Electro Science

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Table of Contents1 Electric Charge and Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Static Electricity and Charge: Conservation of Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Conductors and Insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Electric Field: Concept of a Field Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Electric Field Lines: Multiple Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Electric Forces in Biology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Conductors and Electric Fields in Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Applications of Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Electric Potential and Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Electric Potential Energy: Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Electric Potential in a Uniform Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46Electrical Potential Due to a Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Equipotential Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Capacitors and Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Capacitors in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Energy Stored in Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3 Electric Current, Resistance, and Ohm's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74Ohm’s Law: Resistance and Simple Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79Resistance and Resistivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Electric Power and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Alternating Current versus Direct Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Electric Hazards and the Human Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Nerve Conduction–Electrocardiograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4 Circuits, Bioelectricity, and DC Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Resistors in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Electromotive Force: Terminal Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120Kirchhoff’s Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126DC Voltmeters and Ammeters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130Null Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134DC Circuits Containing Resistors and Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151Magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152Ferromagnets and Electromagnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154Magnetic Fields and Magnetic Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158Force on a Moving Charge in a Magnetic Field: Examples and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159The Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163Magnetic Force on a Current-Carrying Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166Torque on a Current Loop: Motors and Meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168Magnetic Fields Produced by Currents: Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170Magnetic Force between Two Parallel Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174More Applications of Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

6 Electromagnetic Induction, AC Circuits, and Electrical Technologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189Induced Emf and Magnetic Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191Faraday’s Law of Induction: Lenz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192Motional Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195Eddy Currents and Magnetic Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198Electric Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201Back Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204Electrical Safety: Systems and Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215Reactance, Inductive and Capacitive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217RLC Series AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

A Useful Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

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1 ELECTRIC CHARGE AND ELECTRIC FIELD

Figure 1.1 Static electricity from this plastic slide causes the child’s hair to stand on end. The sliding motion stripped electrons away from the child’s body, leaving an excess ofpositive charges, which repel each other along each strand of hair. (credit: Ken Bosma/Wikimedia Commons)

Learning Objectives1.1. Static Electricity and Charge: Conservation of Charge

• Define electric charge, and describe how the two types of charge interact.• Describe three common situations that generate static electricity.• State the law of conservation of charge.

1.2. Conductors and Insulators• Define conductor and insulator, explain the difference, and give examples of each.• Describe three methods for charging an object.• Explain what happens to an electric force as you move farther from the source.• Define polarization.

1.3. Coulomb’s Law• State Coulomb’s law in terms of how the electrostatic force changes with the distance between two objects.• Calculate the electrostatic force between two charged point forces, such as electrons or protons.• Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth.

1.4. Electric Field: Concept of a Field Revisited• Describe a force field and calculate the strength of an electric field due to a point charge.• Calculate the force exerted on a test charge by an electric field.• Explain the relationship between electrical force (F) on a test charge and electrical field strength (E).

1.5. Electric Field Lines: Multiple Charges• Calculate the total force (magnitude and direction) exerted on a test charge from more than one charge• Describe an electric field diagram of a positive point charge; of a negative point charge with twice the magnitude of positive charge• Draw the electric field lines between two points of the same charge; between two points of opposite charge.

1.6. Electric Forces in Biology• Describe how a water molecule is polar.• Explain electrostatic screening by a water molecule within a living cell.

1.7. Conductors and Electric Fields in Static Equilibrium• List the three properties of a conductor in electrostatic equilibrium.• Explain the effect of an electric field on free charges in a conductor.• Explain why no electric field may exist inside a conductor.• Describe the electric field surrounding Earth.• Explain what happens to an electric field applied to an irregular conductor.• Describe how a lightning rod works.• Explain how a metal car may protect passengers inside from the dangerous electric fields caused by a downed line touching the car.

1.8. Applications of Electrostatics• Name several real-world applications of the study of electrostatics.

CHAPTER 1 | ELECTRIC CHARGE AND ELECTRIC FIELD 5

Introduction to Electric Charge and Electric FieldThe image of American politician and scientist Benjamin Franklin (1706–1790) flying a kite in a thunderstorm is familiar to every schoolchild. (SeeFigure 1.2.) In this experiment, Franklin demonstrated a connection between lightning and static electricity. Sparks were drawn from a key hung ona kite string during an electrical storm. These sparks were like those produced by static electricity, such as the spark that jumps from your finger to ametal doorknob after you walk across a wool carpet. What Franklin demonstrated in his dangerous experiment was a connection betweenphenomena on two different scales: one the grand power of an electrical storm, the other an effect of more human proportions. Connections like thisone reveal the underlying unity of the laws of nature, an aspect we humans find particularly appealing.

Figure 1.2 When Benjamin Franklin demonstrated that lightning was related to static electricity, he made a connection that is now part of the evidence that all directlyexperienced forces except the gravitational force are manifestations of the electromagnetic force.

Much has been written about Franklin. His experiments were only part of the life of a man who was a scientist, inventor, revolutionary, statesman, andwriter. Franklin’s experiments were not performed in isolation, nor were they the only ones to reveal connections.

For example, the Italian scientist Luigi Galvani (1737–1798) performed a series of experiments in which static electricity was used to stimulatecontractions of leg muscles of dead frogs, an effect already known in humans subjected to static discharges. But Galvani also found that if he joinedtwo metal wires (say copper and zinc) end to end and touched the other ends to muscles, he produced the same effect in frogs as static discharge.Alessandro Volta (1745–1827), partly inspired by Galvani’s work, experimented with various combinations of metals and developed the battery.

During the same era, other scientists made progress in discovering fundamental connections. The periodic table was developed as the systematicproperties of the elements were discovered. This influenced the development and refinement of the concept of atoms as the basis of matter. Suchsubmicroscopic descriptions of matter also help explain a great deal more.

Atomic and molecular interactions, such as the forces of friction, cohesion, and adhesion, are now known to be manifestations of theelectromagnetic force. Static electricity is just one aspect of the electromagnetic force, which also includes moving electricity and magnetism.

All the macroscopic forces that we experience directly, such as the sensations of touch and the tension in a rope, are due to the electromagneticforce, one of the four fundamental forces in nature. The gravitational force, another fundamental force, is actually sensed through the electromagneticinteraction of molecules, such as between those in our feet and those on the top of a bathroom scale. (The other two fundamental forces, the strongnuclear force and the weak nuclear force, cannot be sensed on the human scale.)

This chapter begins the study of electromagnetic phenomena at a fundamental level. The next several chapters will cover static electricity, movingelectricity, and magnetism—collectively known as electromagnetism. In this chapter, we begin with the study of electric phenomena due to chargesthat are at least temporarily stationary, called electrostatics, or static electricity.

6 CHAPTER 1 | ELECTRIC CHARGE AND ELECTRIC FIELD


1.1 Static Electricity and Charge: Conservation of Charge

Figure 1.3 Borneo amber was mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of silk, the amber gains moreelectrons, giving it a net negative charge. At the same time, the silk, having lost electrons, becomes positively charged. (credit: Sebakoamber, Wikimedia Commons)

What makes plastic wrap cling? Static electricity. Not only are applications of static electricity common these days, its existence has been knownsince ancient times. The first record of its effects dates to ancient Greeks who noted more than 500 years B.C. that polishing amber temporarilyenabled it to attract bits of straw (see Figure 1.3). The very word electric derives from the Greek word for amber (electron).

Many of the characteristics of static electricity can be explored by rubbing things together. Rubbing creates the spark you get from walking across awool carpet, for example. Static cling generated in a clothes dryer and the attraction of straw to recently polished amber also result from rubbing.Similarly, lightning results from air movements under certain weather conditions. You can also rub a balloon on your hair, and the static electricitycreated can then make the balloon cling to a wall. We also have to be cautious of static electricity, especially in dry climates. When we pumpgasoline, we are warned to discharge ourselves (after sliding across the seat) on a metal surface before grabbing the gas nozzle. Attendants inhospital operating rooms must wear booties with aluminum foil on the bottoms to avoid creating sparks which may ignite the oxygen being used.

Some of the most basic characteristics of static electricity include:

• The effects of static electricity are explained by a physical quantity not previously introduced, called electric charge.• There are only two types of charge, one called positive and the other called negative.• Like charges repel, whereas unlike charges attract.• The force between charges decreases with distance.

How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations ofmaterials always produce one type of charge on one material and the opposite type on the other. By convention, we call one type of charge “positive”,and the other type “negative.” For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged.Since the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed withsilk in this manner will repel one another, since each rod has positive charge on it. Similarly, two silk cloths so rubbed will repel, since both clothshave negative charge. Figure 1.4 shows how these simple materials can be used to explore the nature of the force between charges.

Figure 1.4 A glass rod becomes positively charged when rubbed with silk, while the silk becomes negatively charged. (a) The glass rod is attracted to the silk because theircharges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel.

More sophisticated questions arise. Where do these charges come from? Can you create or destroy charge? Is there a smallest unit of charge?Exactly how does the force depend on the amount of charge and the distance between charges? Such questions obviously occurred to BenjaminFranklin and other early researchers, and they interest us even today.

Charge Carried by Electrons and ProtonsFranklin wrote in his letters and books that he could see the effects of electric charge but did not understand what caused the phenomenon. Todaywe have the advantage of knowing that normal matter is made of atoms, and that atoms contain positive and negative charges, usually in equalamounts.

Figure 1.5 shows a simple model of an atom with negative electrons orbiting its positive nucleus. The nucleus is positive due to the presence ofpositively charged protons. Nearly all charge in nature is due to electrons and protons, which are two of the three building blocks of most matter.(The third is the neutron, which is neutral, carrying no charge.) Other charge-carrying particles are observed in cosmic rays and nuclear decay, andare created in particle accelerators. All but the electron and proton survive only a short time and are quite rare by comparison.


Figure 1.5 This simplified (and not to scale) view of an atom is called the planetary model of the atom. Negative electrons orbit a much heavier positive nucleus, as the planetsorbit the much heavier sun. There the similarity ends, because forces in the atom are electromagnetic, whereas those in the planetary system are gravitational. Normalmacroscopic amounts of matter contain immense numbers of atoms and molecules and, hence, even greater numbers of individual negative and positive charges.

The charges of electrons and protons are identical in magnitude but opposite in sign. Furthermore, all charged objects in nature are integral multiplesof this basic quantity of charge, meaning that all charges are made of combinations of a basic unit of charge. Usually, charges are formed bycombinations of electrons and protons. The magnitude of this basic charge is

(1.1)∣ qe ∣ = 1.60×10−19 C.

The symbol q is commonly used for charge and the subscript e indicates the charge of a single electron (or proton).

The SI unit of charge is the coulomb (C). The number of protons needed to make a charge of 1.00 C is

(1.2)1.00 C× 1 proton

1.60×10−19 C= 6.25×1018 protons.

Similarly, 6.25×1018 electrons have a combined charge of −1.00 coulomb. Just as there is a smallest bit of an element (an atom), there is a

smallest bit of charge. There is no directly observed charge smaller than ∣ qe ∣ (see Things Great and Small: The Submicroscopic Origin of

Charge), and all observed charges are integral multiples of ∣ qe ∣ .

Things Great and Small: The Submicroscopic Origin of Charge

With the exception of exotic, short-lived particles, all charge in nature is carried by electrons and protons. Electrons carry the charge we havenamed negative. Protons carry an equal-magnitude charge that we call positive. (See Figure 1.6.) Electron and proton charges are consideredfundamental building blocks, since all other charges are integral multiples of those carried by electrons and protons. Electrons and protons arealso two of the three fundamental building blocks of ordinary matter. The neutron is the third and has zero total charge.

Figure 1.6 shows a person touching a Van de Graaff generator and receiving excess positive charge. The expanded view of a hair shows theexistence of both types of charges but an excess of positive. The repulsion of these positive like charges causes the strands of hair to repel otherstrands of hair and to stand up. The further blowup shows an artist’s conception of an electron and a proton perhaps found in an atom in a strand ofhair.



Figure 1.6 When this person touches a Van de Graaff generator, she receives an excess of positive charge, causing her hair to stand on end. The charges in one hair areshown. An artist’s conception of an electron and a proton illustrate the particles carrying the negative and positive charges. We cannot really see these particles with visiblelight because they are so small (the electron seems to be an infinitesimal point), but we know a great deal about their measurable properties, such as the charges they carry.

The electron seems to have no substructure; in contrast, when the substructure of protons is explored by scattering extremely energetic electronsfrom them, it appears that there are point-like particles inside the proton. These sub-particles, named quarks, have never been directly observed, butthey are believed to carry fractional charges as seen in Figure 1.7. Charges on electrons and protons and all other directly observable particles are

unitary, but these quark substructures carry charges of either −13 or +2

3 . There are continuing attempts to observe fractional charge directly and to

learn of the properties of quarks, which are perhaps the ultimate substructure of matter.

Figure 1.7 Artist’s conception of fractional quark charges inside a proton. A group of three quark charges add up to the single positive charge on the proton:

−13qe + 2

3qe + 23qe = +1qe .

Separation of Charge in AtomsCharges in atoms and molecules can be separated—for example, by rubbing materials together. Some atoms and molecules have a greater affinityfor electrons than others and will become negatively charged by close contact in rubbing, leaving the other material positively charged. (See Figure1.8.) Positive charge can similarly be induced by rubbing. Methods other than rubbing can also separate charges. Batteries, for example, usecombinations of substances that interact in such a way as to separate charges. Chemical interactions may transfer negative charge from onesubstance to the other, making one battery terminal negative and leaving the first one positive.


Figure 1.8 When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber andcloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbedtogether, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, butthe absolute value of the net positive and negative charges will be equal.

No charge is actually created or destroyed when charges are separated as we have been discussing. Rather, existing charges are moved about. Infact, in all situations the total amount of charge is always constant. This universally obeyed law of nature is called the law of conservation ofcharge.

Law of Conservation of Charge

Total charge is constant in any process.

In more exotic situations, such as in particle accelerators, mass, Δm , can be created from energy in the amount Δm = Ec2 . Sometimes, the

created mass is charged, such as when an electron is created. Whenever a charged particle is created, another having an opposite charge is alwayscreated along with it, so that the total charge created is zero. Usually, the two particles are “matter-antimatter” counterparts. For example, anantielectron would usually be created at the same time as an electron. The antielectron has a positive charge (it is called a positron), and so the totalcharge created is zero. (See Figure 1.9.) All particles have antimatter counterparts with opposite signs. When matter and antimatter counterparts arebrought together, they completely annihilate one another. By annihilate, we mean that the mass of the two particles is converted to energy E, again

obeying the relationship Δm = Ec2 . Since the two particles have equal and opposite charge, the total charge is zero before and after the

annihilation; thus, total charge is conserved.

Making Connections: Conservation Laws

Only a limited number of physical quantities are universally conserved. Charge is one—energy, momentum, and angular momentum are others.Because they are conserved, these physical quantities are used to explain more phenomena and form more connections than other, less basicquantities. We find that conserved quantities give us great insight into the rules followed by nature and hints to the organization of nature.Discoveries of conservation laws have led to further discoveries, such as the weak nuclear force and the quark substructure of protons and otherparticles.



Figure 1.9 (a) When enough energy is present, it can be converted into matter. Here the matter created is an electron–antielectron pair. ( me is the electron’s mass.) The total

charge before and after this event is zero. (b) When matter and antimatter collide, they annihilate each other; the total charge is conserved at zero before and after theannihilation.

The law of conservation of charge is absolute—it has never been observed to be violated. Charge, then, is a special physical quantity, joining a veryshort list of other quantities in nature that are always conserved. Other conserved quantities include energy, momentum, and angular momentum.

PhET Explorations: Balloons and Static Electricity

Why does a balloon stick to your sweater? Rub a balloon on a sweater, then let go of the balloon and it flies over and sticks to the sweater. Viewthe charges in the sweater, balloons, and the wall.

Figure 1.10 Balloons and Static Electricity (http://cnx.org/content/m42300/1.5/balloons_en.jar)

1.2 Conductors and Insulators

Figure 1.11 This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to movefreely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don’t allow electric charge to escape outward. (credit: Evan-Amos,Wikimedia Commons)

Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals andsimilar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air movesthrough loose sand. Any substance that has free electrons and allows charge to move relatively freely through it is called a conductor. The movingelectrons may collide with fixed atoms and molecules, losing some energy, but they can move in a conductor. Superconductors allow the movementof charge without any loss of energy. Salty water and other similar conducting materials contain free ions that can move through them. An ion is anatom or molecule having a positive or negative (nonzero) total charge. In other words, the total number of electrons is not equal to the total number ofprotons.


http://cnx.org/content/m42300/1.5/balloons_en.jar

Other substances, such as glass, do not allow charges to move through them. These are called insulators. Electrons and ions in insulators are

bound in the structure and cannot move easily—as much as 1023 times more slowly than in conductors. Pure water and dry table salt areinsulators, for example, whereas molten salt and salty water are conductors.

Figure 1.12 An electroscope is a favorite instrument in physics demonstrations and student laboratories. It is typically made with gold foil leaves hung from a (conducting)metal stem and is insulated from the room air in a glass-walled container. (a) A positively charged glass rod is brought near the tip of the electroscope, attracting electrons tothe top and leaving a net positive charge on the leaves. Like charges in the light flexible gold leaves repel, separating them. (b) When the rod is touched against the ball,electrons are attracted and transferred, reducing the net charge on the glass rod but leaving the electroscope positively charged. (c) The excess charges are evenly distributedin the stem and leaves of the electroscope once the glass rod is removed.

Charging by ContactFigure 1.12 shows an electroscope being charged by touching it with a positively charged glass rod. Because the glass rod is an insulator, it mustactually touch the electroscope to transfer charge to or from it. (Note that the extra positive charges reside on the surface of the glass rod as a resultof rubbing it with silk before starting the experiment.) Since only electrons move in metals, we see that they are attracted to the top of theelectroscope. There, some are transferred to the positive rod by touch, leaving the electroscope with a net positive charge.

Electrostatic repulsion in the leaves of the charged electroscope separates them. The electrostatic force has a horizontal component that results inthe leaves moving apart as well as a vertical component that is balanced by the gravitational force. Similarly, the electroscope can be negativelycharged by contact with a negatively charged object.

Charging by InductionIt is not necessary to transfer excess charge directly to an object in order to charge it. Figure 1.13 shows a method of induction wherein a charge iscreated in a nearby object, without direct contact. Here we see two neutral metal spheres in contact with one another but insulated from the rest ofthe world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged.

This is an example of induced polarization of neutral objects. Polarization is the separation of charges in an object that remains neutral. If thespheres are now separated (before the rod is pulled away), each sphere will have a net charge. Note that the object closest to the charged rodreceives an opposite charge when charged by induction. Note also that no charge is removed from the charged rod, so that this process can berepeated without depleting the supply of excess charge.

Another method of charging by induction is shown in Figure 1.14. The neutral metal sphere is polarized when a charged rod is brought near it. Thesphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since the earth is large and most ground is a goodconductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire,because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with anexcess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction and the charged rod loses none of itsexcess charge.



Figure 1.13 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively chargedglass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod isremoved, thus separating negative and positive charge. (d) The spheres retain net charges after the inducing rod is removed—without ever having been touched by a chargedobject.

Figure 1.14 Charging by induction, using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded,allowing electrons to be attracted from the earth’s ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an inducednegative charge.


Figure 1.15 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules.There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostaticforce decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effectoccurs for a conductor; since the unlike charges are closer, there is a net attraction.

Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plasticcomb through your hair, the charged comb can pick up neutral pieces of paper. Figure 1.15 shows how the polarization of atoms and molecules inneutral objects results in their attraction to a charged object.

When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly.Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, therepulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus a positively charged glass rod attractsneutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural orinherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greaterpolarization effects than molecules with naturally uniform charge distributions.

Check Your Understanding

Can you explain the attraction of water to the charged rod in the figure below?

Figure 1.16

Solution

Water molecules are polarized, giving them slightly positive and slightly negative sides. This makes water even more susceptible to a chargedrod’s attraction. As the water flows downward, due to the force of gravity, the charged conductor exerts a net attraction to the opposite charges inthe stream of water, pulling it closer.



PhET Explorations: John Travoltage

Make sparks fly with John Travoltage. Wiggle Johnnie's foot and he picks up charges from the carpet. Bring his hand close to the door knob andget rid of the excess charge.

Figure 1.17 John Travoltage (http://cnx.org/content/m42306/1.4/travoltage_en.jar)

1.3 Coulomb’s Law

Figure 1.18 This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostaticattraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (credit: NASA/HST)

Through the work of scientists in the late 18th century, the main features of the electrostatic force—the existence of two types of charge, theobservation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as amathematical formula. The mathematical formula for the electrostatic force is called Coulomb’s law after the French physicist Charles Coulomb(1736–1806), who performed experiments and first proposed a formula to calculate it.

Coulomb’s Law(1.3)

F = k |q1 q2|r2 .

Coulomb’s law calculates the magnitude of the force F between two point charges, q1 and q2 , separated by a distance r . In SI units, the

constant k is equal to

(1.4)k = 8.988×109N ⋅ m2

C2 ≈ 8.99×109N ⋅ m2

C2 .

The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the twocharges. (See Figure 1.19.)

Although the formula for Coulomb’s law is simple, it was no mean task to prove it. The experiments Coulomb did, with the primitive equipment thenavailable, were difficult. Modern experiments have verified Coulomb’s law to great precision. For example, it has been shown that the force is

inversely proportional to distance between two objects squared ⎛⎝F ∝ 1 / r2⎞⎠ to an accuracy of 1 part in 1016 . No exceptions have ever been found,

even at the small distances within the atom.

Figure 1.19 The magnitude of the electrostatic force F between point charges q1 and q2 separated by a distance r is given by Coulomb’s law. Note that Newton’s third

law (every force exerted creates an equal and opposite force) applies as usual—the force on q1 is equal in magnitude and opposite in direction to the force it exerts on q2 .

(a) Like charges. (b) Unlike charges.


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Example 1.1 How Strong is the Coulomb Force Relative to the Gravitational Force?

Compare the electrostatic force between an electron and proton separated by 0.530×10−10 m with the gravitational force between them. Thisdistance is their average separation in a hydrogen atom.

Strategy

To compare the two forces, we first compute the electrostatic force using Coulomb’s law, F = k |q1 q2|r2 . We then calculate the gravitational

force using Newton’s universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.

Solution

Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb’s lawyields

(1.5)F = k |q1 q2|

r2

(1.6)= ⎛⎝8.99×109 N ⋅ m2 / C2⎞⎠×

(1.60×10–19 C)(1.60×10–19 C)(0.530×10–10 m)2

Thus the Coulomb force is

(1.7)F = 8.19×10–8 N.The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of

8.99×1022 m / s2 (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as:

(1.8)FG = GmMr2 ,

where G = 6.67×10−11 N ⋅ m2 / kg2 . Here m and M represent the electron and proton masses, which can be found in the appendices.

Entering values for the knowns yields

(1.9)FG = (6.67×10 – 11 N ⋅ m2 / kg2)×(9.11×10–31 kg)(1.67×10–27 kg)

(0.530×10–10 m)2 = 3.61×10–47 N

This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of themagnitude of the electrostatic force to gravitational force in this case is, thus,

(1.10)FFG

= 2.27×1039.

Discussion

This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at anydistance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how muchlarger the Coulomb force is than the gravitational force between two of the most common particles in nature.

As the example implies, gravitational force is completely negligible on a small scale, where the interactions of individual charged particles areimportant. On a large scale, such as between the Earth and a person, the reverse is true. Most objects are nearly electrically neutral, and soattractive and repulsive Coulomb forces nearly cancel. Gravitational force on a large scale dominates interactions between large objects because itis always attractive, while Coulomb forces tend to cancel.

1.4 Electric Field: Concept of a Field RevisitedContact forces, such as between a baseball and a bat, are explained on the small scale by the interaction of the charges in atoms and molecules inclose proximity. They interact through forces that include the Coulomb force. Action at a distance is a force between objects that are not closeenough for their atoms to “touch.” That is, they are separated by more than a few atomic diameters.

For example, a charged rubber comb attracts neutral bits of paper from a distance via the Coulomb force. It is very useful to think of an object beingsurrounded in space by a force field. The force field carries the force to another object (called a test object) some distance away.

Concept of a FieldA field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparentphysical connection. For example, the gravitational field surrounding the earth (and all other masses) represents the gravitational force that would beexperienced if another mass were placed at a given point within the field.



In the same way, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb’s law, F = k|q1q2| / r2 , its magnitude

is given by the equation F = k|qQ| / r2 , for a point charge (a particle having a charge Q ) acting on a test charge q at a distance r (see Figure

1.20). Both the magnitude and direction of the Coulomb force field depend on Q and the test charge q .

Figure 1.20 The Coulomb force field due to a positive charge Q is shown acting on two different charges. Both charges are the same distance from Q . (a) Since q1 is

positive, the force F1 acting on it is repulsive. (b) The charge q2 is negative and greater in magnitude than q1 , and so the force F2 acting on it is attractive and stronger

than F1 . The Coulomb force field is thus not unique at any point in space, because it depends on the test charges q1 and q2 as well as the charge Q .

To simplify things, we would prefer to have a field that depends only on Q and not on the test charge q . The electric field is defined in such a

manner that it represents only the charge creating it and is unique at every point in space. Specifically, the electric field E is defined to be the ratio ofthe Coulomb force to the test charge:

(1.11)E = Fq ,

where F is the electrostatic force (or Coulomb force) exerted on a positive test charge q . It is understood that E is in the same direction as F . It

is also assumed that q is so small that it does not alter the charge distribution creating the electric field. The units of electric field are newtons per

coulomb (N/C). If the electric field is known, then the electrostatic force on any charge q is simply obtained by multiplying charge times electric field,

or F = qE . Consider the electric field due to a point charge Q . According to Coulomb’s law, the force it exerts on a test charge q is

F = k|qQ| / r2 . Thus the magnitude of the electric field, E , for a point charge is

(1.12)E = |Fq | = k| qQ

qr2 | = k |Q|r2 .

Since the test charge cancels, we see that

(1.13)E = k |Q|

r2 .

The electric field is thus seen to depend only on the charge Q and the distance r ; it is completely independent of the test charge q .

Example 1.2 Calculating the Electric Field of a Point Charge

Calculate the strength and direction of the electric field E due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from thecharge.

Strategy

We can find the electric field created by a point charge by using the equation E = kQ / r2 .

Solution

Here Q = 2.00×10−9 C and r = 5.00×10−3 m. Entering those values into the above equation gives

(1.14)E = k Qr2

= (8.99×109 N ⋅ m2/C2 )× (2.00×10−9 C)(5.00×10−3 m)2

= 7.19×105 N/C.


Discussion

This electric field strength is the same at any point 5.00 mm away from the charge Q that creates the field. It is positive, meaning that it has a

direction pointing away from the charge Q .

Example 1.3 Calculating the Force Exerted on a Point Charge by an Electric Field

What force does the electric field found in the previous example exert on a point charge of –0.250 μC ?

Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric fieldE = F / q rearranged to F = qE .

Solution

The magnitude of the force on a charge q = −0.250 μC exerted by a field of strength E = 7.20×105 N/C is thus,

(1.15)F = −qE

= (0.250×10–6 C)(7.20×105 N/C)= 0.180 N.

Because q is negative, the force is directed opposite to the direction of the field.

Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) Thecharges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in staticcling and similar situations.

PhET Explorations: Electric Field of Dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust thedirection and magnitude.

Figure 1.21 Electric Field of Dreams (http://cnx.org/content/m42310/1.6/efield_en.jar)

1.5 Electric Field Lines: Multiple ChargesDrawings using lines to represent electric fields around charged objects are very useful in visualizing field strength and direction. Since the electricfield has both magnitude and direction, it is a vector. Like all vectors, the electric field can be represented by an arrow that has length proportional toits magnitude and that points in the correct direction. (We have used arrows extensively to represent force vectors, for example.)

Figure 1.22 shows two pictorial representations of the same electric field created by a positive point charge Q . Figure 1.22 (b) shows the standard

representation using continuous lines. Figure 1.22 (b) shows numerous individual arrows with each arrow representing the force on a test charge q .

Field lines are essentially a map of infinitesimal force vectors.

Figure 1.22 Two equivalent representations of the electric field due to a positive charge Q . (a) Arrows representing the electric field’s magnitude and direction. (b) In the

standard representation, the arrows are replaced by continuous field lines having the same direction at any point as the electric field. The closeness of the lines is directlyrelated to the strength of the electric field. A test charge placed anywhere will feel a force in the direction of the field line; this force will have a strength proportional to thedensity of the lines (being greater near the charge, for example).



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Note that the electric field is defined for a positive test charge q , so that the field lines point away from a positive charge and toward a negative

charge. (See Figure 1.23.) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the

electric field for a point charge is E = k|Q| / r2 and area is proportional to r2 . This pictorial representation, in which field lines represent the

direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields:electrostatic, gravitational, magnetic, and others.

Figure 1.23 The electric field surrounding three different point charges. (a) A positive charge. (b) A negative charge of equal magnitude. (c) A larger negative charge.

In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created byeach charge. The following example shows how to add electric field vectors.

Example 1.4 Adding Electric Fields

Find the magnitude and direction of the total electric field due to the two point charges, q1 and q2 , at the origin of the coordinate system as

shown in Figure 1.24.

Figure 1.24 The electric fields E1 and E2 at the origin O add to Etot .

Strategy

Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types ofvectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in thisinstance. We pretend that there is a positive test charge, q , at point O, which allows us to determine the direction of the fields E1 and E2 .

Once those fields are found, the total field can be determined using vector addition.

Solution

The electric field strength at the origin due to q1 is labeled E1 and is calculated:

(1.16)E1 = kq1

r12 = ⎛⎝8.99×109 N ⋅ m2/C2⎞⎠

⎛⎝5.00×10−9 C⎞⎠⎛⎝2.00×10−2 m⎞⎠

2

E1 = 1.124×105 N/C.

Similarly, E2 is

(1.17)E2 = kq2

r22 = ⎛⎝8.99×109 N ⋅ m2/C2⎞⎠

⎛⎝10.0×10−9 C⎞⎠⎛⎝4.00×10−2 m⎞⎠

2

E2 = 0.5619×105 N/C.

Four digits have been retained in this solution to illustrate that E1 is exactly twice the magnitude of E2 . Now arrows are drawn to represent the

magnitudes and directions of E1 and E2 . (See Figure 1.24.) The direction of the electric field is that of the force on a positive charge so both


arrows point directly away from the positive charges that create them. The arrow for E1 is exactly twice the length of that for E2 . The arrows

form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field Etot is

(1.18)Etot = (E12 + E2

2 )1/2

= {(1.124×105 N/C)2 + (0.5619×105 N/C)2}1/2

= 1.26×105 N/C.The direction is

(1.19)θ = tan−1⎛⎝

E1E2

⎞⎠

= tan−1⎛⎝ 1.124×105 N/C0.5619×105 N/C

⎞⎠

= 63.4º,

or 63.4º above the x-axis.

Discussion

In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The totalelectric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges overan entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number ofpoints in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

Figure 1.25 shows how the electric field from two point charges can be drawn by finding the total field at representative points and drawing electricfield lines consistent with those points. While the electric fields from multiple charges are more complex than those of single charges, some simplefeatures are easily noticed.

For example, the field is weaker between like charges, as shown by the lines being farther apart in that region. (This is because the fields from eachcharge exert opposing forces on any charge placed between them.) (See Figure 1.25 and Figure 1.26(a).) Furthermore, at a great distance from twolike charges, the field becomes identical to the field from a single, larger charge.

Figure 1.26(b) shows the electric field of two unlike charges. The field is stronger between the charges. In that region, the fields from each chargeare in the same direction, and so their strengths add. The field of two unlike charges is weak at large distances, because the fields of the individualcharges are in opposite directions and so their strengths subtract. At very large distances, the field of two unlike charges looks like that of a smallersingle charge.

Figure 1.25 Two positive point charges q1 and q2 produce the resultant electric field shown. The field is calculated at representative points and then smooth field lines

drawn following the rules outlined in the text.



Figure 1.26 (a) Two negative charges produce the fields shown. It is very similar to the field produced by two positive charges, except that the directions are reversed. Thefield is clearly weaker between the charges. The individual forces on a test charge in that region are in opposite directions. (b) Two opposite charges produce the field shown,which is stronger in the region between the charges.

We use electric field lines to visualize and analyze electric fields (the lines are a pictorial tool, not a physical entity in themselves). The properties ofelectric field lines for any charge distribution can be summarized as follows:

1. Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.2. The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.3. The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area

perpendicular to the lines.4. The direction of the electric field is tangent to the field line at any point in space.5. Field lines can never cross.

The last property means that the field is unique at any point. The field line represents the direction of the field; so if they crossed, the field would havetwo directions at that location (an impossibility if the field is unique).

PhET Explorations: Charges and Fields

Move point charges around on the playing field and then view the electric field, voltages, equipotential lines, and more. It's colorful, it's dynamic,it's free.

Figure 1.27 Charges and Fields (http://cnx.org/content/m42312/1.7/charges-and-fields_en.jar)

1.6 Electric Forces in BiologyClassical electrostatics has an important role to play in modern molecular biology. Large molecules such as proteins, nucleic acids, and so on—soimportant to life—are usually electrically charged. DNA itself is highly charged; it is the electrostatic force that not only holds the molecule together butgives the molecule structure and strength. Figure 1.28 is a schematic of the DNA double helix.

Figure 1.28 DNA is a highly charged molecule. The DNA double helix shows the two coiled strands each containing a row of nitrogenous bases, which “code” the geneticinformation needed by a living organism. The strands are connected by bonds between pairs of bases. While pairing combinations between certain bases are fixed (C-G andA-T), the sequence of nucleotides in the strand varies. (credit: Jerome Walker)


http://cnx.org/content/m42312/1.7/charges-and-fields_en.jar

The four nucleotide bases are given the symbols A (adenine), C (cytosine), G (guanine), and T (thymine). The order of the four bases varies in eachstrand, but the pairing between bases is always the same. C and G are always paired and A and T are always paired, which helps to preserve the

order of bases in cell division (mitosis) so as to pass on the correct genetic information. Since the Coulomb force drops with distance ( F ∝ 1 / r2 ),the distances between the base pairs must be small enough that the electrostatic force is sufficient to hold them together.

DNA is a highly charged molecule, with about 2qe (fundamental charge) per 0.3×10−9 m. The distance separating the two strands that make up

the DNA structure is about 1 nm, while the distance separating the individual atoms within each base is about 0.3 nm.

One might wonder why electrostatic forces do not play a larger role in biology than they do if we have so many charged molecules. The reason is thatthe electrostatic force is “diluted” due to screening between molecules. This is due to the presence of other charges in the cell.

Polarity of Water Molecules

The best example of this charge screening is the water molecule, represented as H2 O . Water is a strongly polar molecule. Its 10 electrons (8 from

the oxygen atom and 2 from the two hydrogen atoms) tend to remain closer to the oxygen nucleus than the hydrogen nuclei. This creates two centersof equal and opposite charges—what is called a dipole, as illustrated in Figure 1.29. The magnitude of the dipole is called the dipole moment.

These two centers of charge will terminate some of the electric field lines coming from a free charge, as on a DNA molecule. This results in areduction in the strength of the Coulomb interaction. One might say that screening makes the Coulomb force a short range force rather than longrange.

Other ions of importance in biology that can reduce or screen Coulomb interactions are Na+ , and K+ , and Cl– . These ions are located both

inside and outside of living cells. The movement of these ions through cell membranes is crucial to the motion of nerve impulses through nerveaxons.

Recent studies of electrostatics in biology seem to show that electric fields in cells can be extended over larger distances, in spite of screening, by“microtubules” within the cell. These microtubules are hollow tubes composed of proteins that guide the movement of chromosomes when cellsdivide, the motion of other organisms within the cell, and provide mechanisms for motion of some cells (as motors).

Figure 1.29 This schematic shows water ( H2 O ) as a polar molecule. Unequal sharing of electrons between the oxygen (O) and hydrogen (H) atoms leads to a net

separation of positive and negative charge—forming a dipole. The symbols δ− and δ+indicate that the oxygen side of the H2 O molecule tends to be more negative,

while the hydrogen ends tend to be more positive. This leads to an attraction of opposite charges between molecules.

1.7 Conductors and Electric Fields in Static EquilibriumConductors contain free charges that move easily. When excess charge is placed on a conductor or the conductor is put into a static electric field,charges in the conductor quickly respond to reach a steady state called electrostatic equilibrium.

Figure 1.30 shows the effect of an electric field on free charges in a conductor. The free charges move until the field is perpendicular to theconductor’s surface. There can be no component of the field parallel to the surface in electrostatic equilibrium, since, if there were, it would producefurther movement of charge. A positive free charge is shown, but free charges can be either positive or negative and are, in fact, negative in metals.The motion of a positive charge is equivalent to the motion of a negative charge in the opposite direction.



Figure 1.30 When an electric field E is applied to a conductor, free charges inside the conductor move until the field is perpendicular to the surface. (a) The electric field is a

vector quantity, with both parallel and perpendicular components. The parallel component ( E∥ ) exerts a force ( F∥ ) on the free charge q , which moves the charge until

F∥ = 0 . (b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor’s surface, leaving electrostatic forces in equilibrium.

A conductor placed in an electric field will be polarized. Figure 1.31 shows the result of placing a neutral conductor in an originally uniform electricfield. The field becomes stronger near the conductor but entirely disappears inside it.

Figure 1.31 This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, untilthe electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive chargeon the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized.

Misconception Alert: Electric Field inside a Conductor

Excess charges placed on a spherical conductor repel and move until they are evenly distributed, as shown in Figure 1.32. Excess charge isforced to the surface until the field inside the conductor is zero. Outside the conductor, the field is exactly the same as if the conductor werereplaced by a point charge at its center equal to the excess charge.

Figure 1.32 The mutual repulsion of excess positive charges on a spherical conductor distributes them uniformly on its surface. The resulting electric field isperpendicular to the surface and zero inside. Outside the conductor, the field is identical to that of a point charge at the center equal to the excess charge.

Properties of a Conductor in Electrostatic Equilibrium1. The electric field is zero inside a conductor.2. Just outside a conductor, the electric field lines are perpendicular to its surface, ending or beginning on charges on the surface.


3. Any excess charge resides entirely on the surface or surfaces of a conductor.

The properties of a conductor are consistent with the situations already discussed and can be used to analyze any conductor in electrostaticequilibrium. This can lead to some interesting new insights, such as described below.

How can a very uniform electric field be created? Consider a system of two metal plates with opposite charges on them, as shown in Figure 1.33.The properties of conductors in electrostatic equilibrium indicate that the electric field between the plates will be uniform in strength and direction.Except near the edges, the excess charges distribute themselves uniformly, producing field lines that are uniformly spaced (hence uniform instrength) and perpendicular to the surfaces (hence uniform in direction, since the plates are flat). The edge effects are less important when the platesare close together.

Figure 1.33 Two metal plates with equal, but opposite, excess charges. The field between them is uniform in strength and direction except near the edges. One use of such afield is to produce uniform acceleration of charges between the plates, such as in the electron gun of a TV tube.

Earth’s Electric FieldA near uniform electric field of approximately 150 N/C, directed downward, surrounds Earth, with the magnitude increasing slightly as we get closer tothe surface. What causes the electric field? At around 100 km above the surface of Earth we have a layer of charged particles, called theionosphere. The ionosphere is responsible for a range of phenomena including the electric field surrounding Earth. In fair weather the ionosphere ispositive and the Earth largely negative, maintaining the electric field (Figure 1.34(a)).

In storm conditions clouds form and localized electric fields can be larger and reversed in direction (Figure 1.34(b)). The exact charge distributionsdepend on the local conditions, and variations of Figure 1.34(b) are possible.

If the electric field is sufficiently large, the insulating properties of the surrounding material break down and it becomes conducting. For air this occurs

at around 3×106 N/C. Air ionizes ions and electrons recombine, and we get discharge in the form of lightning sparks and corona discharge.

Figure 1.34 Earth’s electric field. (a) Fair weather field. Earth and the ionosphere (a layer of charged particles) are both conductors. They produce a uniform electric field ofabout 150 N/C. (credit: D. H. Parks) (b) Storm fields. In the presence of storm clouds, the local electric fields can be larger. At very high fields, the insulating properties of theair break down and lightning can occur. (credit: Jan-Joost Verhoef)

Electric Fields on Uneven SurfacesSo far we have considered excess charges on a smooth, symmetrical conductor surface. What happens if a conductor has sharp corners or ispointed? Excess charges on a nonuniform conductor become concentrated at the sharpest points. Additionally, excess charge may move on or offthe conductor at the sharpest points.

To see how and why this happens, consider the charged conductor in Figure 1.35. The electrostatic repulsion of like charges is most effective inmoving them apart on the flattest surface, and so they become least concentrated there. This is because the forces between identical pairs ofcharges at either end of the conductor are identical, but the components of the forces parallel to the surfaces are different. The component parallel tothe surface is greatest on the flattest surface and, hence, more effective in moving the charge.

The same effect is produced on a conductor by an externally applied electric field, as seen in Figure 1.35 (c). Since the field lines must beperpendicular to the surface, more of them are concentrated on the most curved parts.



Figure 1.35 Excess charge on a nonuniform conductor becomes most concentrated at the location of greatest curvature. (a) The forces between identical pairs of charges ateither end of the conductor are identical, but the components of the forces parallel to the surface are different. It is F∥ that moves the charges apart once they have

reached the surface. (b) F∥ is smallest at the more pointed end, the charges are left closer together, producing the electric field shown. (c) An uncharged conductor in an

originally uniform electric field is polarized, with the most concentrated charge at its most pointed end.

Applications of ConductorsOn a very sharply curved surface, such as shown in Figure 1.36, the charges are so concentrated at the point that the resulting electric field can begreat enough to remove them from the surface. This can be useful.

Lightning rods work best when they are most pointed. The large charges created in storm clouds induce an opposite charge on a building that canresult in a lightning bolt hitting the building. The induced charge is bled away continually by a lightning rod, preventing the more dramatic lightningstrike.

Of course, we sometimes wish to prevent the transfer of charge rather than to facilitate it. In that case, the conductor should be very smooth and haveas large a radius of curvature as possible. (See Figure 1.37.) Smooth surfaces are used on high-voltage transmission lines, for example, to avoidleakage of charge into the air.

Another device that makes use of some of these principles is a Faraday cage. This is a metal shield that encloses a volume. All electrical chargeswill reside on the outside surface of this shield, and there will be no electrical field inside. A Faraday cage is used to prohibit stray electrical fields inthe environment from interfering with sensitive measurements, such as the electrical signals inside a nerve cell.

During electrical storms if you are driving a car, it is best to stay inside the car as its metal body acts as a Faraday cage with zero electrical fieldinside. If in the vicinity of a lightning strike, its effect is felt on the outside of the car and the inside is unaffected, provided you remain totally inside.This is also true if an active (“hot”) electrical wire was broken (in a storm or an accident) and fell on your car.

Figure 1.36 A very pointed conductor has a large charge concentration at the point. The electric field is very strong at the point and can exert a force large enough to transfercharge on or off the conductor. Lightning rods are used to prevent the buildup of large excess charges on structures and, thus, are pointed.


Figure 1.37 (a) A lightning rod is pointed to facilitate the transfer of charge. (credit: Romaine, Wikimedia Commons) (b) This Van de Graaff generator has a smooth surfacewith a large radius of curvature to prevent the transfer of charge and allow a large voltage to be generated. The mutual repulsion of like charges is evident in the person’s hairwhile touching the metal sphere. (credit: Jon ‘ShakataGaNai’ Davis/Wikimedia Commons).

1.8 Applications of ElectrostaticsThe study of electrostatics has proven useful in many areas. This module covers just a few of the many applications of electrostatics.

The Van de Graaff GeneratorVan de Graaff generators (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due to static electricity—they arealso used for serious research. The first was built by Robert Van de Graaff in 1931 (based on original suggestions by Lord Kelvin) for use in nuclearphysics research. Figure 1.38 shows a schematic of a large research version. Van de Graaffs utilize both smooth and pointed surfaces, andconductors and insulators to generate large static charges and, hence, large voltages.

A very large excess charge can be deposited on the sphere, because it moves quickly to the outer surface. Practical limits arise because the largeelectric fields polarize and eventually ionize surrounding materials, creating free charges that neutralize excess charge or allow it to escape.Nevertheless, voltages of 15 million volts are well within practical limits.

Figure 1.38 Schematic of Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a movinginsulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes thecharge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outside surface. An ion source inside the sphereproduces positive ions, which are accelerated away from the positive sphere to high velocities.

Take-Home Experiment: Electrostatics and Humidity

Rub a comb through your hair and use it to lift pieces of paper. It may help to tear the pieces of paper rather than cut them neatly. Repeat theexercise in your bathroom after you have had a long shower and the air in the bathroom is moist. Is it easier to get electrostatic effects in dry ormoist air? Why would torn paper be more attractive to the comb than cut paper? Explain your observations.



XerographyMost copy machines use an electrostatic process called xerography—a word coined from the Greek words xeros for dry and graphos for writing. Theheart of the process is shown in simplified form in Figure 1.39.

A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium is a substance with aninteresting property—it is a photoconductor. That is, selenium is an insulator when in the dark and a conductor when exposed to light.

In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is induced under the thin layer ofuniformly positively charged selenium. In the second stage, the surface of the drum is exposed to the image of whatever is to be copied. Where theimage is light, the selenium becomes conducting, and the positive charge is neutralized. In dark areas, the positive charge remains, and so the imagehas been transferred to the drum.

The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it will be attracted to the positive regions of thedrum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it will pull the toner from the drum. Finally, the paperand electrostatically held toner are passed through heated pressure rollers, which melt and permanently adhere the toner within the fibers of thepaper.

Figure 1.39 Xerography is a dry copying process based on electrostatics. The major steps in the process are the charging of the photoconducting drum, transfer of an imagecreating a positive charge duplicate, attraction of toner to the charged parts of the drum, and transfer of toner to the paper. Not shown are heat treatment of the paper andcleansing of the drum for the next copy.

Laser PrintersLaser printers use the xerographic process to make high-quality images on paper, employing a laser to produce an image on the photoconductingdrum as shown in Figure 1.40. In its most common application, the laser printer receives output from a computer, and it can achieve high-qualityoutput because of the precision with which laser light can be controlled. Many laser printers do significant information processing, such as makingsophisticated letters or fonts, and may contain a computer more powerful than the one giving them the raw data to be printed.

Figure 1.40 In a laser printer, a laser beam is scanned across a photoconducting drum, leaving a positive charge image. The other steps for charging the drum andtransferring the image to paper are the same as in xerography. Laser light can be very precisely controlled, enabling laser printers to produce high-quality images.

Ink Jet Printers and Electrostatic PaintingThe ink jet printer, commonly used to print computer-generated text and graphics, also employs electrostatics. A nozzle makes a fine spray of tinyink droplets, which are then given an electrostatic charge. (See Figure 1.41.)

Once charged, the droplets can be directed, using pairs of charged plates, with great precision to form letters and images on paper. Ink jet printerscan produce color images by using a black jet and three other jets with primary colors, usually cyan, magenta, and yellow, much as a color televisionproduces color. (This is more difficult with xerography, requiring multiple drums and toners.)


Figure 1.41 The nozzle of an ink-jet printer produces small ink droplets, which are sprayed with electrostatic charge. Various computer-driven devices are then used to directthe droplets to the correct positions on a page.

Electrostatic painting employs electrostatic charge to spray paint onto odd-shaped surfaces. Mutual repulsion of like charges causes the paint to flyaway from its source. Surface tension forms drops, which are then attracted by unlike charges to the surface to be painted. Electrostatic painting canreach those hard-to-get at places, applying an even coat in a controlled manner. If the object is a conductor, the electric field is perpendicular to thesurface, tending to bring the drops in perpendicularly. Corners and points on conductors will receive extra paint. Felt can similarly be applied.

Smoke Precipitators and Electrostatic Air CleaningAnother important application of electrostatics is found in air cleaners, both large and small. The electrostatic part of the process places excess(usually positive) charge on smoke, dust, pollen, and other particles in the air and then passes the air through an oppositely charged grid that attractsand retains the charged particles. (See Figure 1.42.)

Large electrostatic precipitators are used industrially to remove over 99% of the particles from stack gas emissions associated with the burning ofcoal and oil. Home precipitators, often in conjunction with the home heating and air conditioning system, are very effective in removing pollutingparticles, irritants, and allergens.

Figure 1.42 (a) Schematic of an electrostatic precipitator. Air is passed through grids of opposite charge. The first grid charges airborne particles, while the second attracts andcollects them. (b) The dramatic effect of electrostatic precipitators is seen by the absence of smoke from this power plant. (credit: Cmdalgleish, Wikimedia Commons)

Problem-Solving Strategies for Electrostatics1. Examine the situation to determine if static electricity is involved. This may concern separated stationary charges, the forces among them,

and the electric fields they create.2. Identify the system of interest. This includes noting the number, locations, and types of charges involved.3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the

Coulomb force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb

force F from the electric field E , for example.5. Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.6. Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?



Integrated ConceptsThe Integrated Concepts exercises for this module involve concepts such as electric charges, electric fields, and several other topics. Physics is mostinteresting when applied to general situations involving more than a narrow set of physical principles. The electric field exerts force on charges, forexample, and hence the relevance of Dynamics: Force and Newton’s Laws of Motion (http://cnx.org/content/m42129/latest/) . The followingtopics are involved in some or all of the problems labeled “Integrated Concepts”:

• Kinematics (http://cnx.org/content/m42122/latest/)• Two-Dimensional Kinematics (http://cnx.org/content/m42126/latest/)• Dynamics: Force and Newton’s Laws of Motion (http://cnx.org/content/m42129/latest/)• Uniform Circular Motion and Gravitation (http://cnx.org/content/m42140/latest/)• Statics and Torque (http://cnx.org/content/m42167/latest/)• Fluid Statics (http://cnx.org/content/m42185/latest/)

The following worked example illustrates how this strategy is applied to an Integrated Concept problem:

Example 1.5 Acceleration of a Charged Drop of Gasoline

If steps are not taken to ground a gasoline pump, static electricity can be placed on gasoline when filling your car’s tank. Suppose a tiny drop of

gasoline has a mass of 4.00×10–15 kg and is given a positive charge of 3.20×10–19 C . (a) Find the weight of the drop. (b) Calculate the

electric force on the drop if there is an upward electric field of strength 3.00×105 N/C due to other static electricity in the vicinity. (c) Calculatethe drop’s acceleration.

Strategy

To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found.Part (a) of this example asks for weight. This is a topic of dynamics and is defined in Dynamics: Force and Newton’s Laws of Motion(http://cnx.org/content/m42129/latest/) . Part (b) deals with electric force on a charge, a topic of Electric Charge and Electric Field. Part (c)asks for acceleration, knowing forces and mass. These are part of Newton’s laws, also found in Dynamics: Force and Newton’s Laws ofMotion (http://cnx.org/content/m42129/latest/) .

The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifyingknowns and unknowns, checking to see if the answer is reasonable, and so on.

Solution for (a)

Weight is mass times the acceleration due to gravity, as first expressed in

(1.20)w = mg.

Entering the given mass and the average acceleration due to gravity yields

(1.21)w = (4.00×10−15 kg)(9.80 m/s2 ) = 3.92×10−14 N.

Discussion for (a)

This is a small weight, consistent with the small mass of the drop.

Solution for (b)

The force an electric field exerts on a charge is given by rearranging the following equation:

(1.22)F = qE.

Here we are given the charge ( 3.20×10–19 C is twice the fundamental unit of charge) and the electric field strength, and so the electric forceis found to be

(1.23)F = (3.20×10−19 C)(3.00×105 N/C) = 9.60×10−14 N.

Discussion for (b)

While this is a small force, it is greater than the weight of the drop.

Solution for (c)

The acceleration can be found using Newton’s second law, provided we can identify all of the external forces acting on the drop. We assumeonly the drop’s weight and the electric force are significant. Since the drop has a positive charge and the electric field is given to be upward, theelectric force is upward. We thus have a one-dimensional (vertical direction) problem, and we can state Newton’s second law as

(1.24)a = Fnet

m .

where Fnet = F − w . Entering this and the known values into the expression for Newton’s second law yields

(1.25)a = F − wm

= 9.60×10−14 N − 3.92×10−14 N4.00×10−15 kg

= 14.2 m/s2.


http://cnx.org/content/m42129/latest/











Coulomb force:

Coulomb interaction:

Coulomb’s law:

conductor:

conductor:

dipole:

electric charge:

electric field lines:

electric field:

electromagnetic force:

electron:

electrostatic equilibrium:

electrostatic force:

electrostatic precipitators:

electrostatic repulsion:

electrostatics:

Faraday cage:

field:

free charge:

free electron:

grounded:

grounded:

Discussion for (c)

This is an upward acceleration great enough to carry the drop to places where you might not wish to have gasoline.

This worked example illustrates how to apply problem-solving strategies to situations that include topics in different chapters. The first step is toidentify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies.These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example,you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course,such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad applicationof physical principles.

Unreasonable Results

The Unreasonable Results exercises for this module have results that are unreasonable because some premise is unreasonable or becausecertain of the premises are inconsistent with one another. Physical principles applied correctly then produce unreasonable results. The purposeof these problems is to give practice in assessing whether nature is being accurately described, and if it is not to trace the source of difficulty.

Problem-Solving Strategy

To determine if an answer is reasonable, and to determine the cause if it is not, do the following.

1. Solve the problem using strategies as outlined above. Use the format followed in the worked examples in the text to solve the problemas usual.

2. Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, and so on?3. If the answer is unreasonable, look for what specifically could cause the identified difficulty. Usually, the manner in which the answer is

unreasonable is an indication of the difficulty. For example, an extremely large Coulomb force could be due to the assumption of anexcessively large separated charge.

Glossaryanother term for the electrostatic force

the interaction between two charged particles generated by the Coulomb forces they exert on one another

the mathematical equation calculating the electrostatic force vector between two charged particles

a material that allows electrons to move separately from their atomic orbits

an object with properties that allow charges to move about freely within it

a molecule’s lack of symmetrical charge distribution, causing one side to be more positive and another to be more negative

a physical property of an object that causes it to be attracted toward or repelled from another charged object; each chargedobject generates and is influenced by a force called an electromagnetic force

a series of lines drawn from a point charge representing the magnitude and direction of force exerted by that charge

a three-dimensional map of the electric force extended out into space from a point charge

one of the four fundamental forces of nature; the electromagnetic force consists of static electricity, moving electricity andmagnetism

a particle orbiting the nucleus of an atom and carrying the smallest unit of negative charge

an electrostatically balanced state in which all free electrical charges have stopped moving about

the amount and direction of attraction or repulsion between two charged bodies

filters that apply charges to particles in the air, then attract those charges to a filter, removing them from the airstream

the phenomenon of two objects with like charges repelling each other

the study of electric forces that are static or slow-moving

a metal shield which prevents electric charge from penetrating its surface

a map of the amount and direction of a force acting on other objects, extending out into space

an electrical charge (either positive or negative) which can move about separately from its base molecule

an electron that is free to move away from its atomic orbit

when a conductor is connected to the Earth, allowing charge to freely flow to and from Earth’s unlimited reservoir

connected to the ground with a conductor, so that charge flows freely to and from the Earth to the grounded object



induction:

ink-jet printer:

insulator:

ionosphere:

laser printer:

law of conservation of charge:

photoconductor:

point charge:

polar molecule:

polarization:

polarized:

proton:

screening:

static electricity:

test charge:

Van de Graaff generator:

vector addition:

vector:

xerography:

the process by which an electrically charged object brought near a neutral object creates a charge in that object

small ink droplets sprayed with an electric charge are controlled by electrostatic plates to create images on paper

a material that holds electrons securely within their atomic orbits

a layer of charged particles located around 100 km above the surface of Earth, which is responsible for a range of phenomenaincluding the electric field surrounding Earth

uses a laser to create a photoconductive image on a drum, which attracts dry ink particles that are then rolled onto a sheet of paperto print a high-quality copy of the image

states that whenever a charge is created, an equal amount of charge with the opposite sign is createdsimultaneously

a substance that is an insulator until it is exposed to light, when it becomes a conductor

A charged particle, designated Q , generating an electric field

a molecule with an asymmetrical distribution of positive and negative charge

slight shifting of positive and negative charges to opposite sides of an atom or molecule

a state in which the positive and negative charges within an object have collected in separate locations

a particle in the nucleus of an atom and carrying a positive charge equal in magnitude and opposite in sign to the amount of negativecharge carried by an electron

the dilution or blocking of an electrostatic force on a charged object by the presence of other charges nearby

a buildup of electric charge on the surface of an object

A particle (designated q ) with either a positive or negative charge set down within an electric field generated by a point charge

a machine that produces a large amount of excess charge, used for experiments with high voltage

mathematical combination of two or more vectors, including their magnitudes, directions, and positions

a quantity with both magnitude and direction

a dry copying process based on electrostatics

Section Summary

1.1 Static Electricity and Charge: Conservation of Charge• There are only two types of charge, which we call positive and negative.• Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance.• The vast majority of positive charge in nature is carried by protons, while the vast majority of negative charge is carried by electrons.• The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton.• An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.• The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of

this basic charge ∣ qe ∣ is

∣ qe ∣ = 1.60×10−19 C.• Whenever charge is created or destroyed, equal amounts of positive and negative are involved.• Most often, existing charges are separated from neutral objects to obtain some net charge.• Both positive and negative charges exist in neutral objects and can be separated by rubbing one object with another. For macroscopic objects,

negatively charged means an excess of electrons and positively charged means a depletion of electrons.• The law of conservation of charge ensures that whenever a charge is created, an equal charge of the opposite sign is created at the same time.

1.2 Conductors and Insulators• Polarization is the separation of positive and negative charges in a neutral object.• A conductor is a substance that allows charge to flow freely through its atomic structure.• An insulator holds charge within its atomic structure.• Objects with like charges repel each other, while those with unlike charges attract each other.• A conducting object is said to be grounded if it is connected to the Earth through a conductor. Grounding allows transfer of charge to and from

the earth’s large reservoir.• Objects can be charged by contact with another charged object and obtain the same sign charge.• If an object is temporarily grounded, it can be charged by induction, and obtains the opposite sign charge.• Polarized objects have their positive and negative charges concentrated in different areas, giving them a non-symmetrical charge.• Polar molecules have an inherent separation of charge.

1.3 Coulomb’s Law• Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects.


• Coulomb’s law gives the magnitude of the force between point charges. It is

F = k |q1 q2|r2 ,

where q1 and q2 are two point charges separated by a distance r , and k ≈ 8.99×109 N · m2 / C2

• This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects andunderlies most macroscopic forces.

• The Coulomb force is extraordinarily strong compared with the gravitational force, another basic force—but unlike gravitational force it cancancel, since it can be either attractive or repulsive.

• The electrostatic force between two subatomic particles is far greater than the gravitational force between the same two particles.

1.4 Electric Field: Concept of a Field Revisited• The electrostatic force field surrounding a charged object extends out into space in all directions.• The electrostatic force exerted by a point charge on a test charge at a distance r depends on the charge of both charges, as well as the

distance between the two.• The electric field E is defined to be

E = Fq,

where F is the Coulomb or electrostatic force exerted on a small positive test charge q . E has units of N/C.

• The magnitude of the electric field E created by a point charge Q is

E = k |Q|r2 .

where r is the distance from Q . The electric field E is a vector and fields due to multiple charges add like vectors.

1.5 Electric Field Lines: Multiple Charges• Drawings of electric field lines are useful visual tools. The properties of electric field lines for any charge distribution are that:• Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.• The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.• The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area

perpendicular to the lines.• The direction of the electric field is tangent to the field line at any point in space.• Field lines can never cross.

1.6 Electric Forces in Biology• Many molecules in living organisms, such as DNA, carry a charge.• An uneven distribution of the positive and negative charges within a polar molecule produces a dipole.• The effect of a Coulomb field generated by a charged object may be reduced or blocked by other nearby charged objects.• Biological systems contain water, and because water molecules are polar, they have a strong effect on other molecules in living systems.

1.7 Conductors and Electric Fields in Static Equilibrium• A conductor allows free charges to move about within it.• The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached.• Any excess charge will collect along the surface of a conductor.• Conductors with sharp corners or points will collect more charge at those points.• A lightning rod is a conductor with sharply pointed ends that collect excess charge on the building caused by an electrical storm and allow it to

dissipate back into the air.• Electrical storms result when the electrical field of Earth’s surface in certain locations becomes more strongly charged, due to changes in the

insulating effect of the air.• A Faraday cage acts like a shield around an object, preventing electric charge from penetrating inside.

1.8 Applications of Electrostatics• Electrostatics is the study of electric fields in static equilibrium.• In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including

photocopiers, laser printers, ink-jet printers and electrostatic air filters.

Conceptual Questions

1.1 Static Electricity and Charge: Conservation of Charge1. There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electricity?

2. Why do most objects tend to contain nearly equal numbers of positive and negative charges?

1.2 Conductors and Insulators3. An eccentric inventor attempts to levitate by first placing a large negative charge on himself and then putting a large positive charge on the ceilingof his workshop. Instead, while attempting to place a large negative charge on himself, his clothes fly off. Explain.

4. If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the charge of otherobjects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob?



5. When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative—yet both attract dust. Does the dust have a third type ofcharge that is attracted to both positive and negative? Explain.

6. Why does a car always attract dust right after it is polished? (Note that car wax and car tires are insulators.)

7. Describe how a positively charged object can be used to give another object a negative charge. What is the name of this process?

8. What is grounding? What effect does it have on a charged conductor? On a charged insulator?

1.3 Coulomb’s Law9. Figure 1.43 shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge.Given water’s polar character, explain what effect humidity has on removing excess charge from objects.

Figure 1.43 Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving apermanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions.

10. Using Figure 1.43, explain, in terms of Coulomb’s law, why a polar molecule (such as in Figure 1.43) is attracted by both positive and negativecharges.

11. Given the polar character of water molecules, explain how ions in the air form nucleation centers for rain droplets.

1.4 Electric Field: Concept of a Field Revisited12. Why must the test charge q in the definition of the electric field be vanishingly small?

13. Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field?

1.5 Electric Field Lines: Multiple Charges14. Compare and contrast the Coulomb force field and the electric field. To do this, make a list of five properties for the Coulomb force field analogousto the five properties listed for electric field lines. Compare each item in your list of Coulomb force field properties with those of the electric field—arethey the same or different? (For example, electric field lines cannot cross. Is the same true for Coulomb field lines?)

15. Figure 1.44 shows an electric field extending over three regions, labeled I, II, and III. Answer the following questions. (a) Are there any isolatedcharges? If so, in what region and what are their signs? (b) Where is the field strongest? (c) Where is it weakest? (d) Where is the field the mostuniform?

Figure 1.44

1.6 Electric Forces in Biology16. A cell membrane is a thin layer enveloping a cell. The thickness of the membrane is much less than the size of the cell. In a static situation the

membrane has a charge distribution of −2.5×10−6 C/m 2 on its inner surface and +2.5×10−6 C/m2 on its outer surface. Draw a diagram of the


cell and the surrounding cell membrane. Include on this diagram the charge distribution and the corresponding electric field. Is there any electric fieldinside the cell? Is there any electric field outside the cell?

1.7 Conductors and Electric Fields in Static Equilibrium17. Is the object in Figure 1.45 a conductor or an insulator? Justify your answer.

Figure 1.45

18. If the electric field lines in the figure above were perpendicular to the object, would it necessarily be a conductor? Explain.

19. The discussion of the electric field between two parallel conducting plates, in this module states that edge effects are less important if the platesare close together. What does close mean? That is, is the actual plate separation crucial, or is the ratio of plate separation to plate area crucial?

20. Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from theconductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field?(The answers to both questions have implications for charge transfer utilizing points.)

21. Why is a golfer with a metal club over her shoulder vulnerable to lightning in an open fairway? Would she be any safer under a tree?

22. Can the belt of a Van de Graaff accelerator be a conductor? Explain.

23. Are you relatively safe from lightning inside an automobile? Give two reasons.

24. Discuss pros and cons of a lightning rod being grounded versus simply being attached to a building.

25. Using the symmetry of the arrangement, show that the net Coulomb force on the charge q at the center of the square below (Figure 1.46) is

zero if the charges on the four corners are exactly equal.

Figure 1.46 Four point charges qa , qb , qc , and qd lie on the corners of a square and q is located at its center.

26. (a) Using the symmetry of the arrangement, show that the electric field at the center of the square in Figure 1.46 is zero if the charges on the fourcorners are exactly equal. (b) Show that this is also true for any combination of charges in which qa = qb and qb = qc

27. (a) What is the direction of the total Coulomb force on q in Figure 1.46 if q is negative, qa = qc and both are negative, and qb = qc and

both are positive? (b) What is the direction of the electric field at the center of the square in this situation?

28. Considering Figure 1.46, suppose that qa = qd and qb = qc . First show that q is in static equilibrium. (You may neglect the gravitational

force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction ofdisplacement of q from the center of the square.

29. If qa = 0 in Figure 1.46, under what conditions will there be no net Coulomb force on q ?

30. In regions of low humidity, one develops a special “grip” when opening car doors, or touching metal door knobs. This involves placing as much ofthe hand on the device as possible, not just the ends of one’s fingers. Discuss the induced charge and explain why this is done.

31. Tollbooth stations on roadways and bridges usually have a piece of wire stuck in the pavement before them that will touch a car as it approaches.Why is this done?

32. Suppose a woman carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? Howwould you discharge her? What are the consequences if she simply walks away?



Problems & Exercises

1.1 Static Electricity and Charge: Conservation ofCharge1. Common static electricity involves charges ranging fromnanocoulombs to microcoulombs. (a) How many electrons are needed toform a charge of –2.00 nC (b) How many electrons must be removed

from a neutral object to leave a net charge of 0.500 µC ?

2. If 1.80×1020 electrons move through a pocket calculator during a fullday’s operation, how many coulombs of charge moved through it?

3. To start a car engine, the car battery moves 3.75×1021 electronsthrough the starter motor. How many coulombs of charge were moved?

4. A certain lightning bolt moves 40.0 C of charge. How manyfundamental units of charge ∣ qe ∣ is this?

1.2 Conductors and Insulators5. Suppose a speck of dust in an electrostatic precipitator has

1.0000×1012 protons in it and has a net charge of –5.00 nC (a verylarge charge for a small speck). How many electrons does it have?

6. An amoeba has 1.00×1016 protons and a net charge of 0.300 pC.(a) How many fewer electrons are there than protons? (b) If you pairedthem up, what fraction of the protons would have no electrons?

7. A 50.0 g ball of copper has a net charge of 2.00 µC . What fraction of

the copper’s electrons has been removed? (Each copper atom has 29protons, and copper has an atomic mass of 63.5.)

8. What net charge would you place on a 100 g piece of sulfur if you put

an extra electron on 1 in 1012 of its atoms? (Sulfur has an atomic massof 32.1.)

9. How many coulombs of positive charge are there in 4.00 kg ofplutonium, given its atomic mass is 244 and that each plutonium atomhas 94 protons?

1.3 Coulomb’s Law10. What is the repulsive force between two pith balls that are 8.00 cmapart and have equal charges of – 30.0 nC?

11. (a) How strong is the attractive force between a glass rod with a0.700 μC charge and a silk cloth with a –0.600 μC charge, which are

12.0 cm apart, using the approximation that they act like point charges?(b) Discuss how the answer to this problem might be affected if thecharges are distributed over some area and do not act like point charges.

12. Two point charges exert a 5.00 N force on each other. What will theforce become if the distance between them is increased by a factor ofthree?

13. Two point charges are brought closer together, increasing the forcebetween them by a factor of 25. By what factor was their separationdecreased?

14. How far apart must two point charges of 75.0 nC (typical of staticelectricity) be to have a force of 1.00 N between them?

15. If two equal charges each of 1 C each are separated in air by adistance of 1 km, what is the magnitude of the force acting betweenthem? You will see that even at a distance as large as 1 km, the repulsiveforce is substantial because 1 C is a very significant amount of charge.

16. A test charge of +2 μC is placed halfway between a charge of

+6 μC and another of +4 μC separated by 10 cm. (a) What is the

magnitude of the force on the test charge? (b) What is the direction ofthis force (away from or toward the +6 μC charge)?

17. Bare free charges do not remain stationary when close together. Toillustrate this, calculate the acceleration of two isolated protons separated

by 2.00 nm (a typical distance between gas atoms). Explicitly show howyou follow the steps in the Problem-Solving Strategy for electrostatics.

18. (a) By what factor must you change the distance between two pointcharges to change the force between them by a factor of 10? (b) Explainhow the distance can either increase or decrease by this factor and stillcause a factor of 10 change in the force.

19. Suppose you have a total charge qtot that you can split in any

manner. Once split, the separation distance is fixed. How do you split thecharge to achieve the greatest force?

20. (a) Common transparent tape becomes charged when pulled from adispenser. If one piece is placed above another, the repulsive force canbe great enough to support the top piece’s weight. Assuming equal pointcharges (only an approximation), calculate the magnitude of the charge ifelectrostatic force is great enough to support the weight of a 10.0 mgpiece of tape held 1.00 cm above another. (b) Discuss whether themagnitude of this charge is consistent with what is typical of staticelectricity.

21. (a) Find the ratio of the electrostatic to gravitational force betweentwo electrons. (b) What is this ratio for two protons? (c) Why is the ratiodifferent for electrons and protons?

22. At what distance is the electrostatic force between two protons equalto the weight of one proton?

23. A certain five cent coin contains 5.00 g of nickel. What fraction of thenickel atoms’ electrons, removed and placed 1.00 m above it, wouldsupport the weight of this coin? The atomic mass of nickel is 58.7, andeach nickel atom contains 28 electrons and 28 protons.

24. (a) Two point charges totaling 8.00 µC exert a repulsive force of

0.150 N on one another when separated by 0.500 m. What is the chargeon each? (b) What is the charge on each if the force is attractive?

25. Point charges of 5.00 µC and –3.00 µC are placed 0.250 m apart.

(a) Where can a third charge be placed so that the net force on it is zero?(b) What if both charges are positive?

26. Two point charges q1 and q2 are 3.00 m apart, and their total

charge is 20 µC . (a) If the force of repulsion between them is 0.075N,

what are magnitudes of the two charges? (b) If one charge attracts theother with a force of 0.525N, what are the magnitudes of the twocharges? Note that you may need to solve a quadratic equation to reachyour answer.

1.4 Electric Field: Concept of a Field Revisited27. What is the magnitude and direction of an electric field that exerts a

2.00×10-5 N upward force on a –1.75 μC charge?

28. What is the magnitude and direction of the force exerted on a3.50 μC charge by a 250 N/C electric field that points due east?

29. Calculate the magnitude of the electric field 2.00 m from a pointcharge of 5.00 mC (such as found on the terminal of a Van de Graaff).

30. (a) What magnitude point charge creates a 10,000 N/C electric fieldat a distance of 0.250 m? (b) How large is the field at 10.0 m?

31. Calculate the initial (from rest) acceleration of a proton in a

5.00×106 N/C electric field (such as created by a research Van deGraaff). Explicitly show how you follow the steps in the Problem-SolvingStrategy for electrostatics.

32. (a) Find the direction and magnitude of an electric field that exerts a

4.80×10−17 N westward force on an electron. (b) What magnitudeand direction force does this field exert on a proton?

1.5 Electric Field Lines: Multiple Charges33. (a) Sketch the electric field lines near a point charge +q . (b) Do the

same for a point charge –3.00q .



34. Sketch the electric field lines a long distance from the chargedistributions shown in Figure 1.26 (a) and (b)

35. Figure 1.47 shows the electric field lines near two charges q1 and

q2 . What is the ratio of their magnitudes? (b) Sketch the electric field

lines a long distance from the charges shown in the figure.

Figure 1.47 The electric field near two charges.

36. Sketch the electric field lines in the vicinity of two opposite charges,where the negative charge is three times greater in magnitude than thepositive. (See Figure 1.47 for a similar situation).

1.7 Conductors and Electric Fields in Static Equilibrium37. Sketch the electric field lines in the vicinity of the conductor in Figure1.48 given the field was originally uniform and parallel to the object’s longaxis. Is the resulting field small near the long side of the object?

Figure 1.48

38. Sketch the electric field lines in the vicinity of the conductor in Figure1.49 given the field was originally uniform and parallel to the object’s longaxis. Is the resulting field small near the long side of the object?

Figure 1.49

39. Sketch the electric field between the two conducting plates shown inFigure 1.50, given the top plate is positive and an equal amount ofnegative charge is on the bottom plate. Be certain to indicate thedistribution of charge on the plates.

Figure 1.50

40. Sketch the electric field lines in the vicinity of the charged insulator inFigure 1.51 noting its nonuniform charge distribution.

Figure 1.51 A charged insulating rod such as might be used in a classroomdemonstration.

41. What is the force on the charge located at x = 8.00 cm in Figure1.52(a) given that q = 1.00 μC ?

Figure 1.52 (a) Point charges located at 3.00, 8.00, and 11.0 cm along the x-axis. (b)Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x-axis.

42. (a) Find the total electric field at x = 1.00 cm in Figure 1.52(b)

given that q = 5.00 nC . (b) Find the total electric field at

x = 11.00 cm in Figure 1.52(b). (c) If the charges are allowed to moveand eventually be brought to rest by friction, what will the final chargeconfiguration be? (That is, will there be a single charge, double charge,etc., and what will its value(s) be?)

43. (a) Find the electric field at x = 5.00 cm in Figure 1.52(a), given

that q = 1.00 μC . (b) At what position between 3.00 and 8.00 cm is the

total electric field the same as that for –2q alone? (c) Can the electric

field be zero anywhere between 0.00 and 8.00 cm? (d) At very largepositive or negative values of x, the electric field approaches zero in both(a) and (b). In which does it most rapidly approach zero and why? (e) Atwhat position to the right of 11.0 cm is the total electric field zero, otherthan at infinity? (Hint: A graphing calculator can yield considerable insightin this problem.)

44. (a) Find the total Coulomb force on a charge of 2.00 nC located atx = 4.00 cm in Figure 1.52 (b), given that q = 1.00 μC . (b) Find the

x-position at which the electric field is zero in Figure 1.52 (b).

45. Using the symmetry of the arrangement, determine the direction ofthe force on q in the figure below, given that qa = qb=+7.50 μC and

qc = qd = −7.50 μC . (b) Calculate the magnitude of the force on the

charge q , given that the square is 10.0 cm on a side and q = 2.00 μC .


Figure 1.53

46. (a) Using the symmetry of the arrangement, determine the directionof the electric field at the center of the square in Figure 1.53, given thatqa = qb = −1.00 μC and qc = qd=+1.00 μC . (b) Calculate the

magnitude of the electric field at the location of q , given that the square

is 5.00 cm on a side.

47. Find the electric field at the location of qa in Figure 1.53 given that

qb = qc = qd=+2.00 nC , q = −1.00 nC , and the square is 20.0 cm

on a side.

48. Find the total Coulomb force on the charge q in Figure 1.53, given

that q = 1.00 μC , qa = 2.00 μC , qb = −3.00 μC ,

qc = −4.00 μC , and qd =+1.00 μC . The square is 50.0 cm on a

side.

49. (a) Find the electric field at the location of qa in Figure 1.54, given

that qb = +10.00 μC and qc = –5.00 μC . (b) What is the force on

qa , given that qa = +1.50 nC ?

Figure 1.54 Point charges located at the corners of an equilateral triangle 25.0 cm ona side.

50. (a) Find the electric field at the center of the triangular configuration ofcharges in Figure 1.54, given that qa =+2.50 nC , qb = −8.00 nC ,

and qc =+1.50 nC . (b) Is there any combination of charges, other than

qa = qb = qc , that will produce a zero strength electric field at the

center of the triangular configuration?

1.8 Applications of Electrostatics51. (a) What is the electric field 5.00 m from the center of the terminal ofa Van de Graaff with a 3.00 mC charge, noting that the field is equivalentto that of a point charge at the center of the terminal? (b) At this distance,what force does the field exert on a 2.00 μC charge on the Van de

Graaff’s belt?

52. (a) What is the direction and magnitude of an electric field thatsupports the weight of a free electron near the surface of Earth? (b)Discuss what the small value for this field implies regarding the relativestrength of the gravitational and electrostatic forces.

53. A simple and common technique for accelerating electrons is shownin Figure 1.55, where there is a uniform electric field between two plates.Electrons are released, usually from a hot filament, near the negativeplate, and there is a small hole in the positive plate that allows theelectrons to continue moving. (a) Calculate the acceleration of the

electron if the field strength is 2.50×104 N/C . (b) Explain why theelectron will not be pulled back to the positive plate once it movesthrough the hole.

Figure 1.55 Parallel conducting plates with opposite charges on them create arelatively uniform electric field used to accelerate electrons to the right. Those that gothrough the hole can be used to make a TV or computer screen glow or to produce X-rays.

54. Earth has a net charge that produces an electric field ofapproximately 150 N/C downward at its surface. (a) What is themagnitude and sign of the excess charge, noting the electric field of aconducting sphere is equivalent to a point charge at its center? (b) Whatacceleration will the field produce on a free electron near Earth’ssurface? (c) What mass object with a single extra electron will have itsweight supported by this field?

55. Point charges of 25.0 μC and 45.0 μC are placed 0.500 m apart.

(a) At what point along the line between them is the electric field zero?(b) What is the electric field halfway between them?

56. What can you say about two charges q1 and q2 , if the electric field

one-fourth of the way from q1 to q2 is zero?

57. Integrated Concepts

Calculate the angular velocity ω of an electron orbiting a proton in the

hydrogen atom, given the radius of the orbit is 0.530×10–10 m . Youmay assume that the proton is stationary and the centripetal force issupplied by Coulomb attraction.


An electron has an initial velocity of 5.00×106 m/s in a uniform

2.00×105 N/C strength electric field. The field accelerates the electronin the direction opposite to its initial velocity. (a) What is the direction ofthe electric field? (b) How far does the electron travel before coming torest? (c) How long does it take the electron to come to rest? (d) What isthe electron’s velocity when it returns to its starting point?


The practical limit to an electric field in air is about 3.00×106 N/C .Above this strength, sparking takes place because air begins to ionize



and charges flow, reducing the field. (a) Calculate the distance a freeproton must travel in this field to reach 3.00% of the speed of light,starting from rest. (b) Is this practical in air, or must it occur in a vacuum?


A 5.00 g charged insulating ball hangs on a 30.0 cm long string in auniform horizontal electric field as shown in Figure 1.56. Given thecharge on the ball is 1.00 μC , find the strength of the field.

Figure 1.56 A horizontal electric field causes the charged ball to hang at an angle of

8.00º .


Figure 1.57 shows an electron passing between two charged metalplates that create an 100 N/C vertical electric field perpendicular to theelectron’s original horizontal velocity. (These can be used to change theelectron’s direction, such as in an oscilloscope.) The initial speed of the

electron is 3.00×106 m/s , and the horizontal distance it travels in theuniform field is 4.00 cm. (a) What is its vertical deflection? (b) What is thevertical component of its final velocity? (c) At what angle does it exit?Neglect any edge effects.

Figure 1.57


The classic Millikan oil drop experiment was the first to obtain anaccurate measurement of the charge on an electron. In it, oil drops weresuspended against the gravitational force by a vertical electric field. (SeeFigure 1.58.) Given the oil drop to be 1.00 μm in radius and have a

density of 920 kg/m3 : (a) Find the weight of the drop. (b) If the drop

has a single excess electron, find the electric field strength needed tobalance its weight.

Figure 1.58 In the Millikan oil drop experiment, small drops can be suspended in anelectric field by the force exerted on a single excess electron. Classically, thisexperiment was used to determine the electron charge qe by measuring the electric

field and mass of the drop.


(a) In Figure 1.59, four equal charges q lie on the corners of a square.

A fifth charge Q is on a mass m directly above the center of the

square, at a height equal to the length d of one side of the square.

Determine the magnitude of q in terms of Q , m , and d , if the

Coulomb force is to equal the weight of m . (b) Is this equilibrium stableor unstable? Discuss.

Figure 1.59 Four equal charges on the corners of a horizontal square support theweight of a fifth charge located directly above the center of the square.

64. Unreasonable Results

(a) Calculate the electric field strength near a 10.0 cm diameterconducting sphere that has 1.00 C of excess charge on it. (b) What isunreasonable about this result? (c) Which assumptions are responsible?


(a) Two 0.500 g raindrops in a thunderhead are 1.00 cm apart when theyeach acquire 1.00 mC charges. Find their acceleration. (b) What isunreasonable about this result? (c) Which premise or assumption isresponsible?


A wrecking yard inventor wants to pick up cars by charging a 0.400 mdiameter ball and inducing an equal and opposite charge on the car. If acar has a 1000 kg mass and the ball is to be able to lift it from a distanceof 1.00 m: (a) What minimum charge must be used? (b) What is theelectric field near the surface of the ball? (c) Why are these resultsunreasonable? (d) Which premise or assumption is responsible?

67. Construct Your Own Problem

Consider two insulating balls with evenly distributed equal and oppositecharges on their surfaces, held with a certain distance between thecenters of the balls. Construct a problem in which you calculate theelectric field (magnitude and direction) due to the balls at various pointsalong a line running through the centers of the balls and extending toinfinity on either side. Choose interesting points and comment on themeaning of the field at those points. For example, at what points mightthe field be just that due to one ball and where does the field become


negligibly small? Among the things to be considered are the magnitudesof the charges and the distance between the centers of the balls. Yourinstructor may wish for you to consider the electric field off axis or for amore complex array of charges, such as those in a water molecule.


Consider identical spherical conducting space ships in deep space wheregravitational fields from other bodies are negligible compared to thegravitational attraction between the ships. Construct a problem in whichyou place identical excess charges on the space ships to exactly countertheir gravitational attraction. Calculate the amount of excess chargeneeded. Examine whether that charge depends on the distance betweenthe centers of the ships, the masses of the ships, or any other factors.Discuss whether this would be an easy, difficult, or even impossible thingto do in practice.



2 ELECTRIC POTENTIAL AND ELECTRIC FIELD

Figure 2.1 Automated external defibrillator unit (AED) (credit: U.S. Defense Department photo/Tech. Sgt. Suzanne M. Day)

Learning Objectives2.1. Electric Potential Energy: Potential Difference

• Define electric potential and electric potential energy.• Describe the relationship between potential difference and electrical potential energy.• Explain electron volt and its usage in submicroscopic process.• Determine electric potential energy given potential difference and amount of charge.

2.2. Electric Potential in a Uniform Electric Field• Describe the relationship between voltage and electric field.• Derive an expression for the electric potential and electric field.• Calculate electric field strength given distance and voltage.

2.3. Electrical Potential Due to a Point Charge• Explain point charges and express the equation for electric potential of a point charge.• Distinguish between electric potential and electric field.• Determine the electric potential of a point charge given charge and distance.

2.4. Equipotential Lines• Explain equipotential lines and equipotential surfaces.• Describe the action of grounding an electrical appliance.• Compare electric field and equipotential lines.

2.5. Capacitors and Dielectrics• Describe the action of a capacitor and define capacitance.• Explain parallel plate capacitors and their capacitances.• Discuss the process of increasing the capacitance of a dielectric.• Determine capacitance given charge and voltage.

2.6. Capacitors in Series and Parallel• Derive expressions for total capacitance in series and in parallel.• Identify series and parallel parts in the combination of connection of capacitors.• Calculate the effective capacitance in series and parallel given individual capacitances.

2.7. Energy Stored in Capacitors• List some uses of capacitors.• Express in equation form the energy stored in a capacitor.• Explain the function of a defibrillator.

CHAPTER 2 | ELECTRIC POTENTIAL AND ELECTRIC FIELD 41

Introduction to Electric Potential and Electric Energy

In Electric Charge and Electric Field, we just scratched the surface (or at least rubbed it) of electrical phenomena. Two of the most familiar aspectsof electricity are its energy and voltage. We know, for example, that great amounts of electrical energy can be stored in batteries, are transmittedcross-country through power lines, and may jump from clouds to explode the sap of trees. In a similar manner, at molecular levels, ions cross cellmembranes and transfer information. We also know about voltages associated with electricity. Batteries are typically a few volts, the outlets in yourhome produce 120 volts, and power lines can be as high as hundreds of thousands of volts. But energy and voltage are not the same thing. Amotorcycle battery, for example, is small and would not be very successful in replacing the much larger car battery, yet each has the same voltage. Inthis chapter, we shall examine the relationship between voltage and electrical energy and begin to explore some of the many applications ofelectricity.

2.1 Electric Potential Energy: Potential DifferenceWhen a free positive charge q is accelerated by an electric field, such as shown in Figure 2.2, it is given kinetic energy. The process is analogous to

an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is convertedto kinetic energy. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric

potential energy.

Figure 2.2 A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by aforce, but since this force is conservative, we can write W = –ΔPE .

The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous

to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energyassociated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the workdirectly.

We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, ΔPE , is crucial, since the

work done by a conservative force is the negative of the change in potential energy; that is, W = –ΔPE . For example, work W done to accelerate

a positive charge from rest is positive and results from a loss in PE, or a negative ΔPE . There must be a minus sign in front of ΔPE to make Wpositive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Potential Energy

W = –ΔPE . For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative

ΔPE. There must be a minus sign in front of ΔPE to make W positive. PE can be found at any point by taking one point as a reference andcalculating the work needed to move a charge to the other point.

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force andgives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, forexample, to use the concept of voltage (related to electric potential energy) than to deal with the Coulomb force directly.

Calculating the work directly is generally difficult, since W = Fd cos θ and the direction and magnitude of F can be complex for multiple charges,

for odd-shaped objects, and along arbitrary paths. But we do know that, since F = qE , the work, and hence ΔPE , is proportional to the test

charge q. To have a physical quantity that is independent of test charge, we define electric potential V (or simply potential, since electric is

understood) to be the potential energy per unit charge:

(2.1)V = PEq .

Electric Potential

This is the electric potential energy per unit charge.

42 CHAPTER 2 | ELECTRIC POTENTIAL AND ELECTRIC FIELD


(2.2)V = PEq

Since PE is proportional to q , the dependence on q cancels. Thus V does not depend on q . The change in potential energy ΔPE is crucial,

and so we are concerned with the difference in potential or potential difference ΔV between two points, where

(2.3)ΔV = VB − VA = ΔPEq .

The potential difference between points A and B, VB – VA , is thus defined to be the change in potential energy of a charge q moved from A to

B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

(2.4)1 V = 1 JC

Potential Difference

The potential difference between points A and B, VB - VA , is defined to be the change in potential energy of a charge q moved from A to B,

divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

(2.5)1 V = 1 JC

The familiar term voltage is the common name for potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be thepotential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. Morefundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero,such as sea level or perhaps a lecture hall floor.

In summary, the relationship between potential difference (or voltage) and electrical potential energy is given by

(2.6)ΔV = ΔPEq and ΔPE = qΔV .

Potential Difference and Electrical Potential Energy

The relationship between potential difference (or voltage) and electrical potential energy is given by

(2.7)ΔV = ΔPEq and ΔPE = qΔV .

The second equation is equivalent to the first.

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus a motorcycle battery and a car battery can both have the samevoltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other sinceΔPE = qΔV . The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

Example 2.1 Calculating Energy

Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge, and a 12.0 V car battery that can move 60,000 C of charge.How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

Strategy

To say we have a 12.0 V battery means that its terminals have a 12.0 V potential difference. When such a battery moves charge, it puts thecharge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to ΔPE = qΔV .

So to find the energy output, we multiply the charge moved by the potential difference.

Solution

For the motorcycle battery, q = 5000 C and ΔV = 12.0 V . The total energy delivered by the motorcycle battery is

(2.8)ΔPEcycle = (5000 C)(12.0 V)= (5000 C)(12.0 J/C)= 6.00×104 J.

Similarly, for the car battery, q = 60,000 C and

(2.9)ΔPEcar = (60,000 C)(12.0 V)

= 7.20×105 J.Discussion


While voltage and energy are related, they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each isquite different. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as whenheadlights dim because of a low car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy isavailable for external use.

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since itloses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons fromtheir negative terminals (A) through whatever circuitry is involved and attract them to their positive terminals (B) as shown in Figure 2.3. The changein potential is ΔV = VB –VA = +12 V and the charge q is negative, so that ΔPE = qΔV is negative, meaning the potential energy of the

battery has decreased when q has moved from A to B.

Figure 2.3 A battery moves negative charge from its negative terminal through a headlight to its positive terminal. Appropriate combinations of chemicals in the batteryseparate charges so that the negative terminal has an excess of negative charge, which is repelled by it and attracted to the excess positive charge on the other terminal. Interms of potential, the positive terminal is at a higher voltage than the negative. Inside the battery, both positive and negative charges move.

Example 2.2 How Many Electrons Move through a Headlight Each Second?

When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second?

Strategy

To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy throughthe equation ΔPE = qΔV . A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have ΔPE = –30.0 J and, since

the electrons are going from the negative terminal to the positive, we see that ΔV = +12.0 V .

Solution

To find the charge q moved, we solve the equation ΔPE = qΔV :

(2.10)q = ΔPEΔV .

Entering the values for ΔPE and ΔV , we get

(2.11)q = –30.0 J+12.0 V = –30.0 J

+12.0 J/C = –2.50 C.

The number of electrons ne is the total charge divided by the charge per electron. That is,

(2.12)ne = –2.50 C–1.60×10–19 C/e– = 1.56×1019 electrons.

Discussion

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinarysystems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances werenegative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determinewhich is moving or whether both are moving.

The Electron VoltThe energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopicscale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enoughfor these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful xrays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects. Figure 2.4 shows a situation related to thedefinition of such an energy unit. An electron is accelerated between two charged metal plates as it might be in an old-model television tube or



oscilloscope. The electron is given kinetic energy that is later converted to another form—light in the television tube, for example. (Note that downhillfor the electron is uphill for a positive charge.) Since energy is related to voltage by ΔPE = qΔV , we can think of the joule as a coulomb-volt.

Figure 2.4 A typical electron gun accelerates electrons using a potential difference between two metal plates. The energy of the electron in electron volts is numerically thesame as the voltage between the plates. For example, a 5000 V potential difference produces 5000 eV electrons.

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to a fundamentalcharge accelerated through a potential difference of 1 V. In equation form,

(2.13)1 eV = ⎛⎝1.60×10–19 C⎞⎠(1 V) = ⎛⎝1.60×10–19 C⎞⎠(1 J/C)

= 1.60×10–19 J.

Electron Volt

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to afundamental charge accelerated through a potential difference of 1 V. In equation form,

(2.14)1 eV = ⎛⎝1.60×10–19 C⎞⎠(1 V) = ⎛⎝1.60×10–19 C⎞⎠(1 J/C)

= 1.60×10–19 J.

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a doublepositive charge accelerated through 100 V will be given 200 eV of energy. These simple relationships between accelerating voltage and particlecharges make the electron volt a simple and convenient energy unit in such circumstances.

Connections: Energy Units

The electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters onmodern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. Thereare, for example, calories for food energy, kilowatt-hours for electrical energy, and therms for natural gas energy.

The electron volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies areamong the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If aproton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as6000 of these molecules ( 30,000 eV ÷ 5 eV per molecule = 6000 molecules ). Nuclear decay energies are on the order of 1 MeV (1,000,000

eV) per event and can, thus, produce significant biological damage.

Conservation of EnergyThe total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as theelectrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, KE + PE = constant . A loss of PE of a chargedparticle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as

(2.15)KE + PE = constant


or

(2.16)KEi + PEi= KEf + PEf ,

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problemsolving.

Example 2.3 Electrical Potential Energy Converted to Kinetic Energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value isaccurate to three significant figures.)

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (wewill check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms

of energy to be KEi = 0, KEf = ½mv2, PEi = qV , and PEf = 0.

Solution

Conservation of energy states that

(2.17)KEi + PEi= KEf + PEf.

Entering the forms identified above, we obtain

(2.18)qV = mv2

2 .

We solve this for v :

(2.19)v = 2qV

m .

Entering values for q, V , and m gives

(2.20)

v =2⎛⎝–1.60×10–19 C⎞⎠(–100 J/C)

9.11×10–31 kg

= 5.93×106 m/s.Discussion

Note that both the charge and the initial voltage are negative, as in Figure 2.4. From the discussions in Electric Charge and Electric Field, weknow that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirmsthat the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltagesbecause of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltagesproduce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in thisexample.

2.2 Electric Potential in a Uniform Electric FieldIn the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage andelectric field. For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates,labeled A and B. (See Figure 2.5.) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal amore fundamental relationship between electric potential and electric field. From a physicist’s point of view, either ΔV or E can be used to describe

any charge distribution. ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar quantity and has no

direction, while E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity,

is represented by E below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge frompoint A to point B. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiringcalculus. We therefore look at a uniform electric field as an interesting special case.



Figure 2.5 The relationship between V and E for parallel conducting plates is E = V / d . (Note that ΔV = VAB in magnitude. For a charge that is moved from plate

A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: –ΔV = VA – VB = VAB . See the text for details.)

The work done by the electric field in Figure 2.5 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate,

lower potential, is

(2.21)W = –ΔPE = – qΔV .

The potential difference between points A and B is

(2.22)–ΔV = – (VB – VA) = VA – VB = VAB.

Entering this into the expression for work yields

(2.23)W = qVAB.

Work is W = Fd cos θ ; here cos θ = 1 , since the path is parallel to the field, and so W = Fd . Since F = qE , we see that W = qEd .

Substituting this expression for work into the previous equation gives

(2.24)qEd = qVAB.

The charge cancels, and so the voltage between points A and B is seen to be

(2.25)VAB = Ed

E = VABd

⎫⎭⎬(uniform E - field only),

where d is the distance from A to B, or the distance between the plates in Figure 2.5. Note that the above equation implies the units for electric fieldare volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:

(2.26)1 N / C = 1 V / m.

Voltage between Points A and B(2.27)VAB = Ed

E = VABd


where d is the distance from A to B, or the distance between the plates.

Example 2.4 What Is the Highest Voltage Possible between Two Plates?

Dry air will support a maximum electric field strength of about 3.0×106 V/m . Above that value, the field creates enough ionization in the air tomake the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallelconducting plates separated by 2.5 cm of dry air?


Strategy

We are given the maximum electric field E between the plates and the distance d between them. The equation VAB = Ed can thus be used

to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

(2.28)VAB = Ed.

Entering the given values for E and d gives

(2.29)VAB = (3.0×106 V/m)(0.025 m) = 7.5×104 V

or

(2.30)VAB = 75 kV.

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Discussion

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark.This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there arepoints on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that asmaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

Figure 2.6 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows aspark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough tocause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

Example 2.5 Field and Force inside an Electron Gun

(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength betweenthe plates? (b) What force would this field exert on a piece of plastic with a 0.500 μC charge that gets between the plates?

Strategy

Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression E = VABd . Once the

electric field strength is known, the force on a charge is found using F = q E . Since the electric field is in only one direction, we can write this

equation in terms of the magnitudes, F = q E .

Solution for (a)

The expression for the magnitude of the electric field between two uniform metal plates is

(2.31)E = VAB

d .

Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for VAB and

the plate separation of 0.0400 m, we obtain



(2.32)E = 25.0 kV0.0400 m = 6.25×105 V/m.

Solution for (b)

The magnitude of the force on a charge in an electric field is obtained from the equation

(2.33)F = qE.

Substituting known values gives

(2.34)F = (0.500×10–6 C)(6.25×105 V/m) = 0.313 N.

Discussion

Note that the units are newtons, since 1 V/m = 1 N/C . The force on the charge is the same no matter where the charge is located betweenthe plates. This is because the electric field is uniform between the plates.

In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on apositive charge is in the direction of E and also in the direction of lower potential V . Furthermore, the magnitude of E equals the rate of decrease

of V with distance. The faster V decreases over distance, the greater the electric field. In equation form, the general relationship between voltageand electric field is

(2.35)E = – ΔVΔs ,

where Δs is the distance over which the change in potential, ΔV , takes place. The minus sign tells us that E points in the direction of decreasingpotential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

Relationship between Voltage and Electric Field

In equation form, the general relationship between voltage and electric field is

(2.36)E = – ΔVΔs ,

where Δs is the distance over which the change in potential, ΔV , takes place. The minus sign tells us that E points in the direction ofdecreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

For continually changing potentials, ΔV and Δs become infinitesimals and differential calculus must be employed to determine the electric field.

2.3 Electrical Potential Due to a Point ChargePoint charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metalsphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider.Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q , and noting

the connection between work and potential ⎛⎝W = – qΔV ⎞⎠ , it can be shown that the electric potential V of a point charge is

(2.37)V = kQr (Point Charge),

where k is a constant equal to 9.0×109 N · m2 /C2 .

Electric Potential V of a Point Charge

The electric potential V of a point charge is given by

(2.38)V = kQr (Point Charge).

The potential at infinity is chosen to be zero. Thus V for a point charge decreases with distance, whereas E for a point charge decreases withdistance squared:

(2.39)E = Fq = kQ

r2 .

Recall that the electric potential V is a scalar and has no direction, whereas the electric field E is a vector. To find the voltage due to a combinationof point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, takingmagnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas E is closelyassociated with force, a vector.


Example 2.6 What Voltage Is Produced by a Small Charge on a Metal Sphere?

Charges in static electricity are typically in the nanocoulomb (nC) to microcoulomb ⎛⎝µC⎞⎠ range. What is the voltage 5.00 cm away from the

center of a 1-cm diameter metal sphere that has a −3.00 nC static charge?

Strategy

As we have discussed in Electric Charge and Electric Field, charge on a metal sphere spreads out uniformly and produces a field like that of apoint charge located at its center. Thus we can find the voltage using the equation V = kQ / r .

Solution

Entering known values into the expression for the potential of a point charge, we obtain

(2.40)V = kQr

= ⎛⎝9.00×109 N · m2 / C2⎞⎠

⎛⎝–3.00×10–9 C

5.00×10–2 m⎞⎠

= –540 V.Discussion

The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative)than at larger distances. Conversely, a negative charge would be repelled, as expected.

Example 2.7 What Is the Excess Charge on a Van de Graaff Generator

A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (See Figure2.7.) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

Figure 2.7 The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as areference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.

Strategy

The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is12.5 cm.) We can thus determine the excess charge using the equation

(2.41)V = kQr .

Solution

Solving for Q and entering known values gives

(2.42)Q = rVk

=(0.125 m)⎛⎝100×103 V⎞⎠9.00×109 N · m2 / C2

= 1.39×10–6 C = 1.39 µC.

Discussion



This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolatedcharges.

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Groundpotential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is ofimportance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted inElectric Potential Energy: Potential Difference, this is analogous to taking sea level as h = 0 when considering gravitational potential energy,

PEg = mgh .

2.4 Equipotential LinesWe can represent electric potentials (voltages) pictorially, just as we drew pictures to illustrate electric fields. Of course, the two are related. ConsiderFigure 2.8, which shows an isolated positive point charge and its electric field lines. Electric field lines radiate out from a positive charge andterminate on negative charges. While we use blue arrows to represent the magnitude and direction of the electric field, we use green lines torepresent places where the electric potential is constant. These are called equipotential lines in two dimensions, or equipotential surfaces in threedimensions. The term equipotential is also used as a noun, referring to an equipotential line or surface. The potential for a point charge is the sameanywhere on an imaginary sphere of radius r surrounding the charge. This is true since the potential for a point charge is given by V = kQ / r and,

thus, has the same value at any point that is a given distance r from the charge. An equipotential sphere is a circle in the two-dimensional view ofFigure 2.8. Since the electric field lines point radially away from the charge, they are perpendicular to the equipotential lines.

Figure 2.8 An isolated point charge Q with its electric field lines in blue and equipotential lines in green. The potential is the same along each equipotential line, meaning that

no work is required to move a charge anywhere along one of those lines. Work is needed to move a charge from one equipotential line to another. Equipotential lines areperpendicular to electric field lines in every case.

It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along anequipotential, since ΔV = 0 . Thus the work is

(2.43)W = –ΔPE = – qΔV = 0.

Work is zero if force is perpendicular to motion. Force is in the same direction as E , so that motion along an equipotential must be perpendicular to

E . More precisely, work is related to the electric field by

(2.44)W = Fd cos θ = qEd cos θ = 0.

Note that in the above equation, E and F symbolize the magnitudes of the electric field strength and force, respectively. Neither q nor E nor d is

zero, and so cos θ must be 0, meaning θ must be 90º . In other words, motion along an equipotential is perpendicular to E .

One of the rules for static electric fields and conductors is that the electric field must be perpendicular to the surface of any conductor. This impliesthat a conductor is an equipotential surface in static situations. There can be no voltage difference across the surface of a conductor, or charges willflow. One of the uses of this fact is that a conductor can be fixed at zero volts by connecting it to the earth with a good conductor—a process calledgrounding. Grounding can be a useful safety tool. For example, grounding the metal case of an electrical appliance ensures that it is at zero voltsrelative to the earth.

Grounding

A conductor can be fixed at zero volts by connecting it to the earth with a good conductor—a process called grounding.

Because a conductor is an equipotential, it can replace any equipotential surface. For example, in Figure 2.8 a charged spherical conductor canreplace the point charge, and the electric field and potential surfaces outside of it will be unchanged, confirming the contention that a spherical chargedistribution is equivalent to a point charge at its center.

Figure 2.9 shows the electric field and equipotential lines for two equal and opposite charges. Given the electric field lines, the equipotential lines canbe drawn simply by making them perpendicular to the electric field lines. Conversely, given the equipotential lines, as in Figure 2.10(a), the electricfield lines can be drawn by making them perpendicular to the equipotentials, as in Figure 2.10(b).


Figure 2.9 The electric field lines and equipotential lines for two equal but opposite charges. The equipotential lines can be drawn by making them perpendicular to the electricfield lines, if those are known. Note that the potential is greatest (most positive) near the positive charge and least (most negative) near the negative charge.

Figure 2.10 (a) These equipotential lines might be measured with a voltmeter in a laboratory experiment. (b) The corresponding electric field lines are found by drawing themperpendicular to the equipotentials. Note that these fields are consistent with two equal negative charges.

One of the most important cases is that of the familiar parallel conducting plates shown in Figure 2.11. Between the plates, the equipotentials areevenly spaced and parallel. The same field could be maintained by placing conducting plates at the equipotential lines at the potentials shown.

Figure 2.11 The electric field and equipotential lines between two metal plates.

An important application of electric fields and equipotential lines involves the heart. The heart relies on electrical signals to maintain its rhythm. Themovement of electrical signals causes the chambers of the heart to contract and relax. When a person has a heart attack, the movement of theseelectrical signals may be disturbed. An artificial pacemaker and a defibrillator can be used to initiate the rhythm of electrical signals. The equipotentiallines around the heart, the thoracic region, and the axis of the heart are useful ways of monitoring the structure and functions of the heart. Anelectrocardiogram (ECG) measures the small electric signals being generated during the activity of the heart. More about the relationship betweenelectric fields and the heart is discussed in Energy Stored in Capacitors.

PhET Explorations: Charges and Fields

Move point charges around on the playing field and then view the electric field, voltages, equipotential lines, and more. It's colorful, it's dynamic,it's free.



Figure 2.12 Charges and Fields (http://cnx.org/content/m42331/1.3/charges-and-fields_en.jar)

2.5 Capacitors and DielectricsA capacitor is a device used to store electric charge. Capacitors have applications ranging from filtering static out of radio reception to energystorage in heart defibrillators. Typically, commercial capacitors have two conducting parts close to one another, but not touching, such as those inFigure 2.13. (Most of the time an insulator is used between the two plates to provide separation—see the discussion on dielectrics below.) Whenbattery terminals are connected to an initially uncharged capacitor, equal amounts of positive and negative charge, +Q and – Q , are separated

into its two plates. The capacitor remains neutral overall, but we refer to it as storing a charge Q in this circumstance.

Capacitor

A capacitor is a device used to store electric charge.

Figure 2.13 Both capacitors shown here were initially uncharged before being connected to a battery. They now have separated charges of +Q and – Q on their two

halves. (a) A parallel plate capacitor. (b) A rolled capacitor with an insulating material between its two conducting sheets.

The amount of charge Q a capacitor can store depends on two major factors—the voltage applied and the capacitor’s physical characteristics, such

as its size.

The Amount of Charge Q a Capacitor Can Store

The amount of charge Q a capacitor can store depends on two major factors—the voltage applied and the capacitor’s physical characteristics,

such as its size.

A system composed of two identical, parallel conducting plates separated by a distance, as in Figure 2.14, is called a parallel plate capacitor. It iseasy to see the relationship between the voltage and the stored charge for a parallel plate capacitor, as shown in Figure 2.14. Each electric field linestarts on an individual positive charge and ends on a negative one, so that there will be more field lines if there is more charge. (Drawing a single fieldline per charge is a convenience, only. We can draw many field lines for each charge, but the total number is proportional to the number of charges.)The electric field strength is, thus, directly proportional to Q .


http://cnx.org/content/m42331/1.3/charges-and-fields_en.jar

Figure 2.14 Electric field lines in this parallel plate capacitor, as always, start on positive charges and end on negative charges. Since the electric field strength is proportionalto the density of field lines, it is also proportional to the amount of charge on the capacitor.

The field is proportional to the charge:

(2.45)E ∝ Q,

where the symbol ∝ means “proportional to.” From the discussion in Electric Potential in a Uniform Electric Field, we know that the voltage

across parallel plates is V = Ed . Thus,

(2.46)V ∝ E.

It follows, then, that V ∝ Q , and conversely,

(2.47)Q ∝ V .

This is true in general: The greater the voltage applied to any capacitor, the greater the charge stored in it.

Different capacitors will store different amounts of charge for the same applied voltage, depending on their physical characteristics. We define theircapacitance C to be such that the charge Q stored in a capacitor is proportional to C . The charge stored in a capacitor is given by

(2.48)Q = CV.

This equation expresses the two major factors affecting the amount of charge stored. Those factors are the physical characteristics of the capacitor,C , and the voltage, V . Rearranging the equation, we see that capacitance C is the amount of charge stored per volt, or

(2.49)C = QV .

Capacitance

Capacitance C is the amount of charge stored per volt, or

(2.50)C = QV .

The unit of capacitance is the farad (F), named for Michael Faraday (1791–1867), an English scientist who contributed to the fields ofelectromagnetism and electrochemistry. Since capacitance is charge per unit voltage, we see that a farad is a coulomb per volt, or

(2.51)1 F = 1 C1 V.

A 1-farad capacitor would be able to store 1 coulomb (a very large amount of charge) with the application of only 1 volt. One farad is, thus, a very

large capacitance. Typical capacitors range from fractions of a picofarad ⎛⎝1 pF = 10–12 F⎞⎠ to millifarads ⎛⎝1 mF = 10–3 F⎞⎠ .



Figure 2.15 shows some common capacitors. Capacitors are primarily made of ceramic, glass, or plastic, depending upon purpose and size.Insulating materials, called dielectrics, are commonly used in their construction, as discussed below.

Figure 2.15 Some typical capacitors. Size and value of capacitance are not necessarily related. (credit: Windell Oskay)

Parallel Plate Capacitor

The parallel plate capacitor shown in Figure 2.16 has two identical conducting plates, each having a surface area A , separated by a distance d(with no material between the plates). When a voltage V is applied to the capacitor, it stores a charge Q , as shown. We can see how its

capacitance depends on A and d by considering the characteristics of the Coulomb force. We know that like charges repel, unlike charges attract,and the force between charges decreases with distance. So it seems quite reasonable that the bigger the plates are, the more charge they canstore—because the charges can spread out more. Thus C should be greater for larger A . Similarly, the closer the plates are together, the greater

the attraction of the opposite charges on them. So C should be greater for smaller d .

Figure 2.16 Parallel plate capacitor with plates separated by a distance d . Each plate has an area A .

It can be shown that for a parallel plate capacitor there are only two factors ( A and d ) that affect its capacitance C . The capacitance of a parallelplate capacitor in equation form is given by

(2.52)C = ε0Ad .

Capacitance of a Parallel Plate Capacitor(2.53)C = ε0

Ad

A is the area of one plate in square meters, and d is the distance between the plates in meters. The constant ε0 is the permittivity of free space;

its numerical value in SI units is ε0 = 8.85×10 – 12 F/m . The units of F/m are equivalent to C2 /N · m2 . The small numerical value of ε0 is

related to the large size of the farad. A parallel plate capacitor must have a large area to have a capacitance approaching a farad. (Note that theabove equation is valid when the parallel plates are separated by air or free space. When another material is placed between the plates, the equationis modified, as discussed below.)


Example 2.8 Capacitance and Charge Stored in a Parallel Plate Capacitor

(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.00 m2 , separated by 1.00 mm? (b) What charge is

stored in this capacitor if a voltage of 3.00×103 V is applied to it?

Strategy

Finding the capacitance C is a straightforward application of the equation C = ε0A / d . Once C is found, the charge stored can be found

using the equation Q = CV .

Solution for (a)

Entering the given values into the equation for the capacitance of a parallel plate capacitor yields

(2.54)C = ε0

Ad = ⎛⎝8.85×10–12 F

m⎞⎠ 1.00 m2

1.00×10–3 m= 8.85×10–9 F = 8.85 nF.

Discussion for (a)

This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as usingvery large area thin foils placed close together.

Solution for (b)

The charge stored in any capacitor is given by the equation Q = CV . Entering the known values into this equation gives

(2.55)Q = CV = ⎛⎝8.85×10–9 F⎞⎠⎛⎝3.00×103 V⎞⎠

= 26.6 μC.

Discussion for (b)

This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about 3.00×106 V/m , more chargecannot be stored on this capacitor by increasing the voltage.

Another interesting biological example dealing with electric potential is found in the cell’s plasma membrane. The membrane sets a cell off from itssurroundings and also allows ions to selectively pass in and out of the cell. There is a potential difference across the membrane of about –70 mV .

This is due to the mainly negatively charged ions in the cell and the predominance of positively charged sodium ( Na+ ) ions outside. Things change

when a nerve cell is stimulated. Na+ ions are allowed to pass through the membrane into the cell, producing a positive membrane potential—thenerve signal. The cell membrane is about 7 to 10 nm thick. An approximate value of the electric field across it is given by

(2.56)E = V

d = –70×10–3 V8×10–9 m

= –9×106 V/m.

This electric field is enough to cause a breakdown in air.

Dielectric

The previous example highlights the difficulty of storing a large amount of charge in capacitors. If d is made smaller to produce a larger capacitance,

then the maximum voltage must be reduced proportionally to avoid breakdown (since E = V / d ). An important solution to this difficulty is to put an

insulating material, called a dielectric, between the plates of a capacitor and allow d to be as small as possible. Not only does the smaller d makethe capacitance greater, but many insulators can withstand greater electric fields than air before breaking down.

There is another benefit to using a dielectric in a capacitor. Depending on the material used, the capacitance is greater than that given by the

equation C = ε0Ad by a factor κ , called the dielectric constant. A parallel plate capacitor with a dielectric between its plates has a capacitance

given by

(2.57)C = κε0Ad (parallel plate capacitor with dielectric).

Values of the dielectric constant κ for various materials are given in Table 2.1. Note that κ for vacuum is exactly 1, and so the above equation isvalid in that case, too. If a dielectric is used, perhaps by placing Teflon between the plates of the capacitor in Example 2.8, then the capacitance isgreater by the factor κ , which for Teflon is 2.1.

Take-Home Experiment: Building a Capacitor

How large a capacitor can you make using a chewing gum wrapper? The plates will be the aluminum foil, and the separation (dielectric) inbetween will be the paper.



Table 2.1 Dielectric Constants and Dielectric Strengths for Various Materialsat 20ºC

Material Dielectric constant κ Dielectric strength (V/m)

Vacuum 1.00000 —

Air 1.00059 3×106

Bakelite 4.9 24×106

Fused quartz 3.78 8×106

Neoprene rubber 6.7 12×106

Nylon 3.4 14×106

Paper 3.7 16×106

Polystyrene 2.56 24×106

Pyrex glass 5.6 14×106

Silicon oil 2.5 15×106

Strontium titanate 233 8×106

Teflon 2.1 60×106

Water 80 —

Note also that the dielectric constant for air is very close to 1, so that air-filled capacitors act much like those with vacuum between their plates exceptthat the air can become conductive if the electric field strength becomes too great. (Recall that E = V / d for a parallel plate capacitor.) Also shownin Table 2.1 are maximum electric field strengths in V/m, called dielectric strengths, for several materials. These are the fields above which thematerial begins to break down and conduct. The dielectric strength imposes a limit on the voltage that can be applied for a given plate separation. Forinstance, in Example 2.8, the separation is 1.00 mm, and so the voltage limit for air is

(2.58)V = E ⋅ d= (3×106 V/m)(1.00×10−3 m)= 3000 V.

However, the limit for a 1.00 mm separation filled with Teflon is 60,000 V, since the dielectric strength of Teflon is 60×106 V/m. So the samecapacitor filled with Teflon has a greater capacitance and can be subjected to a much greater voltage. Using the capacitance we calculated in theabove example for the air-filled parallel plate capacitor, we find that the Teflon-filled capacitor can store a maximum charge of

(2.59)Q = CV= κCair V

= (2.1)(8.85 nF)(6.0×104 V)= 1.1 mC.

This is 42 times the charge of the same air-filled capacitor.

Dielectric Strength

The maximum electric field strength above which an insulating material begins to break down and conduct is called its dielectric strength.

Microscopically, how does a dielectric increase capacitance? Polarization of the insulator is responsible. The more easily it is polarized, the greater itsdielectric constant κ . Water, for example, is a polar molecule because one end of the molecule has a slight positive charge and the other end has aslight negative charge. The polarity of water causes it to have a relatively large dielectric constant of 80. The effect of polarization can be bestexplained in terms of the characteristics of the Coulomb force. Figure 2.17 shows the separation of charge schematically in the molecules of adielectric material placed between the charged plates of a capacitor. The Coulomb force between the closest ends of the molecules and the chargeon the plates is attractive and very strong, since they are very close together. This attracts more charge onto the plates than if the space were emptyand the opposite charges were a distance d away.


Figure 2.17 (a) The molecules in the insulating material between the plates of a capacitor are polarized by the charged plates. This produces a layer of opposite charge on thesurface of the dielectric that attracts more charge onto the plate, increasing its capacitance. (b) The dielectric reduces the electric field strength inside the capacitor, resulting ina smaller voltage between the plates for the same charge. The capacitor stores the same charge for a smaller voltage, implying that it has a larger capacitance because of thedielectric.

Another way to understand how a dielectric increases capacitance is to consider its effect on the electric field inside the capacitor. Figure 2.17(b)shows the electric field lines with a dielectric in place. Since the field lines end on charges in the dielectric, there are fewer of them going from oneside of the capacitor to the other. So the electric field strength is less than if there were a vacuum between the plates, even though the same chargeis on the plates. The voltage between the plates is V = Ed , so it too is reduced by the dielectric. Thus there is a smaller voltage V for the same

charge Q ; since C = Q / V , the capacitance C is greater.

The dielectric constant is generally defined to be κ = E0 / E , or the ratio of the electric field in a vacuum to that in the dielectric material, and is

intimately related to the polarizability of the material.

Things Great and Small

The Submicroscopic Origin of Polarization

Polarization is a separation of charge within an atom or molecule. As has been noted, the planetary model of the atom pictures it as having apositive nucleus orbited by negative electrons, analogous to the planets orbiting the Sun. Although this model is not completely accurate, it isvery helpful in explaining a vast range of phenomena and will be refined elsewhere, such as in Atomic Physics (http://cnx.org/content/m42585/latest/) . The submicroscopic origin of polarization can be modeled as shown in Figure 2.18.

Figure 2.18 Artist’s conception of a polarized atom. The orbits of electrons around the nucleus are shifted slightly by the external charges (shown exaggerated). The resultingseparation of charge within the atom means that it is polarized. Note that the unlike charge is now closer to the external charges, causing the polarization.





We will find in Atomic Physics (http://cnx.org/content/m42585/latest/) that the orbits of electrons are more properly viewed as electron clouds withthe density of the cloud related to the probability of finding an electron in that location (as opposed to the definite locations and paths of planets intheir orbits around the Sun). This cloud is shifted by the Coulomb force so that the atom on average has a separation of charge. Although the atomremains neutral, it can now be the source of a Coulomb force, since a charge brought near the atom will be closer to one type of charge than theother.

Some molecules, such as those of water, have an inherent separation of charge and are thus called polar molecules. Figure 2.19 illustrates theseparation of charge in a water molecule, which has two hydrogen atoms and one oxygen atom ⎛

⎝H2 O⎞⎠ . The water molecule is not symmetric—the

hydrogen atoms are repelled to one side, giving the molecule a boomerang shape. The electrons in a water molecule are more concentrated aroundthe more highly charged oxygen nucleus than around the hydrogen nuclei. This makes the oxygen end of the molecule slightly negative and leavesthe hydrogen ends slightly positive. The inherent separation of charge in polar molecules makes it easier to align them with external fields andcharges. Polar molecules therefore exhibit greater polarization effects and have greater dielectric constants. Those who study chemistry will find thatthe polar nature of water has many effects. For example, water molecules gather ions much more effectively because they have an electric field anda separation of charge to attract charges of both signs. Also, as brought out in the previous chapter, polar water provides a shield or screening of theelectric fields in the highly charged molecules of interest in biological systems.

Figure 2.19 Artist’s conception of a water molecule. There is an inherent separation of charge, and so water is a polar molecule. Electrons in the molecule are attracted to theoxygen nucleus and leave an excess of positive charge near the two hydrogen nuclei. (Note that the schematic on the right is a rough illustration of the distribution of electronsin the water molecule. It does not show the actual numbers of protons and electrons involved in the structure.)

PhET Explorations: Capacitor Lab

Explore how a capacitor works! Change the size of the plates and add a dielectric to see the effect on capacitance. Change the voltage and seecharges built up on the plates. Observe the electric field in the capacitor. Measure the voltage and the electric field.

Figure 2.20 Capacitor Lab (http://cnx.org/content/m42333/1.4/capacitor-lab_en.jar)

2.6 Capacitors in Series and ParallelSeveral capacitors may be connected together in a variety of applications. Multiple connections of capacitors act like a single equivalent capacitor.The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simpleand common types of connections, called series and parallel, for which we can easily calculate the total capacitance. Certain more complicatedconnections can also be related to combinations of series and parallel.

Capacitance in SeriesFigure 2.21(a) shows a series connection of three capacitors with a voltage applied. As for any capacitor, the capacitance of the combination is

related to charge and voltage by C = QV .

Note in Figure 2.21 that opposite charges of magnitude Q flow to either side of the originally uncharged combination of capacitors when the voltage

V is applied. Conservation of charge requires that equal-magnitude charges be created on the plates of the individual capacitors, since charge isonly being separated in these originally neutral devices. The end result is that the combination resembles a single capacitor with an effective plateseparation greater than that of the individual capacitors alone. (See Figure 2.21(b).) Larger plate separation means smaller capacitance. It is ageneral feature of series connections of capacitors that the total capacitance is less than any of the individual capacitances.



http://cnx.org/content/m42333/1.4/capacitor-lab_en.jar

Figure 2.21 (a) Capacitors connected in series. The magnitude of the charge on each plate is Q . (b) An equivalent capacitor has a larger plate separation d . Series

connections produce a total capacitance that is less than that of any of the individual capacitors.

We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown in Figure 2.21. Solving

C = QV for V gives V = Q

C . The voltages across the individual capacitors are thus V1 = QC1

, V2 = QC2

, and V3 = QC3

. The total voltage is

the sum of the individual voltages:

(2.60)V = V1 + V2 + V3.

Now, calling the total capacitance CS for series capacitance, consider that

(2.61)V = QCS

= V1 + V2 + V3 .

Entering the expressions for V1 , V2 , and V3 , we get

(2.62)QCS

= QC1

+ QC2

+ QC3

.

Canceling the Q s, we obtain the equation for the total capacitance in series CS to be

(2.63)1CS

= 1C1

+ 1C2

+ 1C3

+ ...,

where “...” indicates that the expression is valid for any number of capacitors connected in series. An expression of this form always results in a totalcapacitance CS that is less than any of the individual capacitances C1 , C2 , ..., as the next example illustrates.

Total Capacitance in Series, Cs

Total capacitance in series: 1CS

= 1C1

+ 1C2

+ 1C3

+ ...



Example 2.9 What Is the Series Capacitance?

Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1.000, 5.000, and 8.000 µF .

Strategy

With the given information, the total capacitance can be found using the equation for capacitance in series.

Solution

Entering the given capacitances into the expression for 1CS

gives 1CS

= 1C1

+ 1C2

+ 1C3

.

(2.64)1CS

= 11.000 µF + 1

5.000 µF + 18.000 µF = 1.325

µF

Inverting to find CS yields CS = µF1.325 = 0.755 µF .

Discussion

The total series capacitance Cs is less than the smallest individual capacitance, as promised. In series connections of capacitors, the sum is

less than the parts. In fact, it is less than any individual. Note that it is sometimes possible, and more convenient, to solve an equation like theabove by finding the least common denominator, which in this case (showing only whole-number calculations) is 40. Thus,

(2.65)1CS

= 4040 µF + 8

40 µF + 540 µF = 53

40 µF,

so that

(2.66)CS = 40 µF

53 = 0.755 µF.

Capacitors in ParallelFigure 2.22(a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the seriescase. To find the equivalent total capacitance Cp , we first note that the voltage across each capacitor is V , the same as that of the source, since

they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as thatacross the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source.The total charge Q is the sum of the individual charges:

(2.67)Q = Q1 + Q2 + Q3.

Figure 2.22 (a) Capacitors in parallel. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of theindividual capacitances. (b) The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors.

Using the relationship Q = CV , we see that the total charge is Q = CpV , and the individual charges are Q1 = C1V , Q2 = C2V , and

Q3 = C3V . Entering these into the previous equation gives

(2.68)Cp V = C1V + C2V + C3V .

Canceling V from the equation, we obtain the equation for the total capacitance in parallel Cp :

(2.69)Cp = C1 + C2 + C3 + ....


Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “...” indicates the expression is valid for any number ofcapacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

(2.70)Cp = 1.000 µF+5.000 µF+8.000 µF = 14.000 µF.

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in Figure 2.22(b).

Total Capacitance in Parallel, Cp

Total capacitance in parallel Cp = C1 + C2 + C3 + ...

More complicated connections of capacitors can sometimes be combinations of series and parallel. (See Figure 2.23.) To find the total capacitanceof such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Figure 2.23 (a) This circuit contains both series and parallel connections of capacitors. See Example 2.10 for the calculation of the overall capacitance of the circuit. (b) C1and C2 are in series; their equivalent capacitance CS is less than either of them. (c) Note that CS is in parallel with C3 . The total capacitance is, thus, the sum of CSand C3 .

Example 2.10 A Mixture of Series and Parallel Capacitance

Find the total capacitance of the combination of capacitors shown in Figure 2.23. Assume the capacitances in Figure 2.23 are known to threedecimal places ( C1 = 1.000 µF , C2 = 3.000 µF , and C3 = 8.000 µF ), and round your answer to three decimal places.

Strategy

To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors C1 and C2 are in series.

Their combination, labeled CS in the figure, is in parallel with C3 .

Solution

Since C1 and C2 are in series, their total capacitance is given by 1CS

= 1C1

+ 1C2

+ 1C3

. Entering their values into the equation gives

(2.71)1CS

= 1C1

+ 1C2

= 11.000 μF + 1

5.000 μF = 1.200μF .

Inverting gives

(2.72)CS = 0.833 µF.

This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum

(2.73)Ctot = CS + CS= 0.833 μF + 8.000 μF= 8.833 μF.

Discussion

This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations ofcapacitors.

2.7 Energy Stored in CapacitorsMost of us have seen dramatizations in which medical personnel use a defibrillator to pass an electric current through a patient’s heart to get it tobeat normally. (Review Figure 2.24.) Often realistic in detail, the person applying the shock directs another person to “make it 400 joules this time.”The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit the situation. SI units of joules are often employed. Less



dramatic is the use of capacitors in microelectronics, such as certain handheld calculators, to supply energy when batteries are charged. (See Figure2.24.) Capacitors are also used to supply energy for flash lamps on cameras.

Figure 2.24 Energy stored in the large capacitor is used to preserve the memory of an electronic calculator when its batteries are charged. (credit: Kucharek, WikimediaCommons)

Energy stored in a capacitor is electrical potential energy, and it is thus related to the charge Q and voltage V on the capacitor. We must be careful

when applying the equation for electrical potential energy ΔPE = qΔV to a capacitor. Remember that ΔPE is the potential energy of a charge qgoing through a voltage ΔV . But the capacitor starts with zero voltage and gradually comes up to its full voltage as it is charged. The first charge

placed on a capacitor experiences a change in voltage ΔV = 0 , since the capacitor has zero voltage when uncharged. The final charge placed on a

capacitor experiences ΔV = V , since the capacitor now has its full voltage V on it. The average voltage on the capacitor during the charging

process is V / 2 , and so the average voltage experienced by the full charge q is V / 2 . Thus the energy stored in a capacitor, Ecap , is

(2.74)Ecap = QV2 ,

where Q is the charge on a capacitor with a voltage V applied. (Note that the energy is not QV , but QV / 2 .) Charge and voltage are related to

the capacitance C of a capacitor by Q = CV , and so the expression for Ecap can be algebraically manipulated into three equivalent expressions:

(2.75)Ecap = QV

2 = CV 2

2 = Q2

2C ,

where Q is the charge and V the voltage on a capacitor C . The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in

farads.

Energy Stored in Capacitors

The energy stored in a capacitor can be expressed in three ways:

(2.76)Ecap = QV

2 = CV 2

2 = Q2

2C ,

where Q is the charge, V is the voltage, and C is the capacitance of the capacitor. The energy is in joules for a charge in coulombs, voltage

in volts, and capacitance in farads.

In a defibrillator, the delivery of a large charge in a short burst to a set of paddles across a person’s chest can be a lifesaver. The person’s heartattack might have arisen from the onset of fast, irregular beating of the heart—cardiac or ventricular fibrillation. The application of a large shock ofelectrical energy can terminate the arrhythmia and allow the body’s pacemaker to resume normal patterns. Today it is common for ambulances tocarry a defibrillator, which also uses an electrocardiogram to analyze the patient’s heartbeat pattern. Automated external defibrillators (AED) arefound in many public places (Figure 2.25). These are designed to be used by lay persons. The device automatically diagnoses the patient’s heartcondition and then applies the shock with appropriate energy and waveform. CPR is recommended in many cases before use of an AED.


capacitance:

capacitor:

defibrillator:

dielectric strength:

dielectric:

electric potential:

electron volt:

equipotential line:

grounding:

mechanical energy:

Figure 2.25 Automated external defibrillators are found in many public places. These portable units provide verbal instructions for use in the important first few minutes for aperson suffering a cardiac attack. (credit: Owain Davies, Wikimedia Commons)

Example 2.11 Capacitance in a Heart Defibrillator

A heart defibrillator delivers 4.00×102 J of energy by discharging a capacitor initially at 1.00×104 V . What is its capacitance?

Strategy

We are given Ecap and V , and we are asked to find the capacitance C . Of the three expressions in the equation for Ecap , the most

convenient relationship is

(2.77)Ecap = CV 2

2 .

Solution

Solving this expression for C and entering the given values yields

(2.78)C =

2Ecap

V 2 = 2(4.00×102 J)(1.00×104 V)2 = 8.00×10 – 6 F

= 8.00 µF.

Discussion

This is a fairly large, but manageable, capacitance at 1.00×104 V .

Glossaryamount of charge stored per unit volt

a device that stores electric charge

a machine used to provide an electrical shock to a heart attack victim's heart in order to restore the heart's normal rhythmic pattern

the maximum electric field above which an insulating material begins to break down and conduct

an insulating material

potential energy per unit charge

the energy given to a fundamental charge accelerated through a potential difference of one volt

a line along which the electric potential is constant

fixing a conductor at zero volts by connecting it to the earth or ground

sum of the kinetic energy and potential energy of a system; this sum is a constant



parallel plate capacitor:

polar molecule:

potential difference (or voltage):

scalar:

vector:

two identical conducting plates separated by a distance

a molecule with inherent separation of charge

change in potential energy of a charge moved from one point to another, divided by the charge; units ofpotential difference are joules per coulomb, known as volt

physical quantity with magnitude but no direction

physical quantity with both magnitude and direction

Section Summary

2.1 Electric Potential Energy: Potential Difference• Electric potential is potential energy per unit charge.• The potential difference between points A and B, VB – VA , defined to be the change in potential energy of a charge q moved from A to B, is

equal to the change in potential energy divided by the charge, Potential difference is commonly called voltage, represented by the symbol ΔV .

ΔV = ΔPEq and ΔPE = qΔV .

• An electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

1 eV = ⎛⎝1.60×10–19 C⎞⎠(1 V) = ⎛⎝1.60×10–19 C⎞⎠(1 J/C)

= 1.60×10–19 J.• Mechanical energy is the sum of the kinetic energy and potential energy of a system, that is, KE + PE. This sum is a constant.

2.2 Electric Potential in a Uniform Electric Field• The voltage between points A and B is

VAB = Ed

E = VABd


where d is the distance from A to B, or the distance between the plates.• In equation form, the general relationship between voltage and electric field is

E = – ΔVΔs ,

where Δs is the distance over which the change in potential, ΔV , takes place. The minus sign tells us that E points in the direction ofdecreasing potential.) The electric field is said to be the gradient (as in grade or slope) of the electric potential.

2.3 Electrical Potential Due to a Point Charge• Electric potential of a point charge is V = kQ / r .

• Electric potential is a scalar, and electric field is a vector. Addition of voltages as numbers gives the voltage due to a combination of pointcharges, whereas addition of individual fields as vectors gives the total electric field.

2.4 Equipotential Lines• An equipotential line is a line along which the electric potential is constant.• An equipotential surface is a three-dimensional version of equipotential lines.• Equipotential lines are always perpendicular to electric field lines.• The process by which a conductor can be fixed at zero volts by connecting it to the earth with a good conductor is called grounding.

2.5 Capacitors and Dielectrics• A capacitor is a device used to store charge.• The amount of charge Q a capacitor can store depends on two major factors—the voltage applied and the capacitor’s physical characteristics,

such as its size.• The capacitance C is the amount of charge stored per volt, or

C = QV .

• The capacitance of a parallel plate capacitor is C = ε0Ad , when the plates are separated by air or free space. ε0 is called the permittivity of

free space.• A parallel plate capacitor with a dielectric between its plates has a capacitance given by

C = κε0Ad ,

where κ is the dielectric constant of the material.• The maximum electric field strength above which an insulating material begins to break down and conduct is called dielectric strength.

2.6 Capacitors in Series and Parallel


• Total capacitance in series 1CS

= 1C1

+ 1C2

+ 1C3

+ ...

• Total capacitance in parallel Cp = C1 + C2 + C3 + ...• If a circuit contains a combination of capacitors in series and parallel, identify series and parallel parts, compute their capacitances, and then

find the total.

2.7 Energy Stored in Capacitors• Capacitors are used in a variety of devices, including defibrillators, microelectronics such as calculators, and flash lamps, to supply energy.• The energy stored in a capacitor can be expressed in three ways:

Ecap = QV2 = CV 2

2 = Q2

2C ,

where Q is the charge, V is the voltage, and C is the capacitance of the capacitor. The energy is in joules when the charge is in coulombs,

voltage is in volts, and capacitance is in farads.


2.1 Electric Potential Energy: Potential Difference1. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?

2. If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be donewithout exerting a force? Explain.

3. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potentialenergy?

4. Voltages are always measured between two points. Why?

5. How are units of volts and electron volts related? How do they differ?

2.2 Electric Potential in a Uniform Electric Field6. Discuss how potential difference and electric field strength are related. Give an example.

7. What is the strength of the electric field in a region where the electric potential is constant?

8. Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.

2.3 Electrical Potential Due to a Point Charge9. In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from thatof a point charge?

10. Can the potential of a non-uniformly charged sphere be the same as that of a point charge? Explain.

2.4 Equipotential Lines11. What is an equipotential line? What is an equipotential surface?

12. Explain in your own words why equipotential lines and surfaces must be perpendicular to electric field lines.

13. Can different equipotential lines cross? Explain.

2.5 Capacitors and Dielectrics14. Does the capacitance of a device depend on the applied voltage? What about the charge stored in it?

15. Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the plate area of a capacitor. Similarly, explainwhy capacitance should be inversely proportional to the separation between plates.

16. Give the reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. Whatis the independent reason that a dielectric material also allows a greater voltage to be applied to a capacitor? (The dielectric thus increases C and

permits a greater V .)

17. How does the polar character of water molecules help to explain water’s relatively large dielectric constant? (Figure 2.19)

18. Sparks will occur between the plates of an air-filled capacitor at lower voltage when the air is humid than when dry. Explain why, considering thepolar character of water molecules.

19. Water has a large dielectric constant, but it is rarely used in capacitors. Explain why.

20. Membranes in living cells, including those in humans, are characterized by a separation of charge across the membrane. Effectively, themembranes are thus charged capacitors with important functions related to the potential difference across the membrane. Is energy required toseparate these charges in living membranes and, if so, is its source the metabolization of food energy or some other source?



Figure 2.26 The semipermeable membrane of a cell has different concentrations of ions inside and out. Diffusion moves the K+ (potassium) and Cl– (chloride) ions in thedirections shown, until the Coulomb force halts further transfer. This results in a layer of positive charge on the outside, a layer of negative charge on the inside, and thus a

voltage across the cell membrane. The membrane is normally impermeable to Na+ (sodium ions).

2.6 Capacitors in Series and Parallel21. If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.

2.7 Energy Stored in Capacitors22. How does the energy contained in a charged capacitor change when a dielectric is inserted, assuming the capacitor is isolated and its charge isconstant? Does this imply that work was done?

23. What happens to the energy stored in a capacitor connected to a battery when a dielectric is inserted? Was work done in the process?



2.1 Electric Potential Energy: Potential Difference1. Find the ratio of speeds of an electron and a negative hydrogen ion(one having an extra electron) accelerated through the same voltage,assuming non-relativistic final speeds. Take the mass of the hydrogen ion

to be 1.67×10 – 27 kg.

2. An evacuated tube uses an accelerating voltage of 40 kV to accelerateelectrons to hit a copper plate and produce x rays. Non-relativistically,what would be the maximum speed of these electrons?

3. A bare helium nucleus has two positive charges and a mass of

6.64×10–27 kg. (a) Calculate its kinetic energy in joules at 2.00% of

the speed of light. (b) What is this in electron volts? (c) What voltagewould be needed to obtain this energy?


Singly charged gas ions are accelerated from rest through a voltage of13.0 V. At what temperature will the average kinetic energy of gasmolecules be the same as that given these ions?


The temperature near the center of the Sun is thought to be 15 million

degrees Celsius ⎛⎝1.5×107 ºC⎞⎠ . Through what voltage must a singly

charged ion be accelerated to have the same energy as the averagekinetic energy of ions at this temperature?


(a) What is the average power output of a heart defibrillator thatdissipates 400 J of energy in 10.0 ms? (b) Considering the high-poweroutput, why doesn’t the defibrillator produce serious burns?


A lightning bolt strikes a tree, moving 20.0 C of charge through a

potential difference of 1.00×102 MV . (a) What energy was

dissipated? (b) What mass of water could be raised from 15ºC to theboiling point and then boiled by this energy? (c) Discuss the damage thatcould be caused to the tree by the expansion of the boiling steam.


A 12.0 V battery-operated bottle warmer heats 50.0 g of glass,

2.50×102 g of baby formula, and 2.00×102 g of aluminum from

20.0ºC to 90.0ºC . (a) How much charge is moved by the battery? (b)How many electrons per second flow if it takes 5.00 min to warm theformula? (Hint: Assume that the specific heat of baby formula is about thesame as the specific heat of water.)


A battery-operated car utilizes a 12.0 V system. Find the charge thebatteries must be able to move in order to accelerate the 750 kg car from

rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause

it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force foran hour.


Fusion probability is greatly enhanced when appropriate nuclei arebrought close together, but mutual Coulomb repulsion must be overcome.This can be done using the kinetic energy of high-temperature gas ionsor by accelerating the nuclei toward one another. (a) Calculate thepotential energy of two singly charged nuclei separated by

1.00×10–12 m by finding the voltage of one at that distance andmultiplying by the charge of the other. (b) At what temperature will atomsof a gas have an average kinetic energy equal to this needed electricalpotential energy?


(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00C of excess positive charge on it. (b) What is unreasonable about thisresult? (c) Which assumptions are responsible?


Consider a battery used to supply energy to a cellular phone. Construct aproblem in which you determine the energy that must be supplied by thebattery, and then calculate the amount of charge it must be able to movein order to supply this energy. Among the things to be considered are theenergy needs and battery voltage. You may need to look ahead tointerpret manufacturer’s battery ratings in ampere-hours as energy injoules.

2.2 Electric Potential in a Uniform Electric Field13. Show that units of V/m and N/C for electric field strength are indeedequivalent.

14. What is the strength of the electric field between two parallelconducting plates separated by 1.00 cm and having a potential difference

(voltage) between them of 1.50×104 V ?

15. The electric field strength between two parallel conducting plates

separated by 4.00 cm is 7.50×104 V/m . (a) What is the potentialdifference between the plates? (b) The plate with the lowest potential istaken to be at zero volts. What is the potential 1.00 cm from that plate(and 3.00 cm from the other)?

16. How far apart are two conducting plates that have an electric field

strength of 4.50×103 V/m between them, if their potential difference is15.0 kV?

17. (a) Will the electric field strength between two parallel conducting

plates exceed the breakdown strength for air ( 3.0×106 V/m ) if theplates are separated by 2.00 mm and a potential difference of

5.0×103 V is applied? (b) How close together can the plates be withthis applied voltage?

18. The voltage across a membrane forming a cell wall is 80.0 mV andthe membrane is 9.00 nm thick. What is the electric field strength? (Thevalue is surprisingly large, but correct. Membranes are discussed inCapacitors and Dielectrics and NerveConduction—Electrocardiograms.) You may assume a uniform electricfield.

19. Membrane walls of living cells have surprisingly large electric fieldsacross them due to separation of ions. (Membranes are discussed insome detail in Nerve Conduction—Electrocardiograms.) What is thevoltage across an 8.00 nm–thick membrane if the electric field strengthacross it is 5.50 MV/m? You may assume a uniform electric field.

20. Two parallel conducting plates are separated by 10.0 cm, and one ofthem is taken to be at zero volts. (a) What is the electric field strengthbetween them, if the potential 8.00 cm from the zero volt plate (and 2.00cm from the other) is 450 V? (b) What is the voltage between the plates?

21. Find the maximum potential difference between two parallelconducting plates separated by 0.500 cm of air, given the maximum

sustainable electric field strength in air to be 3.0×106 V/m .

22. A doubly charged ion is accelerated to an energy of 32.0 keV by theelectric field between two parallel conducting plates separated by 2.00cm. What is the electric field strength between the plates?

23. An electron is to be accelerated in a uniform electric field having a

strength of 2.00×106 V/m . (a) What energy in keV is given to theelectron if it is accelerated through 0.400 m? (b) Over what distancewould it have to be accelerated to increase its energy by 50.0 GeV?

2.3 Electrical Potential Due to a Point Charge24. A 0.500 cm diameter plastic sphere, used in a static electricitydemonstration, has a uniformly distributed 40.0 pC charge on its surface.What is the potential near its surface?



25. What is the potential 0.530×10–10 m from a proton (the averagedistance between the proton and electron in a hydrogen atom)?

26. (a) A sphere has a surface uniformly charged with 1.00 C. At whatdistance from its center is the potential 5.00 MV? (b) What does youranswer imply about the practical aspect of isolating such a large charge?

27. How far from a 1.00 µC point charge will the potential be 100 V? At

what distance will it be 2.00×102 V?28. What are the sign and magnitude of a point charge that produces apotential of –2.00 V at a distance of 1.00 mm?

29. If the potential due to a point charge is 5.00×102 V at a distance of15.0 m, what are the sign and magnitude of the charge?

30. In nuclear fission, a nucleus splits roughly in half. (a) What is the

potential 2.00×10 – 14 m from a fragment that has 46 protons in it? (b)What is the potential energy in MeV of a similarly charged fragment atthis distance?

31. A research Van de Graaff generator has a 2.00-m-diameter metalsphere with a charge of 5.00 mC on it. (a) What is the potential near itssurface? (b) At what distance from its center is the potential 1.00 MV? (c)An oxygen atom with three missing electrons is released near the Van deGraaff generator. What is its energy in MeV at this distance?

32. An electrostatic paint sprayer has a 0.200-m-diameter metal sphereat a potential of 25.0 kV that repels paint droplets onto a grounded object.(a) What charge is on the sphere? (b) What charge must a 0.100-mgdrop of paint have to arrive at the object with a speed of 10.0 m/s?

33. In one of the classic nuclear physics experiments at the beginning ofthe 20th century, an alpha particle was accelerated toward a goldnucleus, and its path was substantially deflected by the Coulombinteraction. If the energy of the doubly charged alpha nucleus was 5.00MeV, how close to the gold nucleus (79 protons) could it come beforebeing deflected?

34. (a) What is the potential between two points situated 10 cm and 20cm from a 3.0 µC point charge? (b) To what location should the point at

20 cm be moved to increase this potential difference by a factor of two?


(a) What is the final speed of an electron accelerated from rest through avoltage of 25.0 MV by a negatively charged Van de Graaff terminal?

(b) What is unreasonable about this result?

(c) Which assumptions are responsible?

2.4 Equipotential Lines36. (a) Sketch the equipotential lines near a point charge + q . Indicate

the direction of increasing potential. (b) Do the same for a point charge– 3 q .

37. Sketch the equipotential lines for the two equal positive chargesshown in Figure 2.27. Indicate the direction of increasing potential.

Figure 2.27 The electric field near two equal positive charges is directed away fromeach of the charges.

38. Figure 2.28 shows the electric field lines near two charges q1 and

q2 , the first having a magnitude four times that of the second. Sketch

the equipotential lines for these two charges, and indicate the direction ofincreasing potential.

39. Sketch the equipotential lines a long distance from the chargesshown in Figure 2.28. Indicate the direction of increasing potential.

Figure 2.28 The electric field near two charges.

40. Sketch the equipotential lines in the vicinity of two opposite charges,where the negative charge is three times as great in magnitude as thepositive. See Figure 2.28 for a similar situation. Indicate the direction ofincreasing potential.

41. Sketch the equipotential lines in the vicinity of the negatively chargedconductor in Figure 2.29. How will these equipotentials look a longdistance from the object?

Figure 2.29 A negatively charged conductor.

42. Sketch the equipotential lines surrounding the two conducting platesshown in Figure 2.30, given the top plate is positive and the bottom platehas an equal amount of negative charge. Be certain to indicate thedistribution of charge on the plates. Is the field strongest where the platesare closest? Why should it be?

Figure 2.30

43. (a) Sketch the electric field lines in the vicinity of the charged insulatorin Figure 2.31. Note its non-uniform charge distribution. (b) Sketchequipotential lines surrounding the insulator. Indicate the direction ofincreasing potential.

Figure 2.31 A charged insulating rod such as might be used in a classroomdemonstration.


44. The naturally occurring charge on the ground on a fine day out in the

open country is –1.00 nC/m2 . (a) What is the electric field relative toground at a height of 3.00 m? (b) Calculate the electric potential at thisheight. (c) Sketch electric field and equipotential lines for this scenario.

45. The lesser electric ray (Narcine bancroftii) maintains an incrediblecharge on its head and a charge equal in magnitude but opposite in signon its tail (Figure 2.32). (a) Sketch the equipotential lines surrounding theray. (b) Sketch the equipotentials when the ray is near a ship with aconducting surface. (c) How could this charge distribution be of use to theray?

Figure 2.32 Lesser electric ray (Narcine bancroftii) (credit: National Oceanic andAtmospheric Administration, NOAA's Fisheries Collection).

2.5 Capacitors and Dielectrics46. What charge is stored in a 180 µF capacitor when 120 V is applied

to it?

47. Find the charge stored when 5.50 V is applied to an 8.00 pFcapacitor.

48. What charge is stored in the capacitor in Example 2.8?

49. Calculate the voltage applied to a 2.00 µF capacitor when it holds

3.10 µC of charge.

50. What voltage must be applied to an 8.00 nF capacitor to store 0.160mC of charge?

51. What capacitance is needed to store 3.00 µC of charge at a voltage

of 120 V?

52. What is the capacitance of a large Van de Graaff generator’sterminal, given that it stores 8.00 mC of charge at a voltage of 12.0 MV?

53. Find the capacitance of a parallel plate capacitor having plates of

area 5.00 m2 that are separated by 0.100 mm of Teflon.

54. (a)What is the capacitance of a parallel plate capacitor having plates

of area 1.50 m2 that are separated by 0.0200 mm of neoprene rubber?(b) What charge does it hold when 9.00 V is applied to it?


A prankster applies 450 V to an 80.0 µF capacitor and then tosses it to

an unsuspecting victim. The victim’s finger is burned by the discharge ofthe capacitor through 0.200 g of flesh. What is the temperature increaseof the flesh? Is it reasonable to assume no phase change?


(a) A certain parallel plate capacitor has plates of area 4.00 m2 ,separated by 0.0100 mm of nylon, and stores 0.170 C of charge. What isthe applied voltage? (b) What is unreasonable about this result? (c)Which assumptions are responsible or inconsistent?

2.6 Capacitors in Series and Parallel57. Find the total capacitance of the combination of capacitors in Figure2.33.

Figure 2.33 A combination of series and parallel connections of capacitors.

58. Suppose you want a capacitor bank with a total capacitance of 0.750F and you possess numerous 1.50 mF capacitors. What is the smallestnumber you could hook together to achieve your goal, and how wouldyou connect them?

59. What total capacitances can you make by connecting a 5.00 µF and

an 8.00 µF capacitor together?

60. Find the total capacitance of the combination of capacitors shown inFigure 2.34.


61. Find the total capacitance of the combination of capacitors shown inFigure 2.35.



(a) An 8.00 µF capacitor is connected in parallel to another capacitor,

producing a total capacitance of 5.00 µF . What is the capacitance of

the second capacitor? (b) What is unreasonable about this result? (c)Which assumptions are unreasonable or inconsistent?

2.7 Energy Stored in Capacitors63. (a) What is the energy stored in the 10.0 μF capacitor of a heart

defibrillator charged to 9.00×103 V ? (b) Find the amount of storedcharge.

64. In open heart surgery, a much smaller amount of energy willdefibrillate the heart. (a) What voltage is applied to the 8.00 μFcapacitor of a heart defibrillator that stores 40.0 J of energy? (b) Find theamount of stored charge.



65. A 165 µF capacitor is used in conjunction with a motor. How much

energy is stored in it when 119 V is applied?

66. Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a

7.40 μF capacitor. (a) Find the charge and energy stored if the

capacitors are connected to the battery in series. (b) Do the same for aparallel connection.

67. A nervous physicist worries that the two metal shelves of his woodframe bookcase might obtain a high voltage if charged by staticelectricity, perhaps produced by friction. (a) What is the capacitance of

the empty shelves if they have area 1.00×102 m2 and are 0.200 mapart? (b) What is the voltage between them if opposite charges ofmagnitude 2.00 nC are placed on them? (c) To show that this voltageposes a small hazard, calculate the energy stored.

68. Show that for a given dielectric material the maximum energy aparallel plate capacitor can store is directly proportional to the volume ofdielectric ( Volume = A · d ). Note that the applied voltage is limited bythe dielectric strength.


Consider a heart defibrillator similar to that discussed in Example 2.11.Construct a problem in which you examine the charge stored in thecapacitor of a defibrillator as a function of stored energy. Among thethings to be considered are the applied voltage and whether it shouldvary with energy to be delivered, the range of energies involved, and thecapacitance of the defibrillator. You may also wish to consider the muchsmaller energy needed for defibrillation during open-heart surgery as avariation on this problem.


(a) On a particular day, it takes 9.60×103 J of electric energy to start atruck’s engine. Calculate the capacitance of a capacitor that could storethat amount of energy at 12.0 V. (b) What is unreasonable about thisresult? (c) Which assumptions are responsible?


3 ELECTRIC CURRENT, RESISTANCE, AND OHM'SLAW

Figure 3.1 Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India, by themovement of charge—that is, by electric current. (credit: Chintohere, Wikimedia Commons)

CHAPTER 3 | ELECTRIC CURRENT, RESISTANCE, AND OHM'S LAW 73

Learning Objectives3.1. Current

• Define electric current, ampere, and drift velocity• Describe the direction of charge flow in conventional current.• Use drift velocity to calculate current and vice versa.

3.2. Ohm’s Law: Resistance and Simple Circuits• Explain the origin of Ohm’s law.• Calculate voltages, currents, or resistances with Ohm’s law.• Explain what an ohmic material is.• Describe a simple circuit.

3.3. Resistance and Resistivity• Explain the concept of resistivity.• Use resistivity to calculate the resistance of specified configurations of material.• Use the thermal coefficient of resistivity to calculate the change of resistance with temperature.

3.4. Electric Power and Energy• Calculate the power dissipated by a resistor and power supplied by a power supply.• Calculate the cost of electricity under various circumstances.

3.5. Alternating Current versus Direct Current• Explain the differences and similarities between AC and DC current.• Calculate rms voltage, current, and average power.• Explain why AC current is used for power transmission.

3.6. Electric Hazards and the Human Body• Define thermal hazard, shock hazard, and short circuit.• Explain what effects various levels of current have on the human body.

3.7. Nerve Conduction–Electrocardiograms• Explain the process by which electric signals are transmitted along a neuron.• Explain the effects myelin sheaths have on signal propagation.• Explain what the features of an ECG signal indicate.

Introduction to Electric Current, Resistance, and Ohm's LawThe flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an ultrasound device sending a signal to acomputer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling its load over a mountain pass, a hydroelectricplant sending energy to metropolitan and rural users—these and many other examples of electricity involve electric current, the movement of charge.Humankind has indeed harnessed electricity, the basis of technology, to improve our quality of life. Whereas the previous two chapters concentratedon static electricity and the fundamental force underlying its behavior, the next few chapters will be devoted to electric and magnetic phenomenainvolving current. In addition to exploring applications of electricity, we shall gain new insights into nature—in particular, the fact that all magnetismresults from electric current.

3.1 Current

Electric CurrentElectric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount ofcharge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a longperiod of time. In equation form, electric current I is defined to be

(3.1)I = ΔQΔt ,

where ΔQ is the amount of charge passing through a given area in time Δt . (As in previous chapters, initial time is often taken to be zero, in which

case Δt = t .) (See Figure 3.2.) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère (1775–1836). Since

I = ΔQ / Δt , we see that an ampere is one coulomb per second:

(3.2)1 A = 1 C/sNot only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances.

Figure 3.2 The rate of flow of charge is current. An ampere is the flow of one coulomb through an area in one second.

74 CHAPTER 3 | ELECTRIC CURRENT, RESISTANCE, AND OHM'S LAW


Example 3.1 Calculating Currents: Current in a Truck Battery and a Handheld Calculator

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing?

Strategy

We can use the definition of current in the equation I = ΔQ / Δt to find the current in part (a), since charge and time are given. In part (b), we

rearrange the definition of current and use the given values of charge and current to find the time required.

Solution for (a)

Entering the given values for charge and time into the definition of current gives

(3.3)I = ΔQΔt = 720 C

4.00 s = 180 C/s

= 180 A.Discussion for (a)

This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” arefairly large because large frictional forces need to be overcome when setting something in motion.

Solution for (b)

Solving the relationship I = ΔQ / Δt for time Δt , and entering the known values for charge and current gives

(3.4)Δt = ΔQI = 1.00 C

0.300×10-3 C/s= 3.33×103 s.

Discussion for (b)

This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller chargethan the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It’s because calculatorsrequire very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours ofuse out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons,so the technology requires smaller currents.

Figure 3.3 shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are veryuseful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in Figure 3.3 (b), forexample, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to apenlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need tounderstand a few schematics to apply the concepts and analysis to many more situations.

Figure 3.3 (a) A simple electric circuit. A closed path for current to flow through is supplied by conducting wires connecting a load to the terminals of a battery. (b) In thisschematic, the battery is represented by the two parallel red lines, conducting wires are shown as straight lines, and the zigzag represents the load. The schematic representsa wide variety of similar circuits.

Note that the direction of current flow in Figure 3.3 is from positive to negative. The direction of conventional current is the direction that positivecharge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried


by electrons—that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true innerve cells. A Van de Graaff generator used for nuclear research can produce a current of pure positive charges, such as protons. Figure 3.4illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positivecharge would flow can be traced back to American politician and scientist Benjamin Franklin in the 1700s. He named the type of charge associatedwith electrons negative, long before they were known to carry current in so many situations. Franklin, in fact, was totally unaware of the small-scalestructure of electricity.

It is important to realize that there is an electric field in conductors responsible for producing the current, as illustrated in Figure 3.4. Unlike staticelectricity, where a conductor in equilibrium cannot have an electric field in it, conductors carrying a current have an electric field and are not in staticequilibrium. An electric field is needed to supply energy to move the charges.

Making Connections: Take-Home Investigation—Electric Current Illustration

Find a straw and little peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you pop one peain at one end, a different pea should pop out the other end. This demonstration is an analogy for an electric current. Identify what compares tothe electrons and what compares to the supply of energy. What other analogies can you find for an electric current?

Note that the flow of peas is based on the peas physically bumping into each other; electrons flow due to mutually repulsive electrostatic forces.

Figure 3.4 Current I is the rate at which charge moves through an area A , such as the cross-section of a wire. Conventional current is defined to move in the direction ofthe electric field. (a) Positive charges move in the direction of the electric field and the same direction as conventional current. (b) Negative charges move in the directionopposite to the electric field. Conventional current is in the direction opposite to the movement of negative charge. The flow of electrons is sometimes referred to as electronicflow.

Example 3.2 Calculating the Number of Electrons that Move through a Calculator

If the 0.300-mA current through the calculator mentioned in the Example 3.1 example is carried by electrons, how many electrons per secondpass through it?

Strategy

The current calculated in the previous example was defined for the flow of positive charge. For electrons, the magnitude is the same, but the sign

is opposite, Ielectrons = −0.300×10−3 C/s .Since each electron (e−) has a charge of –1.60×10−19 C , we can convert the current in

coulombs per second to electrons per second.

Solution

Starting with the definition of current, we have

(3.5)Ielectrons = ΔQelectrons

Δt = –0.300×10−3 Cs .

We divide this by the charge per electron, so that

(3.6)e–s = –0.300×10 – 3 C

s × 1 e–

–1.60×10−19 C= 1.88×1015 e–

s .

Discussion

There are so many charged particles moving, even in small currents, that individual charges are not noticed, just as individual water moleculesare not noticed in water flow. Even more amazing is that they do not always keep moving forward like soldiers in a parade. Rather they are like a



crowd of people with movement in different directions but a general trend to move forward. There are lots of collisions with atoms in the metalwire and, of course, with other electrons.

Drift VelocityElectrical signals are known to move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable

delays. Lights come on as soon as a switch is flicked. Most electrical signals carried by currents travel at speeds on the order of 108 m/s , asignificant fraction of the speed of light. Interestingly, the individual charges that make up the current move much more slowly on average, typically

drifting at speeds on the order of 10−4 m/s . How do we reconcile these two speeds, and what does it tell us about standard conductors?

The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forcedinto a wire, as in Figure 3.5, the incoming charge pushes other charges ahead of it, which in turn push on charges farther down the line. The densityof charge in a system cannot easily be increased, and so the signal is passed on rapidly. The resulting electrical shock wave moves through thesystem at nearly the speed of light. To be precise, this rapidly moving signal or shock wave is a rapidly propagating change in electric field.

Figure 3.5 When charged particles are forced into this volume of a conductor, an equal number are quickly forced to leave. The repulsion between like charges makes itdifficult to increase the number of charges in a volume. Thus, as one charge enters, another leaves almost immediately, carrying the signal rapidly forward.

Good conductors have large numbers of free charges in them. In metals, the free charges are free electrons. Figure 3.6 shows how free electronsmove through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quitesmall. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electric field in the conductor that causes theelectrons to drift in the direction shown (opposite to the field, since they are negative). The drift velocity vd is the average velocity of the free

charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, wecan calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current.

Figure 3.6 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of the freecharges is called the drift velocity, vd , and it is in the direction opposite to the electric field for electrons. The collisions normally transfer energy to the conductor, requiring a

constant supply of energy to maintain a steady current.

Conduction of Electricity and Heat

Good electrical conductors are often good heat conductors, too. This is because large numbers of free electrons can carry electrical current andcan transport thermal energy.

The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, butthat work does not increase the kinetic energy (nor speed, therefore) of the electrons. The work is transferred to the conductor’s atoms, possiblyincreasing temperature. Thus a continuous power input is required to keep a current flowing. An exception, of course, is found in superconductors, forreasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy—a great energysavings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase thetemperature of the tungsten filament, so that the filament glows.

Making Connections: Take-Home Investigation—Filament Observations

Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected?


We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire,as illustrated in Figure 3.7. The number of free charges per unit volume is given the symbol n and depends on the material. The shaded segment

has a volume Ax , so that the number of free charges in it is nAx . The charge ΔQ in this segment is thus qnAx , where q is the amount of

charge on each carrier. (Recall that for electrons, q is −1.60×10−19 C .) Current is charge moved per unit time; thus, if all the original charges

move out of this segment in time Δt , the current is

(3.7)I = ΔQ

Δt = qnAxΔt .

Note that x / Δt is the magnitude of the drift velocity, vd , since the charges move an average distance x in a time Δt . Rearranging terms gives

(3.8)I = nqAv d,

where I is the current through a wire of cross-sectional area A made of a material with a free charge density n . The carriers of the current eachhave charge q and move with a drift velocity of magnitude vd .

Figure 3.7 All the charges in the shaded volume of this wire move out in a time t , having a drift velocity of magnitude vd = x / t . See text for further discussion.

Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electronsin a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by freeelectrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei thatthey do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a singleatom but can instead move freely among the atoms in a “sea” of electrons. These free electrons respond by accelerating when an electric field isapplied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor getswarmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons.

Example 3.3 Calculating Drift Velocity in a Common Wire

Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that thereis one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is

usually 20 A.) The density of copper is 8.80×103 kg/m3 .

Strategy

We can calculate the drift velocity using the equation I = nqAvd . The current I = 20.0 A is given, and q = – 1.60×10 – 19 C is the

charge of an electron. We can calculate the area of a cross-section of the wire using the formula A = πr2, where r is one-half the given

diameter, 2.053 mm. We are given the density of copper, 8.80×103 kg/m3, and the periodic table shows that the atomic mass of copper is

63.54 g/mol. We can use these two quantities along with Avogadro’s number, 6.02×1023 atoms/mol, to determine n, the number of free

electrons per cubic meter.

Solution

First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of

copper atoms per m3 . We can now find n as follows:

(3.9)n = 1 e−

atom×6.02×1023 atomsmol × 1 mol

63.54 g×1000 gkg ×8.80×103 kg

1 m3

= 8.342×1028 e− /m3 .The cross-sectional area of the wire is

(3.10)A = πr2

= π⎛⎝2.053×10−3 m2

⎞⎠

2

= 3.310×10–6 m2 .



Rearranging I = nqAvd to isolate drift velocity gives

(3.11)vd = InqA

= 20.0 A(8.342×1028/m3)(–1.60×10–19 C)(3.310×10–6 m2)

= –4.53×10–4 m/s.Discussion

The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity

(on the order of 10−4 m/s ) confirms that the signal moves on the order of 1012 times faster (about 108 m/s ) than the charges that carry it.

3.2 Ohm’s Law: Resistance and Simple CircuitsWhat drives current? We can think of various devices—such as batteries, generators, wall outlets, and so on—which are necessary to maintain acurrent. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to aconductor, it applies a potential difference V that creates an electric field. The electric field in turn exerts force on charges, causing current.

Ohm’s LawThe current that flows through most substances is directly proportional to the voltage V applied to it. The German physicist Georg Simon Ohm(1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied:

(3.12)I ∝ V .This important relationship is known as Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect.This is an empirical law like that for friction—an experimentally observed phenomenon. Such a linear relationship doesn’t always occur.

Resistance and Simple CircuitsIf voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is calledresistance R . Collisions of moving charges with atoms and molecules in a substance transfer energy to the substance and limit current. Resistanceis defined as inversely proportional to current, or

(3.13)I ∝ 1R.

Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives

(3.14)I = VR .

This relationship is also called Ohm’s law. Ohm’s law in this form really defines resistance for certain materials. Ohm’s law (like Hooke’s law) is notuniversally valid. The many substances for which Ohm’s law holds are called ohmic. These include good conductors like copper and aluminum, andsome poor conductors under certain circumstances. Ohmic materials have a resistance R that is independent of voltage V and current I . An

object that has simple resistance is called a resistor, even if its resistance is small. The unit for resistance is an ohm and is given the symbol Ω(upper case Greek omega). Rearranging I = V/R gives R = V/I , and so the units of resistance are 1 ohm = 1 volt per ampere:

(3.15)1 Ω = 1VA .

Figure 3.8 shows the schematic for a simple circuit. A simple circuit has a single voltage source and a single resistor. The wires connecting thevoltage source to the resistor can be assumed to have negligible resistance, or their resistance can be included in R .

Figure 3.8 A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually metal wires) connecting a load to the terminals of a battery,represented by the red parallel lines. The zigzag symbol represents the single resistor and includes any resistance in the connections to the voltage source.


Example 3.4 Calculating Resistance: An Automobile Headlight

What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it?

Strategy

We can rearrange Ohm’s law as stated by I = V/R and use it to find the resistance.

Solution

Rearranging I = V/R and substituting known values gives

(3.16)R = VI = 12.0 V

2.50 A = 4.80 Ω.

Discussion

This is a relatively small resistance, but it is larger than the cold resistance of the headlight. As we shall see in Resistance and Resistivity,resistance usually increases with temperature, and so the bulb has a lower resistance when it is first switched on and will draw considerablymore current during its brief warm-up period.

Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of

1012 Ω or more. A dry person may have a hand-to-foot resistance of 105 Ω , whereas the resistance of the human heart is about 103 Ω . A

meter-long piece of large-diameter copper wire may have a resistance of 10−5 Ω , and superconductors have no resistance at all (they are non-ohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity.

Additional insight is gained by solving I = V/R for V , yielding

(3.17)V = IR.

This expression for V can be interpreted as the voltage drop across a resistor produced by the flow of current I . The phrase IR drop is often used

for this voltage. For instance, the headlight in Example 3.4 has an IR drop of 12.0 V. If voltage is measured at various points in a circuit, it will beseen to increase at the voltage source and decrease at the resistor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating apressure difference, causing current—the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance.Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and theresistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the sourceequals the voltage drop across the resistor, since PE = qΔV , and the same q flows through each. Thus the energy supplied by the voltage source

and the energy converted by the resistor are equal. (See Figure 3.9.)

Figure 3.9 The voltage drop across a resistor in a simple circuit equals the voltage output of the battery.

Making Connections: Conservation of Energy

In a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced hereby the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation ofenergy has other important applications in circuits and is a powerful tool in circuit analysis.

PhET Explorations: Ohm's Law

See how the equation form of Ohm's law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according toOhm's law. The sizes of the symbols in the equation change to match the circuit diagram.

Figure 3.10 Ohm's Law (http://cnx.org/content/m42344/1.4/ohms-law_en.jar)



http://cnx.org/content/m42344/1.4/ohms-law_en.jar

3.3 Resistance and Resistivity

Material and Shape Dependence of ResistanceThe resistance of an object depends on its shape and the material of which it is composed. The cylindrical resistor in Figure 3.11 is easy to analyze,and, by so doing, we can gain insight into the resistance of more complicated shapes. As you might expect, the cylinder’s electric resistance R is

directly proportional to its length L , similar to the resistance of a pipe to fluid flow. The longer the cylinder, the more collisions charges will make with

its atoms. The greater the diameter of the cylinder, the more current it can carry (again similar to the flow of fluid through a pipe). In fact, R is

inversely proportional to the cylinder’s cross-sectional area A .

Figure 3.11 A uniform cylinder of length L and cross-sectional area A . Its resistance to the flow of current is similar to the resistance posed by a pipe to fluid flow. The

longer the cylinder, the greater its resistance. The larger its cross-sectional area A , the smaller its resistance.

For a given shape, the resistance depends on the material of which the object is composed. Different materials offer different resistance to the flow ofcharge. We define the resistivity ρ of a substance so that the resistance R of an object is directly proportional to ρ . Resistivity ρ is an intrinsic

property of a material, independent of its shape or size. The resistance R of a uniform cylinder of length L , of cross-sectional area A , and made ofa material with resistivity ρ , is

(3.18)R = ρLA .

Table 3.1 gives representative values of ρ . The materials listed in the table are separated into categories of conductors, semiconductors, and

insulators, based on broad groupings of resistivities. Conductors have the smallest resistivities, and insulators have the largest; semiconductors haveintermediate resistivities. Conductors have varying but large free charge densities, whereas most charges in insulators are bound to atoms and arenot free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number offree charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use inmodern electronics, as will be explored in later chapters.


Table 3.1 Resistivities ρ of Various materials at 20ºC

Material Resistivity ρ ( Ω ⋅m )

Conductors

Silver 1.59×10−8

Copper 1.72×10−8

Gold 2.44×10−8

Aluminum 2.65×10−8

Tungsten 5.6×10−8

Iron 9.71×10−8

Platinum 10.6×10−8

Steel 20×10−8

Lead 22×10−8

Manganin (Cu, Mn, Ni alloy) 44×10−8

Constantan (Cu, Ni alloy) 49×10−8

Mercury 96×10−8

Nichrome (Ni, Fe, Cr alloy) 100×10−8

Semiconductors[1]

Carbon (pure) 3.5×105

Carbon (3.5 − 60)×105

Germanium (pure) 600×10−3

Germanium (1 − 600)×10−3

Silicon (pure) 2300

Silicon 0.1–2300

Insulators

Amber 5×1014

Glass 109 − 1014

Lucite >1013

Mica 1011 − 1015

Quartz (fused) 75×1016

Rubber (hard) 1013 − 1016

Sulfur 1015

Teflon >1013

Wood 108 − 1011

1. Values depend strongly on amounts and types of impurities



Example 3.5 Calculating Resistor Diameter: A Headlight Filament

A car headlight filament is made of tungsten and has a cold resistance of 0.350 Ω . If the filament is a cylinder 4.00 cm long (it may be coiledto save space), what is its diameter?

Strategy

We can rearrange the equation R = ρLA to find the cross-sectional area A of the filament from the given information. Then its diameter can be

found by assuming it has a circular cross-section.

Solution

The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in R = ρLA , is

(3.19)A = ρLR .

Substituting the given values, and taking ρ from Table 3.1, yields

(3.20)A = (5.6×10–8 Ω ⋅ m)(4.00×10–2 m)

1.350 Ω= 6.40×10–9 m2 .

The area of a circle is related to its diameter D by

(3.21)A = πD2

4 .

Solving for the diameter D , and substituting the value found for A , gives

(3.22)

D = 2⎛⎝Ap⎞⎠

12

= 2⎛⎝6.40×10–9 m2

3.14⎞⎠

12

= 9.0×10–5 m.Discussion

The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.

Temperature Variation of ResistanceThe resistivity of all materials depends on temperature. Some even become superconductors (zero resistivity) at very low temperatures. (See Figure3.12.) Conversely, the resistivity of conductors increases with increasing temperature. Since the atoms vibrate more rapidly and over larger distancesat higher temperatures, the electrons moving through a metal make more collisions, effectively making the resistivity higher. Over relatively smalltemperature changes (about 100ºC or less), resistivity ρ varies with temperature change ΔT as expressed in the following equation

(3.23)ρ = ρ0(1 + αΔT),

where ρ0 is the original resistivity and α is the temperature coefficient of resistivity. (See the values of α in Table 3.2 below.) For larger

temperature changes, α may vary or a nonlinear equation may be needed to find ρ . Note that α is positive for metals, meaning their resistivity

increases with temperature. Some alloys have been developed specifically to have a small temperature dependence. Manganin (which is made ofcopper, manganese and nickel), for example, has α close to zero (to three digits on the scale in Table 3.2), and so its resistivity varies only slightlywith temperature. This is useful for making a temperature-independent resistance standard, for example.


Figure 3.12 The resistance of a sample of mercury is zero at very low temperatures—it is a superconductor up to about 4.2 K. Above that critical temperature, its resistancemakes a sudden jump and then increases nearly linearly with temperature.

Table 3.2 Tempature Coefficients of Resistivity α

Material Coefficient α (1/°C)[2]

Conductors

Silver 3.8×10−3

Copper 3.9×10−3

Gold 3.4×10−3

Aluminum 3.9×10−3

Tungsten 4.5×10−3

Iron 5.0×10−3

Platinum 3.93×10−3

Lead 3.9×10−3

Manganin (Cu, Mn, Ni alloy) 0.000×10−3

Constantan (Cu, Ni alloy) 0.002×10−3

Mercury 0.89×10−3

Nichrome (Ni, Fe, Cr alloy) 0.4×10−3

Semiconductors

Carbon (pure) −0.5×10−3

Germanium (pure) −50×10−3

Silicon (pure) −70×10−3

Note also that α is negative for the semiconductors listed in Table 3.2, meaning that their resistivity decreases with increasing temperature. Theybecome better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carrycurrent. This property of decreasing ρ with temperature is also related to the type and amount of impurities present in the semiconductors.

The resistance of an object also depends on temperature, since R0 is directly proportional to ρ . For a cylinder we know R = ρL / A , and so, if Land A do not change greatly with temperature, R will have the same temperature dependence as ρ . (Examination of the coefficients of linear

expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, and so the effect of temperatureon L and A is about two orders of magnitude less than on ρ .) Thus,

(3.24)R = R0(1 + αΔT)

2. Values at 20°C.



is the temperature dependence of the resistance of an object, where R0 is the original resistance and R is the resistance after a temperature

change ΔT . Numerous thermometers are based on the effect of temperature on resistance. (See Figure 3.13.) One of the most common is thethermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The deviceis small, so that it quickly comes into thermal equilibrium with the part of a person it touches.

Figure 3.13 These familiar thermometers are based on the automated measurement of a thermistor’s temperature-dependent resistance. (credit: Biol, Wikimedia Commons)

Example 3.6 Calculating Resistance: Hot-Filament Resistance

Although caution must be used in applying ρ = ρ0(1 + αΔT) and R = R0(1 + αΔT) for temperature changes greater than 100ºC , for

tungsten the equations work reasonably well for very large temperature changes. What, then, is the resistance of the tungsten filament in theprevious example if its temperature is increased from room temperature ( 20ºC ) to a typical operating temperature of 2850ºC ?

Strategy

This is a straightforward application of R = R0(1 + αΔT) , since the original resistance of the filament was given to be R0 = 0.350 Ω , and

the temperature change is ΔT = 2830ºC .

Solution

The hot resistance R is obtained by entering known values into the above equation:

(3.25)R = R0(1 + αΔT)

= (0.350 Ω)[1 + (4.5×10–3 / ºC)(2830ºC)]= 4.8 Ω.

Discussion

This value is consistent with the headlight resistance example in Ohm’s Law: Resistance and Simple Circuits.

PhET Explorations: Resistance in a Wire

Learn about the physics of resistance in a wire. Change its resistivity, length, and area to see how they affect the wire's resistance. The sizes ofthe symbols in the equation change along with the diagram of a wire.

Figure 3.14 Resistance in a Wire (http://cnx.org/content/m42346/1.5/resistance-in-a-wire_en.jar)

3.4 Electric Power and Energy

Power in Electric CircuitsPower is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression forelectric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a25-W bulb with a 60-W bulb. (See Figure 3.15(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greaterpower rating. Thus the 60-W bulb’s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example,when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how arevoltage, current, and resistance related to electric power?


http://cnx.org/content/m42346/1.5/resistance-in-a-wire_en.jar

Figure 3.15 (a) Which of these lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses themost energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; GregWestfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr)

Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qV , where q is the charge

moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, andso electric power is

(3.26)P = PEt = qV

t .

Recognizing that current is I = q / t (note that Δt = t here), the expression for power becomes

(3.27)P = IV.

Electric power ( P ) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the

joule, power has units of joules per second, or watts. Thus, 1 A ⋅ V = 1 W . For example, cars often have one or more auxiliary power outlets withwhich you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum powerP = IV = (20 A)(12 V) = 240 W . In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (

1 kA ⋅ V = 1 kW ).

To see the relationship of power to resistance, we combine Ohm’s law with P = IV . Substituting I = V/R gives P = (V / R)V = V 2 /R . Similarly,

substituting V = IR gives P = I(IR) = I 2R . Three expressions for electric power are listed together here for convenience:

(3.28)P = IV(3.29)

P = V 2

R(3.30)P = I 2R.

Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and asingle resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, P can be thepower dissipated by a single device and not the total power in the circuit.)

Different insights can be gained from the three different expressions for electric power. For example, P = V 2 / R implies that the lower the

resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P = V 2 / R , the effect ofapplying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance ishigher, too.

Example 3.7 Calculating Power Dissipation and Current: Hot and Cold Power

(a) Consider the examples given in Ohm’s Law: Resistance and Simple Circuits and Resistance and Resistivity. Then find the powerdissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold?

Strategy for (a)



For the hot headlight, we know voltage and current, so we can use P = IV to find the power. For the cold headlight, we know the voltage and

resistance, so we can use P = V 2 / R to find the power.

Solution for (a)

Entering the known values of current and voltage for the hot headlight, we obtain

(3.31)P = IV = (2.50 A)(12.0 V) = 30.0 W.

The cold resistance was 0.350 Ω , and so the power it uses when first switched on is

(3.32)P = V 2

R = (12.0 V)2

0.350 Ω = 411 W.

Discussion for (a)

The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as thebulb’s temperature increases and its resistance increases.

Strategy and Solution for (b)

The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, P = I 2R , and enter knownvalues, obtaining

(3.33)I = PR = 411 W

0.350 Ω = 34.3 A.

Discussion for (b)

The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb’stemperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly asa device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow”fuses.

The Cost of ElectricityThe more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationshipbetween energy and power. You pay for the energy used. Since P = E / t , we see that

(3.34)E = Pt

is the energy used by a device using power P for a time interval t . For example, the more lightbulbs burning, the greater P used; the longer they

are on, the greater t is. The energy unit on electric bills is the kilowatt-hour ( kW ⋅ h ), consistent with the relationship E = Pt . It is easy toestimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on inhours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be

converted to joules. You can prove to yourself that 1 kW ⋅ h = 3.6×106 J .

The electrical energy ( E ) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture.This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest waysto reduce the electrical energy used in a home or business. About 20% of a home’s use of energy goes to lighting, while the number for commercialestablishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubesand the compact fluorescent lights (CFL). (See Figure 3.15(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has thesame brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standardincandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.)The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in thenext example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer thanCFLs. However, their cost is still high.

Making Connections: Energy, Power, and Time

The relationship E = Pt is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the powerlevel and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Eventhe radiation dose of an X-ray image is related to the power and time of exposure.

Example 3.8 Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL)

If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same lightoutput, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be?

Strategy

To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour.


Solution for (a)

The energy used in kilowatt-hours is found by entering the power and time into the expression for energy:

(3.35)E = Pt = (60 W)(1000 h) = 60,000 W ⋅ h.

In kilowatt-hours, this is

(3.36)E = 60.0 kW ⋅ h.Now the electricity cost is

(3.37)cost = (60.0 kW ⋅ h)($0.12/kW ⋅ h) = $7.20.

The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day).

Solution for (b)

Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent,so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95for 1000 hours.

Discussion

Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business mustinclude for replacing the incandescent bulbs more often has not been figured in here.

Making Connections: Take-Home Experiment—Electrical Energy Use Inventory

1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating thana digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances mightonly state the operating current. If the household voltage is 120 V, then use P = IV . 2) Check out the total wattage used in the rest rooms ofyour school’s floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building wasclosed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for anentire year of weekends?

3.5 Alternating Current versus Direct Current

Alternating CurrentMost of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it isthus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Mostwell-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodicallyreverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include thecommercial and residential power that serves so many of our needs. Figure 3.16 shows graphs of voltage and current versus time for typical DC andAC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world.

Figure 3.16 (a) DC voltage and current are constant in time, once the current is established. (b) A graph of voltage and current versus time for 60-Hz AC power. The voltageand current are sinusoidal and are in phase for a simple resistance circuit. The frequencies and peak voltages of AC sources differ greatly.



Figure 3.17 The potential difference V between the terminals of an AC voltage source fluctuates as shown. The mathematical expression for V is given by

V = V0 sin 2π ft .

Figure 3.17 shows a schematic of a simple circuit with an AC voltage source. The voltage between the terminals fluctuates as shown, with the ACvoltage given by

(3.38)V = V0 sin 2π ft,

where V is the voltage at time t , V0 is the peak voltage, and f is the frequency in hertz. For this simple resistance circuit, I = V/R , and so the

AC current is

(3.39)I = I0 sin 2π ft,

where I is the current at time t , and I0 = V0/R is the peak current. For this example, the voltage and current are said to be in phase, as seen in

Figure 3.16(b).

Current in the resistor alternates back and forth just like the driving voltage, since I = V/R . If the resistor is a fluorescent light bulb, for example, itbrightens and dims 120 times per second as the current repeatedly goes through zero. A 120-Hz flicker is too rapid for your eyes to detect, but if youwave your hand back and forth between your face and a fluorescent light, you will see a stroboscopic effect evidencing AC. The fact that the lightoutput fluctuates means that the power is fluctuating. The power supplied is P = IV . Using the expressions for I and V above, we see that the

time dependence of power is P = I0V0 sin2 2π ft , as shown in Figure 3.18.

Making Connections: Take-Home Experiment—AC/DC Lights

Wave your hand back and forth between your face and a fluorescent light bulb. Do you observe the same thing with the headlights on your car?Explain what you observe. Warning: Do not look directly at very bright light.

Figure 3.18 AC power as a function of time. Since the voltage and current are in phase here, their product is non-negative and fluctuates between zero and I0 V0 . Average

power is (1 / 2)I0 V0 .

We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average powerconsumption of 60 W, for example. As illustrated in Figure 3.18, the average power Pave is

(3.40)Pave = 12I0 V0.

This is evident from the graph, since the areas above and below the (1 / 2)I0 V0 line are equal, but it can also be proven using trigonometric

identities. Similarly, we define an average or rms current Irms and average or rms voltage Vrms to be, respectively,


(3.41)Irms = I0

2and

(3.42)Vrms = V0

2.

where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, itsmean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now,

(3.43)Pave = IrmsVrms,

which gives

(3.44)Pave = I0

2⋅ V0

2= 1

2I0 V0,

as stated above. It is standard practice to quote Irms , Vrms , and Pave rather than the peak values. For example, most household electricity is 120

V AC, which means that Vrms is 120 V. The common 10-A circuit breaker will interrupt a sustained Irms greater than 10 A. Your 1.0-kW microwave

oven consumes Pave = 1.0 kW , and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive

circuit.

To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average valuesare used for AC. Thus, for AC, Ohm’s law is written

(3.45)Irms = VrmsR .

The various expressions for AC power Pave are

(3.46)Pave = IrmsVrms,(3.47)

Pave = Vrms2

R ,

and

(3.48)Pave = Irms2 R.

Example 3.9 Peak Voltage and Power for AC

(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?

Strategy

We are told that Vrms is 120 V and Pave is 60.0 W. We can use Vrms = V02

to find the peak voltage, and we can manipulate the definition of

power to find the peak power from the given average power.

Solution for (a)

Solving the equation Vrms = V02

for the peak voltage V0 and substituting the known value for Vrms gives

(3.49)V0 = 2Vrms = 1.414(120 V) = 170 V.

Discussion for (a)

This means that the AC voltage swings from 170 V to –170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V.

Solution for (b)

Peak power is peak current times peak voltage. Thus,

(3.50)P0 = I0V0 = 2⎛⎝12I0 V0⎞⎠ = 2Pave.

We know the average power is 60.0 W, and so

(3.51)P0 = 2(60.0 W) = 120 W.

Discussion

So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.



Why Use AC for Power Distribution?Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts ofthe world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than tobuild numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized.High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See Figure 3.19.) For safety reasons, thevoltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC isused in most large power distribution systems.

Figure 3.19 Power is distributed over large distances at high voltage to reduce power loss in the transmission lines. The voltages generated at the power plant are stepped upby passive devices called transformers (see Transformers) to 330,000 volts (or more in some places worldwide). At the point of use, the transformers reduce the voltagetransmitted for safe residential and commercial use. (Credit: GeorgHH, Wikimedia Commons)

Example 3.10 Power Losses Are Less for High-Voltage Transmission

(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have aresistance of 1.00 Ω ? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given Pave = 100 MW , Vrms = 200 kV , and the resistance of the lines is R = 1.00 Ω . Using these givens, we can find the

current flowing (from P = IV ) and then the power dissipated in the lines ( P = I 2R ), and we take the ratio to the total power transmitted.

Solution

To find the current, we rearrange the relationship Pave = IrmsVrms and substitute known values. This gives

(3.52)Irms = Pave

Vrms= 100×106 W

200×103 V= 500 A.

Solution

Knowing the current and given the resistance of the lines, the power dissipated in them is found from Pave = Irms2 R . Substituting the known

values gives

(3.53)Pave = Irms2 R = (500 A)2(1.00 Ω ) = 250 kW.

Solution

The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:

(3.54)% loss= 250 kW100 MW×100 = 0.250 %.

Discussion

One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would havebeen needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current isneeded, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requireslarger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines atall. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical fortransmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.

It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated withcommon static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are oftenmore harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in NewYork City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocatingthe use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltagetransmission.


PhET Explorations: Generator

Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make abulb light.

Figure 3.20 Generator (http://cnx.org/content/m42348/1.4/generator_en.jar)

3.6 Electric Hazards and the Human BodyThere are two known hazards of electricity—thermal and shock. A thermal hazard is one where excessive electric power causes undesired thermaleffects, such as starting a fire in the wall of a house. A shock hazard occurs when electric current passes through a person. Shocks range in severityfrom painful, but otherwise harmless, to heart-stopping lethality. This section considers these hazards and the various factors affecting them in aquantitative manner. Electrical Safety: Systems and Devices will consider systems and devices for preventing electrical hazards.

Thermal HazardsElectric power causes undesired heating effects whenever electric energy is converted to thermal energy at a rate faster than it can be safelydissipated. A classic example of this is the short circuit, a low-resistance path between terminals of a voltage source. An example of a short circuit isshown in Figure 3.21. Insulation on wires leading to an appliance has worn through, allowing the two wires to come into contact. Such an undesired

contact with a high voltage is called a short. Since the resistance of the short, r , is very small, the power dissipated in the short, P = V 2 / r , is very

large. For example, if V is 120 V and r is 0.100 Ω , then the power is 144 kW, much greater than that used by a typical household appliance.Thermal energy delivered at this rate will very quickly raise the temperature of surrounding materials, melting or perhaps igniting them.

Figure 3.21 A short circuit is an undesired low-resistance path across a voltage source. (a) Worn insulation on the wires of a toaster allow them to come into contact with a low

resistance r . Since P = V 2 / r , thermal power is created so rapidly that the cord melts or burns. (b) A schematic of the short circuit.

One particularly insidious aspect of a short circuit is that its resistance may actually be decreased due to the increase in temperature. This can

happen if the short creates ionization. These charged atoms and molecules are free to move and, thus, lower the resistance r . Since P = V 2 / r ,the power dissipated in the short rises, possibly causing more ionization, more power, and so on. High voltages, such as the 480-V AC used in someindustrial applications, lend themselves to this hazard, because higher voltages create higher initial power production in a short.

Another serious, but less dramatic, thermal hazard occurs when wires supplying power to a user are overloaded with too great a current. As

discussed in the previous section, the power dissipated in the supply wires is P = I 2Rw , where Rw is the resistance of the wires and I the

current flowing through them. If either I or Rw is too large, the wires overheat. For example, a worn appliance cord (with some of its braided wires

broken) may have Rw = 2.00 Ω rather than the 0.100 Ω it should be. If 10.0 A of current passes through the cord, then

P = I 2Rw = 200 W is dissipated in the cord—much more than is safe. Similarly, if a wire with a 0.100 - Ω resistance is meant to carry a few

amps, but is instead carrying 100 A, it will severely overheat. The power dissipated in the wire will in that case be P = 1000 W . Fuses and circuitbreakers are used to limit excessive currents. (See Figure 3.22 and Figure 3.23.) Each device opens the circuit automatically when a sustainedcurrent exceeds safe limits.



http://cnx.org/content/m42348/1.4/generator_en.jar

Figure 3.22 (a) A fuse has a metal strip with a low melting point that, when overheated by an excessive current, permanently breaks the connection of a circuit to a voltagesource. (b) A circuit breaker is an automatic but restorable electric switch. The one shown here has a bimetallic strip that bends to the right and into the notch if overheated.The spring then forces the metal strip downward, breaking the electrical connection at the points.

Figure 3.23 Schematic of a circuit with a fuse or circuit breaker in it. Fuses and circuit breakers act like automatic switches that open when sustained current exceeds desiredlimits.

Fuses and circuit breakers for typical household voltages and currents are relatively simple to produce, but those for large voltages and currentsexperience special problems. For example, when a circuit breaker tries to interrupt the flow of high-voltage electricity, a spark can jump across itspoints that ionizes the air in the gap and allows the current to continue flowing. Large circuit breakers found in power-distribution systems employinsulating gas and even use jets of gas to blow out such sparks. Here AC is safer than DC, since AC current goes through zero 120 times persecond, giving a quick opportunity to extinguish these arcs.

Shock HazardsElectrical currents through people produce tremendously varied effects. An electrical current can be used to block back pain. The possibility of usingelectrical current to stimulate muscle action in paralyzed limbs, perhaps allowing paraplegics to walk, is under study. TV dramatizations in whichelectrical shocks are used to bring a heart attack victim out of ventricular fibrillation (a massively irregular, often fatal, beating of the heart) are morethan common. Yet most electrical shock fatalities occur because a current put the heart into fibrillation. A pacemaker uses electrical shocks tostimulate the heart to beat properly. Some fatal shocks do not produce burns, but warts can be safely burned off with electric current (though freezingusing liquid nitrogen is now more common). Of course, there are consistent explanations for these disparate effects. The major factors upon whichthe effects of electrical shock depend are

1. The amount of current I2. The path taken by the current3. The duration of the shock4. The frequency f of the current ( f = 0 for DC)

Table 3.3 gives the effects of electrical shocks as a function of current for a typical accidental shock. The effects are for a shock that passes throughthe trunk of the body, has a duration of 1 s, and is caused by 60-Hz power.


Figure 3.24 An electric current can cause muscular contractions with varying effects. (a) The victim is “thrown” backward by involuntary muscle contractions that extend thelegs and torso. (b) The victim can’t let go of the wire that is stimulating all the muscles in the hand. Those that close the fingers are stronger than those that open them.

Table 3.3 Effects of Electrical Shock as a Function of Current[3]

Current(mA) Effect

1 Threshold of sensation

5 Maximum harmless current

10–20 Onset of sustained muscular contraction; cannot let go for duration of shock; contraction of chest muscles may stop breathing duringshock

50 Onset of pain

100–300+ Ventricular fibrillation possible; often fatal

300 Onset of burns depending on concentration of current

6000 (6 A) Onset of sustained ventricular contraction and respiratory paralysis; both cease when shock ends; heartbeat may return to normal;used to defibrillate the heart

Our bodies are relatively good conductors due to the water in our bodies. Given that larger currents will flow through sections with lower resistance(to be further discussed in the next chapter), electric currents preferentially flow through paths in the human body that have a minimum resistance ina direct path to earth. The earth is a natural electron sink. Wearing insulating shoes, a requirement in many professions, prohibits a pathway forelectrons by providing a large resistance in that path. Whenever working with high-power tools (drills), or in risky situations, ensure that you do notprovide a pathway for current flow (especially through the heart).

Very small currents pass harmlessly and unfelt through the body. This happens to you regularly without your knowledge. The threshold of sensation isonly 1 mA and, although unpleasant, shocks are apparently harmless for currents less than 5 mA. A great number of safety rules take the 5-mA valuefor the maximum allowed shock. At 10 to 20 mA and above, the current can stimulate sustained muscular contractions much as regular nerveimpulses do. People sometimes say they were knocked across the room by a shock, but what really happened was that certain muscles contracted,propelling them in a manner not of their own choosing. (See Figure 3.24(a).) More frightening, and potentially more dangerous, is the “can’t let go”effect illustrated in Figure 3.24(b). The muscles that close the fingers are stronger than those that open them, so the hand closes involuntarily on thewire shocking it. This can prolong the shock indefinitely. It can also be a danger to a person trying to rescue the victim, because the rescuer’s handmay close about the victim’s wrist. Usually the best way to help the victim is to give the fist a hard knock/blow/jar with an insulator or to throw aninsulator at the fist. Modern electric fences, used in animal enclosures, are now pulsed on and off to allow people who touch them to get free,rendering them less lethal than in the past.

Greater currents may affect the heart. Its electrical patterns can be disrupted, so that it beats irregularly and ineffectively in a condition called“ventricular fibrillation.” This condition often lingers after the shock and is fatal due to a lack of blood circulation. The threshold for ventricularfibrillation is between 100 and 300 mA. At about 300 mA and above, the shock can cause burns, depending on the concentration of current—themore concentrated, the greater the likelihood of burns.

Very large currents cause the heart and diaphragm to contract for the duration of the shock. Both the heart and breathing stop. Interestingly, bothoften return to normal following the shock. The electrical patterns on the heart are completely erased in a manner that the heart can start afresh withnormal beating, as opposed to the permanent disruption caused by smaller currents that can put the heart into ventricular fibrillation. The latter issomething like scribbling on a blackboard, whereas the former completely erases it. TV dramatizations of electric shock used to bring a heart attackvictim out of ventricular fibrillation also show large paddles. These are used to spread out current passed through the victim to reduce the likelihood ofburns.

Current is the major factor determining shock severity (given that other conditions such as path, duration, and frequency are fixed, such as in thetable and preceding discussion). A larger voltage is more hazardous, but since I = V/R , the severity of the shock depends on the combination of

voltage and resistance. For example, a person with dry skin has a resistance of about 200 k Ω . If he comes into contact with 120-V AC, a current

I = (120 V) / (200 k Ω )= 0.6 mA passes harmlessly through him. The same person soaking wet may have a resistance of 10.0 k Ω and the

same 120 V will produce a current of 12 mA—above the “can’t let go” threshold and potentially dangerous.

Most of the body’s resistance is in its dry skin. When wet, salts go into ion form, lowering the resistance significantly. The interior of the body has amuch lower resistance than dry skin because of all the ionic solutions and fluids it contains. If skin resistance is bypassed, such as by an intravenous

3. For an average male shocked through trunk of body for 1 s by 60-Hz AC. Values for females are 60–80% of those listed.



infusion, a catheter, or exposed pacemaker leads, a person is rendered microshock sensitive. In this condition, currents about 1/1000 those listedin Table 3.3 produce similar effects. During open-heart surgery, currents as small as 20 μA can be used to still the heart. Stringent electrical safety

requirements in hospitals, particularly in surgery and intensive care, are related to the doubly disadvantaged microshock-sensitive patient. The breakin the skin has reduced his resistance, and so the same voltage causes a greater current, and a much smaller current has a greater effect.

Figure 3.25 Graph of average values for the threshold of sensation and the “can’t let go” current as a function of frequency. The lower the value, the more sensitive the body isat that frequency.

Factors other than current that affect the severity of a shock are its path, duration, and AC frequency. Path has obvious consequences. For example,the heart is unaffected by an electric shock through the brain, such as may be used to treat manic depression. And it is a general truth that the longerthe duration of a shock, the greater its effects. Figure 3.25 presents a graph that illustrates the effects of frequency on a shock. The curves show theminimum current for two different effects, as a function of frequency. The lower the current needed, the more sensitive the body is at that frequency.Ironically, the body is most sensitive to frequencies near the 50- or 60-Hz frequencies in common use. The body is slightly less sensitive for DC (f = 0 ), mildly confirming Edison’s claims that AC presents a greater hazard. At higher and higher frequencies, the body becomes progressively less

sensitive to any effects that involve nerves. This is related to the maximum rates at which nerves can fire or be stimulated. At very high frequencies,electrical current travels only on the surface of a person. Thus a wart can be burned off with very high frequency current without causing the heart tostop. (Do not try this at home with 60-Hz AC!) Some of the spectacular demonstrations of electricity, in which high-voltage arcs are passed throughthe air and over people’s bodies, employ high frequencies and low currents. (See Figure 3.26.) Electrical safety devices and techniques arediscussed in detail in Electrical Safety: Systems and Devices.

Figure 3.26 Is this electric arc dangerous? The answer depends on the AC frequency and the power involved. (credit: Khimich Alex, Wikimedia Commons)

3.7 Nerve Conduction–Electrocardiograms

Nerve ConductionElectric currents in the vastly complex system of billions of nerves in our body allow us to sense the world, control parts of our body, and think. Theseare representative of the three major functions of nerves. First, nerves carry messages from our sensory organs and others to the central nervoussystem, consisting of the brain and spinal cord. Second, nerves carry messages from the central nervous system to muscles and other organs. Third,nerves transmit and process signals within the central nervous system. The sheer number of nerve cells and the incredibly greater number ofconnections between them makes this system the subtle wonder that it is. Nerve conduction is a general term for electrical signals carried by nervecells. It is one aspect of bioelectricity, or electrical effects in and created by biological systems.

Nerve cells, properly called neurons, look different from other cells—they have tendrils, some of them many centimeters long, connecting them withother cells. (See Figure 3.27.) Signals arrive at the cell body across synapses or through dendrites, stimulating the neuron to generate its own signal,sent along its long axon to other nerve or muscle cells. Signals may arrive from many other locations and be transmitted to yet others, conditioningthe synapses by use, giving the system its complexity and its ability to learn.


Figure 3.27 A neuron with its dendrites and long axon. Signals in the form of electric currents reach the cell body through dendrites and across synapses, stimulating theneuron to generate its own signal sent down the axon. The number of interconnections can be far greater than shown here.

The method by which these electric currents are generated and transmitted is more complex than the simple movement of free charges in aconductor, but it can be understood with principles already discussed in this text. The most important of these are the Coulomb force and diffusion.

Figure 3.28 illustrates how a voltage (potential difference) is created across the cell membrane of a neuron in its resting state. This thin membrane

separates electrically neutral fluids having differing concentrations of ions, the most important varieties being Na+ , K+ , and Cl- (these aresodium, potassium, and chlorine ions with single plus or minus charges as indicated). As discussed in Molecular Transport Phenomena: Diffusion,Osmosis, and Related Processes (http://cnx.org/content/m42212/latest/) , free ions will diffuse from a region of high concentration to one of lowconcentration. But the cell membrane is semipermeable, meaning that some ions may cross it while others cannot. In its resting state, the cell

membrane is permeable to K+ and Cl- , and impermeable to Na+ . Diffusion of K+ and Cl- thus creates the layers of positive and negativecharge on the outside and inside of the membrane. The Coulomb force prevents the ions from diffusing across in their entirety. Once the charge layerhas built up, the repulsion of like charges prevents more from moving across, and the attraction of unlike charges prevents more from leaving eitherside. The result is two layers of charge right on the membrane, with diffusion being balanced by the Coulomb force. A tiny fraction of the chargesmove across and the fluids remain neutral (other ions are present), while a separation of charge and a voltage have been created across themembrane.

Figure 3.28 The semipermeable membrane of a cell has different concentrations of ions inside and out. Diffusion moves the K+and Cl- ions in the direction shown, until

the Coulomb force halts further transfer. This results in a layer of positive charge on the outside, a layer of negative charge on the inside, and thus a voltage across the cell

membrane. The membrane is normally impermeable to Na+.





Figure 3.29 An action potential is the pulse of voltage inside a nerve cell graphed here. It is caused by movements of ions across the cell membrane as shown. Depolarization

occurs when a stimulus makes the membrane permeable to Na+ions. Repolarization follows as the membrane again becomes impermeable to Na+, and K+

moves

from high to low concentration. In the long term, active transport slowly maintains the concentration differences, but the cell may fire hundreds of times in rapid successionwithout seriously depleting them.

The separation of charge creates a potential difference of 70 to 90 mV across the cell membrane. While this is a small voltage, the resulting electricfield ( E = V / d ) across the only 8-nm-thick membrane is immense (on the order of 11 MV/m!) and has fundamental effects on its structure andpermeability. Now, if the exterior of a neuron is taken to be at 0 V, then the interior has a resting potential of about –90 mV. Such voltages are createdacross the membranes of almost all types of animal cells but are largest in nerve and muscle cells. In fact, fully 25% of the energy used by cells goestoward creating and maintaining these potentials.

Electric currents along the cell membrane are created by any stimulus that changes the membrane’s permeability. The membrane thus temporarily

becomes permeable to Na+ , which then rushes in, driven both by diffusion and the Coulomb force. This inrush of Na+ first neutralizes the inside

membrane, or depolarizes it, and then makes it slightly positive. The depolarization causes the membrane to again become impermeable to Na+ ,

and the movement of K+ quickly returns the cell to its resting potential, or repolarizes it. This sequence of events results in a voltage pulse, calledthe action potential. (See Figure 3.29.) Only small fractions of the ions move, so that the cell can fire many hundreds of times without depleting the

excess concentrations of Na+ and K+ . Eventually, the cell must replenish these ions to maintain the concentration differences that createbioelectricity. This sodium-potassium pump is an example of active transport, wherein cell energy is used to move ions across membranes againstdiffusion gradients and the Coulomb force.

The action potential is a voltage pulse at one location on a cell membrane. How does it get transmitted along the cell membrane, and in particulardown an axon, as a nerve impulse? The answer is that the changing voltage and electric fields affect the permeability of the adjacent cell membrane,so that the same process takes place there. The adjacent membrane depolarizes, affecting the membrane further down, and so on, as illustrated inFigure 3.30. Thus the action potential stimulated at one location triggers a nerve impulse that moves slowly (about 1 m/s) along the cell membrane.


Figure 3.30 A nerve impulse is the propagation of an action potential along a cell membrane. A stimulus causes an action potential at one location, which changes thepermeability of the adjacent membrane, causing an action potential there. This in turn affects the membrane further down, so that the action potential moves slowly (in

electrical terms) along the cell membrane. Although the impulse is due to Na+and K+

going across the membrane, it is equivalent to a wave of charge moving along theoutside and inside of the membrane.

Some axons, like that in Figure 3.27, are sheathed with myelin, consisting of fat-containing cells. Figure 3.31 shows an enlarged view of an axonhaving myelin sheaths characteristically separated by unmyelinated gaps (called nodes of Ranvier). This arrangement gives the axon a number ofinteresting properties. Since myelin is an insulator, it prevents signals from jumping between adjacent nerves (cross talk). Additionally, the myelinatedregions transmit electrical signals at a very high speed, as an ordinary conductor or resistor would. There is no action potential in the myelinatedregions, so that no cell energy is used in them. There is an IR signal loss in the myelin, but the signal is regenerated in the gaps, where the voltagepulse triggers the action potential at full voltage. So a myelinated axon transmits a nerve impulse faster, with less energy consumption, and is betterprotected from cross talk than an unmyelinated one. Not all axons are myelinated, so that cross talk and slow signal transmission are a characteristicof the normal operation of these axons, another variable in the nervous system.

The degeneration or destruction of the myelin sheaths that surround the nerve fibers impairs signal transmission and can lead to numerousneurological effects. One of the most prominent of these diseases comes from the body’s own immune system attacking the myelin in the centralnervous system—multiple sclerosis. MS symptoms include fatigue, vision problems, weakness of arms and legs, loss of balance, and tingling ornumbness in one’s extremities (neuropathy). It is more apt to strike younger adults, especially females. Causes might come from infection,environmental or geographic affects, or genetics. At the moment there is no known cure for MS.



Most animal cells can fire or create their own action potential. Muscle cells contract when they fire and are often induced to do so by a nerve impulse.In fact, nerve and muscle cells are physiologically similar, and there are even hybrid cells, such as in the heart, that have characteristics of bothnerves and muscles. Some animals, like the infamous electric eel (see Figure 3.32), use muscles ganged so that their voltages add in order to createa shock great enough to stun prey.

Figure 3.31 Propagation of a nerve impulse down a myelinated axon, from left to right. The signal travels very fast and without energy input in the myelinated regions, but itloses voltage. It is regenerated in the gaps. The signal moves faster than in unmyelinated axons and is insulated from signals in other nerves, limiting cross talk.

Figure 3.32 An electric eel flexes its muscles to create a voltage that stuns prey. (credit: chrisbb, Flickr)

ElectrocardiogramsJust as nerve impulses are transmitted by depolarization and repolarization of adjacent membrane, the depolarization that causes muscle contractioncan also stimulate adjacent muscle cells to depolarize (fire) and contract. Thus, a depolarization wave can be sent across the heart, coordinating itsrhythmic contractions and enabling it to perform its vital function of propelling blood through the circulatory system. Figure 3.33 is a simplified graphicof a depolarization wave spreading across the heart from the sinoarterial (SA) node, the heart’s natural pacemaker.

Figure 3.33 The outer surface of the heart changes from positive to negative during depolarization. This wave of depolarization is spreading from the top of the heart and isrepresented by a vector pointing in the direction of the wave. This vector is a voltage (potential difference) vector. Three electrodes, labeled RA, LA, and LL, are placed on thepatient. Each pair (called leads I, II, and III) measures a component of the depolarization vector and is graphed in an ECG.

An electrocardiogram (ECG) is a record of the voltages created by the wave of depolarization and subsequent repolarization in the heart. Voltagesbetween pairs of electrodes placed on the chest are vector components of the voltage wave on the heart. Standard ECGs have 12 or moreelectrodes, but only three are shown in Figure 3.33 for clarity. Decades ago, three-electrode ECGs were performed by placing electrodes on the leftand right arms and the left leg. The voltage between the right arm and the left leg is called the lead II potential and is the most often graphed. Weshall examine the lead II potential as an indicator of heart-muscle function and see that it is coordinated with arterial blood pressure as well.

Heart function and its four-chamber action are explored in Viscosity and Laminar Flow; Poiseuille’s Law (http://cnx.org/content/m42209/latest/). Basically, the right and left atria receive blood from the body and lungs, respectively, and pump the blood into the ventricles. The right and leftventricles, in turn, pump blood through the lungs and the rest of the body, respectively. Depolarization of the heart muscle causes it to contract. After



AC current:

AC voltage:

alternating current:

contraction it is repolarized to ready it for the next beat. The ECG measures components of depolarization and repolarization of the heart muscle andcan yield significant information on the functioning and malfunctioning of the heart.

Figure 3.34 shows an ECG of the lead II potential and a graph of the corresponding arterial blood pressure. The major features are labeled P, Q, R,S, and T. The P wave is generated by the depolarization and contraction of the atria as they pump blood into the ventricles. The QRS complex iscreated by the depolarization of the ventricles as they pump blood to the lungs and body. Since the shape of the heart and the path of thedepolarization wave are not simple, the QRS complex has this typical shape and time span. The lead II QRS signal also masks the repolarization ofthe atria, which occur at the same time. Finally, the T wave is generated by the repolarization of the ventricles and is followed by the next P wave inthe next heartbeat. Arterial blood pressure varies with each part of the heartbeat, with systolic (maximum) pressure occurring closely after the QRScomplex, which signals contraction of the ventricles.

Figure 3.34 A lead II ECG with corresponding arterial blood pressure. The QRS complex is created by the depolarization and contraction of the ventricles and is followedshortly by the maximum or systolic blood pressure. See text for further description.

Taken together, the 12 leads of a state-of-the-art ECG can yield a wealth of information about the heart. For example, regions of damaged hearttissue, called infarcts, reflect electrical waves and are apparent in one or more lead potentials. Subtle changes due to slight or gradual damage to theheart are most readily detected by comparing a recent ECG to an older one. This is particularly the case since individual heart shape, size, andorientation can cause variations in ECGs from one individual to another. ECG technology has advanced to the point where a portable ECG monitorwith a liquid crystal instant display and a printer can be carried to patients' homes or used in emergency vehicles. See Figure 3.35.

Figure 3.35 This NASA scientist and NEEMO 5 aquanaut’s heart rate and other vital signs are being recorded by a portable device while living in an underwater habitat.(credit: NASA, Life Sciences Data Archive at Johnson Space Center, Houston, Texas)

PhET Explorations: Neuron

Figure 3.36 Neuron (http://cnx.org/content/m42352/1.3/neuron_en.jar)

Stimulate a neuron and monitor what happens. Pause, rewind, and move forward in time in order to observe the ions as they move across theneuron membrane.

Glossarycurrent that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f

is the frequency in hertz

voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage,and f is the frequency in hertz

(AC) the flow of electric charge that periodically reverses direction



http://cnx.org/content/m42352/1.3/neuron_en.jar

ampere:

bioelectricity:

direct current:

drift velocity:

electric current:

electric power:

electrocardiogram (ECG):

microshock sensitive:

nerve conduction:

Ohm’s law:

ohmic:

ohm:

resistance:

resistivity:

rms current:

rms voltage:

semipermeable:

shock hazard:

short circuit:

simple circuit:

temperature coefficient of resistivity:

thermal hazard:

(amp) the SI unit for current; 1 A = 1 C/s

electrical effects in and created by biological systems

(DC) the flow of electric charge in only one direction

the average velocity at which free charges flow in response to an electric field

the rate at which charge flows, I = ΔQ/Δt

the rate at which electrical energy is supplied by a source or dissipated by a device; it is the product of current times voltage

usually abbreviated ECG, a record of voltages created by depolarization and repolarization, especially in the heart

a condition in which a person’s skin resistance is bypassed, possibly by a medical procedure, rendering the personvulnerable to electrical shock at currents about 1/1000 the normally required level

the transport of electrical signals by nerve cells

an empirical relation stating that the current I is proportional to the potential difference V, ∝ V; it is often written as I = V/R, where R isthe resistance

a type of a material for which Ohm's law is valid

the unit of resistance, given by 1Ω = 1 V/A

the electric property that impedes current; for ohmic materials, it is the ratio of voltage to current, R = V/I

an intrinsic property of a material, independent of its shape or size, directly proportional to the resistance, denoted by ρ

the root mean square of the current, Irms = I0 / 2 , where I0 is the peak current, in an AC system

the root mean square of the voltage, Vrms = V0 / 2 , where V0 is the peak voltage, in an AC system

property of a membrane that allows only certain types of ions to cross it

when electric current passes through a person

also known as a “short,” a low-resistance path between terminals of a voltage source

a circuit with a single voltage source and a single resistor

an empirical quantity, denoted by α, which describes the change in resistance or resistivity of a materialwith temperature

a hazard in which electric current causes undesired thermal effects

Section Summary

3.1 Current• Electric current I is the rate at which charge flows, given by

I = ΔQΔt ,

where ΔQ is the amount of charge passing through an area in time Δt .

• The direction of conventional current is taken as the direction in which positive charge moves.• The SI unit for current is the ampere (A), where 1 A = 1 C/s.• Current is the flow of free charges, such as electrons and ions.• Drift velocity vd is the average speed at which these charges move.

• Current I is proportional to drift velocity vd , as expressed in the relationship I = nqAvd . Here, I is the current through a wire of cross-

sectional area A . The wire’s material has a free-charge density n , and each carrier has charge q and a drift velocity vd .

• Electrical signals travel at speeds about 1012 times greater than the drift velocity of free electrons.

3.2 Ohm’s Law: Resistance and Simple Circuits• A simple circuit is one in which there is a single voltage source and a single resistance.

• One statement of Ohm’s law gives the relationship between current I , voltage V , and resistance R in a simple circuit to be I = VR .

• Resistance has units of ohms ( Ω ), related to volts and amperes by 1 Ω = 1 V/A .

• There is a voltage or IR drop across a resistor, caused by the current flowing through it, given by V = IR .

3.3 Resistance and Resistivity


• The resistance R of a cylinder of length L and cross-sectional area A is R = ρLA , where ρ is the resistivity of the material.

• Values of ρ in Table 3.1 show that materials fall into three groups—conductors, semiconductors, and insulators.

• Temperature affects resistivity; for relatively small temperature changes ΔT , resistivity is ρ = ρ0(1 + αΔT) , where ρ0 is the original

resistivity and α is the temperature coefficient of resistivity.• Table 3.2 gives values for α , the temperature coefficient of resistivity.

• The resistance R of an object also varies with temperature: R = R0(1 + αΔT) , where R0 is the original resistance, and R is the

resistance after the temperature change.

3.4 Electric Power and Energy• Electric power P is the rate (in watts) that energy is supplied by a source or dissipated by a device.• Three expressions for electrical power are

P = IV,

P = V 2

R ,

and

P = I 2R.• The energy used by a device with a power P over a time t is E = Pt .

3.5 Alternating Current versus Direct Current• Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant.• The voltage source of an alternating current (AC) system puts out V = V0 sin 2π ft , where V is the voltage at time t , V0 is the peak

voltage, and f is the frequency in hertz.

• In a simple circuit, I = V/R and AC current is I = I0 sin 2π ft , where I is the current at time t , and I0 = V0/R is the peak current.

• The average AC power is Pave = 12I0 V0 .

• Average (rms) current Irms and average (rms) voltage Vrms are Irms = I02

and Vrms = V02

, where rms stands for root mean square.

• Thus, Pave = IrmsVrms .

• Ohm’s law for AC is Irms = VrmsR .

• Expressions for the average power of an AC circuit are Pave = IrmsVrms , Pave = Vrms2

R , and Pave = Irms2 R , analogous to the expressions

for DC circuits.

3.6 Electric Hazards and the Human Body• The two types of electric hazards are thermal (excessive power) and shock (current through a person).• Shock severity is determined by current, path, duration, and AC frequency.• Table 3.3 lists shock hazards as a function of current.• Figure 3.25 graphs the threshold current for two hazards as a function of frequency.

3.7 Nerve Conduction–Electrocardiograms• Electric potentials in neurons and other cells are created by ionic concentration differences across semipermeable membranes.• Stimuli change the permeability and create action potentials that propagate along neurons.• Myelin sheaths speed this process and reduce the needed energy input.• This process in the heart can be measured with an electrocardiogram (ECG).


3.1 Current1. Can a wire carry a current and still be neutral—that is, have a total charge of zero? Explain.

2. Car batteries are rated in ampere-hours ( A ⋅ h ). To what physical quantity do ampere-hours correspond (voltage, charge, . . .), and whatrelationship do ampere-hours have to energy content?

3. If two different wires having identical cross-sectional areas carry the same current, will the drift velocity be higher or lower in the better conductor?

Explain in terms of the equation vd = InqA , by considering how the density of charge carriers n relates to whether or not a material is a good

conductor.

4. Why are two conducting paths from a voltage source to an electrical device needed to operate the device?

5. In cars, one battery terminal is connected to the metal body. How does this allow a single wire to supply current to electrical devices rather thantwo wires?



6. Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneouslywith its wings.

3.2 Ohm’s Law: Resistance and Simple Circuits7. The IR drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passesthrough a resistor? Explain.

8. How is the IR drop in a resistor similar to the pressure drop in a fluid flowing through a pipe?

3.3 Resistance and Resistivity9. In which of the three semiconducting materials listed in Table 3.1 do impurities supply free charges? (Hint: Examine the range of resistivity for eachand determine whether the pure semiconductor has the higher or lower conductivity.)

10. Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar—is its resistance thesame along its length as across its width? (See Figure 3.37.)

Figure 3.37 Does current taking two different paths through the same object encounter different resistance?

11. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why?

12. Explain why R = R0(1 + αΔT) for the temperature variation of the resistance R of an object is not as accurate as ρ = ρ0(1 + αΔT) , which

gives the temperature variation of resistivity ρ .

3.4 Electric Power and Energy13. Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break?

14. The power dissipated in a resistor is given by P = V 2 / R , which means power decreases if resistance increases. Yet this power is also given by

P = I 2R , which means power increases if resistance increases. Explain why there is no contradiction here.

3.5 Alternating Current versus Direct Current15. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied bybatteries.

16. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity?

17. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights makedashed streaks. Explain.

3.6 Electric Hazards and the Human Body18. Using an ohmmeter, a student measures the resistance between various points on his body. He finds that the resistance between two points onthe same finger is about the same as the resistance between two points on opposite hands—both are several hundred thousand ohms. Furthermore,the resistance decreases when more skin is brought into contact with the probes of the ohmmeter. Finally, there is a dramatic drop in resistance (to afew thousand ohms) when the skin is wet. Explain these observations and their implications regarding skin and internal resistance of the humanbody.

19. What are the two major hazards of electricity?

20. Why isn’t a short circuit a shock hazard?

21. What determines the severity of a shock? Can you say that a certain voltage is hazardous without further information?

22. An electrified needle is used to burn off warts, with the circuit being completed by having the patient sit on a large butt plate. Why is this platelarge?

23. Some surgery is performed with high-voltage electricity passing from a metal scalpel through the tissue being cut. Considering the nature ofelectric fields at the surface of conductors, why would you expect most of the current to flow from the sharp edge of the scalpel? Do you think high- orlow-frequency AC is used?

24. Some devices often used in bathrooms, such as hairdryers, often have safety messages saying “Do not use when the bathtub or basin is full ofwater.” Why is this so?

25. We are often advised to not flick electric switches with wet hands, dry your hand first. We are also advised to never throw water on an electric fire.Why is this so?

26. Before working on a power transmission line, linemen will touch the line with the back of the hand as a final check that the voltage is zero. Whythe back of the hand?

27. Why is the resistance of wet skin so much smaller than dry, and why do blood and other bodily fluids have low resistances?


28. Could a person on intravenous infusion (an IV) be microshock sensitive?

29. In view of the small currents that cause shock hazards and the larger currents that circuit breakers and fuses interrupt, how do they play a role inpreventing shock hazards?

3.7 Nerve Conduction–Electrocardiograms

30. Note that in Figure 3.28, both the concentration gradient and the Coulomb force tend to move Na+ ions into the cell. What prevents this?

31. Define depolarization, repolarization, and the action potential.

32. Explain the properties of myelinated nerves in terms of the insulating properties of myelin.




3.1 Current1. What is the current in milliamperes produced by the solar cells of apocket calculator through which 4.00 C of charge passes in 4.00 h?

2. A total of 600 C of charge passes through a flashlight in 0.500 h. Whatis the average current?

3. What is the current when a typical static charge of 0.250 μC moves

from your finger to a metal doorknob in 1.00 μs ?

4. Find the current when 2.00 nC jumps between your comb and hairover a 0.500 - μs time interval.

5. A large lightning bolt had a 20,000-A current and moved 30.0 C ofcharge. What was its duration?

6. The 200-A current through a spark plug moves 0.300 mC of charge.How long does the spark last?

7. (a) A defibrillator sends a 6.00-A current through the chest of a patientby applying a 10,000-V potential as in the figure below. What is theresistance of the path? (b) The defibrillator paddles make contact with thepatient through a conducting gel that greatly reduces the path resistance.Discuss the difficulties that would ensue if a larger voltage were used toproduce the same current through the patient, but with the path havingperhaps 50 times the resistance. (Hint: The current must be about thesame, so a higher voltage would imply greater power. Use this equation

for power: P = I 2R .)

Figure 3.38 The capacitor in a defibrillation unit drives a current through the heart of apatient.

8. During open-heart surgery, a defibrillator can be used to bring a patientout of cardiac arrest. The resistance of the path is 500 Ω and a10.0-mA current is needed. What voltage should be applied?

9. (a) A defibrillator passes 12.0 A of current through the torso of aperson for 0.0100 s. How much charge moves? (b) How many electronspass through the wires connected to the patient? (See figure twoproblems earlier.)

10. A clock battery wears out after moving 10,000 C of charge throughthe clock at a rate of 0.500 mA. (a) How long did the clock run? (b) Howmany electrons per second flowed?

11. The batteries of a submerged non-nuclear submarine supply 1000 Aat full speed ahead. How long does it take to move Avogadro’s number (

6.02×1023 ) of electrons at this rate?

12. Electron guns are used in X-ray tubes. The electrons are acceleratedthrough a relatively large voltage and directed onto a metal target,producing X-rays. (a) How many electrons per second strike the target ifthe current is 0.500 mA? (b) What charge strikes the target in 0.750 s?

13. A large cyclotron directs a beam of He++ nuclei onto a target with a

beam current of 0.250 mA. (a) How many He++ nuclei per second isthis? (b) How long does it take for 1.00 C to strike the target? (c) How

long before 1.00 mol of He++ nuclei strike the target?

14. Repeat the above example on Example 3.3, but for a wire made ofsilver and given there is one free electron per silver atom.

15. Using the results of the above example on Example 3.3, find the driftvelocity in a copper wire of twice the diameter and carrying 20.0 A.

16. A 14-gauge copper wire has a diameter of 1.628 mm. Whatmagnitude current flows when the drift velocity is 1.00 mm/s? (See aboveexample on Example 3.3 for useful information.)

17. SPEAR, a storage ring about 72.0 m in diameter at the StanfordLinear Accelerator (closed in 2009), has a 20.0-A circulating beam ofelectrons that are moving at nearly the speed of light. (See Figure 3.39.)How many electrons are in the beam?

Figure 3.39 Electrons circulating in the storage ring called SPEAR constitute a 20.0-Acurrent. Because they travel close to the speed of light, each electron completesmany orbits in each second.

3.2 Ohm’s Law: Resistance and Simple Circuits18. What current flows through the bulb of a 3.00-V flashlight when its hotresistance is 3.60 Ω ?

19. Calculate the effective resistance of a pocket calculator that has a1.35-V battery and through which 0.200 mA flows.

20. What is the effective resistance of a car’s starter motor when 150 Aflows through it as the car battery applies 11.0 V to the motor?

21. How many volts are supplied to operate an indicator light on a DVDplayer that has a resistance of 140 Ω , given that 25.0 mA passesthrough it?

22. (a) Find the voltage drop in an extension cord having a 0.0600- Ωresistance and through which 5.00 A is flowing. (b) A cheaper cordutilizes thinner wire and has a resistance of 0.300 Ω . What is thevoltage drop in it when 5.00 A flows? (c) Why is the voltage to whateverappliance is being used reduced by this amount? What is the effect onthe appliance?

23. A power transmission line is hung from metal towers with glass

insulators having a resistance of 1.00×109 Ω . What current flowsthrough the insulator if the voltage is 200 kV? (Some high-voltage linesare DC.)

3.3 Resistance and Resistivity24. What is the resistance of a 20.0-m-long piece of 12-gauge copperwire having a 2.053-mm diameter?

25. The diameter of 0-gauge copper wire is 8.252 mm. Find theresistance of a 1.00-km length of such wire used for power transmission.

26. If the 0.100-mm diameter tungsten filament in a light bulb is to have aresistance of 0.200 Ω at 20.0ºC , how long should it be?

27. Find the ratio of the diameter of aluminum to copper wire, if they havethe same resistance per unit length (as they might in household wiring).


28. What current flows through a 2.54-cm-diameter rod of pure silicon

that is 20.0 cm long, when 1.00 × 103 V is applied to it? (Such a rodmay be used to make nuclear-particle detectors, for example.)

29. (a) To what temperature must you raise a copper wire, originally at20.0ºC , to double its resistance, neglecting any changes indimensions? (b) Does this happen in household wiring under ordinarycircumstances?

30. A resistor made of Nichrome wire is used in an application where itsresistance cannot change more than 1.00% from its value at 20.0ºC .Over what temperature range can it be used?

31. Of what material is a resistor made if its resistance is 40.0% greaterat 100ºC than at 20.0ºC ?

32. An electronic device designed to operate at any temperature in therange from –10.0ºC to 55.0ºC contains pure carbon resistors. By whatfactor does their resistance increase over this range?

33. (a) Of what material is a wire made, if it is 25.0 m long with a 0.100mm diameter and has a resistance of 77.7 Ω at 20.0ºC ? (b) What is

its resistance at 150ºC ?

34. Assuming a constant temperature coefficient of resistivity, what is themaximum percent decrease in the resistance of a constantan wirestarting at 20.0ºC ?

35. A wire is drawn through a die, stretching it to four times its originallength. By what factor does its resistance increase?

36. A copper wire has a resistance of 0.500 Ω at 20.0ºC , and an

iron wire has a resistance of 0.525 Ω at the same temperature. Atwhat temperature are their resistances equal?

37. (a) Digital medical thermometers determine temperature bymeasuring the resistance of a semiconductor device called a thermistor(which has α = – 0.0600 / ºC ) when it is at the same temperature asthe patient. What is a patient’s temperature if the thermistor’s resistanceat that temperature is 82.0% of its value at 37.0ºC (normal bodytemperature)? (b) The negative value for α may not be maintained forvery low temperatures. Discuss why and whether this is the case here.(Hint: Resistance can’t become negative.)


(a) Redo Exercise 3.25 taking into account the thermal expansion of thetungsten filament. You may assume a thermal expansion coefficient of

12×10−6 / ºC . (b) By what percentage does your answer differ fromthat in the example?


(a) To what temperature must you raise a resistor made of constantan todouble its resistance, assuming a constant temperature coefficient ofresistivity? (b) To cut it in half? (c) What is unreasonable about theseresults? (d) Which assumptions are unreasonable, or which premises areinconsistent?

3.4 Electric Power and Energy

40. What is the power of a 1.00×102 MV lightning bolt having a

current of 2.00 × 104 A ?

41. What power is supplied to the starter motor of a large truck that draws250 A of current from a 24.0-V battery hookup?

42. A charge of 4.00 C of charge passes through a pocket calculator’ssolar cells in 4.00 h. What is the power output, given the calculator’svoltage output is 3.00 V? (See Figure 3.40.)

Figure 3.40 The strip of solar cells just above the keys of this calculator convert lightto electricity to supply its energy needs. (credit: Evan-Amos, Wikimedia Commons)

43. How many watts does a flashlight that has 6.00×102 C passthrough it in 0.500 h use if its voltage is 3.00 V?

44. Find the power dissipated in each of these extension cords: (a) anextension cord having a 0.0600 - Ω resistance and through which5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with aresistance of 0.300 Ω .45. Verify that the units of a volt-ampere are watts, as implied by theequation P = IV .

46. Show that the units 1 V2 / Ω = 1W , as implied by the equation

P = V 2 / R .

47. Show that the units 1 A2 ⋅ Ω = 1 W , as implied by the equation

P = I 2R .

48. Verify the energy unit equivalence that 1 kW ⋅ h = 3.60×106 J .

49. Electrons in an X-ray tube are accelerated through 1.00×102 kVand directed toward a target to produce X-rays. Calculate the power ofthe electron beam in this tube if it has a current of 15.0 mA.

50. An electric water heater consumes 5.00 kW for 2.00 h per day. Whatis the cost of running it for one year if electricity costs12.0 cents/kW ⋅ h ? See Figure 3.41.

Figure 3.41 On-demand electric hot water heater. Heat is supplied to water only whenneeded. (credit: aviddavid, Flickr)

51. With a 1200-W toaster, how much electrical energy is needed tomake a slice of toast (cooking time = 1 minute)? At 9.0 cents/kW · h ,how much does this cost?

52. What would be the maximum cost of a CFL such that the total cost(investment plus operating) would be the same for both CFL andincandescent 60-W bulbs? Assume the cost of the incandescent bulb is25 cents and that electricity costs 10 cents/kWh . Calculate the cost for1000 hours, as in the cost effectiveness of CFL example.

53. Some makes of older cars have 6.00-V electrical systems. (a) What isthe hot resistance of a 30.0-W headlight in such a car? (b) What currentflows through it?



54. Alkaline batteries have the advantage of putting out constant voltageuntil very nearly the end of their life. How long will an alkaline batteryrated at 1.00 A ⋅ h and 1.58 V keep a 1.00-W flashlight bulb burning?

55. A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at15.0 kV. (a) What is its power output? (b) What is the resistance of thepath?

56. The average television is said to be on 6 hours per day. Estimate theyearly cost of electricity to operate 100 million TVs, assuming their powerconsumption averages 150 W and the cost of electricity averages12.0 cents/kW ⋅ h .

57. An old lightbulb draws only 50.0 W, rather than its original 60.0 W,due to evaporative thinning of its filament. By what factor is its diameterreduced, assuming uniform thinning along its length? Neglect any effectscaused by temperature differences.

58. 00-gauge copper wire has a diameter of 9.266 mm. Calculate the

power loss in a kilometer of such wire when it carries 1.00×102 A .


Cold vaporizers pass a current through water, evaporating it with only asmall increase in temperature. One such home device is rated at 3.50 Aand utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporizationrate in grams per minute? (b) How much water must you put into thevaporizer for 8.00 h of overnight operation? (See Figure 3.42.)

Figure 3.42 This cold vaporizer passes current directly through water, vaporizing itdirectly with relatively little temperature increase.


(a) What energy is dissipated by a lightning bolt having a 20,000-A

current, a voltage of 1.00×102 MV , and a length of 1.00 ms? (b) What

mass of tree sap could be raised from 18.0ºC to its boiling point andthen evaporated by this energy, assuming sap has the same thermalcharacteristics as water?


What current must be produced by a 12.0-V battery-operated bottlewarmer in order to heat 75.0 g of glass, 250 g of baby formula, and

3.00×102 g of aluminum from 20.0ºC to 90.0ºC in 5.00 min?


How much time is needed for a surgical cauterizer to raise thetemperature of 1.00 g of tissue from 37.0ºC to 100ºC and then boilaway 0.500 g of water, if it puts out 2.00 mA at 15.0 kV? Ignore heattransfer to the surroundings.


Hydroelectric generators (see Figure 3.43) at Hoover Dam produce a

maximum current of 8.00×103 A at 250 kV. (a) What is the poweroutput? (b) The water that powers the generators enters and leaves thesystem at low speed (thus its kinetic energy does not change) but loses

160 m in altitude. How many cubic meters per second are needed,assuming 85.0% efficiency?

Figure 3.43 Hydroelectric generators at the Hoover dam. (credit: Jon Sullivan)


(a) Assuming 95.0% efficiency for the conversion of electrical power bythe motor, what current must the 12.0-V batteries of a 750-kg electric carbe able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b)

To climb a 2.00×102 -m -high hill in 2.00 min at a constant 25.0-m/s

speed while exerting 5.00×102 N of force to overcome air resistanceand friction? (c) To travel at a constant 25.0-m/s speed, exerting a

5.00×102 N force to overcome air resistance and friction? See Figure3.44.

Figure 3.44 This REVAi, an electric car, gets recharged on a street in London. (credit:Frank Hebbert)


A light-rail commuter train draws 630 A of 650-V DC electricity whenaccelerating. (a) What is its power consumption rate in kilowatts? (b) Howlong does it take to reach 20.0 m/s starting from rest if its loaded mass is

5.30×104 kg , assuming 95.0% efficiency and constant power? (c) Find

its average acceleration. (d) Discuss how the acceleration you found forthe light-rail train compares to what might be typical for an automobile.


(a) An aluminum power transmission line has a resistance of0.0580 Ω / km . What is its mass per kilometer? (b) What is the massper kilometer of a copper line having the same resistance? A lowerresistance would shorten the heating time. Discuss the practical limits tospeeding the heating by lowering the resistance.


(a) An immersion heater utilizing 120 V can raise the temperature of a

1.00×102 -g aluminum cup containing 350 g of water from 20.0ºC to

95.0ºC in 2.00 min. Find its resistance, assuming it is constant duringthe process. (b) A lower resistance would shorten the heating time.Discuss the practical limits to speeding the heating by lowering theresistance.



(a) What is the cost of heating a hot tub containing 1500 kg of water from10.0ºC to 40.0ºC , assuming 75.0% efficiency to account for heat

transfer to the surroundings? The cost of electricity is 12 cents/kW ⋅ h .(b) What current was used by the 220-V AC electric heater, if this took4.00 h?


(a) What current is needed to transmit 1.00×102 MW of power at 480V? (b) What power is dissipated by the transmission lines if they have a1.00 - Ω resistance? (c) What is unreasonable about this result? (d)Which assumptions are unreasonable, or which premises areinconsistent?


(a) What current is needed to transmit 1.00×102 MW of power at 10.0kV? (b) Find the resistance of 1.00 km of wire that would cause a0.0100% power loss. (c) What is the diameter of a 1.00-km-long copperwire having this resistance? (d) What is unreasonable about theseresults? (e) Which assumptions are unreasonable, or which premises areinconsistent?


Consider an electric immersion heater used to heat a cup of water tomake tea. Construct a problem in which you calculate the neededresistance of the heater so that it increases the temperature of the waterand cup in a reasonable amount of time. Also calculate the cost of theelectrical energy used in your process. Among the things to beconsidered are the voltage used, the masses and heat capacitiesinvolved, heat losses, and the time over which the heating takes place.Your instructor may wish for you to consider a thermal safety switch(perhaps bimetallic) that will halt the process before damagingtemperatures are reached in the immersion unit.

3.5 Alternating Current versus Direct Current72. (a) What is the hot resistance of a 25-W light bulb that runs on 120-VAC? (b) If the bulb’s operating temperature is 2700ºC , what is its

resistance at 2600ºC ?

73. Certain heavy industrial equipment uses AC power that has a peakvoltage of 679 V. What is the rms voltage?

74. A certain circuit breaker trips when the rms current is 15.0 A. What isthe corresponding peak current?

75. Military aircraft use 400-Hz AC power, because it is possible todesign lighter-weight equipment at this higher frequency. What is the timefor one complete cycle of this power?

76. A North American tourist takes his 25.0-W, 120-V AC razor toEurope, finds a special adapter, and plugs it into 240 V AC. Assumingconstant resistance, what power does the razor consume as it is ruined?

77. In this problem, you will verify statements made at the end of thepower losses for Example 3.10. (a) What current is needed to transmit100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a1.00 - Ω transmission line. (c) What percent loss does this represent?

78. A small office-building air conditioner operates on 408-V AC andconsumes 50.0 kW. (a) What is its effective resistance? (b) What is thecost of running the air conditioner during a hot summer month when it ison 8.00 h per day for 30 days and electricity costs 9.00 cents/kW ⋅ h ?

79. What is the peak power consumption of a 120-V AC microwave oventhat draws 10.0 A?

80. What is the peak current through a 500-W room heater that operateson 120-V AC power?

81. Two different electrical devices have the same power consumption,but one is meant to be operated on 120-V AC and the other on 240-VAC. (a) What is the ratio of their resistances? (b) What is the ratio of theircurrents? (c) Assuming its resistance is unaffected, by what factor will thepower increase if a 120-V AC device is connected to 240-V AC?

82. Nichrome wire is used in some radiative heaters. (a) Find theresistance needed if the average power output is to be 1.00 kW utilizing120-V AC. (b) What length of Nichrome wire, having a cross-sectional

area of 5.00mm2 , is needed if the operating temperature is 500º C ?(c) What power will it draw when first switched on?

83. Find the time after t = 0 when the instantaneous voltage of 60-Hz

AC first reaches the following values: (a) V0 / 2 (b) V0 (c) 0.

84. (a) At what two times in the first period following t = 0 does the

instantaneous voltage in 60-Hz AC equal Vrms ? (b) −Vrms ?

3.6 Electric Hazards and the Human Body85. (a) How much power is dissipated in a short circuit of 240-V ACthrough a resistance of 0.250 Ω ? (b) What current flows?

86. What voltage is involved in a 1.44-kW short circuit through a0.100 - Ω resistance?

87. Find the current through a person and identify the likely effect on herif she touches a 120-V AC source: (a) if she is standing on a rubber matand offers a total resistance of 300 k Ω ; (b) if she is standing barefoot

on wet grass and has a resistance of only 4000 k Ω .

88. While taking a bath, a person touches the metal case of a radio. Thepath through the person to the drainpipe and ground has a resistance of4000 Ω . What is the smallest voltage on the case of the radio thatcould cause ventricular fibrillation?

89. Foolishly trying to fish a burning piece of bread from a toaster with ametal butter knife, a man comes into contact with 120-V AC. He does noteven feel it since, luckily, he is wearing rubber-soled shoes. What is theminimum resistance of the path the current follows through the person?

90. (a) During surgery, a current as small as 20.0 μA applied directly to

the heart may cause ventricular fibrillation. If the resistance of theexposed heart is 300 Ω , what is the smallest voltage that poses thisdanger? (b) Does your answer imply that special electrical safetyprecautions are needed?

91. (a) What is the resistance of a 220-V AC short circuit that generates apeak power of 96.8 kW? (b) What would the average power be if thevoltage was 120 V AC?

92. A heart defibrillator passes 10.0 A through a patient’s torso for 5.00ms in an attempt to restore normal beating. (a) How much chargepassed? (b) What voltage was applied if 500 J of energy was dissipated?(c) What was the path’s resistance? (d) Find the temperature increasecaused in the 8.00 kg of affected tissue.


A short circuit in a 120-V appliance cord has a 0.500- Ω resistance.Calculate the temperature rise of the 2.00 g of surrounding materials,assuming their specific heat capacity is 0.200 cal/g⋅ºC and that it takes

0.0500 s for a circuit breaker to interrupt the current. Is this likely to bedamaging?


Consider a person working in an environment where electric currentsmight pass through her body. Construct a problem in which you calculatethe resistance of insulation needed to protect the person from harm.Among the things to be considered are the voltage to which the personmight be exposed, likely body resistance (dry, wet, …), and acceptablecurrents (safe but sensed, safe and unfelt, …).

3.7 Nerve Conduction–Electrocardiograms95. Integrated Concepts

Use the ECG in Figure 3.34 to determine the heart rate in beats perminute assuming a constant time between beats.




(a) Referring to Figure 3.34, find the time systolic pressure lags behindthe middle of the QRS complex. (b) Discuss the reasons for the time lag.


4 CIRCUITS, BIOELECTRICITY, AND DCINSTRUMENTS

Figure 4.1 The complexity of the electric circuits in a computer is surpassed by those in the human brain. (credit: Airman 1st Class Mike Meares, United States Air Force)

Learning Objectives4.1. Resistors in Series and Parallel

• Draw a circuit with resistors in parallel and in series.• Calculate the voltage drop of a current across a resistor using Ohm’s law.• Contrast the way total resistance is calculated for resistors in series and in parallel.• Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit.• Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel.

4.2. Electromotive Force: Terminal Voltage• Compare and contrast the voltage and the electromagnetic force of an electric power source.• Describe what happens to the terminal voltage, current, and power delivered to a load as internal resistance of the voltage source

increases (due to aging of batteries, for example).• Explain why it is beneficial to use more than one voltage source connected in parallel.

4.3. Kirchhoff’s Rules• Analyze a complex circuit using Kirchhoff’s rules, using the conventions for determining the correct signs of various terms.

4.4. DC Voltmeters and Ammeters• Explain why a voltmeter must be connected in parallel with the circuit.• Draw a diagram showing an ammeter correctly connected in a circuit.• Describe how a galvanometer can be used as either a voltmeter or an ammeter.• Find the resistance that must be placed in series with a galvanometer to allow it to be used as a voltmeter with a given reading.• Explain why measuring the voltage or current in a circuit can never be exact.

4.5. Null Measurements• Explain why a null measurement device is more accurate than a standard voltmeter or ammeter.• Demonstrate how a Wheatstone bridge can be used to accurately calculate the resistance in a circuit.

4.6. DC Circuits Containing Resistors and Capacitors• Explain the importance of the time constant, τ , and calculate the time constant for a given resistance and capacitance.• Explain why batteries in a flashlight gradually lose power and the light dims over time.• Describe what happens to a graph of the voltage across a capacitor over time as it charges.• Explain how a timing circuit works and list some applications.• Calculate the necessary speed of a strobe flash needed to “stop” the movement of an object over a particular length.

CHAPTER 4 | CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS 111

Introduction to Circuits, Bioelectricity, and DC InstrumentsElectric circuits are commonplace. Some are simple, such as those in flashlights. Others, such as those used in supercomputers, are extremelycomplex.

This collection of modules takes the topic of electric circuits a step beyond simple circuits. When the circuit is purely resistive, everything in thismodule applies to both DC and AC. Matters become more complex when capacitance is involved. We do consider what happens when capacitorsare connected to DC voltage sources, but the interaction of capacitors and other nonresistive devices with AC is left for a later chapter. Finally, anumber of important DC instruments, such as meters that measure voltage and current, are covered in this chapter.

4.1 Resistors in Series and ParallelMost circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow iscalled resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 4.2. The total resistance of acombination of resistors depends on both their individual values and how they are connected.

Figure 4.2 (a) A series connection of resistors. (b) A parallel connection of resistors.

Resistors in SeriesWhen are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. Forexample, if current flows through a person holding a screwdriver and into the Earth, then R1 in Figure 4.2(a) could be the resistance of the

screwdriver’s shaft, R2 the resistance of its handle, R3 the person’s body resistance, and R4 the resistance of her shoes.

Figure 4.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individualresistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing toavoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of theresistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Figure 4.3 Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 4.3.

According to Ohm’s law, the voltage drop, V , across a resistor when a current flows through it is calculated using the equation V = IR , where Iequals the current in amps (A) and R is the resistance in ohms ( Ω ) . Another way to think of this is that V is the voltage necessary to make a

current I flow through a resistance R .

So the voltage drop across R1 is V1 = IR1 , that across R2 is V2 = IR2 , and that across R3 is V3 = IR3 . The sum of these voltages equals

the voltage output of the source; that is,

(4.1)V = V1 + V2 + V3.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equationPE = qV , where q is the electric charge and V is the voltage. Thus the energy supplied by the source is qV , while that dissipated by the

resistors is

(4.2)qV1 + qV2 + qV3.

112 CHAPTER 4 | CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS


Connections: Conservation Laws

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge,which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena andwill be invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, qV = qV1 + qV2 + qV3 .

The charge q cancels, yielding V = V1 + V2 + V3 , as stated. (Note that the same amount of charge passes through the battery and each resistor

in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

(4.3)V = IR1 + IR2 + IR3 = I(R1 + R2 + R3).

Note that for the equivalent single series resistance Rs , we have

(4.4)V = IRs.

This implies that the total or equivalent series resistance Rs of three resistors is Rs = R1 + R2 + R3 .

This logic is valid in general for any number of resistors in series; thus, the total resistance Rs of a series connection is

(4.5)Rs = R1 + R2 + R3 + ...,

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

Example 4.1 Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a SeriesCircuit

Suppose the voltage output of the battery in Figure 4.3 is 12.0 V , and the resistances are R1 = 1.00 Ω , R2 = 6.00 Ω , and

R3 = 13.0 Ω . (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to

equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and showthat it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

(4.6)Rs = R1 + R2 + R3= 1.00 Ω + 6.00 Ω + 13.0 Ω= 20.0 Ω.


The current is found using Ohm’s law, V = IR . Entering the value of the applied voltage and the total resistance yields the current for thecircuit:

(4.7)I = VRs

= 12.0 V20.0 Ω = 0.600 A.

Strategy and Solution for (c)

The voltage—or IR drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

(4.8)V1 = IR1 = (0.600 A)(1.0 Ω ) = 0.600 V.

Similarly,

(4.9)V2 = IR2 = (0.600 A)(6.0 Ω ) = 3.60 V

and

(4.10)V3 = IR3 = (0.600 A)(13.0 Ω ) = 7.80 V.

Discussion for (c)

The three IR drops add to 12.0 V , as predicted:

(4.11)V1 + V2 + V3 = (0.600 + 3.60 + 7.80) V = 12.0 V.

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law, P = IV , where P is electric

power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law V = IR into Joule’s law, we get thepower dissipated by the first resistor as


(4.12)P1 = I 2R1 = (0.600 A)2(1.00 Ω ) = 0.360 W.

Similarly,

(4.13)P2 = I 2R2 = (0.600 A)2(6.00 Ω ) = 2.16 W

and

(4.14)P3 = I 2R3 = (0.600 A)2(13.0 Ω ) = 4.68 W.

Discussion for (d)

Power can also be calculated using either P = IV or P = V 2

R , where V is the voltage drop across the resistor (not the full voltage of the

source). The same values will be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use P = IV , where V is the source voltage. This gives

(4.15)P = (0.600 A)(12.0 V) = 7.20 W.

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is,

(4.16)P1 + P2 + P3 = (0.360 + 2.16 + 4.68) W = 7.20 W.

Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total powerdissipated by the resistors.

Major Features of Resistors in Series1. Series resistances add: Rs = R1 + R2 + R3 + ....2. The same current flows through each resistor in series.3. Individual resistors in series do not get the total source voltage, but divide it.

Resistors in ParallelFigure 4.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltagesource by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it.

Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). Forexample, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operatecompletely independently. The same is true in your house, or any building. (See Figure 4.4(b).)



Figure 4.4 (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G,Wikimedia Commons)

To find an expression for the equivalent parallel resistance Rp , let us consider the currents that flow and how they are related to resistance. Since

each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are I1 = VR1

, I2 = VR2

, and I3 = VR3

.

Conservation of charge implies that the total current I produced by the source is the sum of these currents:

(4.17)I = I1 + I2 + I3.

Substituting the expressions for the individual currents gives

(4.18)I = V

R1+ V

R2+ V

R3= V⎛⎝ 1

R1+ 1

R2+ 1

R3

⎞⎠.

Note that Ohm’s law for the equivalent single resistance gives

(4.19)I = V

Rp= V⎛⎝ 1

Rp

⎞⎠.

The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance Rp of a

parallel connection is related to the individual resistances by

(4.20)1Rp

= 1R1

+ 1R2

+ 1R .3

+ ....

This relationship results in a total resistance Rp that is less than the smallest of the individual resistances. (This is seen in the next example.) When

resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance islower.


Example 4.2 Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a ParallelCircuit

Let the voltage output of the battery and resistances in the parallel connection in Figure 4.4 be the same as the previously considered seriesconnection: V = 12.0 V , R1 = 1.00 Ω , R2 = 6.00 Ω , and R3 = 13.0 Ω . (a) What is the total resistance? (b) Find the total current.

(c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipatedby each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.


The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives

(4.21)1Rp

= 1R1

+ 1R2

+ 1R3

= 11.00 Ω + 1

6.00 Ω + 113.0 Ω .

Thus,

(4.22)1Rp

= 1.00Ω + 0.1667

Ω + 0.07692Ω = 1.2436

Ω .

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistance Rp . This yields

(4.23)Rp = 11.2436 Ω = 0.8041 Ω .

The total resistance with the correct number of significant digits is Rp = 0.804 Ω .

Discussion for (a)

Rp is, as predicted, less than the smallest individual resistance.


The total current can be found from Ohm’s law, substituting Rp for the total resistance. This gives

(4.24)I = VRp

= 12.0 V0.8041 Ω = 14.92 A.

Discussion for (b)

Current I for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallelconnections has a smaller total resistance than the resistors connected in series.


The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

(4.25)I1 = VR1

= 12.0 V1.00 Ω = 12.0 A.

Similarly,

(4.26)I2 = VR2

= 12.0 V6.00 Ω = 2.00 A

and

(4.27)I3 = VR3

= 12.0 V13.0 Ω = 0.92 A.

Discussion for (c)

The total current is the sum of the individual currents:

(4.28)I1 + I2 + I3 = 14.92 A.

This is consistent with conservation of charge.


The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three

are known. Let us use P = V 2

R , since each resistor gets full voltage. Thus,

(4.29)P1 = V 2

R1= (12.0 V)2

1.00 Ω = 144 W.

Similarly,



(4.30)P2 = V 2

R2= (12.0 V)2

6.00 Ω = 24.0 W

and

(4.31)P3 = V 2

R3= (12.0 V)2

13.0 Ω = 11.1 W.

Discussion for (d)

The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source.

Strategy and Solution for (e)

The total power can also be calculated in several ways. Choosing P = IV , and entering the total current, yields

(4.32)P = IV = (14.92 A)(12.0 V) = 179 W.

Discussion for (e)

Total power dissipated by the resistors is also 179 W:

(4.33)P1 + P2 + P3 = 144 W + 24.0 W + 11.1 W = 179 W.

This is consistent with the law of conservation of energy.

Overall Discussion

Note that both the currents and powers in parallel connections are greater than for the same devices in series.

Major Features of Resistors in Parallel

1. Parallel resistance is found from 1Rp

= 1R1

+ 1R2

+ 1R3

+ ... , and it is smaller than any individual resistance in the combination.

2. Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallelconnections to supply the myriad devices served with the same voltage and to allow them to operate independently.)

3. Parallel resistors do not each get the total current; they divide it.

Combinations of Series and ParallelMore complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially whenwire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel.

Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 4.5. Various parts areidentified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more timeconsuming than difficult.

Figure 4.5 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduceduntil a single equivalent resistance is reached.

The simplest combination of series and parallel resistance, shown in Figure 4.6, is also the most instructive, since it is found in many applications.For example, R1 could be the resistance of wires from a car battery to its electrical devices, which are in parallel. R2 and R3 could be the starter


motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects,as the next example indicates.

Example 4.3 Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and ParallelCircuits

Figure 4.6 shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can considerR1 to be the resistance of wires leading to R2 and R3 . (a) Find the total resistance. (b) What is the IR drop in R1 ? (c) Find the current I2through R2 . (d) What power is dissipated by R2 ?

Figure 4.6 These three resistors are connected to a voltage source so that R2 and R3 are in parallel with one another and that combination is in series with R1 .


To find the total resistance, we note that R2 and R3 are in parallel and their combination Rp is in series with R1 . Thus the total (equivalent)

resistance of this combination is

(4.34)Rtot = R1 + Rp.

First, we find Rp using the equation for resistors in parallel and entering known values:

(4.35)1Rp

= 1R2

+ 1R3

= 16.00 Ω + 1

13.0 Ω = 0.2436Ω .

Inverting gives

(4.36)Rp = 10.2436 Ω = 4.11 Ω .

So the total resistance is

(4.37)Rtot = R1 + Rp = 1.00 Ω + 4.11 Ω = 5.11 Ω .

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω , respectively)found for the same resistors in the two previous examples.


To find the IR drop in R1 , we note that the full current I flows through R1 . Thus its IR drop is

(4.38)V1 = IR1.

We must find I before we can calculate V1 . The total current I is found using Ohm’s law for the circuit. That is,

(4.39)I = VRtot

= 12.0 V5.11 Ω = 2.35 A.

Entering this into the expression above, we get

(4.40)V1 = IR1 = (2.35 A)(1.00 Ω ) = 2.35 V.

Discussion for (b)

The voltage applied to R2 and R3 is less than the total voltage by an amount V1 . When wire resistance is large, it can significantly affect the

operation of the devices represented by R2 and R3 .


To find the current through R2 , we must first find the voltage applied to it. We call this voltage Vp , because it is applied to a parallel

combination of resistors. The voltage applied to both R2 and R3 is reduced by the amount V1 , and so it is



(4.41)Vp = V − V1 = 12.0 V − 2.35 V = 9.65 V.

Now the current I2 through resistance R2 is found using Ohm’s law:

(4.42)I2 =

VpR2

= 9.65 V6.00 Ω = 1.61 A.

Discussion for (c)

The current is less than the 2.00 A that flowed through R2 when it was connected in parallel to the battery in the previous parallel circuit

example.


The power dissipated by R2 is given by

(4.43)P2 = (I2 )2R2 = (1.61 A)2(6.00 Ω ) = 15.5 W.

Discussion for (d)

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

Practical ImplicationsOne implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relativelylarge, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the IR drop in the wires can also besignificant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see thepassenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in Figure 4.7. The device represented by R3 has a very low resistance, and so when

it is switched on, a large current flows. This increased current causes a larger IR drop in the wires represented by R1 , reducing the voltage across

the light bulb (which is R2 ), which then dims noticeably.

Figure 4.7 Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant IR drop in the wiresand reduces the voltage across the light.


Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram ofresistors that cannot be broken down into combinations of series and parallel.

SolutionNo, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such casesKirchhoff’s rules, to be introduced in Kirchhoff’s Rules, will allow you to analyze the circuit.

Problem-Solving Strategies for Series and Parallel Resistors1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they

are labeled in your circuit diagram.2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this

assessment. Resistors are in series if the same current must pass sequentially through them.


4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series andanother for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series orparallel connections, as done in this module and the examples. Special note: When finding R , the reciprocal must be taken with care.

5. Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistanceshould be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices inparallel compared with series, and so on.

4.2 Electromotive Force: Terminal VoltageWhen you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they simply blink off when the battery’s energy isgone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted.

Furthermore, if you connect an excessive number of 12-V lights in parallel to a car battery, they will be dim even when the battery is fresh and even ifthe wires to the lights have very low resistance. This implies that the battery’s output voltage is reduced by the overload.

The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a sourceof electrical energy and an internal resistance. Let us examine both.

Electromotive ForceYou can think of many different types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electricalgenerators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectricdevices create voltage from temperature differences.

A few voltage sources are shown in Figure 4.8. All such devices create a potential difference and can supply current if connected to a resistance.On the small scale, the potential difference creates an electric field that exerts force on charges, causing current. We thus use the nameelectromotive force, abbreviated emf.

Emf is not a force at all; it is a special type of potential difference. To be precise, the electromotive force (emf) is the potential difference of a sourcewhen no current is flowing. Units of emf are volts.

Figure 4.8 A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Leaflet, Wikimedia Commons); the Krasnoyarsk Dam inRussia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of eachdepends on its construction and load, and equals emf only if there is no load.

Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery. However, emfdiffers from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf whenthe battery supplies current, and it declines further as the battery is depleted or loaded down. However, if the device’s output voltage can bemeasured without drawing current, then output voltage will equal emf (even for a very depleted battery).

Internal ResistanceAs noted before, a 12-V truck battery is physically larger, contains more charge and energy, and can deliver a larger current than a 12-V motorcyclebattery. Both are lead-acid batteries with identical emf, but, because of its size, the truck battery has a smaller internal resistance r . Internalresistance is the inherent resistance to the flow of current within the source itself.

Figure 4.9 is a schematic representation of the two fundamental parts of any voltage source. The emf (represented by a script E in the figure) andinternal resistance r are in series. The smaller the internal resistance for a given emf, the more current and the more power the source can supply.



Figure 4.9 Any voltage source (in this case, a carbon-zinc dry cell) has an emf related to its source of potential difference, and an internal resistance r related to its

construction. (Note that the script E stands for emf.). Also shown are the output terminals across which the terminal voltage V is measured. Since V = emf − Ir ,terminal voltage equals emf only if there is no current flowing.

The internal resistance r can behave in complex ways. As noted, r increases as a battery is depleted. But internal resistance may also depend onthe magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeablenickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted.

Things Great and Small: The Submicroscopic Origin of Battery Potential

Various types of batteries are available, with emfs determined by the combination of chemicals involved. We can view this as a molecularreaction (what much of chemistry is about) that separates charge.

The lead-acid battery used in cars and other vehicles is one of the most common types. A single cell (one of six) of this battery is seen in Figure4.10. The cathode (positive) terminal of the cell is connected to a lead oxide plate, while the anode (negative) terminal is connected to a leadplate. Both plates are immersed in sulfuric acid, the electrolyte for the system.

Figure 4.10 Artist’s conception of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is connected tothe lead plates. The lead oxide plates are connected to the positive or cathode terminal of the cell. Sulfuric acid conducts the charge as well as participating in thechemical reaction.

The details of the chemical reaction are left to the reader to pursue in a chemistry text, but their results at the molecular level help explain thepotential created by the battery. Figure 4.11 shows the result of a single chemical reaction. Two electrons are placed on the anode, making itnegative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. Inshort, a separation of charge has been driven by a chemical reaction.

Note that the reaction will not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under manycircumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note also that since the chemicalreactions involve substances with resistance, it is not possible to create the emf without an internal resistance.

Figure 4.11 Artist’s conception of two electrons being forced onto the anode of a cell and two electrons being removed from the cathode of the cell. The chemical reactionin a lead-acid battery places two electrons on the anode and removes two from the cathode. It requires a closed circuit to proceed, since the two electrons must besupplied to the cathode.


Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into account thetypes of atoms and numbers of electrons in them, are able to predict the energy states they can have and the energies of reactions between them.

In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy

divided by charge: V = PEq . An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is

given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a differentenergy and, hence, a different voltage.

Terminal VoltageThe voltage output of a device is measured across its terminals and, thus, is called its terminal voltage V . Terminal voltage is given by

(4.44)V = emf − Ir,

where r is the internal resistance and I is the current flowing at the time of the measurement.

I is positive if current flows away from the positive terminal, as shown in Figure 4.9. You can see that the larger the current, the smaller the terminalvoltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage.

Suppose a load resistance Rload is connected to a voltage source, as in Figure 4.12. Since the resistances are in series, the total resistance in the

circuit is Rload + r . Thus the current is given by Ohm’s law to be

(4.45)I = emfRload + r .

Figure 4.12 Schematic of a voltage source and its load Rload . Since the internal resistance r is in series with the load, it can significantly affect the terminal voltage and

current delivered to the load. (Note that the script E stands for emf.)

We see from this expression that the smaller the internal resistance r , the greater the current the voltage source supplies to its load Rload . As

batteries are depleted, r increases. If r becomes a significant fraction of the load resistance, then the current is significantly reduced, as thefollowing example illustrates.

Example 4.4 Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage andLoad

A certain battery has a 12.0-V emf and an internal resistance of 0.100 Ω . (a) Calculate its terminal voltage when connected to a 10.0- Ωload. (b) What is the terminal voltage when connected to a 0.500- Ω load? (c) What power does the 0.500- Ω load dissipate? (d) If the

internal resistance grows to 0.500 Ω , find the current, terminal voltage, and power dissipated by a 0.500- Ω load.

Strategy

The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltagecan be calculated using the equation V = emf − Ir . Once current is found, the power dissipated by a resistor can also be found.

Solution for (a)

Entering the given values for the emf, load resistance, and internal resistance into the expression above yields

(4.46)I = emfRload + r = 12.0 V

10.1 Ω = 1.188 A.

Enter the known values into the equation V = emf − Ir to get the terminal voltage:

(4.47)V = emf − Ir = 12.0 V − (1.188 A)(0.100 Ω)= 11.9 V.

Discussion for (a)

The terminal voltage here is only slightly lower than the emf, implying that 10.0 Ω is a light load for this particular battery.

Solution for (b)



Similarly, with Rload = 0.500 Ω , the current is

(4.48)I = emfRload + r = 12.0 V

0.600 Ω = 20.0 A.

The terminal voltage is now

(4.49)V = emf − Ir = 12.0 V − (20.0 A)(0.100 Ω)= 10.0 V.

Discussion for (b)

This terminal voltage exhibits a more significant reduction compared with emf, implying 0.500 Ω is a heavy load for this battery.

Solution for (c)

The power dissipated by the 0.500 - Ω load can be found using the formula P = I 2R . Entering the known values gives

(4.50)Pload = I 2Rload = (20.0 A)2(0.500 Ω) = 2.00×102 W.

Discussion for (c)

Note that this power can also be obtained using the expressions V 2

R or IV , where V is the terminal voltage (10.0 V in this case).

Solution for (d)

Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. Asbefore, we first find the current by entering the known values into the expression, yielding

(4.51)I = emfRload + r = 12.0 V

1.00 Ω = 12.0 A.

Now the terminal voltage is

(4.52)V = emf − Ir = 12.0 V − (12.0 A)(0.500 Ω)= 6.00 V,

and the power dissipated by the load is

(4.53)Pload = I 2Rload = (12.0 A)2(0.500 Ω ) = 72.0 W.

Discussion for (d)

We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.

Battery testers, such as those in Figure 4.13, use small load resistors to intentionally draw current to determine whether the terminal voltage dropsbelow an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by itslow terminal voltage.

Figure 4.13 These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy electronicstechnician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo byPhotographer’s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage.(credit: Keith Williamson)

Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is doneroutinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in Figure 4.14. The voltage outputof the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to begreater than the emf, since V = emf − Ir , and I is now negative.


Figure 4.14 A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential.

Multiple Voltage SourcesThere are two voltage sources when a battery charger is used. Voltage sources connected in series are relatively simple. When voltage sources arein series, their internal resistances add and their emfs add algebraically. (See Figure 4.15.) Series connections of voltage sources are common—forexample, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total emf.

But if the cells oppose one another, such as when one is put into an appliance backward, the total emf is less, since it is the algebraic sum of theindividual emfs.

A battery is a multiple connection of voltaic cells, as shown in Figure 4.16. The disadvantage of series connections of cells is that their internalresistances add. One of the authors once owned a 1957 MGA that had two 6-V batteries in series, rather than a single 12-V battery. Thisarrangement produced a large internal resistance that caused him many problems in starting the engine.

Figure 4.15 A series connection of two voltage sources. The emfs (each labeled with a script E) and internal resistances add, giving a total emf of emf1 + emf2 and a

total internal resistance of r1 + r2 .

Figure 4.16 Batteries are multiple connections of individual cells, as shown in this modern rendition of an old print. Single cells, such as AA or C cells, are commonly calledbatteries, although this is technically incorrect.

If the series connection of two voltage sources is made into a complete circuit with the emfs in opposition, then a current of magnitude

I =⎛⎝emf1 – emf2

⎞⎠

r1 + r2flows. See Figure 4.17, for example, which shows a circuit exactly analogous to the battery charger discussed above. If two

voltage sources in series with emfs in the same sense are connected to a load Rload , as in Figure 4.18, then I =⎛⎝emf1 + emf2

⎞⎠

r1 + r2 + Rloadflows.



Figure 4.17 These two voltage sources are connected in series with their emfs in opposition. Current flows in the direction of the greater emf and is limited to

I =⎛⎝emf1 − emf2

⎞⎠

r1 + r2by the sum of the internal resistances. (Note that each emf is represented by script E in the figure.) A battery charger connected to a battery is an

example of such a connection. The charger must have a larger emf than the battery to reverse current through it.

Figure 4.18 This schematic represents a flashlight with two cells (voltage sources) and a single bulb (load resistance) in series. The current that flows is

I =⎛⎝emf1 + emf2

⎞⎠

r1 + r2 + Rload. (Note that each emf is represented by script E in the figure.)

Take-Home Experiment: Flashlight Batteries

Find a flashlight that uses several batteries and find new and old batteries. Based on the discussions in this module, predict the brightness of theflashlight when different combinations of batteries are used. Do your predictions match what you observe? Now place new batteries in theflashlight and leave the flashlight switched on for several hours. Is the flashlight still quite bright? Do the same with the old batteries. Is theflashlight as bright when left on for the same length of time with old and new batteries? What does this say for the case when you are limited inthe number of available new batteries?

Figure 4.19 shows two voltage sources with identical emfs in parallel and connected to a load resistance. In this simple case, the total emf is thesame as the individual emfs. But the total internal resistance is reduced, since the internal resistances are in parallel. The parallel connection thuscan produce a larger current.

Here, I = emf⎛⎝rtot + Rload

⎞⎠

flows through the load, and rtot is less than those of the individual batteries. For example, some diesel-powered cars

use two 12-V batteries in parallel; they produce a total emf of 12 V but can deliver the larger current needed to start a diesel engine.

Figure 4.19 Two voltage sources with identical emfs (each labeled by script E) connected in parallel produce the same emf but have a smaller total internal resistance than the

individual sources. Parallel combinations are often used to deliver more current. Here I = emf⎛⎝rtot + Rload

⎞⎠

flows through the load.


Animals as Electrical DetectorsA number of animals both produce and detect electrical signals. Fish, sharks, platypuses, and echidnas (spiny anteaters) all detect electric fieldsgenerated by nerve activity in prey. Electric eels produce their own emf through biological cells (electric organs) called electroplaques, which arearranged in both series and parallel as a set of batteries.

Electroplaques are flat, disk-like cells; those of the electric eel have a voltage of 0.15 V across each one. These cells are usually located toward thehead or tail of the animal, although in the case of the electric eel, they are found along the entire body. The electroplaques in the South American eelare arranged in 140 rows, with each row stretching horizontally along the body and containing 5,000 electroplaques. This can yield an emf ofapproximately 600 V, and a current of 1 A—deadly.

The mechanism for detection of external electric fields is similar to that for producing nerve signals in the cell through depolarization andrepolarization—the movement of ions across the cell membrane. Within the fish, weak electric fields in the water produce a current in a gel-filledcanal that runs from the skin to sensing cells, producing a nerve signal. The Australian platypus, one of the very few mammals that lay eggs, can

detect fields of 30 mVm , while sharks have been found to be able to sense a field in their snouts as small as 100 mV

m (Figure 4.20). Electric eels

use their own electric fields produced by the electroplaques to stun their prey or enemies.

Figure 4.20 Sand tiger sharks (Carcharias taurus), like this one at the Minnesota Zoo, use electroreceptors in their snouts to locate prey. (credit: Jim Winstead, Flickr)

Solar Cell ArraysAnother example dealing with multiple voltage sources is that of combinations of solar cells—wired in both series and parallel combinations to yield adesired voltage and current. Photovoltaic generation (PV), the conversion of sunlight directly into electricity, is based upon the photoelectric effect, inwhich photons hitting the surface of a solar cell create an electric current in the cell.

Most solar cells are made from pure silicon—either as single-crystal silicon, or as a thin film of silicon deposited upon a glass or metal backing. Mostsingle cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight upon the cell (the incident solar

radiation—the insolation). Under bright noon sunlight, a current of about 100 mA/cm2 of cell surface area is produced by typical single-crystalcells.

Individual solar cells are connected electrically in modules to meet electrical-energy needs. They can be wired together in series or inparallel—connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power outputof 50 W to 140 W.

The output of the solar cells is direct current. For most uses in a home, AC is required, so a device called an inverter must be used to convert the DCto AC. Any extra output can then be passed on to the outside electrical grid for sale to the utility.

Take-Home Experiment: Virtual Solar Cells

One can assemble a “virtual” solar cell array by using playing cards, or business or index cards, to represent a solar cell. Combinations of thesecards in series and/or parallel can model the required array output. Assume each card has an output of 0.5 V and a current (under bright light) of2 A. Using your cards, how would you arrange them to produce an output of 6 A at 3 V (18 W)?

Suppose you were told that you needed only 18 W (but no required voltage). Would you need more cards to make this arrangement?

4.3 Kirchhoff’s RulesMany complex circuits, such as the one in Figure 4.21, cannot be analyzed with the series-parallel techniques developed in Resistors in Series andParallel and Electromotive Force: Terminal Voltage. There are, however, two circuit analysis rules that can be used to analyze any circuit, simpleor complex. These rules are special cases of the laws of conservation of charge and conservation of energy. The rules are known as Kirchhoff’srules, after their inventor Gustav Kirchhoff (1824–1887).



Figure 4.21 This circuit cannot be reduced to a combination of series and parallel connections. Kirchhoff’s rules, special applications of the laws of conservation of charge andenergy, can be used to analyze it. (Note: The script E in the figure represents electromotive force, emf.)

Kirchhoff’s Rules• Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction.• Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.

Explanations of the two rules will now be given, followed by problem-solving hints for applying Kirchhoff’s rules, and a worked example that usesthem.

Kirchhoff’s First RuleKirchhoff’s first rule (the junction rule) is an application of the conservation of charge to a junction; it is illustrated in Figure 4.22. Current is the flowof charge, and charge is conserved; thus, whatever charge flows into the junction must flow out. Kirchhoff’s first rule requires that I1 = I2 + I3 (see

figure). Equations like this can and will be used to analyze circuits and to solve circuit problems.

Making Connections: Conservation Laws

Kirchhoff’s rules for circuit analysis are applications of conservation laws to circuits. The first rule is the application of conservation of charge,while the second rule is the application of conservation of energy. Conservation laws, even used in a specific application, such as circuit analysis,are so basic as to form the foundation of that application.

Figure 4.22 The junction rule. The diagram shows an example of Kirchhoff’s first rule where the sum of the currents into a junction equals the sum of the currents out of ajunction. In this case, the current going into the junction splits and comes out as two currents, so that I1 = I2 + I3 . Here I1 must be 11 A, since I2 is 7 A and I3 is 4

A.

Kirchhoff’s Second RuleKirchhoff’s second rule (the loop rule) is an application of conservation of energy. The loop rule is stated in terms of potential, V , rather than

potential energy, but the two are related since PEelec = qV . Recall that emf is the potential difference of a source when no current is flowing. In a

closed loop, whatever energy is supplied by emf must be transferred into other forms by devices in the loop, since there are no other ways in whichenergy can be transferred into or out of the circuit. Figure 4.23 illustrates the changes in potential in a simple series circuit loop.


Kirchhoff’s second rule requires emf − Ir − IR1 − IR2 = 0 . Rearranged, this is emf = Ir + IR1 + IR2 , which means the emf equals the sum of

the IR (voltage) drops in the loop.

Figure 4.23 The loop rule. An example of Kirchhoff’s second rule where the sum of the changes in potential around a closed loop must be zero. (a) In this standard schematicof a simple series circuit, the emf supplies 18 V, which is reduced to zero by the resistances, with 1 V across the internal resistance, and 12 V and 5 V across the two loadresistances, for a total of 18 V. (b) This perspective view represents the potential as something like a roller coaster, where charge is raised in potential by the emf and loweredby the resistances. (Note that the script E stands for emf.)

Applying Kirchhoff’s RulesBy applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, orresistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can besolved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in theequations you obtain from applying the rules.

1. When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. Forexample, in Figure 4.21, Figure 4.22, and Figure 4.23, currents are labeled I1 , I2 , I3 , and I , and arrows indicate their directions. There is

no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative.2. When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or

counterclockwise. For example, in Figure 4.23 the loop was traversed in the same direction as the current (clockwise). Again, there is no risk;going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of theequation by –1.

Figure 4.24 and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs aretraversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to beconsistent for the sign of the change in potential. (See Example 4.5.)

Figure 4.24 Each of these resistors and voltage sources is traversed from a to b. The potential changes are shown beneath each element and are explained in the text. (Notethat the script E stands for emf.)

• When a resistor is traversed in the same direction as the current, the change in potential is −IR . (See Figure 4.24.)

• When a resistor is traversed in the direction opposite to the current, the change in potential is +IR . (See Figure 4.24.)• When an emf is traversed from – to + (the same direction it moves positive charge), the change in potential is +emf. (See Figure 4.24.)• When an emf is traversed from + to – (opposite to the direction it moves positive charge), the change in potential is − emf. (See Figure

4.24.)



Example 4.5 Calculating Current: Using Kirchhoff’s Rules

Find the currents flowing in the circuit in Figure 4.25.

Figure 4.25 This circuit is similar to that in Figure 4.21, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch arelabeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents.

Strategy

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to useKirchhoff’s rules. Currents have been labeled I1 , I2 , and I3 in the figure and assumptions have been made about their directions. Locations

on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independentequations to allow us to solve for the three unknown currents.

Solution

We begin by applying Kirchhoff’s first or junction rule at point a. This gives

(4.54)I1 = I2 + I3,

since I1 flows into the junction, while I2 and I3 flow out. Applying the junction rule at e produces exactly the same equation, so that no new

information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must beapplied.

Now we consider the loop abcdea. Going from a to b, we traverse R2 in the same (assumed) direction of the current I2 , and so the change in

potential is −I2R2 . Then going from b to c, we go from – to +, so that the change in potential is +emf1 . Traversing the internal resistance

r1 from c to d gives −I2r1 . Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a

change in potential of −I1R1 .

The loop rule states that the changes in potential sum to zero. Thus,

(4.55)−I2R2 + emf1 − I2r1 − I1R1 = −I2(R2 + r1) + emf1 − I1R1 = 0.

Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives

(4.56)−3I 2 + 18 − 6I 1 = 0.

Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives

(4.57)+ I1R1 + I3R3 + I3r2 − emf2= +I1 R1 + I3⎛⎝R3 + r2

⎞⎠− emf2 = 0.

Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered,this becomes

(4.58)+ 6I 1 + 2I 3 − 45 = 0.

These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for I2 :

(4.59)I2 = 6 − 2I 1.

Now solve the third equation for I3 :

(4.60)I3 = 22.5 − 3I 1.


Substituting these two new equations into the first one allows us to find a value for I1 :

(4.61)I1 = I2 + I3 = (6 − 2I 1) + (22.5 − 3I 1) = 28.5 − 5I 1.

Combining terms gives

(4.62)6I 1 = 28.5, and(4.63)I1 = 4.75 A.

Substituting this value for I1 back into the fourth equation gives

(4.64)I2 = 6 − 2I 1 = 6 − 9.50(4.65)I2 = −3.50 A.

The minus sign means I2 flows in the direction opposite to that assumed in Figure 4.25.

Finally, substituting the value for I1 into the fifth equation gives

(4.66)I3 = 22.5−3I 1 = 22.5 − 14.25(4.67)I3 = 8.25 A.

Discussion

Just as a check, we note that indeed I1 = I2 + I3 . The results could also have been checked by entering all of the values into the equation for

the abcdefgha loop.

Problem-Solving Strategies for Kirchhoff’s Rules1. Make certain there is a clear circuit diagram on which you can label all known and unknown resistances, emfs, and currents. If a current is

unknown, you must assign it a direction. This is necessary for determining the signs of potential changes. If you assign the directionincorrectly, the current will be found to have a negative value—no harm done.

2. Apply the junction rule to any junction in the circuit. Each time the junction rule is applied, you should get an equation with a current thatdoes not appear in a previous application—if not, then the equation is redundant.

3. Apply the loop rule to as many loops as needed to solve for the unknowns in the problem. (There must be as many independent equationsas unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine thesigns of the potential changes for each element using the four bulleted points discussed above in conjunction with Figure 4.24.

4. Solve the simultaneous equations for the unknowns. This may involve many algebraic steps, requiring careful checking and rechecking.5. Check to see whether the answers are reasonable and consistent. The numbers should be of the correct order of magnitude, neither

exceedingly large nor vanishingly small. The signs should be reasonable—for example, no resistance should be negative. Check to seethat the values obtained satisfy the various equations obtained from applying the rules. The currents should satisfy the junction rule, forexample.

The material in this section is correct in theory. We should be able to verify it by making measurements of current and voltage. In fact, some of thedevices used to make such measurements are straightforward applications of the principles covered so far and are explored in the next modules. Aswe shall see, a very basic, even profound, fact results—making a measurement alters the quantity being measured.


Can Kirchhoff’s rules be applied to simple series and parallel circuits or are they restricted for use in more complicated circuits that are notcombinations of series and parallel?

SolutionKirchhoff's rules can be applied to any circuit since they are applications to circuits of two conservation laws. Conservation laws are the mostbroadly applicable principles in physics. It is usually mathematically simpler to use the rules for series and parallel in simpler circuits so weemphasize Kirchhoff’s rules for use in more complicated situations. But the rules for series and parallel can be derived from Kirchhoff’s rules.Moreover, Kirchhoff’s rules can be expanded to devices other than resistors and emfs, such as capacitors, and are one of the basic analysisdevices in circuit analysis.

4.4 DC Voltmeters and AmmetersVoltmeters measure voltage, whereas ammeters measure current. Some of the meters in automobile dashboards, digital cameras, cell phones, andtuner-amplifiers are voltmeters or ammeters. (See Figure 4.26.) The internal construction of the simplest of these meters and how they areconnected to the system they monitor give further insight into applications of series and parallel connections.



Figure 4.26 The fuel and temperature gauges (far right and far left, respectively) in this 1996 Volkswagen are voltmeters that register the voltage output of “sender” units,which are hopefully proportional to the amount of gasoline in the tank and the engine temperature. (credit: Christian Giersing)

Voltmeters are connected in parallel with whatever device’s voltage is to be measured. A parallel connection is used because objects in parallelexperience the same potential difference. (See Figure 4.27, where the voltmeter is represented by the symbol V.)

Ammeters are connected in series with whatever device’s current is to be measured. A series connection is used because objects in series have thesame current passing through them. (See Figure 4.28, where the ammeter is represented by the symbol A.)

Figure 4.27 (a) To measure potential differences in this series circuit, the voltmeter (V) is placed in parallel with the voltage source or either of the resistors. Note that terminalvoltage is measured between points a and b. It is not possible to connect the voltmeter directly across the emf without including its internal resistance, r . (b) A digitalvoltmeter in use. (credit: Messtechniker, Wikimedia Commons)


Figure 4.28 An ammeter (A) is placed in series to measure current. All of the current in this circuit flows through the meter. The ammeter would have the same reading iflocated between points d and e or between points f and a as it does in the position shown. (Note that the script capital E stands for emf, and r stands for the internalresistance of the source of potential difference.)

Analog Meters: GalvanometersAnalog meters have a needle that swivels to point at numbers on a scale, as opposed to digital meters, which have numerical readouts similar to ahand-held calculator. The heart of most analog meters is a device called a galvanometer, denoted by G. Current flow through a galvanometer, IG ,

produces a proportional needle deflection. (This deflection is due to the force of a magnetic field upon a current-carrying wire.)

The two crucial characteristics of a given galvanometer are its resistance and current sensitivity. Current sensitivity is the current that gives a full-scale deflection of the galvanometer’s needle, the maximum current that the instrument can measure. For example, a galvanometer with a currentsensitivity of 50 μA has a maximum deflection of its needle when 50 μA flows through it, reads half-scale when 25 μA flows through it, and so

on.

If such a galvanometer has a 25- Ω resistance, then a voltage of only V = IR = ⎛⎝50 μA⎞⎠(25 Ω) = 1.25 mV produces a full-scale reading. By

connecting resistors to this galvanometer in different ways, you can use it as either a voltmeter or ammeter that can measure a broad range ofvoltages or currents.

Galvanometer as Voltmeter

Figure 4.29 shows how a galvanometer can be used as a voltmeter by connecting it in series with a large resistance, R . The value of the resistance

R is determined by the maximum voltage to be measured. Suppose you want 10 V to produce a full-scale deflection of a voltmeter containing a

25-Ω galvanometer with a 50-μA sensitivity. Then 10 V applied to the meter must produce a current of 50 μA . The total resistance must be

(4.68)Rtot = R + r = VI = 10 V

50 μA = 200 kΩ, or

(4.69)R = Rtot − r = 200 kΩ − 25 Ω ≈ 200 k Ω .

( R is so large that the galvanometer resistance, r , is nearly negligible.) Note that 5 V applied to this voltmeter produces a half-scale deflection by

producing a 25-μA current through the meter, and so the voltmeter’s reading is proportional to voltage as desired.

This voltmeter would not be useful for voltages less than about half a volt, because the meter deflection would be small and difficult to readaccurately. For other voltage ranges, other resistances are placed in series with the galvanometer. Many meters have a choice of scales. That choiceinvolves switching an appropriate resistance into series with the galvanometer.

Figure 4.29 A large resistance R placed in series with a galvanometer G produces a voltmeter, the full-scale deflection of which depends on the choice of R . The larger the

voltage to be measured, the larger R must be. (Note that r represents the internal resistance of the galvanometer.)

Galvanometer as Ammeter

The same galvanometer can also be made into an ammeter by placing it in parallel with a small resistance R , often called the shunt resistance, asshown in Figure 4.30. Since the shunt resistance is small, most of the current passes through it, allowing an ammeter to measure currents muchgreater than those producing a full-scale deflection of the galvanometer.

Suppose, for example, an ammeter is needed that gives a full-scale deflection for 1.0 A, and contains the same 25- Ω galvanometer with its

50-μA sensitivity. Since R and r are in parallel, the voltage across them is the same.

These IR drops are IR = IGr so that IR = IGI = R

r . Solving for R , and noting that IG is 50 μA and I is 0.999950 A, we have

(4.70)R = rIG

I = (25 Ω ) 50 μA0.999950 A = 1.25×10−3 Ω .



Figure 4.30 A small shunt resistance R placed in parallel with a galvanometer G produces an ammeter, the full-scale deflection of which depends on the choice of R . The

larger the current to be measured, the smaller R must be. Most of the current ( I ) flowing through the meter is shunted through R to protect the galvanometer. (Note that rrepresents the internal resistance of the galvanometer.) Ammeters may also have multiple scales for greater flexibility in application. The various scales are achieved byswitching various shunt resistances in parallel with the galvanometer—the greater the maximum current to be measured, the smaller the shunt resistance must be.

Taking Measurements Alters the CircuitWhen you use a voltmeter or ammeter, you are connecting another resistor to an existing circuit and, thus, altering the circuit. Ideally, voltmeters andammeters do not appreciably affect the circuit, but it is instructive to examine the circumstances under which they do or do not interfere.

First, consider the voltmeter, which is always placed in parallel with the device being measured. Very little current flows through the voltmeter if itsresistance is a few orders of magnitude greater than the device, and so the circuit is not appreciably affected. (See Figure 4.31(a).) (A largeresistance in parallel with a small one has a combined resistance essentially equal to the small one.) If, however, the voltmeter’s resistance iscomparable to that of the device being measured, then the two in parallel have a smaller resistance, appreciably affecting the circuit. (See Figure4.31(b).) The voltage across the device is not the same as when the voltmeter is out of the circuit.

Figure 4.31 (a) A voltmeter having a resistance much larger than the device ( RVoltmeter >>R ) with which it is in parallel produces a parallel resistance essentially the

same as the device and does not appreciably affect the circuit being measured. (b) Here the voltmeter has the same resistance as the device ( RVoltmeter ≅ R ), so that the

parallel resistance is half of what it is when the voltmeter is not connected. This is an example of a significant alteration of the circuit and is to be avoided.

An ammeter is placed in series in the branch of the circuit being measured, so that its resistance adds to that branch. Normally, the ammeter’sresistance is very small compared with the resistances of the devices in the circuit, and so the extra resistance is negligible. (See Figure 4.32(a).)However, if very small load resistances are involved, or if the ammeter is not as low in resistance as it should be, then the total series resistance issignificantly greater, and the current in the branch being measured is reduced. (See Figure 4.32(b).)

A practical problem can occur if the ammeter is connected incorrectly. If it was put in parallel with the resistor to measure the current in it, you couldpossibly damage the meter; the low resistance of the ammeter would allow most of the current in the circuit to go through the galvanometer, and thiscurrent would be larger since the effective resistance is smaller.

Figure 4.32 (a) An ammeter normally has such a small resistance that the total series resistance in the branch being measured is not appreciably increased. The circuit isessentially unaltered compared with when the ammeter is absent. (b) Here the ammeter’s resistance is the same as that of the branch, so that the total resistance is doubledand the current is half what it is without the ammeter. This significant alteration of the circuit is to be avoided.

One solution to the problem of voltmeters and ammeters interfering with the circuits being measured is to use galvanometers with greater sensitivity.This allows construction of voltmeters with greater resistance and ammeters with smaller resistance than when less sensitive galvanometers areused.

There are practical limits to galvanometer sensitivity, but it is possible to get analog meters that make measurements accurate to a few percent. Notethat the inaccuracy comes from altering the circuit, not from a fault in the meter.

Connections: Limits to Knowledge

Making a measurement alters the system being measured in a manner that produces uncertainty in the measurement. For macroscopic systems,such as the circuits discussed in this module, the alteration can usually be made negligibly small, but it cannot be eliminated entirely. Forsubmicroscopic systems, such as atoms, nuclei, and smaller particles, measurement alters the system in a manner that cannot be made


arbitrarily small. This actually limits knowledge of the system—even limiting what nature can know about itself. We shall see profoundimplications of this when the Heisenberg uncertainty principle is discussed in the modules on quantum mechanics.

There is another measurement technique based on drawing no current at all and, hence, not altering the circuit at all. These are called nullmeasurements and are the topic of Null Measurements. Digital meters that employ solid-state electronics and null measurements can attain

accuracies of one part in 106 .


Digital meters are able to detect smaller currents than analog meters employing galvanometers. How does this explain their ability to measurevoltage and current more accurately than analog meters?

SolutionSince digital meters require less current than analog meters, they alter the circuit less than analog meters. Their resistance as a voltmeter can befar greater than an analog meter, and their resistance as an ammeter can be far less than an analog meter. Consult Figure 4.27 and Figure 4.28and their discussion in the text.

PhET Explorations: Circuit Construction Kit (DC Only), Virtual Lab

Stimulate a neuron and monitor what happens. Pause, rewind, and move forward in time in order to observe the ions as they move across theneuron membrane.

Figure 4.33 Circuit Construction Kit (DC Only), Virtual Lab (http://cnx.org/content/m42360/1.6/circuit-construction-kit-dc-virtual-lab_en.jar)

4.5 Null MeasurementsStandard measurements of voltage and current alter the circuit being measured, introducing uncertainties in the measurements. Voltmeters drawsome extra current, whereas ammeters reduce current flow. Null measurements balance voltages so that there is no current flowing through themeasuring device and, therefore, no alteration of the circuit being measured.

Null measurements are generally more accurate but are also more complex than the use of standard voltmeters and ammeters, and they still havelimits to their precision. In this module, we shall consider a few specific types of null measurements, because they are common and interesting, andthey further illuminate principles of electric circuits.

The PotentiometerSuppose you wish to measure the emf of a battery. Consider what happens if you connect the battery directly to a standard voltmeter as shown inFigure 4.34. (Once we note the problems with this measurement, we will examine a null measurement that improves accuracy.) As discussed before,the actual quantity measured is the terminal voltage V , which is related to the emf of the battery by V = emf − Ir , where I is the current thatflows and r is the internal resistance of the battery.

The emf could be accurately calculated if r were very accurately known, but it is usually not. If the current I could be made zero, then V = emf ,and so emf could be directly measured. However, standard voltmeters need a current to operate; thus, another technique is needed.

Figure 4.34 An analog voltmeter attached to a battery draws a small but nonzero current and measures a terminal voltage that differs from the emf of the battery. (Note thatthe script capital E symbolizes electromotive force, or emf.) Since the internal resistance of the battery is not known precisely, it is not possible to calculate the emf precisely.

A potentiometer is a null measurement device for measuring potentials (voltages). (See Figure 4.35.) A voltage source is connected to a resistorR, say, a long wire, and passes a constant current through it. There is a steady drop in potential (an IR drop) along the wire, so that a variable

potential can be obtained by making contact at varying locations along the wire.

Figure 4.35(b) shows an unknown emfx (represented by script Ex in the figure) connected in series with a galvanometer. Note that emfxopposes the other voltage source. The location of the contact point (see the arrow on the drawing) is adjusted until the galvanometer reads zero.



http://cnx.org/content/m42360/1.6/circuit-construction-kit-dc-virtual-lab_en.jar

When the galvanometer reads zero, emfx = IRx , where Rx is the resistance of the section of wire up to the contact point. Since no current flows

through the galvanometer, none flows through the unknown emf, and so emfx is directly sensed.

Now, a very precisely known standard emfs is substituted for emfx , and the contact point is adjusted until the galvanometer again reads zero, so

that emfs = IRs . In both cases, no current passes through the galvanometer, and so the current I through the long wire is the same. Upon taking

the ratioemfxemfs

, I cancels, giving

(4.71)emfxemfs

= IRxIRs

= RxRs

.

Solving for emfx gives

(4.72)emfx = emfsRxRs

.

Figure 4.35 The potentiometer, a null measurement device. (a) A voltage source connected to a long wire resistor passes a constant current I through it. (b) An unknown emf

(labeled script Ex in the figure) is connected as shown, and the point of contact along R is adjusted until the galvanometer reads zero. The segment of wire has a

resistance Rx and script Ex = IRx , where I is unaffected by the connection since no current flows through the galvanometer. The unknown emf is thus proportional to

the resistance of the wire segment.

Because a long uniform wire is used for R , the ratio of resistances Rx / Rs is the same as the ratio of the lengths of wire that zero the galvanometer

for each emf. The three quantities on the right-hand side of the equation are now known or measured, and emfx can be calculated. The uncertainty

in this calculation can be considerably smaller than when using a voltmeter directly, but it is not zero. There is always some uncertainty in the ratio ofresistances Rx / Rs and in the standard emfs . Furthermore, it is not possible to tell when the galvanometer reads exactly zero, which introduces

error into both Rx and Rs , and may also affect the current I .

Resistance Measurements and the Wheatstone BridgeThere is a variety of so-called ohmmeters that purport to measure resistance. What the most common ohmmeters actually do is to apply a voltage toa resistance, measure the current, and calculate the resistance using Ohm’s law. Their readout is this calculated resistance. Two configurations forohmmeters using standard voltmeters and ammeters are shown in Figure 4.36. Such configurations are limited in accuracy, because the meters alterboth the voltage applied to the resistor and the current that flows through it.


Figure 4.36 Two methods for measuring resistance with standard meters. (a) Assuming a known voltage for the source, an ammeter measures current, and resistance is

calculated as R = VI . (b) Since the terminal voltage V varies with current, it is better to measure it. V is most accurately known when I is small, but I itself is most

accurately known when it is large.

The Wheatstone bridge is a null measurement device for calculating resistance by balancing potential drops in a circuit. (See Figure 4.37.) Thedevice is called a bridge because the galvanometer forms a bridge between two branches. A variety of bridge devices are used to make nullmeasurements in circuits.

Resistors R1 and R2 are precisely known, while the arrow through R3 indicates that it is a variable resistance. The value of R3 can be precisely

read. With the unknown resistance Rx in the circuit, R3 is adjusted until the galvanometer reads zero. The potential difference between points b

and d is then zero, meaning that b and d are at the same potential. With no current running through the galvanometer, it has no effect on the rest ofthe circuit. So the branches abc and adc are in parallel, and each branch has the full voltage of the source. That is, the IR drops along abc and adc

are the same. Since b and d are at the same potential, the IR drop along ad must equal the IR drop along ab. Thus,

(4.73)I1 R1 = I2R3.

Again, since b and d are at the same potential, the IR drop along dc must equal the IR drop along bc. Thus,

(4.74)I1 R2 = I2Rx.

Taking the ratio of these last two expressions gives

(4.75)I1 R1I1 R2

= I2 R3I2 Rx

.

Canceling the currents and solving for Rx yields

(4.76)Rx = R3

R2R1

.

Figure 4.37 The Wheatstone bridge is used to calculate unknown resistances. The variable resistance R3 is adjusted until the galvanometer reads zero with the switch

closed. This simplifies the circuit, allowing Rx to be calculated based on the IR drops as discussed in the text.

This equation is used to calculate the unknown resistance when current through the galvanometer is zero. This method can be very accurate (often tofour significant digits), but it is limited by two factors. First, it is not possible to get the current through the galvanometer to be exactly zero. Second,there are always uncertainties in R1 , R2 , and R3 , which contribute to the uncertainty in Rx .


Identify other factors that might limit the accuracy of null measurements. Would the use of a digital device that is more sensitive than agalvanometer improve the accuracy of null measurements?



SolutionOne factor would be resistance in the wires and connections in a null measurement. These are impossible to make zero, and they can changeover time. Another factor would be temperature variations in resistance, which can be reduced but not completely eliminated by choice ofmaterial. Digital devices sensitive to smaller currents than analog devices do improve the accuracy of null measurements because they allow youto get the current closer to zero.

4.6 DC Circuits Containing Resistors and CapacitorsWhen you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tinyfraction of a second. Why does charging take longer than discharging? This question and a number of other phenomena that involve charging anddischarging capacitors are discussed in this module.

RC Circuits

An RC circuit is one containing a resistor R and a capacitor C . The capacitor is an electrical component that stores electric charge.

Figure 4.38 shows a simple RC circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switchis closed, current flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to theflow of charge by the repulsion of like charges on each plate.

In terms of voltage, this is because voltage across the capacitor is given by Vc = Q / C , where Q is the amount of charge stored on each plate and

C is the capacitance. This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases

from its initial value of I0 = emfR to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no

IR drop, and so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule(the loop rule), discussed in Kirchhoff’s Rules, which says that the algebraic sum of changes in potential around any closed loop must be zero.

The initial current is I0 = emfR , because all of the IR drop is in the resistance. Therefore, the smaller the resistance, the faster a given capacitor

will be charged. Note that the internal resistance of the voltage source is included in R , as are the resistances of the capacitor and the connectingwires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing thecurrent and lengthening the time it takes to get ready for the next flash.

Figure 4.38 (a) An RC circuit with an initially uncharged capacitor. Current flows in the direction shown (opposite of electron flow) as soon as the switch is closed. Mutualrepulsion of like charges in the capacitor progressively slows the flow as the capacitor is charged, stopping the current when the capacitor is fully charged andQ = C ⋅ emf . (b) A graph of voltage across the capacitor versus time, with the switch closing at time t = 0 . (Note that in the two parts of the figure, the capital script E

stands for emf, q stands for the charge stored on the capacitor, and τ is the RC time constant.)

Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. Figure 4.38(b) shows a graph of capacitorvoltage versus time ( t ) starting when the switch is closed at t = 0 . The voltage approaches emf asymptotically, since the closer it gets to emf the

less current flows. The equation for voltage versus time when charging a capacitor C through a resistor R , derived using calculus, is

(4.77)V = emf(1 − e−t / RC) (charging),

where V is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the

natural logarithm. Note that the units of RC are seconds. We define

(4.78)τ = RC,

where τ (the Greek letter tau) is called the time constant for an RC circuit. As noted before, a small resistance R allows the capacitor to charge

faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor C , the less

time needed to charge it. Both factors are contained in τ = RC .

More quantitatively, consider what happens when t = τ = RC . Then the voltage on the capacitor is

(4.79)V = emf⎛⎝1 − e−1⎞⎠ = emf(1 − 0.368) = 0.632 ⋅ emf.


This means that in the time τ = RC , the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time τ . It isa characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, τ .In just a few multiples of the time constant τ , then, the final value is very nearly achieved, as the graph in Figure 4.38(b) illustrates.

Discharging a Capacitor

Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure 4.39 illustrates. Initially, the current is I0 = V0R , driven by the

initial voltage V0 on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential

formula for V . Using calculus, the voltage V on a capacitor C being discharged through a resistor R is found to be

(4.80)V = V e−t / RC(discharging).

Figure 4.39 (a) Closing the switch discharges the capacitor C through the resistor R . Mutual repulsion of like charges on each plate drives the current. (b) A graph of

voltage across the capacitor versus time, with V = V0 at t = 0 . The voltage decreases exponentially, falling a fixed fraction of the way to zero in each subsequent time

constant τ .

The graph in Figure 4.39(b) is an example of this exponential decay. Again, the time constant is τ = RC . A small resistance R allows the capacitorto discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In

the first time interval τ = RC after the switch is closed, the voltage falls to 0.368 of its initial value, since V = V0 ⋅ e−1 = 0.368V0 .

During each successive time τ , the voltage falls to 0.368 of its preceding value. In a few multiples of τ , the voltage becomes very close to zero, asindicated by the graph in Figure 4.39(b).

Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging issignificantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the batteryages, the increasing internal resistance makes the charging process even slower. (You may have noticed this.)

The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure 4.40,can capture a brief instant of a rapid motion because the flash can be less than a microsecond in duration. Such flashes can be made extremelyintense.

During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer ofenemy territory. The brevity of the flash eliminated blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps isto pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to reemit the energy in another form.

Figure 4.40 This stop-motion photograph of a rufous hummingbird (Selasphorus rufus) feeding on a flower was obtained with an extremely brief and intense flash of lightpowered by the discharge of a capacitor through a gas. (credit: Dean E. Biggins, U.S. Fish and Wildlife Service)

Example 4.6 Integrated Concept Problem: Calculating Capacitor Size—Strobe Lights

High-speed flash photography was pioneered by Doc Edgerton in the 1930s, while he was a professor of electrical engineering at MIT. You mighthave seen examples of his work in the amazing shots of hummingbirds in motion, a drop of milk splattering on a table, or a bullet penetrating anapple (see Figure 4.40). To stop the motion and capture these pictures, one needs a high-intensity, very short pulsed flash, as mentioned earlierin this module.



Suppose one wished to capture the picture of a bullet (moving at 5.0×102 m/s ) that was passing through an apple. The duration of the flash is

related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was

10.0 Ω ? Assume the apple is a sphere with a diameter of 8.0×10–2 m.Strategy

We begin by identifying the physical principles involved. This example deals with the strobe light, as discussed above. Figure 4.39 shows thecircuit for this probe. The characteristic time τ of the strobe is given as τ = RC .

Solution

We wish to find C , but we don’t know τ . We want the flash to be on only while the bullet traverses the apple. So we need to use the kinematicequations that describe the relationship between distance x , velocity v , and time t :

(4.81)x = vt or t = xv .

The bullet’s velocity is given as 5.0×102 m/s , and the distance x is 8.0×10–2 m. The traverse time, then, is

(4.82)t = x

v = 8.0×10–2 m5.0×102 m/s

= 1.6×10−4 s.

We set this value for the crossing time t equal to τ . Therefore,

(4.83)C = t

R = 1.6×10−4 s10.0 Ω = 16 μF.

(Note: Capacitance C is typically measured in farads, F , defined as Coulombs per volt. From the equation, we see that C can also be statedin units of seconds per ohm.)

Discussion

The flash interval of 160 μs (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from

science to entertainment. The information from the picture of the apple and bullet was used in the Warren Commission Report on theassassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.

RC Circuits for Timing

RC circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern

cars. The time between wipes is varied by adjusting the resistance in an RC circuit. Another example of an RC circuit is found in novelty jewelry,Halloween costumes, and various toys that have battery-powered flashing lights. (See Figure 4.41 for a timing circuit.)

A more crucial use of RC circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled byelectrical signals generated by the sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract andpump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.

The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant.Pacemakers have sensors that detect body motion and breathing to increase the heart rate during exercise to meet the body’s increased needs forblood and oxygen.

Figure 4.41 (a) The lamp in this RC circuit ordinarily has a very high resistance, so that the battery charges the capacitor as if the lamp were not there. When the voltagereaches a threshold value, a current flows through the lamp that dramatically reduces its resistance, and the capacitor discharges through the lamp as if the battery andcharging resistor were not there. Once discharged, the process starts again, with the flash period determined by the RC constant τ . (b) A graph of voltage versus time forthis circuit.


ammeter:

analog meter:

bridge device:

Example 4.7 Calculating Time: RC Circuit in a Heart Defibrillator

A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of thecircuit is seen in Figure 4.39. (a) What is the time constant if an 8.00-μF capacitor is used and the path resistance through her body is

1.00×103 Ω ? (b) If the initial voltage is 10.0 kV, how long does it take to decline to 5.00×102 V ?

Strategy

Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the

time for the voltage to decline to 5.00×102 V , we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to

5.00×102 V is obtained. Each multiplication corresponds to a time of τ seconds.

Solution for (a)

The time constant τ is given by the equation τ = RC . Entering the given values for resistance and capacitance (and remembering that units

for a farad can be expressed as s / Ω ) gives

(4.84)τ = RC = (1.00×103 Ω )(8.00 μF) = 8.00 ms.

Solution for (b)

In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:

(4.85)V = 0.368V0 = 3.680×103 V at t = 8.00 ms.

(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is

(4.86)V′ = 0.368V= (0.368)⎛⎝3.680×103 V⎞⎠= 1.354×103 V at t = 16.0 ms.

Similarly, after another 8.00 ms, the voltage is

(4.87)V′′ = 0.368V′ = (0.368)(1.354×103 V)= 498 V at t = 24.0 ms.

Discussion

So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value.Such brief times are useful in heart defibrillation, because thebrief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex thanthe one in Figure 4.39, to compensate for magnetic and AC effects that will be covered in Magnetism.


When is the potential difference across a capacitor an emf?

SolutionOnly when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their outputvoltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But thesource of potential difference in a capacitor is fundamental and it is an emf.

PhET Explorations: Circuit Construction Kit (DC only)

An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realisticammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view.

Figure 4.42 Circuit Construction Kit (DC only) (http://cnx.org/content/m42363/1.5/circuit-construction-kit-dc_en.jar)

Glossaryan instrument that measures current

a measuring instrument that gives a readout in the form of a needle movement over a marked gauge

a device that forms a bridge between two branches of a circuit; some bridge devices are used to make null measurements incircuits



http://cnx.org/content/m42363/1.5/circuit-construction-kit-dc_en.jar

capacitance:

capacitor:

conservation laws:

current sensitivity:

current:

digital meter:

electromotive force (emf):

full-scale deflection:

galvanometer:

internal resistance:

Joule’s law:

junction rule:

Kirchhoff’s rules:

loop rule:

null measurements:

Ohm’s law:

ohmmeter:

parallel:

potential difference:

potentiometer:

RC circuit:

resistance:

resistor:

series:

shunt resistance:

terminal voltage:

voltage drop:

voltage:

voltmeter:

Wheatstone bridge:

the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential

an electrical component used to store energy by separating electric charge on two opposing plates

require that energy and charge be conserved in a system

the maximum current that a galvanometer can read

the flow of charge through an electric circuit past a given point of measurement

a measuring instrument that gives a readout in a digital form

the potential difference of a source of electricity when no current is flowing; measured in volts

the maximum deflection of a galvanometer needle, also known as current sensitivity; a galvanometer with a full-scaledeflection of 50 μA has a maximum deflection of its needle when 50 μA flows through it

an analog measuring device, denoted by G, that measures current flow using a needle deflection caused by a magnetic field forceacting upon a current-carrying wire

the amount of resistance within the voltage source

the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: Pe = IV

Kirchhoff’s first rule, which applies the conservation of charge to a junction; current is the flow of charge; thus, whatever chargeflows into the junction must flow out; the rule can be stated I1 = I2 + I3

a set of two rules, based on conservation of charge and energy, governing current and changes in potential in an electric circuit

Kirchhoff’s second rule, which states that in a closed loop, whatever energy is supplied by emf must be transferred into other forms bydevices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Thus, the emf equals the sumof the IR (voltage) drops in the loop and can be stated: emf = Ir + IR1 + IR2

methods of measuring current and voltage more accurately by balancing the circuit so that no current flows through themeasurement device

the relationship between current, voltage, and resistance within an electrical circuit: V = IR

an instrument that applies a voltage to a resistance, measures the current, calculates the resistance using Ohm’s law, and provides areadout of this calculated resistance

the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the powersource; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder

the difference in electric potential between two points in an electric circuit, measured in volts

a null measurement device for measuring potentials (voltages)

a circuit that contains both a resistor and a capacitor

causing a loss of electrical power in a circuit

a component that provides resistance to the current flowing through an electrical circuit

a sequence of resistors or other components wired into a circuit one after the other

a small resistance R placed in parallel with a galvanometer G to produce an ammeter; the larger the current to be measured,

the smaller R must be; most of the current flowing through the meter is shunted through R to protect the galvanometer

the voltage measured across the terminals of a source of potential difference

the loss of electrical power as a current travels through a resistor, wire or other component

the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery

an instrument that measures voltage

a null measurement device for calculating resistance by balancing potential drops in a circuit

Section Summary

4.1 Resistors in Series and Parallel• The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: Rs = R1 + R2 + R3 + ....• Each resistor in a series circuit has the same amount of current flowing through it.


• The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the powersource input.

• The total resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can bedetermined using the formula:

1Rp

= 1R1

+ 1R2

+ 1R3

+ ....

• Each resistor in a parallel circuit has the same full voltage of the source applied to it.• The current flowing through each resistor in a parallel circuit is different, depending on the resistance.• If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by

identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached.

4.2 Electromotive Force: Terminal Voltage• All voltage sources have two fundamental parts—a source of electrical energy that has a characteristic electromotive force (emf), and an

internal resistance r .• The emf is the potential difference of a source when no current is flowing.• The numerical value of the emf depends on the source of potential difference.• The internal resistance r of a voltage source affects the output voltage when a current flows.

• The voltage output of a device is called its terminal voltage V and is given by V = emf − Ir , where I is the electric current and is positivewhen flowing away from the positive terminal of the voltage source.

• When multiple voltage sources are in series, their internal resistances add and their emfs add algebraically.• Solar cells can be wired in series or parallel to provide increased voltage or current, respectively.

4.3 Kirchhoff’s Rules• Kirchhoff’s rules can be used to analyze any circuit, simple or complex.• Kirchhoff’s first rule—the junction rule: The sum of all currents entering a junction must equal the sum of all currents leaving the junction.• Kirchhoff’s second rule—the loop rule: The algebraic sum of changes in potential around any closed circuit path (loop) must be zero.• The two rules are based, respectively, on the laws of conservation of charge and energy.• When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various

terms.• The simpler series and parallel rules are special cases of Kirchhoff’s rules.

4.4 DC Voltmeters and Ammeters• Voltmeters measure voltage, and ammeters measure current.• A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit.• An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit.• Both can be based on the combination of a resistor and a galvanometer, a device that gives an analog reading of current.• Standard voltmeters and ammeters alter the circuit being measured and are thus limited in accuracy.

4.5 Null Measurements• Null measurement techniques achieve greater accuracy by balancing a circuit so that no current flows through the measuring device.• One such device, for determining voltage, is a potentiometer.• Another null measurement device, for determining resistance, is the Wheatstone bridge.• Other physical quantities can also be measured with null measurement techniques.

4.6 DC Circuits Containing Resistors and Capacitors• An RC circuit is one that has both a resistor and a capacitor.

• The time constant τ for an RC circuit is τ = RC .

• When an initially uncharged ( V0 = 0 at t = 0 ) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises,

asymptotically approaching the emf of the voltage source; as a function of time,

V = emf(1 − e−t / RC)(charging).• Within the span of each time constant τ , the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically.

• If a capacitor with an initial voltage V0 is discharged through a resistor starting at t = 0 , then its voltage decreases exponentially as given by

V = V0e−t / RC (discharging).• In each time constant τ , the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically.


4.1 Resistors in Series and Parallel1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device itcontrols. Explain the effect the switch in Figure 4.43 has on current when open and when closed.



Figure 4.43 A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely largeresistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.)

2. What is the voltage across the open switch in Figure 4.43?

3. There is a voltage across an open switch, such as in Figure 4.43. Why, then, is the power dissipated by the open switch small?

4. Why is the power dissipated by a closed switch, such as in Figure 4.43, small?

5. A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 4.44. Explain why the bulb is on when the switch isopen, and off when the switch is closed. (Do not try this—it is hard on the battery!)

Figure 4.44 A wiring mistake put this switch in parallel with the device represented by R . (Note that in this diagram, the script E represents the voltage (or electromotive

force) of the battery.)

6. Knowing that the severity of a shock depends on the magnitude of the current through your body, would you prefer to be in series or parallel with aresistance, such as the heating element of a toaster, if shocked by it? Explain.

7. Would your headlights dim when you start your car’s engine if the wires in your automobile were superconductors? (Do not neglect the battery’sinternal resistance.) Explain.

8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like anopen switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identicalbulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one suchbulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operatingvoltage of each?

9. If two household lightbulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? Explain.

10. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than therequested value. How would you connect the available resistances to attempt to get the smaller value asked for?

11. Before World War II, some radios got power through a “resistance cord” that had a significant resistance. Such a resistance cord reduces thevoltage to a desired level for the radio’s tubes and the like, and it saves the expense of a transformer. Explain why resistance cords become warmand waste energy when the radio is on.

12. Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired inparallel. What is the minimum number of filaments needed for three power settings?

4.2 Electromotive Force: Terminal Voltage13. Is every emf a potential difference? Is every potential difference an emf? Explain.

14. Explain which battery is doing the charging and which is being charged in Figure 4.45.

Figure 4.45


15. Given a battery, an assortment of resistors, and a variety of voltage and current measuring devices, describe how you would determine theinternal resistance of the battery.

16. Two different 12-V automobile batteries on a store shelf are rated at 600 and 850 “cold cranking amps.” Which has the smallest internalresistance?

17. What are the advantages and disadvantages of connecting batteries in series? In parallel?

18. Semitractor trucks use four large 12-V batteries. The starter system requires 24 V, while normal operation of the truck’s other electricalcomponents utilizes 12 V. How could the four batteries be connected to produce 24 V? To produce 12 V? Why is 24 V better than 12 V for starting thetruck’s engine (a very heavy load)?

4.3 Kirchhoff’s Rules19. Can all of the currents going into the junction in Figure 4.46 be positive? Explain.

Figure 4.46

20. Apply the junction rule to junction b in Figure 4.47. Is any new information gained by applying the junction rule at e? (In the figure, each emf isrepresented by script E.)

Figure 4.47

21. (a) What is the potential difference going from point a to point b in Figure 4.47? (b) What is the potential difference going from c to b? (c) From eto g? (d) From e to d?

22. Apply the loop rule to loop afedcba in Figure 4.47.

23. Apply the loop rule to loops abgefa and cbgedc in Figure 4.47.

4.4 DC Voltmeters and Ammeters24. Why should you not connect an ammeter directly across a voltage source as shown in Figure 4.48? (Note that script E in the figure stands foremf.)

Figure 4.48

25. Suppose you are using a multimeter (one designed to measure a range of voltages, currents, and resistances) to measure current in a circuit andyou inadvertently leave it in a voltmeter mode. What effect will the meter have on the circuit? What would happen if you were measuring voltage butaccidentally put the meter in the ammeter mode?



26. Specify the points to which you could connect a voltmeter to measure the following potential differences in Figure 4.49: (a) the potentialdifference of the voltage source; (b) the potential difference across R1 ; (c) across R2 ; (d) across R3 ; (e) across R2 and R3 . Note that there may

be more than one answer to each part.

Figure 4.49

27. To measure currents in Figure 4.49, you would replace a wire between two points with an ammeter. Specify the points between which you wouldplace an ammeter to measure the following: (a) the total current; (b) the current flowing through R1 ; (c) through R2 ; (d) through R3 . Note that

there may be more than one answer to each part.

4.5 Null Measurements28. Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of nullmeasurements?

29. If a potentiometer is used to measure cell emfs on the order of a few volts, why is it most accurate for the standard emfs to be the same order of

magnitude and the resistances to be in the range of a few ohms?

4.6 DC Circuits Containing Resistors and Capacitors30. Regarding the units involved in the relationship τ = RC , verify that the units of resistance times capacitance are time, that is, Ω ⋅ F = s .

31. The RC time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how

would you manipulate resistance in the circuit to adjust the RC constant τ ? Would an adjustment of the applied voltage also be needed to ensurethat the current delivered has an appropriate value?

32. When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the RCconstant of the circuit—it is not possible to measure time variations shorter than RC . How would you manipulate R and C in the circuit to allow thenecessary measurements?

33. Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in thecircuit in Figure 4.38, starting from t = 0 . Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 4.39, starting at

t = 0 , with an initial charge Q0 . Show at least two intervals of τ .

34. When charging a capacitor, as discussed in conjunction with Figure 4.38, how long does it take for the voltage on the capacitor to reach emf? Isthis a problem?

35. When discharging a capacitor, as discussed in conjunction with Figure 4.39, how long does it take for the voltage on the capacitor to reach zero?Is this a problem?

36. Referring to Figure 4.38, draw a graph of potential difference across the resistor versus time, showing at least two intervals of τ . Also draw agraph of current versus time for this situation.

37. A long, inexpensive extension cord is connected from inside the house to a refrigerator outside. The refrigerator doesn’t run as it should. Whatmight be the problem?

38. In Figure 4.41, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have lowresistance? How would you adjust R to get a longer time between flashes? Would adjusting R affect the discharge time?

39. An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when theapparatus is switched off. A “bleeder resistor” is therefore placed across such a capacitor, as shown schematically in Figure 4.50, to bleed the chargefrom it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How doesthis affect the time constant for discharging the capacitor?


Figure 4.50 A bleeder resistor Rbl discharges the capacitor in this electronic device once it is switched off.




4.1 Resistors in Series and ParallelNote: Data taken from figures can be assumed to be accurate tothree significant digits.

1. (a) What is the resistance of ten 275-Ω resistors connected inseries? (b) In parallel?

2. (a) What is the resistance of a 1.00×102 -Ω , a 2.50-kΩ , and a

4.00-k Ω resistor connected in series? (b) In parallel?

3. What are the largest and smallest resistances you can obtain byconnecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?

4. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp areplugged into the same outlet in a 15-A, 120-V circuit. (The three devicesare in parallel when plugged into the same socket.). (a) What current isdrawn by each device? (b) Will this combination blow the 15-A fuse?

5. Your car’s 30.0-W headlight and 2.40-kW starter are ordinarilyconnected in parallel in a 12.0-V system. What power would oneheadlight and the starter consume if connected in series to a 12.0-Vbattery? (Neglect any other resistance in the circuit and any change inresistance in the two devices.)

6. (a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, findthe current and power for each when connected in series. (b) Repeatwhen the resistances are in parallel.

7. Referring to the example combining series and parallel circuits andFigure 4.6, calculate I3 in the following two different ways: (a) from the

known values of I and I2 ; (b) using Ohm’s law for R3 . In both parts

explicitly show how you follow the steps in the Problem-SolvingStrategies for Series and Parallel Resistors.

8. Referring to Figure 4.6: (a) Calculate P3 and note how it compares

with P3 found in the first two example problems in this module. (b) Find

the total power supplied by the source and compare it with the sum of thepowers dissipated by the resistors.

9. Refer to Figure 4.7 and the discussion of lights dimming when a heavyappliance comes on. (a) Given the voltage source is 120 V, the wireresistance is 0.800 Ω , and the bulb is nominally 75.0 W, what powerwill the bulb dissipate if a total of 15.0 A passes through the wires whenthe motor comes on? Assume negligible change in bulb resistance. (b)What power is consumed by the motor?

10. A 240-kV power transmission line carrying 5.00×102 A is hungfrom grounded metal towers by ceramic insulators, each having a

1.00×109 -Ω resistance. Figure 4.51. (a) What is the resistance toground of 100 of these insulators? (b) Calculate the power dissipated by100 of them. (c) What fraction of the power carried by the line is this?Explicitly show how you follow the steps in the Problem-SolvingStrategies for Series and Parallel Resistors.

Figure 4.51 High-voltage (240-kV) transmission line carrying 5.00×102 A is

hung from a grounded metal transmission tower. The row of ceramic insulators

provide 1.00×109 Ω of resistance each.

11. Show that if two resistors R1 and R2 are combined and one is

much greater than the other ( R1 >>R2 ): (a) Their series resistance is

very nearly equal to the greater resistance R1 . (b) Their parallel

resistance is very nearly equal to smaller resistance R2 .


Two resistors, one having a resistance of 145 Ω , are connected in

parallel to produce a total resistance of 150 Ω . (a) What is the value ofthe second resistance? (b) What is unreasonable about this result? (c)Which assumptions are unreasonable or inconsistent?


Two resistors, one having a resistance of 900 kΩ , are connected in

series to produce a total resistance of 0.500 MΩ . (a) What is the valueof the second resistance? (b) What is unreasonable about this result? (c)Which assumptions are unreasonable or inconsistent?

4.2 Electromotive Force: Terminal Voltage14. Standard automobile batteries have six lead-acid cells in series,creating a total emf of 12.0 V. What is the emf of an individual lead-acidcell?

15. Carbon-zinc dry cells (sometimes referred to as non-alkaline cells)have an emf of 1.54 V, and they are produced as single cells or in variouscombinations to form other voltages. (a) How many 1.54-V cells areneeded to make the common 9-V battery used in many small electronicdevices? (b) What is the actual emf of the approximately 9-V battery? (c)Discuss how internal resistance in the series connection of cells willaffect the terminal voltage of this approximately 9-V battery.

16. What is the output voltage of a 3.0000-V lithium cell in a digitalwristwatch that draws 0.300 mA, if the cell’s internal resistance is2.00 Ω ?

17. (a) What is the terminal voltage of a large 1.54-V carbon-zinc dry cellused in a physics lab to supply 2.00 A to a circuit, if the cell’s internalresistance is 0.100 Ω ? (b) How much electrical power does the cellproduce? (c) What power goes to its load?

18. What is the internal resistance of an automobile battery that has anemf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A ischarging it?

19. (a) Find the terminal voltage of a 12.0-V motorcycle battery having a0.600-Ω internal resistance, if it is being charged by a current of 10.0 A.(b) What is the output voltage of the battery charger?

20. A car battery with a 12-V emf and an internal resistance of0.050 Ω is being charged with a current of 60 A. Note that in thisprocess the battery is being charged. (a) What is the potential differenceacross its terminals? (b) At what rate is thermal energy being dissipatedin the battery? (c) At what rate is electric energy being converted tochemical energy? (d) What are the answers to (a) and (b) when thebattery is used to supply 60 A to the starter motor?

21. The hot resistance of a flashlight bulb is 2.30 Ω , and it is run by a

1.58-V alkaline cell having a 0.100-Ω internal resistance. (a) Whatcurrent flows? (b) Calculate the power supplied to the bulb using

I 2 Rbulb . (c) Is this power the same as calculated using V 2

Rbulb?

22. The label on a portable radio recommends the use of rechargeablenickel-cadmium cells (nicads), although they have a 1.25-V emf whilealkaline cells have a 1.58-V emf. The radio has a 3.20-Ω resistance. (a)Draw a circuit diagram of the radio and its batteries. Now, calculate thepower delivered to the radio. (b) When using Nicad cells each having an


internal resistance of 0.0400 Ω . (c) When using alkaline cells each

having an internal resistance of 0.200 Ω . (d) Does this difference seemsignificant, considering that the radio’s effective resistance is loweredwhen its volume is turned up?

23. An automobile starter motor has an equivalent resistance of0.0500 Ω and is supplied by a 12.0-V battery with a 0.0100-Ωinternal resistance. (a) What is the current to the motor? (b) What voltageis applied to it? (c) What power is supplied to the motor? (d) Repeatthese calculations for when the battery connections are corroded and add0.0900 Ω to the circuit. (Significant problems are caused by evensmall amounts of unwanted resistance in low-voltage, high-currentapplications.)

24. A child’s electronic toy is supplied by three 1.58-V alkaline cellshaving internal resistances of 0.0200 Ω in series with a 1.53-V

carbon-zinc dry cell having a 0.100-Ω internal resistance. The load

resistance is 10.0 Ω . (a) Draw a circuit diagram of the toy and itsbatteries. (b) What current flows? (c) How much power is supplied to theload? (d) What is the internal resistance of the dry cell if it goes bad,resulting in only 0.500 W being supplied to the load?

25. (a) What is the internal resistance of a voltage source if its terminalvoltage drops by 2.00 V when the current supplied increases by 5.00 A?(b) Can the emf of the voltage source be found with the informationsupplied?

26. A person with body resistance between his hands of 10.0 k Ωaccidentally grasps the terminals of a 20.0-kV power supply. (Do NOT dothis!) (a) Draw a circuit diagram to represent the situation. (b) If theinternal resistance of the power supply is 2000 Ω , what is the currentthrough his body? (c) What is the power dissipated in his body? (d) If thepower supply is to be made safe by increasing its internal resistance,what should the internal resistance be for the maximum current in thissituation to be 1.00 mA or less? (e) Will this modification compromise theeffectiveness of the power supply for driving low-resistance devices?Explain your reasoning.

27. Electric fish generate current with biological cells calledelectroplaques, which are physiological emf devices. The electroplaquesin the South American eel are arranged in 140 rows, each row stretchinghorizontally along the body and each containing 5000 electroplaques.Each electroplaque has an emf of 0.15 V and internal resistance of0.25 Ω . If the water surrounding the fish has resistance of 800 Ω ,how much current can the eel produce in water from near its head to nearits tail?


A 12.0-V emf automobile battery has a terminal voltage of 16.0 V whenbeing charged by a current of 10.0 A. (a) What is the battery’s internalresistance? (b) What power is dissipated inside the battery? (c) At whatrate (in ºC/min ) will its temperature increase if its mass is 20.0 kg and it

has a specific heat of 0.300 kcal/kg ⋅ ºC , assuming no heat escapes?


A 1.58-V alkaline cell with a 0.200-Ω internal resistance is supplying8.50 A to a load. (a) What is its terminal voltage? (b) What is the value ofthe load resistance? (c) What is unreasonable about these results? (d)Which assumptions are unreasonable or inconsistent?


(a) What is the internal resistance of a 1.54-V dry cell that supplies 1.00W of power to a 15.0-Ω bulb? (b) What is unreasonable about thisresult? (c) Which assumptions are unreasonable or inconsistent?

4.3 Kirchhoff’s Rules31. Apply the loop rule to loop abcdefgha in Figure 4.25.

32. Apply the loop rule to loop aedcba in Figure 4.25.

33. Verify the second equation in Example 4.5 by substituting the valuesfound for the currents I1 and I2 .

34. Verify the third equation in Example 4.5 by substituting the valuesfound for the currents I1 and I3 .

35. Apply the junction rule at point a in Figure 4.52.

Figure 4.52

36. Apply the loop rule to loop abcdefghija in Figure 4.52.

37. Apply the loop rule to loop akledcba in Figure 4.52.

38. Find the currents flowing in the circuit in Figure 4.52. Explicitly showhow you follow the steps in the Problem-Solving Strategies for Seriesand Parallel Resistors.

39. Solve Example 4.5, but use loop abcdefgha instead of loop akledcba.Explicitly show how you follow the steps in the Problem-SolvingStrategies for Series and Parallel Resistors.

40. Find the currents flowing in the circuit in Figure 4.47.


Consider the circuit in Figure 4.53, and suppose that the emfs areunknown and the currents are given to be I1 = 5.00 A , I2 = 3.0 A ,

and I3 = –2.00 A . (a) Could you find the emfs? (b) What is wrong with

the assumptions?

Figure 4.53

4.4 DC Voltmeters and Ammeters42. What is the sensitivity of the galvanometer (that is, what current givesa full-scale deflection) inside a voltmeter that has a 1.00-M Ωresistance on its 30.0-V scale?

43. What is the sensitivity of the galvanometer (that is, what current givesa full-scale deflection) inside a voltmeter that has a 25.0-k Ωresistance on its 100-V scale?



44. Find the resistance that must be placed in series with a 25.0-Ωgalvanometer having a 50.0-μA sensitivity (the same as the one

discussed in the text) to allow it to be used as a voltmeter with a 0.100-Vfull-scale reading.

45. Find the resistance that must be placed in series with a 25.0-Ωgalvanometer having a 50.0-μA sensitivity (the same as the one

discussed in the text) to allow it to be used as a voltmeter with a 3000-Vfull-scale reading. Include a circuit diagram with your solution.

46. Find the resistance that must be placed in parallel with a 25.0-Ωgalvanometer having a 50.0-μA sensitivity (the same as the one

discussed in the text) to allow it to be used as an ammeter with a 10.0-Afull-scale reading. Include a circuit diagram with your solution.

47. Find the resistance that must be placed in parallel with a 25.0-Ωgalvanometer having a 50.0-μA sensitivity (the same as the one

discussed in the text) to allow it to be used as an ammeter with a 300-mAfull-scale reading.

48. Find the resistance that must be placed in series with a 10.0-Ωgalvanometer having a 100-μA sensitivity to allow it to be used as a

voltmeter with: (a) a 300-V full-scale reading, and (b) a 0.300-V full-scalereading.

49. Find the resistance that must be placed in parallel with a 10.0-Ωgalvanometer having a 100-μA sensitivity to allow it to be used as an

ammeter with: (a) a 20.0-A full-scale reading, and (b) a 100-mA full-scalereading.

50. Suppose you measure the terminal voltage of a 1.585-V alkaline cellhaving an internal resistance of 0.100 Ω by placing a 1.00-k Ωvoltmeter across its terminals. (See Figure 4.54.) (a) What current flows?(b) Find the terminal voltage. (c) To see how close the measured terminalvoltage is to the emf, calculate their ratio.

Figure 4.54

51. Suppose you measure the terminal voltage of a 3.200-V lithium cellhaving an internal resistance of 5.00 Ω by placing a 1.00-k Ωvoltmeter across its terminals. (a) What current flows? (b) Find theterminal voltage. (c) To see how close the measured terminal voltage isto the emf, calculate their ratio.

52. A certain ammeter has a resistance of 5.00×10−5 Ω on its

3.00-A scale and contains a 10.0-Ω galvanometer. What is thesensitivity of the galvanometer?

53. A 1.00-MΩ voltmeter is placed in parallel with a 75.0-k Ωresistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Whatis the resistance of the combination? (c) If the voltage across thecombination is kept the same as it was across the 75.0-k Ω resistoralone, what is the percent increase in current? (d) If the current throughthe combination is kept the same as it was through the 75.0-k Ωresistor alone, what is the percentage decrease in voltage? (e) Are thechanges found in parts (c) and (d) significant? Discuss.

54. A 0.0200-Ω ammeter is placed in series with a 10.00-Ω resistor ina circuit. (a) Draw a circuit diagram of the connection. (b) Calculate theresistance of the combination. (c) If the voltage is kept the same acrossthe combination as it was through the 10.00-Ω resistor alone, what isthe percent decrease in current? (d) If the current is kept the samethrough the combination as it was through the 10.00-Ω resistor alone,

what is the percent increase in voltage? (e) Are the changes found inparts (c) and (d) significant? Discuss.


Suppose you have a 40.0-Ω galvanometer with a 25.0-μA sensitivity.

(a) What resistance would you put in series with it to allow it to be usedas a voltmeter that has a full-scale deflection for 0.500 mV? (b) What isunreasonable about this result? (c) Which assumptions are responsible?


(a) What resistance would you put in parallel with a 40.0-Ωgalvanometer having a 25.0-μA sensitivity to allow it to be used as an

ammeter that has a full-scale deflection for 10.0-μA ? (b) What is

unreasonable about this result? (c) Which assumptions are responsible?

4.5 Null Measurements57. What is the emfx of a cell being measured in a potentiometer, if the

standard cell’s emf is 12.0 V and the potentiometer balances forRx = 5.000 Ω and Rs = 2.500 Ω ?

58. Calculate the emfx of a dry cell for which a potentiometer is

balanced when Rx = 1.200 Ω , while an alkaline standard cell with an

emf of 1.600 V requires Rs = 1.247 Ω to balance the potentiometer.

59. When an unknown resistance Rx is placed in a Wheatstone bridge,

it is possible to balance the bridge by adjusting R3 to be 2500 Ω .

What is Rx ifR2R1

= 0.625 ?

60. To what value must you adjust R3 to balance a Wheatstone bridge,

if the unknown resistance Rx is 100 Ω , R1 is 50.0 Ω , and R2 is

175 Ω ?

61. (a) What is the unknown emfx in a potentiometer that balances

when Rx is 10.0 Ω , and balances when Rs is 15.0 Ω for a

standard 3.000-V emf? (b) The same emfx is placed in the same

potentiometer, which now balances when Rs is 15.0 Ω for a standard

emf of 3.100 V. At what resistance Rx will the potentiometer balance?

62. Suppose you want to measure resistances in the range from10.0 Ω to 10.0 kΩ using a Wheatstone bridge that has

R2R1

= 2.000 . Over what range should R3 be adjustable?

4.6 DC Circuits Containing Resistors and Capacitors63. The timing device in an automobile’s intermittent wiper system isbased on an RC time constant and utilizes a 0.500-μF capacitor and

a variable resistor. Over what range must R be made to vary to achievetime constants from 2.00 to 15.0 s?

64. A heart pacemaker fires 72 times a minute, each time a 25.0-nFcapacitor is charged (by a battery in series with a resistor) to 0.632 of itsfull voltage. What is the value of the resistance?

65. The duration of a photographic flash is related to an RC time

constant, which is 0.100 μs for a certain camera. (a) If the resistance of

the flash lamp is 0.0400 Ω during discharge, what is the size of thecapacitor supplying its energy? (b) What is the time constant for chargingthe capacitor, if the charging resistance is 800 kΩ ?

66. A 2.00- and a 7.50-μF capacitor can be connected in series or

parallel, as can a 25.0- and a 100-kΩ resistor. Calculate the four RC


time constants possible from connecting the resulting capacitance andresistance in series.

67. After two time constants, what percentage of the final voltage, emf, ison an initially uncharged capacitor C , charged through a resistance R?

68. A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V

emf are connected in series. (a) What is the initial current? (b) What isthe RC time constant? (c) What is the current after one time constant?(d) What is the voltage on the capacitor after one time constant?

69. A heart defibrillator being used on a patient has an RC timeconstant of 10.0 ms due to the resistance of the patient and thecapacitance of the defibrillator. (a) If the defibrillator has an 8.00-μFcapacitance, what is the resistance of the path through the patient? (Youmay neglect the capacitance of the patient and the resistance of thedefibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to

decline to 6.00×102 V ?

70. An ECG monitor must have an RC time constant less than

1.00×102 μs to be able to measure variations in voltage over small

time intervals. (a) If the resistance of the circuit (due mostly to that of thepatient’s chest) is 1.00 kΩ , what is the maximum capacitance of thecircuit? (b) Would it be difficult in practice to limit the capacitance to lessthan the value found in (a)?

71. Figure 4.55 shows how a bleeder resistor is used to discharge acapacitor after an electronic device is shut off, allowing a person to workon the electronics with less risk of shock. (a) What is the time constant?(b) How long will it take to reduce the voltage on the capacitor to 0.250%(5% of 5%) of its full value once discharge begins? (c) If the capacitor ischarged to a voltage V0 through a 100-Ω resistance, calculate the

time it takes to rise to 0.865V0 (This is about two time constants.)

Figure 4.55

72. Using the exact exponential treatment, find how much time isrequired to discharge a 250-μF capacitor through a 500-Ω resistor

down to 1.00% of its original voltage.

73. Using the exact exponential treatment, find how much time isrequired to charge an initially uncharged 100-pF capacitor through a75.0-M Ω resistor to 90.0% of its final voltage.


If you wish to take a picture of a bullet traveling at 500 m/s, then a verybrief flash of light produced by an RC discharge through a flash tube

can limit blurring. Assuming 1.00 mm of motion during one RC constant

is acceptable, and given that the flash is driven by a 600-μF capacitor,

what is the resistance in the flash tube?


A flashing lamp in a Christmas earring is based on an RC discharge ofa capacitor through its resistance. The effective duration of the flash is0.250 s, during which it produces an average 0.500 W from an average3.00 V. (a) What energy does it dissipate? (b) How much charge movesthrough the lamp? (c) Find the capacitance. (d) What is the resistance ofthe lamp?


A 160-μF capacitor charged to 450 V is discharged through a

31.2-k Ω resistor. (a) Find the time constant. (b) Calculate thetemperature increase of the resistor, given that its mass is 2.50 g and its

specific heat is 1.67 kJkg ⋅ ºC , noting that most of the thermal energy is

retained in the short time of the discharge. (c) Calculate the newresistance, assuming it is pure carbon. (d) Does this change in resistanceseem significant?


(a) Calculate the capacitance needed to get an RC time constant of

1.00×103 s with a 0.100-Ω resistor. (b) What is unreasonable aboutthis result? (c) Which assumptions are responsible?


Consider a camera’s flash unit. Construct a problem in which youcalculate the size of the capacitor that stores energy for the flash lamp.Among the things to be considered are the voltage applied to thecapacitor, the energy needed in the flash and the associated chargeneeded on the capacitor, the resistance of the flash lamp duringdischarge, and the desired RC time constant.


Consider a rechargeable lithium cell that is to be used to power acamcorder. Construct a problem in which you calculate the internalresistance of the cell during normal operation. Also, calculate theminimum voltage output of a battery charger to be used to recharge yourlithium cell. Among the things to be considered are the emf and usefulterminal voltage of a lithium cell and the current it should be able tosupply to a camcorder.



5 MAGNETISM

Figure 5.1 The magnificent spectacle of the Aurora Borealis, or northern lights, glows in the northern sky above Bear Lake near Eielson Air Force Base, Alaska. Shaped by theEarth’s magnetic field, this light is produced by radiation spewed from solar storms. (credit: Senior Airman Joshua Strang, via Flickr)

Learning Objectives5.1. Magnets

• Describe the difference between the north and south poles of a magnet.• Describe how magnetic poles interact with each other.

5.2. Ferromagnets and Electromagnets• Define ferromagnet.• Describe the role of magnetic domains in magnetization.• Explain the significance of the Curie temperature.• Describe the relationship between electricity and magnetism.

5.3. Magnetic Fields and Magnetic Field Lines• Define magnetic field and describe the magnetic field lines of various magnetic fields.

5.4. Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field• Describe the effects of magnetic fields on moving charges.• Use the right hand rule 1 to determine the velocity of a charge, the direction of the magnetic field, and the direction of the magnetic force

on a moving charge.• Calculate the magnetic force on a moving charge.

5.5. Force on a Moving Charge in a Magnetic Field: Examples and Applications• Describe the effects of a magnetic field on a moving charge.• Calculate the radius of curvature of the path of a charge that is moving in a magnetic field.

5.6. The Hall Effect• Describe the Hall effect.• Calculate the Hall emf across a current-carrying conductor.

5.7. Magnetic Force on a Current-Carrying Conductor• Describe the effects of a magnetic force on a current-carrying conductor.• Calculate the magnetic force on a current-carrying conductor.

5.8. Torque on a Current Loop: Motors and Meters• Describe how motors and meters work in terms of torque on a current loop.• Calculate the torque on a current-carrying loop in a magnetic field.

5.9. Magnetic Fields Produced by Currents: Ampere’s Law• Calculate current that produces a magnetic field.• Use the right hand rule 2 to determine the direction of current or the direction of magnetic field loops.

5.10. Magnetic Force between Two Parallel Conductors• Describe the effects of the magnetic force between two conductors.• Calculate the force between two parallel conductors.

5.11. More Applications of Magnetism• Describe some applications of magnetism.

CHAPTER 5 | MAGNETISM 151

Introduction to MagnetismOne evening, an Alaskan sticks a note to his refrigerator with a small magnet. Through the kitchen window, the Aurora Borealis glows in the night sky.This grand spectacle is shaped by the same force that holds the note to the refrigerator.

People have been aware of magnets and magnetism for thousands of years. The earliest records date to well before the time of Christ, particularly ina region of Asia Minor called Magnesia (the name of this region is the source of words like magnetic). Magnetic rocks found in Magnesia, which isnow part of western Turkey, stimulated interest during ancient times. A practical application for magnets was found later, when they were employedas navigational compasses. The use of magnets in compasses resulted not only in improved long-distance sailing, but also in the names of “north”and “south” being given to the two types of magnetic poles.

Today magnetism plays many important roles in our lives. Physicists’ understanding of magnetism has enabled the development of technologies thataffect our everyday lives. The iPod in your purse or backpack, for example, wouldn’t have been possible without the applications of magnetism andelectricity on a small scale.

The discovery that weak changes in a magnetic field in a thin film of iron and chromium could bring about much larger changes in electricalresistance was one of the first large successes of nanotechnology. The 2007 Nobel Prize in Physics went to Albert Fert from France and PeterGrunberg from Germany for this discovery of giant magnetoresistance and its applications to computer memory.

All electric motors, with uses as diverse as powering refrigerators, starting cars, and moving elevators, contain magnets. Generators, whetherproducing hydroelectric power or running bicycle lights, use magnetic fields. Recycling facilities employ magnets to separate iron from other refuse.Hundreds of millions of dollars are spent annually on magnetic containment of fusion as a future energy source. Magnetic resonance imaging (MRI)has become an important diagnostic tool in the field of medicine, and the use of magnetism to explore brain activity is a subject of contemporaryresearch and development. The list of applications also includes computer hard drives, tape recording, detection of inhaled asbestos, and levitation ofhigh-speed trains. Magnetism is used to explain atomic energy levels, cosmic rays, and charged particles trapped in the Van Allen belts. Once again,we will find all these disparate phenomena are linked by a small number of underlying physical principles.

Figure 5.2 Engineering of technology like iPods would not be possible without a deep understanding magnetism. (credit: Jesse! S?, Flickr)

5.1 Magnets

Figure 5.3 Magnets come in various shapes, sizes, and strengths. All have both a north pole and a south pole. There is never an isolated pole (a monopole).

All magnets attract iron, such as that in a refrigerator door. However, magnets may attract or repel other magnets. Experimentation shows that allmagnets have two poles. If freely suspended, one pole will point toward the north. The two poles are thus named the north magnetic pole and thesouth magnetic pole (or more properly, north-seeking and south-seeking poles, for the attractions in those directions).

Universal Characteristics of Magnets and Magnetic Poles

It is a universal characteristic of all magnets that like poles repel and unlike poles attract. (Note the similarity with electrostatics: unlike chargesattract and like charges repel.)

Further experimentation shows that it is impossible to separate north and south poles in the manner that + and − charges can be separated.

152 CHAPTER 5 | MAGNETISM


Figure 5.4 One end of a bar magnet is suspended from a thread that points toward north. The magnet’s two poles are labeled N and S for north-seeking and south-seekingpoles, respectively.

Misconception Alert: Earth’s Geographic North Pole Hides an S

The Earth acts like a very large bar magnet with its south-seeking pole near the geographic North Pole. That is why the north pole of yourcompass is attracted toward the geographic north pole of the Earth—because the magnetic pole that is near the geographic North Pole isactually a south magnetic pole! Confusion arises because the geographic term “North Pole” has come to be used (incorrectly) for the magneticpole that is near the North Pole. Thus, “North magnetic pole” is actually a misnomer—it should be called the South magnetic pole.

Figure 5.5 Unlike poles attract, whereas like poles repel.

Figure 5.6 North and south poles always occur in pairs. Attempts to separate them result in more pairs of poles. If we continue to split the magnet, we will eventually get downto an iron atom with a north pole and a south pole—these, too, cannot be separated.


The fact that magnetic poles always occur in pairs of north and south is true from the very large scale—for example, sunspots always occur in pairsthat are north and south magnetic poles—all the way down to the very small scale. Magnetic atoms have both a north pole and a south pole, as domany types of subatomic particles, such as electrons, protons, and neutrons.

Making Connections: Take-Home Experiment—Refrigerator Magnets

We know that like magnetic poles repel and unlike poles attract. See if you can show this for two refrigerator magnets. Will the magnets stick ifyou turn them over? Why do they stick to the door anyway? What can you say about the magnetic properties of the door next to the magnet? Dorefrigerator magnets stick to metal or plastic spoons? Do they stick to all types of metal?

5.2 Ferromagnets and Electromagnets

FerromagnetsOnly certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are called ferromagnetic, afterthe Latin word for iron, ferrum. A group of materials made from the alloys of the rare earth elements are also used as strong and permanent magnets;a popular one is neodymium. Other materials exhibit weak magnetic effects, which are detectable only with sensitive instruments. Not only doferromagnetic materials respond strongly to magnets (the way iron is attracted to magnets), they can also be magnetized themselves—that is, theycan be induced to be magnetic or made into permanent magnets.

Figure 5.7 An unmagnetized piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron becomes a permanent magnet withthe poles aligned as shown: its south pole is adjacent to the north pole of the original magnet, and its north pole is adjacent to the south pole of the original magnet. Note thatthere are attractive forces between the magnets.

When a magnet is brought near a previously unmagnetized ferromagnetic material, it causes local magnetization of the material with unlike polesclosest, as in Figure 5.7. (This results in the attraction of the previously unmagnetized material to the magnet.) What happens on a microscopic scaleis illustrated in Figure 5.8. The regions within the material called domains act like small bar magnets. Within domains, the poles of individual atomsare aligned. Each atom acts like a tiny bar magnet. Domains are small and randomly oriented in an unmagnetized ferromagnetic object. In responseto an external magnetic field, the domains may grow to millimeter size, aligning themselves as shown in Figure 5.8(b). This induced magnetizationcan be made permanent if the material is heated and then cooled, or simply tapped in the presence of other magnets.

Figure 5.8 (a) An unmagnetized piece of iron (or other ferromagnetic material) has randomly oriented domains. (b) When magnetized by an external field, the domains showgreater alignment, and some grow at the expense of others. Individual atoms are aligned within domains; each atom acts like a tiny bar magnet.

Conversely, a permanent magnet can be demagnetized by hard blows or by heating it in the absence of another magnet. Increased thermal motion athigher temperature can disrupt and randomize the orientation and the size of the domains. There is a well-defined temperature for ferromagneticmaterials, which is called the Curie temperature, above which they cannot be magnetized. The Curie temperature for iron is 1043 K (770ºC) ,

which is well above room temperature. There are several elements and alloys that have Curie temperatures much lower than room temperature andare ferromagnetic only below those temperatures.

ElectromagnetsEarly in the 19th century, it was discovered that electrical currents cause magnetic effects. The first significant observation was by the Danishscientist Hans Christian Oersted (1777–1851), who found that a compass needle was deflected by a current-carrying wire. This was the firstsignificant evidence that the movement of charges had any connection with magnets. Electromagnetism is the use of electric current to makemagnets. These temporarily induced magnets are called electromagnets. Electromagnets are employed for everything from a wrecking yard cranethat lifts scrapped cars to controlling the beam of a 90-km-circumference particle accelerator to the magnets in medical imaging machines (SeeFigure 5.9).



Figure 5.9 Instrument for magnetic resonance imaging (MRI). The device uses a superconducting cylindrical coil for the main magnetic field. The patient goes into this “tunnel”on the gurney. (credit: Bill McChesney, Flickr)

Figure 5.10 shows that the response of iron filings to a current-carrying coil and to a permanent bar magnet. The patterns are similar. In fact,electromagnets and ferromagnets have the same basic characteristics—for example, they have north and south poles that cannot be separated andfor which like poles repel and unlike poles attract.

Figure 5.10 Iron filings near (a) a current-carrying coil and (b) a magnet act like tiny compass needles, showing the shape of their fields. Their response to a current-carryingcoil and a permanent magnet is seen to be very similar, especially near the ends of the coil and the magnet.

Combining a ferromagnet with an electromagnet can produce particularly strong magnetic effects. (See Figure 5.11.) Whenever strong magneticeffects are needed, such as lifting scrap metal, or in particle accelerators, electromagnets are enhanced by ferromagnetic materials. Limits to howstrong the magnets can be made are imposed by coil resistance (it will overheat and melt at sufficiently high current), and so superconductingmagnets may be employed. These are still limited, because superconducting properties are destroyed by too great a magnetic field.

Figure 5.11 An electromagnet with a ferromagnetic core can produce very strong magnetic effects. Alignment of domains in the core produces a magnet, the poles of whichare aligned with the electromagnet.

Figure 5.12 shows a few uses of combinations of electromagnets and ferromagnets. Ferromagnetic materials can act as memory devices, becausethe orientation of the magnetic fields of small domains can be reversed or erased. Magnetic information storage on videotapes and computer harddrives are among the most common applications. This property is vital in our digital world.


Figure 5.12 An electromagnet induces regions of permanent magnetism on a floppy disk coated with a ferromagnetic material. The information stored here is digital (a regionis either magnetic or not); in other applications, it can be analog (with a varying strength), such as on audiotapes.

Current: The Source of All MagnetismAn electromagnet creates magnetism with an electric current. In later sections we explore this more quantitatively, finding the strength and directionof magnetic fields created by various currents. But what about ferromagnets? Figure 5.13 shows models of how electric currents create magnetismat the submicroscopic level. (Note that we cannot directly observe the paths of individual electrons about atoms, and so a model or visual image,consistent with all direct observations, is made. We can directly observe the electron’s orbital angular momentum, its spin momentum, andsubsequent magnetic moments, all of which are explained with electric-current-creating subatomic magnetism.) Currents, including those associatedwith other submicroscopic particles like protons, allow us to explain ferromagnetism and all other magnetic effects. Ferromagnetism, for example,results from an internal cooperative alignment of electron spins, possible in some materials but not in others.

Crucial to the statement that electric current is the source of all magnetism is the fact that it is impossible to separate north and south magnetic poles.(This is far different from the case of positive and negative charges, which are easily separated.) A current loop always produces a magneticdipole—that is, a magnetic field that acts like a north pole and south pole pair. Since isolated north and south magnetic poles, called magneticmonopoles, are not observed, currents are used to explain all magnetic effects. If magnetic monopoles did exist, then we would have to modify thisunderlying connection that all magnetism is due to electrical current. There is no known reason that magnetic monopoles should not exist—they aresimply never observed—and so searches at the subnuclear level continue. If they do not exist, we would like to find out why not. If they do exist, wewould like to see evidence of them.

Electric Currents and Magnetism

Electric current is the source of all magnetism.

Figure 5.13 (a) In the planetary model of the atom, an electron orbits a nucleus, forming a closed-current loop and producing a magnetic field with a north pole and a southpole. (b) Electrons have spin and can be crudely pictured as rotating charge, forming a current that produces a magnetic field with a north pole and a south pole. Neither theplanetary model nor the image of a spinning electron is completely consistent with modern physics. However, they do provide a useful way of understanding phenomena.



PhET Explorations: Magnets and Electromagnets

Explore the interactions between a compass and bar magnet. Discover how you can use a battery and wire to make a magnet! Can you make ita stronger magnet? Can you make the magnetic field reverse?

Figure 5.14 Magnets and Electromagnets (http://cnx.org/content/m42368/1.4/magnets-and-electromagnets_en.jar)

5.3 Magnetic Fields and Magnetic Field LinesEinstein is said to have been fascinated by a compass as a child, perhaps musing on how the needle felt a force without direct physical contact. Hisability to think deeply and clearly about action at a distance, particularly for gravitational, electric, and magnetic forces, later enabled him to create hisrevolutionary theory of relativity. Since magnetic forces act at a distance, we define a magnetic field to represent magnetic forces. The pictorialrepresentation of magnetic field lines is very useful in visualizing the strength and direction of the magnetic field. As shown in Figure 5.15, thedirection of magnetic field lines is defined to be the direction in which the north end of a compass needle points. The magnetic field is traditionallycalled the B-field.

Figure 5.15 Magnetic field lines are defined to have the direction that a small compass points when placed at a location. (a) If small compasses are used to map the magneticfield around a bar magnet, they will point in the directions shown: away from the north pole of the magnet, toward the south pole of the magnet. (Recall that the Earth’s northmagnetic pole is really a south pole in terms of definitions of poles on a bar magnet.) (b) Connecting the arrows gives continuous magnetic field lines. The strength of the fieldis proportional to the closeness (or density) of the lines. (c) If the interior of the magnet could be probed, the field lines would be found to form continuous closed loops.

Small compasses used to test a magnetic field will not disturb it. (This is analogous to the way we tested electric fields with a small test charge. Inboth cases, the fields represent only the object creating them and not the probe testing them.) Figure 5.16 shows how the magnetic field appears fora current loop and a long straight wire, as could be explored with small compasses. A small compass placed in these fields will align itself parallel tothe field line at its location, with its north pole pointing in the direction of B. Note the symbols used for field into and out of the paper.

Figure 5.16 Small compasses could be used to map the fields shown here. (a) The magnetic field of a circular current loop is similar to that of a bar magnet. (b) A long andstraight wire creates a field with magnetic field lines forming circular loops. (c) When the wire is in the plane of the paper, the field is perpendicular to the paper. Note that thesymbols used for the field pointing inward (like the tail of an arrow) and the field pointing outward (like the tip of an arrow).

Making Connections: Concept of a Field

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. Thefield represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields mapmagnetic forces.

Extensive exploration of magnetic fields has revealed a number of hard-and-fast rules. We use magnetic field lines to represent the field (the lines area pictorial tool, not a physical entity in and of themselves). The properties of magnetic field lines can be summarized by these rules:

1. The direction of the magnetic field is tangent to the field line at any point in space. A small compass will point in the direction of the field line.


http://cnx.org/content/m42368/1.4/magnets-and-electromagnets_en.jar

2. The strength of the field is proportional to the closeness of the lines. It is exactly proportional to the number of lines per unit area perpendicularto the lines (called the areal density).

3. Magnetic field lines can never cross, meaning that the field is unique at any point in space.4. Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the south pole.

The last property is related to the fact that the north and south poles cannot be separated. It is a distinct difference from electric field lines, whichbegin and end on the positive and negative charges. If magnetic monopoles existed, then magnetic field lines would begin and end on them.

5.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic FieldWhat is the mechanism by which one magnet exerts a force on another? The answer is related to the fact that all magnetism is caused by current,the flow of charge. Magnetic fields exert forces on moving charges, and so they exert forces on other magnets, all of which have moving charges.

Right Hand Rule 1The magnetic force on a moving charge is one of the most fundamental known. Magnetic force is as important as the electrostatic or Coulomb force.Yet the magnetic force is more complex, in both the number of factors that affects it and in its direction, than the relatively simple Coulomb force. Themagnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

(5.1)F = qvB sin θ,

where θ is the angle between the directions of v and B. This force is often called the Lorentz force. In fact, this is how we define the magnetic

field strength B —in terms of the force on a charged particle moving in a magnetic field. The SI unit for magnetic field strength B is called the tesla(T) after the eccentric but brilliant inventor Nikola Tesla (1856–1943). To determine how the tesla relates to other SI units, we solve F = qvB sin θfor B .

(5.2)B = Fqv sin θ

Because sin θ is unitless, the tesla is

(5.3)1 T = 1 NC ⋅ m/s = 1 N

A ⋅ m(note that C/s = A).

Another smaller unit, called the gauss (G), where 1 G = 10−4 T , is sometimes used. The strongest permanent magnets have fields near 2 T;

superconducting electromagnets may attain 10 T or more. The Earth’s magnetic field on its surface is only about 5×10−5 T , or 0.5 G.

The direction of the magnetic force F is perpendicular to the plane formed by v and B , as determined by the right hand rule 1 (or RHR-1), whichis illustrated in Figure 5.17. RHR-1 states that, to determine the direction of the magnetic force on a positive moving charge, you point the thumb ofthe right hand in the direction of v , the fingers in the direction of B , and a perpendicular to the palm points in the direction of F . One way toremember this is that there is one velocity, and so the thumb represents it. There are many field lines, and so the fingers represent them. The force isin the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Figure 5.17 Magnetic fields exert forces on moving charges. This force is one of the most basic known. The direction of the magnetic force on a moving charge isperpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1) as shown. The magnitude of the force is proportional to q , v , B , and the sine of

the angle between v and B .



Making Connections: Charges and Magnets

There is no magnetic force on static charges. However, there is a magnetic force on moving charges. When charges are stationary, their electricfields do not affect magnets. But, when charges move, they produce magnetic fields that exert forces on other magnets. When there is relativemotion, a connection between electric and magnetic fields emerges—each affects the other.

Example 5.1 Calculating Magnetic Force: Earth’s Magnetic Field on a Charged Glass Rod

With the exception of compasses, you seldom see or personally experience forces due to the Earth’s small magnetic field. To illustrate this,suppose that in a physics lab you rub a glass rod with silk, placing a 20-nC positive charge on it. Calculate the force on the rod due to the Earth’smagnetic field, if you throw it with a horizontal velocity of 10 m/s due west in a place where the Earth’s field is due north parallel to the ground.(The direction of the force is determined with right hand rule 1 as shown in Figure 5.18.)

Figure 5.18 A positively charged object moving due west in a region where the Earth’s magnetic field is due north experiences a force that is straight down as shown. Anegative charge moving in the same direction would feel a force straight up.

Strategy

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation F = qvB sin θ to find the

force.

Solution

The magnetic force is

(5.4)F = qvb sin θ.

We see that sin θ = 1 , since the angle between the velocity and the direction of the field is 90º . Entering the other given quantities yields

(5.5)F = ⎛⎝20×10–9 C⎞⎠(10 m/s)⎛⎝5×10–5 T⎞⎠

= 1×10–11 (C ⋅ m/s)⎛⎝ NC ⋅ m/s

⎞⎠ = 1×10–11 N.

Discussion

This force is completely negligible on any macroscopic object, consistent with experience. (It is calculated to only one digit, since the Earth’s fieldvaries with location and is given to only one digit.) The Earth’s magnetic field, however, does produce very important effects, particularly onsubmicroscopic particles. Some of these are explored in Force on a Moving Charge in a Magnetic Field: Examples and Applications.

5.5 Force on a Moving Charge in a Magnetic Field: Examples and ApplicationsMagnetic force can cause a charged particle to move in a circular or spiral path. Cosmic rays are energetic charged particles in outer space, some ofwhich approach the Earth. They can be forced into spiral paths by the Earth’s magnetic field. Protons in giant accelerators are kept in a circular pathby magnetic force. The bubble chamber photograph in Figure 5.19 shows charged particles moving in such curved paths. The curved paths ofcharged particles in magnetic fields are the basis of a number of phenomena and can even be used analytically, such as in a mass spectrometer.


Figure 5.19 Trails of bubbles are produced by high-energy charged particles moving through the superheated liquid hydrogen in this artist’s rendition of a bubble chamber.There is a strong magnetic field perpendicular to the page that causes the curved paths of the particles. The radius of the path can be used to find the mass, charge, andenergy of the particle.

So does the magnetic force cause circular motion? Magnetic force is always perpendicular to velocity, so that it does no work on the charged particle.The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected, but not the speed. This is typical of uniform circularmotion. The simplest case occurs when a charged particle moves perpendicular to a uniform B -field, such as shown in Figure 5.20. (If this takesplace in a vacuum, the magnetic field is the dominant factor determining the motion.) Here, the magnetic force supplies the centripetal force

Fc = mv2 / r . Noting that sin θ = 1 , we see that F = qvB .

Figure 5.20 A negatively charged particle moves in the plane of the page in a region where the magnetic field is perpendicular into the page (represented by the small circleswith x’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, and so velocity changes in direction but not magnitude. Uniform circular motion results.

Because the magnetic force F supplies the centripetal force Fc , we have

(5.6)qvB = mv2

r .

Solving for r yields

(5.7)r = mvqB .

Here, r is the radius of curvature of the path of a charged particle with mass m and charge q , moving at a speed v perpendicular to a magnetic

field of strength B . If the velocity is not perpendicular to the magnetic field, then v is the component of the velocity perpendicular to the field. Thecomponent of the velocity parallel to the field is unaffected, since the magnetic force is zero for motion parallel to the field. This produces a spiralmotion rather than a circular one.



Example 5.2 Calculating the Curvature of the Path of an Electron Moving in a Magnetic Field: A Magnet on a TVScreen

A magnet brought near an old-fashioned TV screen such as in Figure 5.21 (TV sets with cathode ray tubes instead of LCD screens) severelydistorts its picture by altering the path of the electrons that make its phosphors glow. (Don’t try this at home, as it will permanently magnetizeand ruin the TV.) To illustrate this, calculate the radius of curvature of the path of an electron having a velocity of 6.00×107 m/s(corresponding to the accelerating voltage of about 10.0 kV used in some TVs) perpendicular to a magnetic field of strength B = 0.500 T(obtainable with permanent magnets).

Figure 5.21 Side view showing what happens when a magnet comes in contact with a computer monitor or TV screen. Electrons moving toward the screen spiral aboutmagnetic field lines, maintaining the component of their velocity parallel to the field lines. This distorts the image on the screen.

Strategy

We can find the radius of curvature r directly from the equation r = mvqB , since all other quantities in it are given or known.

Solution

Using known values for the mass and charge of an electron, along with the given values of v and B gives us

(5.8)r = mv

qB =⎛⎝9.11×10−31 kg⎞⎠

⎛⎝6.00×107 m/s⎞⎠

⎛⎝1.60×10−19 C⎞⎠(0.500 T)

= 6.83×10−4 mor

(5.9)r = 0.683 mm.Discussion

The small radius indicates a large effect. The electrons in the TV picture tube are made to move in very tight circles, greatly altering their pathsand distorting the image.

Figure 5.22 shows how electrons not moving perpendicular to magnetic field lines follow the field lines. The component of velocity parallel to the linesis unaffected, and so the charges spiral along the field lines. If field strength increases in the direction of motion, the field will exert a force to slow thecharges, forming a kind of magnetic mirror, as shown below.

Figure 5.22 When a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences a force that reduces thecomponent of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a “magnetic mirror.”


The properties of charged particles in magnetic fields are related to such different things as the Aurora Australis or Aurora Borealis and particleaccelerators. Charged particles approaching magnetic field lines may get trapped in spiral orbits about the lines rather than crossing them, as seenabove. Some cosmic rays, for example, follow the Earth’s magnetic field lines, entering the atmosphere near the magnetic poles and causing thesouthern or northern lights through their ionization of molecules in the atmosphere. This glow of energized atoms and molecules is seen in Figure5.1. Those particles that approach middle latitudes must cross magnetic field lines, and many are prevented from penetrating the atmosphere.Cosmic rays are a component of background radiation; consequently, they give a higher radiation dose at the poles than at the equator.

Figure 5.23 Energetic electrons and protons, components of cosmic rays, from the Sun and deep outer space often follow the Earth’s magnetic field lines rather than crossthem. (Recall that the Earth’s north magnetic pole is really a south pole in terms of a bar magnet.)

Some incoming charged particles become trapped in the Earth’s magnetic field, forming two belts above the atmosphere known as the Van Allenradiation belts after the discoverer James A. Van Allen, an American astrophysicist. (See Figure 5.24.) Particles trapped in these belts form radiationfields (similar to nuclear radiation) so intense that manned space flights avoid them and satellites with sensitive electronics are kept out of them. Inthe few minutes it took lunar missions to cross the Van Allen radiation belts, astronauts received radiation doses more than twice the allowed annualexposure for radiation workers. Other planets have similar belts, especially those having strong magnetic fields like Jupiter.

Figure 5.24 The Van Allen radiation belts are two regions in which energetic charged particles are trapped in the Earth’s magnetic field. One belt lies about 300 km above theEarth’s surface, the other about 16,000 km. Charged particles in these belts migrate along magnetic field lines and are partially reflected away from the poles by the strongerfields there. The charged particles that enter the atmosphere are replenished by the Sun and sources in deep outer space.

Back on Earth, we have devices that employ magnetic fields to contain charged particles. Among them are the giant particle accelerators that havebeen used to explore the substructure of matter. (See Figure 5.25.) Magnetic fields not only control the direction of the charged particles, they alsoare used to focus particles into beams and overcome the repulsion of like charges in these beams.



Figure 5.25 The Fermilab facility in Illinois has a large particle accelerator (the most powerful in the world until 2008) that employs magnetic fields (magnets seen here inorange) to contain and direct its beam. This and other accelerators have been in use for several decades and have allowed us to discover some of the laws underlying allmatter. (credit: ammcrim, Flickr)

Thermonuclear fusion (like that occurring in the Sun) is a hope for a future clean energy source. One of the most promising devices is the tokamak,which uses magnetic fields to contain (or trap) and direct the reactive charged particles. (See Figure 5.26.) Less exotic, but more immediatelypractical, amplifiers in microwave ovens use a magnetic field to contain oscillating electrons. These oscillating electrons generate the microwavessent into the oven.

Figure 5.26 Tokamaks such as the one shown in the figure are being studied with the goal of economical production of energy by nuclear fusion. Magnetic fields in thedoughnut-shaped device contain and direct the reactive charged particles. (credit: David Mellis, Flickr)

Mass spectrometers have a variety of designs, and many use magnetic fields to measure mass. The curvature of a charged particle’s path in the fieldis related to its mass and is measured to obtain mass information. (See More Applications of Magnetism.) Historically, such techniques wereemployed in the first direct observations of electron charge and mass. Today, mass spectrometers (sometimes coupled with gas chromatographs) areused to determine the make-up and sequencing of large biological molecules.

5.6 The Hall EffectWe have seen effects of a magnetic field on free-moving charges. The magnetic field also affects charges moving in a conductor. One result is theHall effect, which has important implications and applications.

Figure 5.27 shows what happens to charges moving through a conductor in a magnetic field. The field is perpendicular to the electron drift velocityand to the width of the conductor. Note that conventional current is to the right in both parts of the figure. In part (a), electrons carry the current andmove to the left. In part (b), positive charges carry the current and move to the right. Moving electrons feel a magnetic force toward one side of theconductor, leaving a net positive charge on the other side. This separation of charge creates a voltage ε , known as the Hall emf, across theconductor. The creation of a voltage across a current-carrying conductor by a magnetic field is known as the Hall effect, after Edwin Hall, theAmerican physicist who discovered it in 1879.


Figure 5.27 The Hall effect. (a) Electrons move to the left in this flat conductor (conventional current to the right). The magnetic field is directly out of the page, represented bycircled dots; it exerts a force on the moving charges, causing a voltage ε , the Hall emf, across the conductor. (b) Positive charges moving to the right (conventional currentalso to the right) are moved to the side, producing a Hall emf of the opposite sign, –ε . Thus, if the direction of the field and current are known, the sign of the charge carrierscan be determined from the Hall effect.

One very important use of the Hall effect is to determine whether positive or negative charges carries the current. Note that in Figure 5.27(b), wherepositive charges carry the current, the Hall emf has the sign opposite to when negative charges carry the current. Historically, the Hall effect wasused to show that electrons carry current in metals and it also shows that positive charges carry current in some semiconductors. The Hall effect isused today as a research tool to probe the movement of charges, their drift velocities and densities, and so on, in materials. In 1980, it wasdiscovered that the Hall effect is quantized, an example of quantum behavior in a macroscopic object.

The Hall effect has other uses that range from the determination of blood flow rate to precision measurement of magnetic field strength. To examinethese quantitatively, we need an expression for the Hall emf, ε , across a conductor. Consider the balance of forces on a moving charge in a situation

where B , v , and l are mutually perpendicular, such as shown in Figure 5.28. Although the magnetic force moves negative charges to one side,

they cannot build up without limit. The electric field caused by their separation opposes the magnetic force, F = qvB , and the electric force,

Fe = qE , eventually grows to equal it. That is,

(5.10)qE = qvB

or

(5.11)E = vB.

Note that the electric field E is uniform across the conductor because the magnetic field B is uniform, as is the conductor. For a uniform electric

field, the relationship between electric field and voltage is E = ε / l , where l is the width of the conductor and ε is the Hall emf. Entering this intothe last expression gives

(5.12)εl = vB.

Solving this for the Hall emf yields

(5.13)ε = Blv (B, v, and l, mutually perpendicular),

where ε is the Hall effect voltage across a conductor of width l through which charges move at a speed v .



Figure 5.28 The Hall emf ε produces an electric force that balances the magnetic force on the moving charges. The magnetic force produces charge separation, which buildsup until it is balanced by the electric force, an equilibrium that is quickly reached.

One of the most common uses of the Hall effect is in the measurement of magnetic field strength B . Such devices, called Hall probes, can be madevery small, allowing fine position mapping. Hall probes can also be made very accurate, usually accomplished by careful calibration. Anotherapplication of the Hall effect is to measure fluid flow in any fluid that has free charges (most do). (See Figure 5.29.) A magnetic field appliedperpendicular to the flow direction produces a Hall emf ε as shown. Note that the sign of ε depends not on the sign of the charges, but only on the

directions of B and v . The magnitude of the Hall emf is ε = Blv , where l is the pipe diameter, so that the average velocity v can be determinedfrom ε providing the other factors are known.

Figure 5.29 The Hall effect can be used to measure fluid flow in any fluid having free charges, such as blood. The Hall emf ε is measured across the tube perpendicular tothe applied magnetic field and is proportional to the average velocity v .

Example 5.3 Calculating the Hall emf: Hall Effect for Blood Flow

A Hall effect flow probe is placed on an artery, applying a 0.100-T magnetic field across it, in a setup similar to that in Figure 5.29. What is theHall emf, given the vessel’s inside diameter is 4.00 mm and the average blood velocity is 20.0 cm/s?

Strategy

Because B , v , and l are mutually perpendicular, the equation ε = Blv can be used to find ε .

Solution

Entering the given values for B , v , and l gives

(5.14)ε = Blv = (0.100 T)⎛⎝4.00×10−3 m⎞⎠(0.200 m/s)= 80.0 μV

Discussion

This is the average voltage output. Instantaneous voltage varies with pulsating blood flow. The voltage is small in this type of measurement. ε isparticularly difficult to measure, because there are voltages associated with heart action (ECG voltages) that are on the order of millivolts. In


practice, this difficulty is overcome by applying an AC magnetic field, so that the Hall emf is AC with the same frequency. An amplifier can bevery selective in picking out only the appropriate frequency, eliminating signals and noise at other frequencies.

5.7 Magnetic Force on a Current-Carrying ConductorBecause charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.

Figure 5.30 The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual movingcharges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces addbecause they are in the same direction.) The force on an individual charge moving at the drift velocity vd is given by F = qvdB sin θ . Taking B to

be uniform over a length of wire l and zero elsewhere, the total magnetic force on the wire is then F = (qvdB sin θ)(N) , where N is the number

of charge carriers in the section of wire of length l . Now, N = nV , where n is the number of charge carriers per unit volume and V is the volume

of wire in the field. Noting that V = Al , where A is the cross-sectional area of the wire, then the force on the wire is F = (qvdB sin θ)(nAl) .

Gathering terms,

(5.15)F = (nqAvd)lB sin θ.

Because nqAvd = I (see Current),

(5.16)F = IlB sin θ

is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B , as shown in Figure 5.31. If we divide

both sides of this expression by l , we find that the magnetic force per unit length of wire in a uniform field is Fl = IB sin θ . The direction of this

force is given by RHR-1, with the thumb in the direction of the current I . Then, with the fingers in the direction of B , a perpendicular to the palm

points in the direction of F , as in Figure 5.31.

Figure 5.31 The force on a current-carrying wire in a magnetic field is F = IlB sin θ . Its direction is given by RHR-1.



Example 5.4 Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field

Calculate the force on the wire shown in Figure 5.30, given B = 1.50 T , l = 5.00 cm , and I = 20.0 A .

Strategy

The force can be found with the given information by using F = IlB sin θ and noting that the angle θ between I and B is 90º , so that

sin θ = 1 .

Solution

Entering the given values into F = IlB sin θ yields

(5.17)F = IlB sin θ = (20.0 A)(0.0500 m)(1.50 T)(1).

The units for tesla are 1 T = NA ⋅ m ; thus,

(5.18)F = 1.50 N.Discussion

This large magnetic field creates a significant force on a small length of wire.

Magnetic force on current-carrying conductors is used to convert electric energy to work. (Motors are a prime example—they employ loops of wireand are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to a clever application where magnetic forcepumps fluids without moving mechanical parts. (See Figure 5.32.)

Figure 5.32 Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.

A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluidparallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as theliquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhapscircumventing the adverse effects of mechanical pumps. (Cell membranes, however, are affected by the large fields needed in MHD, delaying itspractical application in humans.) MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter thanconventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike.As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (SeeFigure 5.33.) Existing MHD drives are heavy and inefficient—much development work is needed.

Figure 5.33 An MHD propulsion system in a nuclear submarine could produce significantly less turbulence than propellers and allow it to run more silently. The development ofa silent drive submarine was dramatized in the book and the film The Hunt for Red October.


5.8 Torque on a Current Loop: Motors and MetersMotors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magnetic field. When current ispassed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work inthe process. (See Figure 5.34.)

Figure 5.34 Torque on a current loop. A current-carrying loop of wire attached to a vertically rotating shaft feels magnetic forces that produce a clockwise torque as viewedfrom above.

Let us examine the force on each segment of the loop in Figure 5.34 to find the torques produced about the axis of the vertical shaft. (This will lead toa useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width w and height l .First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those verticalforces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. Figure 5.35 shows views of the loop fromabove. Torque is defined as τ = rF sin θ , where F is the force, r is the distance from the pivot that the force is applied, and θ is the angle

between r and F . As seen in Figure 5.35(a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in direction, so

that the net force is again zero. However, each force produces a clockwise torque. Since r = w / 2 , the torque on each vertical segment is

(w / 2)F sin θ , and the two add to give a total torque.

(5.19)τ = w2 F sin θ + w

2 F sin θ = wF sin θ



Figure 5.35 Top views of a current-carrying loop in a magnetic field. (a) The equation for torque is derived using this view. Note that the perpendicular to the loop makes anangle θ with the field that is the same as the angle between w / 2 and F . (b) The maximum torque occurs when θ is a right angle and sin θ = 1 . (c) Zero (minimum)

torque occurs when θ is zero and sin θ = 0 . (d) The torque reverses once the loop rotates past θ = 0 .

Now, each vertical segment has a length l that is perpendicular to B , so that the force on each is F = IlB . Entering F into the expression fortorque yields

(5.20)τ = wIlB sin θ.

If we have a multiple loop of N turns, we get N times the torque of one loop. Finally, note that the area of the loop is A = wl ; the expression forthe torque becomes

(5.21)τ = NIAB sin θ.This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. The loopcarries a current I , has N turns, each of area A , and the perpendicular to the loop makes an angle θ with the field B . The net force on the loopis zero.

Example 5.5 Calculating Torque on a Current-Carrying Loop in a Strong Magnetic Field

Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-T field.

Strategy

Torque on the loop can be found using τ = NIAB sin θ . Maximum torque occurs when θ = 90º and sin θ = 1 .

Solution

For sin θ = 1 , the maximum torque is

(5.22)τmax = NIAB.

Entering known values yields

(5.23)τmax = (100)(15.0 A)⎛⎝0.100 m2⎞⎠(2.00 T)= 30.0 N ⋅ m.


Discussion

This torque is large enough to be useful in a motor.

The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at θ = 0 . The torque then reverses its

direction once the coil rotates past θ = 0 . (See Figure 5.35(d).) This means that, unless we do something, the coil will oscillate back and forth about

equilibrium at θ = 0 . To get the coil to continue rotating in the same direction, we can reverse the current as it passes through θ = 0 withautomatic switches called brushes. (See Figure 5.36.)

Figure 5.36 (a) As the angular momentum of the coil carries it through θ = 0 , the brushes reverse the current to keep the torque clockwise. (b) The coil will rotatecontinuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque.

Meters, such as those in analog fuel gauges on a car, are another common application of magnetic torque on a current-carrying loop. Figure 5.37shows that a meter is very similar in construction to a motor. The meter in the figure has its magnets shaped to limit the effect of θ by making Bperpendicular to the loop over a large angular range. Thus the torque is proportional to I and not θ . A linear spring exerts a counter-torque that

balances the current-produced torque. This makes the needle deflection proportional to I . If an exact proportionality cannot be achieved, the gaugereading can be calibrated. To produce a galvanometer for use in analog voltmeters and ammeters that have a low resistance and respond to smallcurrents, we use a large loop area A , high magnetic field B , and low-resistance coils.

Figure 5.37 Meters are very similar to motors but only rotate through a part of a revolution. The magnetic poles of this meter are shaped to keep the component of Bperpendicular to the loop constant, so that the torque does not depend on θ and the deflection against the return spring is proportional only to the current I .

5.9 Magnetic Fields Produced by Currents: Ampere’s LawHow much current is needed to produce a significant magnetic field, perhaps as strong as the Earth’s field? Surveyors will tell you that overheadelectric power lines create magnetic fields that interfere with their compass readings. Indeed, when Oersted discovered in 1820 that a current in awire affected a compass needle, he was not dealing with extremely large currents. How does the shape of wires carrying current affect the shape ofthe magnetic field created? We noted earlier that a current loop created a magnetic field similar to that of a bar magnet, but what about a straight wireor a toroid (doughnut)? How is the direction of a current-created field related to the direction of the current? Answers to these questions are exploredin this section, together with a brief discussion of the law governing the fields created by currents.



Magnetic Field Created by a Long Straight Current-Carrying Wire: Right Hand Rule 2Magnetic fields have both direction and magnitude. As noted before, one way to explore the direction of a magnetic field is with compasses, asshown for a long straight current-carrying wire in Figure 5.38. Hall probes can determine the magnitude of the field. The field around a long straightwire is found to be in circular loops. The right hand rule 2 (RHR-2) emerges from this exploration and is valid for any current segment—point thethumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.

Figure 5.38 (a) Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. (b) Right hand rule 2 states that, ifthe right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire andis valid for any current segment.

The magnetic field strength (magnitude) produced by a long straight current-carrying wire is found by experiment to be

(5.24)B = μ0 I

2πr (long straight wire),

where I is the current, r is the shortest distance to the wire, and the constant μ0 = 4π × 10−7 T ⋅ m/A is the permeability of free space. (μ0is one of the basic constants in nature. We will see later that μ0 is related to the speed of light.) Since the wire is very long, the magnitude of the

field depends only on distance from the wire r , not on position along the wire.

Example 5.6 Calculating Current that Produces a Magnetic Field

Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth’s at a distance of 5.0 cm from the wire.

Strategy

The Earth’s field is about 5.0×10−5 T , and so here B due to the wire is taken to be 1.0×10−4 T . The equation B = μ0 I2πr can be used to

find I , since all other quantities are known.

Solution

Solving for I and entering known values gives

(5.25)I = 2πrB

μ0=

2π⎛⎝5.0×10−2 m⎞⎠⎛⎝1.0×10−4 T⎞⎠

4π×10−7 T ⋅ m/A= 25 A.

Discussion

So a moderately large current produces a significant magnetic field at a distance of 5.0 cm from a long straight wire. Note that the answer isstated to only two digits, since the Earth’s field is specified to only two digits in this example.


Ampere’s Law and OthersThe magnetic field of a long straight wire has more implications than you might at first suspect. Each segment of current produces a magnetic fieldlike that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement ofthe direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for anarbitrary shape current. This results in a more complete law, called Ampere’s law, which relates magnetic field and current in a general way.Ampere’s law in turn is a part of Maxwell’s equations, which give a complete theory of all electromagnetic phenomena. Considerations of howMaxwell’s equations appear to different observers led to the modern theory of relativity, and the realization that electric and magnetic fields aredifferent manifestations of the same thing. Most of this is beyond the scope of this text in both mathematical level, requiring calculus, and in theamount of space that can be devoted to it. But for the interested student, and particularly for those who continue in physics, engineering, or similarpursuits, delving into these matters further will reveal descriptions of nature that are elegant as well as profound. In this text, we shall keep thegeneral features in mind, such as RHR-2 and the rules for magnetic field lines listed in Magnetic Fields and Magnetic Field Lines, whileconcentrating on the fields created in certain important situations.

Making Connections: Relativity

Hearing all we do about Einstein, we sometimes get the impression that he invented relativity out of nothing. On the contrary, one of Einstein’smotivations was to solve difficulties in knowing how different observers see magnetic and electric fields.

Magnetic Field Produced by a Current-Carrying Circular LoopThe magnetic field near a current-carrying loop of wire is shown in Figure 5.39. Both the direction and the magnitude of the magnetic field producedby a current-carrying loop are complex. RHR-2 can be used to give the direction of the field near the loop, but mapping with compasses and the rulesabout field lines given in Magnetic Fields and Magnetic Field Lines are needed for more detail. There is a simple formula for the magnetic fieldstrength at the center of a circular loop. It is

(5.26)B = μ0 I

2R (at center of loop),

where R is the radius of the loop. This equation is very similar to that for a straight wire, but it is valid only at the center of a circular loop of wire. The

similarity of the equations does indicate that similar field strength can be obtained at the center of a loop. One way to get a larger field is to have Nloops; then, the field is B = Nμ0I / (2R) . Note that the larger the loop, the smaller the field at its center, because the current is farther away.

Figure 5.39 (a) RHR-2 gives the direction of the magnetic field inside and outside a current-carrying loop. (b) More detailed mapping with compasses or with a Hall probecompletes the picture. The field is similar to that of a bar magnet.

Magnetic Field Produced by a Current-Carrying SolenoidA solenoid is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside a solenoid can be veryuniform, and also very strong. The field just outside the coils is nearly zero. Figure 5.40 shows how the field looks and how its direction is given byRHR-2.



Figure 5.40 (a) Because of its shape, the field inside a solenoid of length l is remarkably uniform in magnitude and direction, as indicated by the straight and uniformlyspaced field lines. The field outside the coils is nearly zero. (b) This cutaway shows the magnetic field generated by the current in the solenoid.

The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does it begin to weaken andchange direction. The field outside has similar complexities to flat loops and bar magnets, but the magnetic field strength inside a solenoid issimply

(5.27)B = μ0nI (inside a solenoid),

where n is the number of loops per unit length of the solenoid (n = N / l , with N being the number of loops and l the length). Note that B is the

field strength anywhere in the uniform region of the interior and not just at the center. Large uniform fields spread over a large volume are possiblewith solenoids, as Example 5.7 implies.

Example 5.7 Calculating Field Strength inside a Solenoid

What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?

Strategy

To find the field strength inside a solenoid, we use B = μ0nI . First, we note the number of loops per unit length is

(5.28)n−1 = Nl = 2000

2.00 m = 1000 m−1 = 10 cm−1 .

Solution

Substituting known values gives

(5.29)B = μ0nI = ⎛⎝4π×10−7 T ⋅ m/A⎞⎠⎛⎝1000 m−1⎞⎠(1600 A)

= 2.01 T.Discussion

This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging(MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires,although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.

There are interesting variations of the flat coil and solenoid. For example, the toroidal coil used to confine the reactive particles in tokamaks is muchlike a solenoid bent into a circle. The field inside a toroid is very strong but circular. Charged particles travel in circles, following the field lines, andcollide with one another, perhaps inducing fusion. But the charged particles do not cross field lines and escape the toroid. A whole range of coilshapes are used to produce all sorts of magnetic field shapes. Adding ferromagnetic materials produces greater field strengths and can have asignificant effect on the shape of the field. Ferromagnetic materials tend to trap magnetic fields (the field lines bend into the ferromagnetic material,leaving weaker fields outside it) and are used as shields for devices that are adversely affected by magnetic fields, including the Earth’s magneticfield.





5.10 Magnetic Force between Two Parallel ConductorsYou might expect that there are significant forces between current-carrying wires, since ordinary currents produce significant magnetic fields andthese fields exert significant forces on ordinary currents. But you might not expect that the force between wires is used to define the ampere. It mightalso surprise you to learn that this force has something to do with why large circuit breakers burn up when they attempt to interrupt large currents.

The force between two long straight and parallel conductors separated by a distance r can be found by applying what we have developed inpreceding sections. Figure 5.42 shows the wires, their currents, the fields they create, and the subsequent forces they exert on one another. Let usconsider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2 ). The field due to I1 at a distance r is given to be

(5.30)B1 = μ0 I1

2πr .

Figure 5.42 (a) The magnetic field produced by a long straight conductor is perpendicular to a parallel conductor, as indicated by RHR-2. (b) A view from above of the twowires shown in (a), with one magnetic field line shown for each wire. RHR-1 shows that the force between the parallel conductors is attractive when the currents are in thesame direction. A similar analysis shows that the force is repulsive between currents in opposite directions.

This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by F = IlB sin θ with sin θ = 1 :

(5.31)F2 = I2lB1.

By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2 . (Note that F1 = −F2 .)

Since the wires are very long, it is convenient to think in terms of F / l , the force per unit length. Substituting the expression for B1 into the last

equation and rearranging terms gives

(5.32)Fl = μ0 I1 I2

2πr .

F / l is the force per unit length between two parallel currents I1 and I2 separated by a distance r . The force is attractive if the currents are in the

same direction and repulsive if they are in opposite directions.

This force is responsible for the pinch effect in electric arcs and plasmas. The force exists whether the currents are in wires or not. In an electric arc,where currents are moving parallel to one another, there is an attraction that squeezes currents into a smaller tube. In large circuit breakers, likethose used in neighborhood power distribution systems, the pinch effect can concentrate an arc between plates of a switch trying to break a largecurrent, burn holes, and even ignite the equipment. Another example of the pinch effect is found in the solar plasma, where jets of ionized material,such as solar flares, are shaped by magnetic forces.

The operational definition of the ampere is based on the force between current-carrying wires. Note that for parallel wires separated by 1 meter witheach carrying 1 ampere, the force per meter is

(5.33)Fl =

⎛⎝4π×10−7 T ⋅ m/A⎞⎠(1 A)2

(2π)(1 m) = 2×10−7 N/m.

Since μ0 is exactly 4π×10−7 T ⋅ m/A by definition, and because 1 T = 1 N/(A ⋅ m) , the force per meter is exactly 2×10−7 N/m . This is the

basis of the operational definition of the ampere.




The Ampere

The official definition of the ampere is:

One ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic

fields, causes a force of exactly 2×10−7 N/m on each conductor.

Infinite-length straight wires are impractical and so, in practice, a current balance is constructed with coils of wire separated by a few centimeters.Force is measured to determine current. This also provides us with a method for measuring the coulomb. We measure the charge that flows for acurrent of one ampere in one second. That is, 1 C = 1 A ⋅ s . For both the ampere and the coulomb, the method of measuring force betweenconductors is the most accurate in practice.

5.11 More Applications of Magnetism

Mass SpectrometryThe curved paths followed by charged particles in magnetic fields can be put to use. A charged particle moving perpendicular to a magnetic fieldtravels in a circular path having a radius r .

(5.34)r = mvqB

It was noted that this relationship could be used to measure the mass of charged particles such as ions. A mass spectrometer is a device thatmeasures such masses. Most mass spectrometers use magnetic fields for this purpose, although some of them have extremely sophisticateddesigns. Since there are five variables in the relationship, there are many possibilities. However, if v , q , and B can be fixed, then the radius of the

path r is simply proportional to the mass m of the charged particle. Let us examine one such mass spectrometer that has a relatively simpledesign. (See Figure 5.43.) The process begins with an ion source, a device like an electron gun. The ion source gives ions their charge, acceleratesthem to some velocity v , and directs a beam of them into the next stage of the spectrometer. This next region is a velocity selector that only allowsparticles with a particular value of v to get through.

Figure 5.43 This mass spectrometer uses a velocity selector to fix v so that the radius of the path is proportional to mass.

The velocity selector has both an electric field and a magnetic field, perpendicular to one another, producing forces in opposite directions on the ions.Only those ions for which the forces balance travel in a straight line into the next region. If the forces balance, then the electric force F = qE equals

the magnetic force F = qvB , so that qE = qvB . Noting that q cancels, we see that

(5.35)v = EB

is the velocity particles must have to make it through the velocity selector, and further, that v can be selected by varying E and B . In the finalregion, there is only a uniform magnetic field, and so the charged particles move in circular arcs with radii proportional to particle mass. The pathsalso depend on charge q , but since q is in multiples of electron charges, it is easy to determine and to discriminate between ions in different charge

states.


Mass spectrometry today is used extensively in chemistry and biology laboratories to identify chemical and biological substances according to theirmass-to-charge ratios. In medicine, mass spectrometers are used to measure the concentration of isotopes used as tracers. Usually, biologicalmolecules such as proteins are very large, so they are broken down into smaller fragments before analyzing. Recently, large virus particles havebeen analyzed as a whole on mass spectrometers. Sometimes a gas chromatograph or high-performance liquid chromatograph provides an initialseparation of the large molecules, which are then input into the mass spectrometer.

Cathode Ray Tubes—CRTs—and the LikeWhat do non-flat-screen TVs, old computer monitors, x-ray machines, and the 2-mile-long Stanford Linear Accelerator have in common? All of themaccelerate electrons, making them different versions of the electron gun. Many of these devices use magnetic fields to steer the acceleratedelectrons. Figure 5.44 shows the construction of the type of cathode ray tube (CRT) found in some TVs, oscilloscopes, and old computer monitors.Two pairs of coils are used to steer the electrons, one vertically and the other horizontally, to their desired destination.

Figure 5.44 The cathode ray tube (CRT) is so named because rays of electrons originate at the cathode in the electron gun. Magnetic coils are used to steer the beam inmany CRTs. In this case, the beam is moved down. Another pair of horizontal coils would steer the beam horizontally.

Magnetic Resonance ImagingMagnetic resonance imaging (MRI) is one of the most useful and rapidly growing medical imaging tools. It non-invasively produces two-dimensionaland three-dimensional images of the body that provide important medical information with none of the hazards of x-rays. MRI is based on an effectcalled nuclear magnetic resonance (NMR) in which an externally applied magnetic field interacts with the nuclei of certain atoms, particularly thoseof hydrogen (protons). These nuclei possess their own small magnetic fields, similar to those of electrons and the current loops discussed earlier inthis chapter.

When placed in an external magnetic field, such nuclei experience a torque that pushes or aligns the nuclei into one of two new energystates—depending on the orientation of its spin (analogous to the N pole and S pole in a bar magnet). Transitions from the lower to higher energystate can be achieved by using an external radio frequency signal to “flip” the orientation of the small magnets. (This is actually a quantummechanical process. The direction of the nuclear magnetic field is quantized as is energy in the radio waves. We will return to these topics in laterchapters.) The specific frequency of the radio waves that are absorbed and reemitted depends sensitively on the type of nucleus, the chemicalenvironment, and the external magnetic field strength. Therefore, this is a resonance phenomenon in which nuclei in a magnetic field act likeresonators (analogous to those discussed in the treatment of sound in Oscillatory Motion and Waves (http://cnx.org/content/m42239/latest/) )that absorb and reemit only certain frequencies. Hence, the phenomenon is named nuclear magnetic resonance (NMR).

NMR has been used for more than 50 years as an analytical tool. It was formulated in 1946 by F. Bloch and E. Purcell, with the 1952 Nobel Prize inPhysics going to them for their work. Over the past two decades, NMR has been developed to produce detailed images in a process now calledmagnetic resonance imaging (MRI), a name coined to avoid the use of the word “nuclear” and the concomitant implication that nuclear radiation isinvolved. (It is not.) The 2003 Nobel Prize in Medicine went to P. Lauterbur and P. Mansfield for their work with MRI applications.

The largest part of the MRI unit is a superconducting magnet that creates a magnetic field, typically between 1 and 2 T in strength, over a relativelylarge volume. MRI images can be both highly detailed and informative about structures and organ functions. It is helpful that normal and non-normaltissues respond differently for slight changes in the magnetic field. In most medical images, the protons that are hydrogen nuclei are imaged. (About2/3 of the atoms in the body are hydrogen.) Their location and density give a variety of medically useful information, such as organ function, thecondition of tissue (as in the brain), and the shape of structures, such as vertebral disks and knee-joint surfaces. MRI can also be used to follow themovement of certain ions across membranes, yielding information on active transport, osmosis, dialysis, and other phenomena. With excellent spatialresolution, MRI can provide information about tumors, strokes, shoulder injuries, infections, etc.

An image requires position information as well as the density of a nuclear type (usually protons). By varying the magnetic field slightly over thevolume to be imaged, the resonant frequency of the protons is made to vary with position. Broadcast radio frequencies are swept over an appropriaterange and nuclei absorb and reemit them only if the nuclei are in a magnetic field with the correct strength. The imaging receiver gathers informationthrough the body almost point by point, building up a tissue map. The reception of reemitted radio waves as a function of frequency thus givesposition information. These “slices” or cross sections through the body are only several mm thick. The intensity of the reemitted radio waves isproportional to the concentration of the nuclear type being flipped, as well as information on the chemical environment in that area of the body.Various techniques are available for enhancing contrast in images and for obtaining more information. Scans called T1, T2, or proton density scansrely on different relaxation mechanisms of nuclei. Relaxation refers to the time it takes for the protons to return to equilibrium after the external field isturned off. This time depends upon tissue type and status (such as inflammation).

While MRI images are superior to x rays for certain types of tissue and have none of the hazards of x rays, they do not completely supplant x-rayimages. MRI is less effective than x rays for detecting breaks in bone, for example, and in imaging breast tissue, so the two diagnostic toolscomplement each other. MRI images are also expensive compared to simple x-ray images and tend to be used most often where they supplyinformation not readily obtained from x rays. Another disadvantage of MRI is that the patient is totally enclosed with detectors close to the body forabout 30 minutes or more, leading to claustrophobia. It is also difficult for the obese patient to be in the magnet tunnel. New “open-MRI” machines arenow available in which the magnet does not completely surround the patient.

Over the last decade, the development of much faster scans, called “functional MRI” (fMRI), has allowed us to map the functioning of various regionsin the brain responsible for thought and motor control. This technique measures the change in blood flow for activities (thought, experiences, action)in the brain. The nerve cells increase their consumption of oxygen when active. Blood hemoglobin releases oxygen to active nerve cells and has




B-field:

Ampere’s law:

Biot-Savart law:

Curie temperature:

direction of magnetic field lines:

domains:

electromagnet:

electromagnetism:

ferromagnetic:

gauss:

Hall effect:

Hall emf:

Lorentz force:

Maxwell’s equations:

magnetic field lines:

somewhat different magnetic properties when oxygenated than when deoxygenated. With MRI, we can measure this and detect a blood oxygen-dependent signal. Most of the brain scans today use fMRI.

Other Medical Uses of Magnetic FieldsCurrents in nerve cells and the heart create magnetic fields like any other currents. These can be measured but with some difficulty since their

strengths are about 10−6 to 10−8 less than the Earth’s magnetic field. Recording of the heart’s magnetic field as it beats is called amagnetocardiogram (MCG), while measurements of the brain’s magnetic field is called a magnetoencephalogram (MEG). Both give informationthat differs from that obtained by measuring the electric fields of these organs (ECGs and EEGs), but they are not yet of sufficient importance tomake these difficult measurements common.

In both of these techniques, the sensors do not touch the body. MCG can be used in fetal studies, and is probably more sensitive thanechocardiography. MCG also looks at the heart’s electrical activity whose voltage output is too small to be recorded by surface electrodes as in EKG.It has the potential of being a rapid scan for early diagnosis of cardiac ischemia (obstruction of blood flow to the heart) or problems with the fetus.

MEG can be used to identify abnormal electrical discharges in the brain that produce weak magnetic signals. Therefore, it looks at brain activity, notjust brain structure. It has been used for studies of Alzheimer’s disease and epilepsy. Advances in instrumentation to measure very small magneticfields have allowed these two techniques to be used more in recent years. What is used is a sensor called a SQUID, for superconducting quantuminterference device. This operates at liquid helium temperatures and can measure magnetic fields thousands of times smaller than the Earth’s.

Finally, there is a burgeoning market for magnetic cures in which magnets are applied in a variety of ways to the body, from magnetic bracelets tomagnetic mattresses. The best that can be said for such practices is that they are apparently harmless, unless the magnets get close to the patient’scomputer or magnetic storage disks. Claims are made for a broad spectrum of benefits from cleansing the blood to giving the patient more energy,but clinical studies have not verified these claims, nor is there an identifiable mechanism by which such benefits might occur.

PhET Explorations: Magnet and Compass

Ever wonder how a compass worked to point you to the Arctic? Explore the interactions between a compass and bar magnet, and then add theEarth and find the surprising answer! Vary the magnet's strength, and see how things change both inside and outside. Use the field meter tomeasure how the magnetic field changes.

Figure 5.45 Magnet and Compass (http://cnx.org/content/m42388/1.3/magnet-and-compass_en.jar)

Glossaryanother term for magnetic field

the physical law that states that the magnetic field around an electric current is proportional to the current; each segment of currentproduces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to eachsegment

a physical law that describes the magnetic field generated by an electric current in terms of a specific equation

the temperature above which a ferromagnetic material cannot be magnetized

the direction that the north end of a compass needle points

regions within a material that behave like small bar magnets

an object that is temporarily magnetic when an electrical current is passed through it

the use of electrical currents to induce magnetism

materials, such as iron, cobalt, nickel, and gadolinium, that exhibit strong magnetic effects

G, the unit of the magnetic field strength; 1 G = 10–4 T

the creation of voltage across a current-carrying conductor by a magnetic field

the electromotive force created by a current-carrying conductor by a magnetic field, ε = Blv

the force on a charge moving in a magnetic field

a set of four equations that describe electromagnetic phenomena

the pictorial representation of the strength and the direction of a magnetic field


http://cnx.org/content/m42388/1.3/magnet-and-compass_en.jar

magnetic field strength (magnitude) produced by a long straight current-carrying wire:

magnetic field strength at the center of a circular loop:

magnetic field strength inside a solenoid:

magnetic field:

magnetic force:

magnetic monopoles:

magnetic resonance imaging (MRI):

magnetized:

magnetocardiogram (MCG):

magnetoencephalogram (MEG):

meter:

motor:

north magnetic pole:

nuclear magnetic resonance (NMR):

permeability of free space:

right hand rule 1 (RHR-1):

right hand rule 2 (RHR-2):

solenoid:

south magnetic pole:

tesla:

defined as B = μ0 I2πr , where I is the current, r is

the shortest distance to the wire, and μ0 is the permeability of free space

defined as B = μ0 I2R where R is the radius of the loop

defined as B = μ0nI where n is the number of loops per unit length of the solenoid (n = N / l ,

with N being the number of loops and l the length)

the representation of magnetic forces

the force on a charge produced by its motion through a magnetic field; the Lorentz force

an isolated magnetic pole; a south pole without a north pole, or vice versa (no magnetic monopole has ever beenobserved)

a medical imaging technique that uses magnetic fields create detailed images of internal tissues and organs

to be turned into a magnet; to be induced to be magnetic

a recording of the heart’s magnetic field as it beats

a measurement of the brain’s magnetic field

common application of magnetic torque on a current-carrying loop that is very similar in construction to a motor; by design, the torque isproportional to I and not θ , so the needle deflection is proportional to the current

loop of wire in a magnetic field; when current is passed through the loops, the magnetic field exerts torque on the loops, which rotates ashaft; electrical energy is converted to mechanical work in the process

the end or the side of a magnet that is attracted toward Earth’s geographic north pole

a phenomenon in which an externally applied magnetic field interacts with the nuclei of certain atoms

the measure of the ability of a material, in this case free space, to support a magnetic field; the constant

μ0 = 4π×10−7 T ⋅ m/A

the rule to determine the direction of the magnetic force on a positive moving charge: when the thumb of the right handpoints in the direction of the charge’s velocity v and the fingers point in the direction of the magnetic field B , then the force on the charge isperpendicular and away from the palm; the force on a negative charge is perpendicular and into the palm

a rule to determine the direction of the magnetic field induced by a current-carrying wire: Point the thumb of the righthand in the direction of current, and the fingers curl in the direction of the magnetic field loops

a thin wire wound into a coil that produces a magnetic field when an electric current is passed through it

the end or the side of a magnet that is attracted toward Earth’s geographic south pole

T, the SI unit of the magnetic field strength; 1 T = 1 NA ⋅ m

Section Summary

5.1 Magnets• Magnetism is a subject that includes the properties of magnets, the effect of the magnetic force on moving charges and currents, and the

creation of magnetic fields by currents.• There are two types of magnetic poles, called the north magnetic pole and south magnetic pole.• North magnetic poles are those that are attracted toward the Earth’s geographic north pole.• Like poles repel and unlike poles attract.• Magnetic poles always occur in pairs of north and south—it is not possible to isolate north and south poles.

5.2 Ferromagnets and Electromagnets• Magnetic poles always occur in pairs of north and south—it is not possible to isolate north and south poles.• All magnetism is created by electric current.• Ferromagnetic materials, such as iron, are those that exhibit strong magnetic effects.• The atoms in ferromagnetic materials act like small magnets (due to currents within the atoms) and can be aligned, usually in millimeter-sized

regions called domains.• Domains can grow and align on a larger scale, producing permanent magnets. Such a material is magnetized, or induced to be magnetic.• Above a material’s Curie temperature, thermal agitation destroys the alignment of atoms, and ferromagnetism disappears.• Electromagnets employ electric currents to make magnetic fields, often aided by induced fields in ferromagnetic materials.



5.3 Magnetic Fields and Magnetic Field Lines• Magnetic fields can be pictorially represented by magnetic field lines, the properties of which are as follows:

1. The field is tangent to the magnetic field line.2. Field strength is proportional to the line density.3. Field lines cannot cross.4. Field lines are continuous loops.

5.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field• Magnetic fields exert a force on a moving charge q, the magnitude of which is

F = qvB sin θ,where θ is the angle between the directions of v and B .

• The SI unit for magnetic field strength B is the tesla (T), which is related to other units by

1 T = 1 NC ⋅ m/s = 1 N

A ⋅ m.

• The direction of the force on a moving charge is given by right hand rule 1 (RHR-1): Point the thumb of the right hand in the direction of v , the

fingers in the direction of B , and a perpendicular to the palm points in the direction of F .

• The force is perpendicular to the plane formed by v and B . Since the force is zero if v is parallel to B , charged particles often followmagnetic field lines rather than cross them.

5.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications• Magnetic force can supply centripetal force and cause a charged particle to move in a circular path of radius

r = mvqB,

where v is the component of the velocity perpendicular to B for a charged particle with mass m and charge q .

5.6 The Hall Effect• The Hall effect is the creation of voltage ε , known as the Hall emf, across a current-carrying conductor by a magnetic field.• The Hall emf is given by

ε = Blv (B, v, and l, mutually perpendicular)for a conductor of width l through which charges move at a speed v .

5.7 Magnetic Force on a Current-Carrying Conductor• The magnetic force on current-carrying conductors is given by

F = IlB sin θ,where I is the current, l is the length of a straight conductor in a uniform magnetic field B , and θ is the angle between I and B . The force

follows RHR-1 with the thumb in the direction of I .

5.8 Torque on a Current Loop: Motors and Meters• The torque τ on a current-carrying loop of any shape in a uniform magnetic field. is

τ = NIAB sin θ,where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between theperpendicular to the loop and the magnetic field.

5.9 Magnetic Fields Produced by Currents: Ampere’s Law• The strength of the magnetic field created by current in a long straight wire is given by

B = μ0 I2πr (long straight wire),

where I is the current, r is the shortest distance to the wire, and the constant μ0 = 4π × 10−7 T ⋅ m/A is the permeability of free space.

• The direction of the magnetic field created by a long straight wire is given by right hand rule 2 (RHR-2): Point the thumb of the right hand in thedirection of current, and the fingers curl in the direction of the magnetic field loops created by it.

• The magnetic field created by current following any path is the sum (or integral) of the fields due to segments along the path (magnitude anddirection as for a straight wire), resulting in a general relationship between current and field known as Ampere’s law.

• The magnetic field strength at the center of a circular loop is given by

B = μ0 I2R (at center of loop),

where R is the radius of the loop. This equation becomes B = μ0nI / (2R) for a flat coil of N loops. RHR-2 gives the direction of the field

about the loop. A long coil is called a solenoid.• The magnetic field strength inside a solenoid is

B = μ0nI (inside a solenoid),where n is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.

5.10 Magnetic Force between Two Parallel Conductors


• The force between two parallel currents I1 and I2 , separated by a distance r , has a magnitude per unit length given by

Fl = μ0 I1 I2

2πr .

• The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

5.11 More Applications of Magnetism• Crossed (perpendicular) electric and magnetic fields act as a velocity filter, giving equal and opposite forces on any charge with velocity

perpendicular to the fields and of magnitude

v = EB.


5.1 Magnets1. Volcanic and other such activity at the mid-Atlantic ridge extrudes material to fill the gap between separating tectonic plates associated withcontinental drift. The magnetization of rocks is found to reverse in a coordinated manner with distance from the ridge. What does this imply about theEarth’s magnetic field and how could the knowledge of the spreading rate be used to give its historical record?

5.3 Magnetic Fields and Magnetic Field Lines2. Explain why the magnetic field would not be unique (that is, not have a single value) at a point in space where magnetic field lines might cross.(Consider the direction of the field at such a point.)

3. List the ways in which magnetic field lines and electric field lines are similar. For example, the field direction is tangent to the line at any point inspace. Also list the ways in which they differ. For example, electric force is parallel to electric field lines, whereas magnetic force on moving chargesis perpendicular to magnetic field lines.

4. Noting that the magnetic field lines of a bar magnet resemble the electric field lines of a pair of equal and opposite charges, do you expect themagnetic field to rapidly decrease in strength with distance from the magnet? Is this consistent with your experience with magnets?

5. Is the Earth’s magnetic field parallel to the ground at all locations? If not, where is it parallel to the surface? Is its strength the same at all locations?If not, where is it greatest?

5.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field6. If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region is necessarily zero?

5.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications7. How can the motion of a charged particle be used to distinguish between a magnetic and an electric field?

8. High-velocity charged particles can damage biological cells and are a component of radiation exposure in a variety of locations ranging fromresearch facilities to natural background. Describe how you could use a magnetic field to shield yourself.

9. If a cosmic ray proton approaches the Earth from outer space along a line toward the center of the Earth that lies in the plane of the equator, inwhat direction will it be deflected by the Earth’s magnetic field? What about an electron? A neutron?

10. What are the signs of the charges on the particles in Figure 5.46?

Figure 5.46

11. Which of the particles in Figure 5.47 has the greatest velocity, assuming they have identical charges and masses?

Figure 5.47

12. Which of the particles in Figure 5.47 has the greatest mass, assuming all have identical charges and velocities?



13. While operating, a high-precision TV monitor is placed on its side during maintenance. The image on the monitor changes color and blurs slightly.Discuss the possible relation of these effects to the Earth’s magnetic field.

5.6 The Hall Effect14. Discuss how the Hall effect could be used to obtain information on free charge density in a conductor. (Hint: Consider how drift velocity andcurrent are related.)

5.7 Magnetic Force on a Current-Carrying Conductor15. Draw a sketch of the situation in Figure 5.30 showing the direction of electrons carrying the current, and use RHR-1 to verify the direction of theforce on the wire.

16. Verify that the direction of the force in an MHD drive, such as that in Figure 5.32, does not depend on the sign of the charges carrying the currentacross the fluid.

17. Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconducting magnets bedesirable?

18. Which is more likely to interfere with compass readings, AC current in your refrigerator or DC current when you start your car? Explain.

5.8 Torque on a Current Loop: Motors and Meters19. Draw a diagram and use RHR-1 to show that the forces on the top and bottom segments of the motor’s current loop in Figure 5.34 are verticaland produce no torque about the axis of rotation.

5.9 Magnetic Fields Produced by Currents: Ampere’s Law20. Make a drawing and use RHR-2 to find the direction of the magnetic field of a current loop in a motor (such as in Figure 5.34). Then show thatthe direction of the torque on the loop is the same as produced by like poles repelling and unlike poles attracting.

5.10 Magnetic Force between Two Parallel Conductors21. Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?

22. If you have three parallel wires in the same plane, as in Figure 5.48, with currents in the outer two running in opposite directions, is it possible forthe middle wire to be repelled by both? Attracted by both? Explain.

Figure 5.48 Three parallel coplanar wires with currents in the outer two in opposite directions.

23. Suppose two long straight wires run perpendicular to one another without touching. Does one exert a net force on the other? If so, what is itsdirection? Does one exert a net torque on the other? If so, what is its direction? Justify your responses by using the right hand rules.

24. Use the right hand rules to show that the force between the two loops in Figure 5.49 is attractive if the currents are in the same direction andrepulsive if they are in opposite directions. Is this consistent with like poles of the loops repelling and unlike poles of the loops attracting? Drawsketches to justify your answers.


Figure 5.49 Two loops of wire carrying currents can exert forces and torques on one another.

25. If one of the loops in Figure 5.49 is tilted slightly relative to the other and their currents are in the same direction, what are the directions of thetorques they exert on each other? Does this imply that the poles of the bar magnet-like fields they create will line up with each other if the loops areallowed to rotate?

26. Electric field lines can be shielded by the Faraday cage effect. Can we have magnetic shielding? Can we have gravitational shielding?

5.11 More Applications of Magnetism27. Measurements of the weak and fluctuating magnetic fields associated with brain activity are called magnetoencephalograms (MEGs). Do thebrain’s magnetic fields imply coordinated or uncoordinated nerve impulses? Explain.

28. Discuss the possibility that a Hall voltage would be generated on the moving heart of a patient during MRI imaging. Also discuss the same effecton the wires of a pacemaker. (The fact that patients with pacemakers are not given MRIs is significant.)

29. A patient in an MRI unit turns his head quickly to one side and experiences momentary dizziness and a strange taste in his mouth. Discuss thepossible causes.

30. You are told that in a certain region there is either a uniform electric or magnetic field. What measurement or observation could you make todetermine the type? (Ignore the Earth’s magnetic field.)

31. An example of magnetohydrodynamics (MHD) comes from the flow of a river (salty water). This fluid interacts with the Earth’s magnetic field toproduce a potential difference between the two river banks. How would you go about calculating the potential difference?

32. Draw gravitational field lines between 2 masses, electric field lines between a positive and a negative charge, electric field lines between 2positive charges and magnetic field lines around a magnet. Qualitatively describe the differences between the fields and the entities responsible forthe field lines.




5.4 Magnetic Field Strength: Force on a Moving Chargein a Magnetic Field1. What is the direction of the magnetic force on a positive charge thatmoves as shown in each of the six cases shown in Figure 5.50?

Figure 5.50

2. Repeat Exercise 5.1 for a negative charge.

3. What is the direction of the velocity of a negative charge thatexperiences the magnetic force shown in each of the three cases inFigure 5.51, assuming it moves perpendicular to B?

Figure 5.51

4. Repeat Exercise 5.3 for a positive charge.

5. What is the direction of the magnetic field that produces the magneticforce on a positive charge as shown in each of the three cases in thefigure below, assuming B is perpendicular to v ?

Figure 5.52

6. Repeat Exercise 5.5 for a negative charge.

7. What is the maximum force on an aluminum rod with a 0.100-μCcharge that you pass between the poles of a 1.50-T permanent magnetat a speed of 5.00 m/s? In what direction is the force?

8. (a) Aircraft sometimes acquire small static charges. Suppose asupersonic jet has a 0.500-μC charge and flies due west at a speed of

660 m/s over the Earth’s south magnetic pole, where the 8.00×10−5-Tmagnetic field points straight up. What are the direction and themagnitude of the magnetic force on the plane? (b) Discuss whether thevalue obtained in part (a) implies this is a significant or negligible effect.

9. (a) A cosmic ray proton moving toward the Earth at 5.00×107 m/sexperiences a magnetic force of 1.70×10−16 N . What is the strength

of the magnetic field if there is a 45º angle between it and the proton’svelocity? (b) Is the value obtained in part (a) consistent with the knownstrength of the Earth’s magnetic field on its surface? Discuss.

10. An electron moving at 4.00×103 m/s in a 1.25-T magnetic field

experiences a magnetic force of 1.40×10−16 N . What angle does thevelocity of the electron make with the magnetic field? There are twoanswers.

11. (a) A physicist performing a sensitive measurement wants to limit themagnetic force on a moving charge in her equipment to less than

1.00×10−12 N . What is the greatest the charge can be if it moves at amaximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether itwould be difficult to limit the charge to less than the value found in (a) bycomparing it with typical static electricity and noting that static is oftenabsent.

5.5 Force on a Moving Charge in a Magnetic Field:Examples and ApplicationsIf you need additional support for these problems, see MoreApplications of Magnetism.

12. A cosmic ray electron moves at 7.50×106 m/s perpendicular to theEarth’s magnetic field at an altitude where field strength is

1.00×10−5 T . What is the radius of the circular path the electronfollows?

13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field.The field causes the proton to travel in a circular path of radius 0.800 m.What is the field strength?

14. (a) Viewers of Star Trek hear of an antimatter drive on the StarshipEnterprise. One possibility for such a futuristic energy source is to storeantimatter charged particles in a vacuum chamber, circulating in amagnetic field, and then extract them as needed. Antimatter annihilateswith normal matter, producing pure energy. What strength magnetic field

is needed to hold antiprotons, moving at 5.00×107 m/s in a circularpath 2.00 m in radius? Antiprotons have the same mass as protons butthe opposite (negative) charge. (b) Is this field strength obtainable withtoday’s technology or is it a futuristic possibility?

15. (a) An oxygen-16 ion with a mass of 2.66×10−26 kg travels at

5.00×106 m/s perpendicular to a 1.20-T magnetic field, which makes itmove in a circular arc with a 0.231-m radius. What positive charge is onthe ion? (b) What is the ratio of this charge to the charge of an electron?(c) Discuss why the ratio found in (b) should be an integer.

16. What radius circular path does an electron travel if it moves at thesame speed and in the same magnetic field as the proton in Exercise5.13?

17. A velocity selector in a mass spectrometer uses a 0.100-T magneticfield. (a) What electric field strength is needed to select a speed of

4.00×106 m/s ? (b) What is the voltage between the plates if they areseparated by 1.00 cm?

18. An electron in a TV CRT moves with a speed of 6.00×107 m/s , in adirection perpendicular to the Earth’s field, which has a strength of

5.00×10−5 T . (a) What strength electric field must be applied


perpendicular to the Earth’s field to make the electron moves in a straightline? (b) If this is done between plates separated by 1.00 cm, what is thevoltage applied? (Note that TVs are usually surrounded by aferromagnetic material to shield against external magnetic fields andavoid the need for such a correction.)

19. (a) At what speed will a proton move in a circular path of the sameradius as the electron in Exercise 5.12? (b) What would the radius of thepath be if the proton had the same speed as the electron? (c) Whatwould the radius be if the proton had the same kinetic energy as theelectron? (d) The same momentum?

20. A mass spectrometer is being used to separate common oxygen-16from the much rarer oxygen-18, taken from a sample of old glacial ice.(The relative abundance of these oxygen isotopes is related to climatictemperature at the time the ice was deposited.) The ratio of the massesof these two ions is 16 to 18, the mass of oxygen-16 is

2.66×10−26 kg, and they are singly charged and travel at

5.00×106 m/s in a 1.20-T magnetic field. What is the separationbetween their paths when they hit a target after traversing a semicircle?

21. (a) Triply charged uranium-235 and uranium-238 ions are beingseparated in a mass spectrometer. (The much rarer uranium-235 is used

as reactor fuel.) The masses of the ions are 3.90×10−25 kg and

3.95×10−25 kg , respectively, and they travel at 3.00×105 m/s in a

0.250-T field. What is the separation between their paths when they hit atarget after traversing a semicircle? (b) Discuss whether this distancebetween their paths seems to be big enough to be practical in theseparation of uranium-235 from uranium-238.

5.6 The Hall Effect22. A large water main is 2.50 m in diameter and the average watervelocity is 6.00 m/s. Find the Hall voltage produced if the pipe runs

perpendicular to the Earth’s 5.00×10−5-T field.

23. What Hall voltage is produced by a 0.200-T field applied across a2.60-cm-diameter aorta when blood velocity is 60.0 cm/s?

24. (a) What is the speed of a supersonic aircraft with a 17.0-mwingspan, if it experiences a 1.60-V Hall voltage between its wing tipswhen in level flight over the north magnetic pole, where the Earth’s field

strength is 8.00×10−5 T? (b) Explain why very little current flows as aresult of this Hall voltage.

25. A nonmechanical water meter could utilize the Hall effect by applyinga magnetic field across a metal pipe and measuring the Hall voltageproduced. What is the average fluid velocity in a 3.00-cm-diameter pipe,if a 0.500-T field across it creates a 60.0-mV Hall voltage?

26. Calculate the Hall voltage induced on a patient’s heart while beingscanned by an MRI unit. Approximate the conducting path on the heartwall by a wire 7.50 cm long that moves at 10.0 cm/s perpendicular to a1.50-T magnetic field.

27. A Hall probe calibrated to read 1.00 μV when placed in a 2.00-T

field is placed in a 0.150-T field. What is its output voltage?

28. Using information in Example 3.6, what would the Hall voltage be if a2.00-T field is applied across a 10-gauge copper wire (2.588 mm indiameter) carrying a 20.0-A current?

29. Show that the Hall voltage across wires made of the same material,carrying identical currents, and subjected to the same magnetic field isinversely proportional to their diameters. (Hint: Consider how drift velocitydepends on wire diameter.)

30. A patient with a pacemaker is mistakenly being scanned for an MRIimage. A 10.0-cm-long section of pacemaker wire moves at a speed of10.0 cm/s perpendicular to the MRI unit’s magnetic field and a 20.0-mVHall voltage is induced. What is the magnetic field strength?

5.7 Magnetic Force on a Current-Carrying Conductor

31. What is the direction of the magnetic force on the current in each ofthe six cases in Figure 5.53?

Figure 5.53

32. What is the direction of a current that experiences the magnetic forceshown in each of the three cases in Figure 5.54, assuming the currentruns perpendicular to B ?

Figure 5.54

33. What is the direction of the magnetic field that produces the magneticforce shown on the currents in each of the three cases in Figure 5.55,assuming B is perpendicular to I ?

Figure 5.55

34. (a) What is the force per meter on a lightning bolt at the equator that

carries 20,000 A perpendicular to the Earth’s 3.00×10−5-T field? (b)What is the direction of the force if the current is straight up and theEarth’s field direction is due north, parallel to the ground?

35. (a) A DC power line for a light-rail system carries 1000 A at an angle

of 30.0º to the Earth’s 5.00×10−5 -T field. What is the force on a100-m section of this line? (b) Discuss practical concerns this presents, ifany.

36. What force is exerted on the water in an MHD drive utilizing a25.0-cm-diameter tube, if 100-A current is passed across the tube that isperpendicular to a 2.00-T magnetic field? (The relatively small size of thisforce indicates the need for very large currents and magnetic fields tomake practical MHD drives.)



37. A wire carrying a 30.0-A current passes between the poles of astrong magnet that is perpendicular to its field and experiences a 2.16-Nforce on the 4.00 cm of wire in the field. What is the average fieldstrength?

38. (a) A 0.750-m-long section of cable carrying current to a car starter

motor makes an angle of 60º with the Earth’s 5.50×10−5 T field.

What is the current when the wire experiences a force of 7.00×10−3 N? (b) If you run the wire between the poles of a strong horseshoe magnet,subjecting 5.00 cm of it to a 1.75-T field, what force is exerted on thissegment of wire?

39. (a) What is the angle between a wire carrying an 8.00-A current andthe 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic forceof 2.40 N? (b) What is the force on the wire if it is rotated to make anangle of 90º with the field?

40. The force on the rectangular loop of wire in the magnetic field inFigure 5.56 can be used to measure field strength. The field is uniform,and the plane of the loop is perpendicular to the field. (a) What is thedirection of the magnetic force on the loop? Justify the claim that theforces on the sides of the loop are equal and opposite, independent ofhow much of the loop is in the field and do not affect the net force on theloop. (b) If a current of 5.00 A is used, what is the force per tesla on the20.0-cm-wide loop?

Figure 5.56 A rectangular loop of wire carrying a current is perpendicular to amagnetic field. The field is uniform in the region shown and is zero outside that region.

5.8 Torque on a Current Loop: Motors and Meters41. (a) By how many percent is the torque of a motor decreased if itspermanent magnets lose 5.0% of their strength? (b) How many percentwould the current need to be increased to return the torque to originalvalues?

42. (a) What is the maximum torque on a 150-turn square loop of wire18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) Whatis the torque when θ is 10.9º?43. Find the current through a loop needed to create a maximum torqueof 9.00 N ⋅ m. The loop has 50 square turns that are 15.0 cm on a sideand is in a uniform 0.800-T magnetic field.

44. Calculate the magnetic field strength needed on a 200-turn squareloop 20.0 cm on a side to create a maximum torque of 300 N ⋅ m if theloop is carrying 25.0 A.

45. Since the equation for torque on a current-carrying loop is

τ = NIAB sin θ , the units of N ⋅ m must equal units of A ⋅ m2 T .Verify this.

46. (a) At what angle θ is the torque on a current loop 90.0% ofmaximum? (b) 50.0% of maximum? (c) 10.0% of maximum?

47. A proton has a magnetic field due to its spin on its axis. The field is

similar to that created by a circular current loop 0.650×10−15 m in

radius with a current of 1.05×104 A (no kidding). Find the maximumtorque on a proton in a 2.50-T field. (This is a significant torque on asmall particle.)

48. (a) A 200-turn circular loop of radius 50.0 cm is vertical, with its axison an east-west line. A current of 100 A circulates clockwise in the loopwhen viewed from the east. The Earth’s field here is due north, parallel to

the ground, with a strength of 3.00×10−5 T . What are the direction andmagnitude of the torque on the loop? (b) Does this device have anypractical applications as a motor?

49. Repeat Exercise 5.41, but with the loop lying flat on the ground withits current circulating counterclockwise (when viewed from above) in alocation where the Earth’s field is north, but at an angle 45.0º below the

horizontal and with a strength of 6.00×10−5 T .

5.10 Magnetic Force between Two Parallel Conductors50. (a) The hot and neutral wires supplying DC power to a light-railcommuter train carry 800 A and are separated by 75.0 cm. What is themagnitude and direction of the force between 50.0 m of these wires? (b)Discuss the practical consequences of this force, if any.

51. The force per meter between the two wires of a jumper cable beingused to start a stalled car is 0.225 N/m. (a) What is the current in thewires, given they are separated by 2.00 cm? (b) Is the force attractive orrepulsive?

52. A 2.50-m segment of wire supplying current to the motor of asubmerged submarine carries 1000 A and feels a 4.00-N repulsive forcefrom a parallel wire 5.00 cm away. What is the direction and magnitude ofthe current in the other wire?

53. The wire carrying 400 A to the motor of a commuter train feels an

attractive force of 4.00×10−3 N/m due to a parallel wire carrying 5.00A to a headlight. (a) How far apart are the wires? (b) Are the currents inthe same direction?

54. An AC appliance cord has its hot and neutral wires separated by 3.00mm and carries a 5.00-A current. (a) What is the average force per meterbetween the wires in the cord? (b) What is the maximum force per meterbetween the wires? (c) Are the forces attractive or repulsive? (d) Doappliance cords need any special design features to compensate forthese forces?

55. Figure 5.57 shows a long straight wire near a rectangular currentloop. What is the direction and magnitude of the total force on the loop?

Figure 5.57

56. Find the direction and magnitude of the force that each wireexperiences in Figure 5.58(a) by, using vector addition.


Figure 5.58

57. Find the direction and magnitude of the force that each wireexperiences in Figure 5.58(b), using vector addition.

5.11 More Applications of Magnetism58. Indicate whether the magnetic field created in each of the threesituations shown in Figure 5.59 is into or out of the page on the left andright of the current.

Figure 5.59

59. What are the directions of the fields in the center of the loop and coilsshown in Figure 5.60?

Figure 5.60

60. What are the directions of the currents in the loop and coils shown inFigure 5.61?

Figure 5.61

61. To see why an MRI utilizes iron to increase the magnetic field createdby a coil, calculate the current needed in a 400-loop-per-meter circularcoil 0.660 m in radius to create a 1.20-T field (typical of an MRIinstrument) at its center with no iron present. The magnetic field of aproton is approximately like that of a circular current loop

0.650×10−15 m in radius carrying 1.05×104 A . What is the field atthe center of such a loop?

62. Inside a motor, 30.0 A passes through a 250-turn circular loop that is10.0 cm in radius. What is the magnetic field strength created at itscenter?

63. Nonnuclear submarines use batteries for power when submerged. (a)Find the magnetic field 50.0 cm from a straight wire carrying 1200 A fromthe batteries to the drive mechanism of a submarine. (b) What is the fieldif the wires to and from the drive mechanism are side by side? (c)Discuss the effects this could have for a compass on the submarine thatis not shielded.

64. How strong is the magnetic field inside a solenoid with 10,000 turnsper meter that carries 20.0 A?

65. What current is needed in the solenoid described in Exercise 5.58 to

produce a magnetic field 104 times the Earth’s magnetic field of

5.00×10−5 T ?

66. How far from the starter cable of a car, carrying 150 A, must you be to

experience a field less than the Earth’s (5.00×10−5 T)? Assume a

long straight wire carries the current. (In practice, the body of your carshields the dashboard compass.)

67. Measurements affect the system being measured, such as thecurrent loop in Figure 5.56. (a) Estimate the field the loop creates bycalculating the field at the center of a circular loop 20.0 cm in diametercarrying 5.00 A. (b) What is the smallest field strength this loop can beused to measure, if its field must alter the measured field by less than0.0100%?

68. Figure 5.62 shows a long straight wire just touching a loop carrying acurrent I1 . Both lie in the same plane. (a) What direction must the

current I2 in the straight wire have to create a field at the center of the

loop in the direction opposite to that created by the loop? (b) What is theratio of I1 / I2 that gives zero field strength at the center of the loop? (c)

What is the direction of the field directly above the loop under thiscircumstance?

Figure 5.62

69. Find the magnitude and direction of the magnetic field at the pointequidistant from the wires in Figure 5.58(a), using the rules of vectoraddition to sum the contributions from each wire.

70. Find the magnitude and direction of the magnetic field at the pointequidistant from the wires in Figure 5.58(b), using the rules of vectoraddition to sum the contributions from each wire.

71. What current is needed in the top wire in Figure 5.58(a) to produce afield of zero at the point equidistant from the wires, if the currents in thebottom two wires are both 10.0 A into the page?

72. Calculate the size of the magnetic field 20 m below a high voltagepower line. The line carries 450 MW at a voltage of 300,000 V.


(a) A pendulum is set up so that its bob (a thin copper disk) swingsbetween the poles of a permanent magnet as shown in Figure 5.63.What is the magnitude and direction of the magnetic force on the bob atthe lowest point in its path, if it has a positive 0.250 μC charge and is

released from a height of 30.0 cm above its lowest point? The magneticfield strength is 1.50 T. (b) What is the acceleration of the bob at thebottom of its swing if its mass is 30.0 grams and it is hung from a flexiblestring? Be certain to include a free-body diagram as part of your analysis.



Figure 5.63


(a) What voltage will accelerate electrons to a speed of

6.00×10−7 m/s ? (b) Find the radius of curvature of the path of aproton accelerated through this potential in a 0.500-T field and comparethis with the radius of curvature of an electron accelerated through thesame potential.


Find the radius of curvature of the path of a 25.0-MeV proton movingperpendicularly to the 1.20-T field of a cyclotron.


To construct a nonmechanical water meter, a 0.500-T magnetic field isplaced across the supply water pipe to a home and the Hall voltage isrecorded. (a) Find the flow rate in liters per second through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV. (b) What would the Hallvoltage be for the same flow rate through a 10.0-cm-diameter pipe withthe same field applied?


(a) Using the values given for an MHD drive in Exercise 5.59, andassuming the force is uniformly applied to the fluid, calculate the pressure

created in N/m2 . (b) Is this a significant fraction of an atmosphere?


(a) Calculate the maximum torque on a 50-turn, 1.50 cm radius circularcurrent loop carrying 50 μA in a 0.500-T field. (b) If this coil is to be

used in a galvanometer that reads 50 μA full scale, what force constant

spring must be used, if it is attached 1.00 cm from the axis of rotation andis stretched by the 60º arc moved?


A current balance used to define the ampere is designed so that thecurrent through it is constant, as is the distance between wires. Even so,if the wires change length with temperature, the force between them willchange. What percent change in force per degree will occur if the wiresare copper?


(a) Show that the period of the circular orbit of a charged particle movingperpendicularly to a uniform magnetic field is T = 2πm / (qB) . (b) What

is the frequency f ? (c) What is the angular velocity ω ? Note that these

results are independent of the velocity and radius of the orbit and, hence,of the energy of the particle. (Figure 5.64.)

Figure 5.64 Cyclotrons accelerate charged particles orbiting in a magnetic field byplacing an AC voltage on the metal Dees, between which the particles move, so thatenergy is added twice each orbit. The frequency is constant, since it is independent ofthe particle energy—the radius of the orbit simply increases with energy until theparticles approach the edge and are extracted for various experiments andapplications.


A cyclotron accelerates charged particles as shown in Figure 5.64. Usingthe results of the previous problem, calculate the frequency of theaccelerating voltage needed for a proton in a 1.20-T field.


(a) A 0.140-kg baseball, pitched at 40.0 m/s horizontally and

perpendicular to the Earth’s horizontal 5.00×10−5 T field, has a100-nC charge on it. What distance is it deflected from its path by themagnetic force, after traveling 30.0 m horizontally? (b) Would yousuggest this as a secret technique for a pitcher to throw curve balls?


(a) What is the direction of the force on a wire carrying a current due eastin a location where the Earth’s field is due north? Both are parallel to theground. (b) Calculate the force per meter if the wire carries 20.0 A and

the field strength is 3.00×10−5 T . (c) What diameter copper wire wouldhave its weight supported by this force? (d) Calculate the resistance permeter and the voltage per meter needed.


One long straight wire is to be held directly above another by repulsionbetween their currents. The lower wire carries 100 A and the wire 7.50cm above it is 10-gauge (2.588 mm diameter) copper wire. (a) Whatcurrent must flow in the upper wire, neglecting the Earth’s field? (b) What

is the smallest current if the Earth’s 3.00×10−5 T field is parallel to theground and is not neglected? (c) Is the supported wire in a stable orunstable equilibrium if displaced vertically? If displaced horizontally?


(a) Find the charge on a baseball, thrown at 35.0 m/s perpendicular to

the Earth’s 5.00×10−5 T field, that experiences a 1.00-N magneticforce. (b) What is unreasonable about this result? (c) Which assumptionor premise is responsible?


A charged particle having mass 6.64×10−27 kg (that of a helium atom)

moving at 8.70×105 m/s perpendicular to a 1.50-T magnetic fieldtravels in a circular path of radius 16.0 mm. (a) What is the charge of theparticle? (b) What is unreasonable about this result? (c) Whichassumptions are responsible?


An inventor wants to generate 120-V power by moving a 1.00-m-long

wire perpendicular to the Earth’s 5.00×10−5 T field. (a) Find the speed


with which the wire must move. (b) What is unreasonable about thisresult? (c) Which assumption is responsible?


Frustrated by the small Hall voltage obtained in blood flowmeasurements, a medical physicist decides to increase the appliedmagnetic field strength to get a 0.500-V output for blood moving at 30.0cm/s in a 1.50-cm-diameter vessel. (a) What magnetic field strength isneeded? (b) What is unreasonable about this result? (c) Which premiseis responsible?


A surveyor 100 m from a long straight 200-kV DC power line suspectsthat its magnetic field may equal that of the Earth and affect compassreadings. (a) Calculate the current in the wire needed to create a

5.00×10−5 T field at this distance. (b) What is unreasonable about thisresult? (c) Which assumption or premise is responsible?


Consider a mass separator that applies a magnetic field perpendicular tothe velocity of ions and separates the ions based on the radius ofcurvature of their paths in the field. Construct a problem in which youcalculate the magnetic field strength needed to separate two ions thatdiffer in mass, but not charge, and have the same initial velocity. Amongthe things to consider are the types of ions, the velocities they can begiven before entering the magnetic field, and a reasonable value for theradius of curvature of the paths they follow. In addition, calculate theseparation distance between the ions at the point where they aredetected.


Consider using the torque on a current-carrying coil in a magnetic field todetect relatively small magnetic fields (less than the field of the Earth, forexample). Construct a problem in which you calculate the maximumtorque on a current-carrying loop in a magnetic field. Among the things tobe considered are the size of the coil, the number of loops it has, thecurrent you pass through the coil, and the size of the field you wish todetect. Discuss whether the torque produced is large enough to beeffectively measured. Your instructor may also wish for you to considerthe effects, if any, of the field produced by the coil on the surroundingsthat could affect detection of the small field.



6 ELECTROMAGNETIC INDUCTION, AC CIRCUITS,AND ELECTRICAL TECHNOLOGIES

Figure 6.1 This wind turbine in the Thames Estuary in the UK is an example of induction at work. Wind pushes the blades of the turbine, spinning a shaft attached to magnets.The magnets spin around a conductive coil, inducing an electric current in the coil, and eventually feeding the electrical grid. (credit: phault, Flickr)

CHAPTER 6 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES 189

Learning Objectives6.1. Induced Emf and Magnetic Flux

• Calculate the flux of a uniform magnetic field through a loop of arbitrary orientation.• Describe methods to produce an electromotive force (emf) with a magnetic field or magnet and a loop of wire.

6.2. Faraday’s Law of Induction: Lenz’s Law• Calculate emf, current, and magnetic fields using Faraday’s Law.• Explain the physical results of Lenz’s Law

6.3. Motional Emf• Calculate emf, force, magnetic field, and work due to the motion of an object in a magnetic field.

6.4. Eddy Currents and Magnetic Damping• Explain the magnitude and direction of an induced eddy current, and the effect this will have on the object it is induced in.• Describe several applications of magnetic damping.

6.5. Electric Generators• Calculate the emf induced in a generator.• Calculate the peak emf which can be induced in a particular generator system.

6.6. Back Emf• Explain what back emf is and how it is induced.

6.7. Transformers• Explain how a transformer works.• Calculate voltage, current, and/or number of turns given the other quantities.

6.8. Electrical Safety: Systems and Devices• Explain how various modern safety features in electric circuits work, with an emphasis on how induction is employed.

6.9. Inductance• Calculate the inductance of an inductor.• Calculate the energy stored in an inductor.• Calculate the emf generated in an inductor.

6.10. RL Circuits• Calculate the current in an RL circuit after a specified number of characteristic time steps.• Calculate the characteristic time of an RL circuit.• Sketch the current in an RL circuit over time.

6.11. Reactance, Inductive and Capacitive• Sketch voltage and current versus time in simple inductive, capacitive, and resistive circuits.• Calculate inductive and capacitive reactance.• Calculate current and/or voltage in simple inductive, capacitive, and resistive circuits.

6.12. RLC Series AC Circuits• Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and/or current in a RLC series circuit.• Draw the circuit diagram for an RLC series circuit.• Explain the significance of the resonant frequency.

Introduction to Electromagnetic Induction, AC Circuits and Electrical TechnologiesNature’s displays of symmetry are beautiful and alluring. A butterfly’s wings exhibit an appealing symmetry in a complex system. (See Figure 6.2.)The laws of physics display symmetries at the most basic level—these symmetries are a source of wonder and imply deeper meaning. Since weplace a high value on symmetry, we look for it when we explore nature. The remarkable thing is that we find it.

Figure 6.2 Physics, like this butterfly, has inherent symmetries. (credit: Thomas Bresson)

The hint of symmetry between electricity and magnetism found in the preceding chapter will be elaborated upon in this chapter. Specifically, we knowthat a current creates a magnetic field. If nature is symmetric here, then perhaps a magnetic field can create a current. The Hall effect is a voltagecaused by a magnetic force. That voltage could drive a current. Historically, it was very shortly after Oersted discovered currents cause magneticfields that other scientists asked the following question: Can magnetic fields cause currents? The answer was soon found by experiment to be yes. In1831, some 12 years after Oersted’s discovery, the English scientist Michael Faraday (1791–1862) and the American scientist Joseph Henry(1797–1878) independently demonstrated that magnetic fields can produce currents. The basic process of generating emfs (electromotive force) and,hence, currents with magnetic fields is known as induction; this process is also called magnetic induction to distinguish it from charging by induction,which utilizes the Coulomb force.

Today, currents induced by magnetic fields are essential to our technological society. The ubiquitous generator—found in automobiles, on bicycles, innuclear power plants, and so on—uses magnetism to generate current. Other devices that use magnetism to induce currents include pickup coils inelectric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances.Not so familiar perhaps, but important nevertheless, is that the behavior of AC circuits depends strongly on the effect of magnetic fields on currents.

190 CHAPTER 6 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES


6.1 Induced Emf and Magnetic FluxThe apparatus used by Faraday to demonstrate that magnetic fields can create currents is illustrated in Figure 6.3. When the switch is closed, amagnetic field is produced in the coil on the top part of the iron ring and transmitted to the coil on the bottom part of the ring. The galvanometer isused to detect any current induced in the coil on the bottom. It was found that each time the switch is closed, the galvanometer detects a current inone direction in the coil on the bottom. (You can also observe this in a physics lab.) Each time the switch is opened, the galvanometer detects acurrent in the opposite direction. Interestingly, if the switch remains closed or open for any length of time, there is no current through thegalvanometer. Closing and opening the switch induces the current. It is the change in magnetic field that creates the current. More basic than thecurrent that flows is the emfthat causes it. The current is a result of an emf induced by a changing magnetic field, whether or not there is a path forcurrent to flow.

Figure 6.3 Faraday’s apparatus for demonstrating that a magnetic field can produce a current. A change in the field produced by the top coil induces an emf and, hence, acurrent in the bottom coil. When the switch is opened and closed, the galvanometer registers currents in opposite directions. No current flows through the galvanometer whenthe switch remains closed or open.

An experiment easily performed and often done in physics labs is illustrated in Figure 6.4. An emf is induced in the coil when a bar magnet is pushedin and out of it. Emfs of opposite signs are produced by motion in opposite directions, and the emfs are also reversed by reversing poles. The sameresults are produced if the coil is moved rather than the magnet—it is the relative motion that is important. The faster the motion, the greater the emf,and there is no emf when the magnet is stationary relative to the coil.

Figure 6.4 Movement of a magnet relative to a coil produces emfs as shown. The same emfs are produced if the coil is moved relative to the magnet. The greater the speed,the greater the magnitude of the emf, and the emf is zero when there is no motion.

The method of inducing an emf used in most electric generators is shown in Figure 6.5. A coil is rotated in a magnetic field, producing an alternatingcurrent emf, which depends on rotation rate and other factors that will be explored in later sections. Note that the generator is remarkably similar inconstruction to a motor (another symmetry).


Figure 6.5 Rotation of a coil in a magnetic field produces an emf. This is the basic construction of a generator, where work done to turn the coil is converted to electric energy.Note the generator is very similar in construction to a motor.

So we see that changing the magnitude or direction of a magnetic field produces an emf. Experiments revealed that there is a crucial quantity calledthe magnetic flux, Φ , given by

(6.1)Φ = BA cos θ,

where B is the magnetic field strength over an area A , at an angle θ with the perpendicular to the area as shown in Figure 6.6. Any change in

magnetic flux Φ induces an emf. This process is defined to be electromagnetic induction. Units of magnetic flux Φ are T ⋅ m2 . As seen in

Figure 6.6, B cos θ = B ⊥, which is the component of B perpendicular to the area A . Thus magnetic flux is Φ = B⊥A , the product of the area

and the component of the magnetic field perpendicular to it.

Figure 6.6 Magnetic flux Φ is related to the magnetic field and the area over which it exists. The flux Φ = BA cos θ is related to induction; any change in Φ induces anemf.

All induction, including the examples given so far, arises from some change in magnetic flux Φ . For example, Faraday changed B and hence Φwhen opening and closing the switch in his apparatus (shown in Figure 6.3). This is also true for the bar magnet and coil shown in Figure 6.4. Whenrotating the coil of a generator, the angle θ and, hence, Φ is changed. Just how great an emf and what direction it takes depend on the change in

Φ and how rapidly the change is made, as examined in the next section.

6.2 Faraday’s Law of Induction: Lenz’s Law

Faraday’s and Lenz’s LawFaraday’s experiments showed that the emf induced by a change in magnetic flux depends on only a few factors. First, emf is directly proportional tothe change in flux ΔΦ . Second, emf is greatest when the change in time Δt is smallest—that is, emf is inversely proportional to Δt . Finally, if a



coil has N turns, an emf will be produced that is N times greater than for a single coil, so that emf is directly proportional to N . The equation forthe emf induced by a change in magnetic flux is

(6.2)emf = −N ΔΦΔt .

This relationship is known as Faraday’s law of induction. The units for emf are volts, as is usual.

The minus sign in Faraday’s law of induction is very important. The minus means that the emf creates a current I and magnetic field B that opposethe change in flux ΔΦ —this is known as Lenz’s law. The direction (given by the minus sign) of the emfis so important that it is called Lenz’s lawafter the Russian Heinrich Lenz (1804–1865), who, like Faraday and Henry,independently investigated aspects of induction. Faraday was aware ofthe direction, but Lenz stated it so clearly that he is credited for its discovery. (See Figure 6.7.)

Figure 6.7 (a) When this bar magnet is thrust into the coil, the strength of the magnetic field increases in the coil. The current induced in the coil creates another field, in theopposite direction of the bar magnet’s to oppose the increase. This is one aspect of Lenz’s law—induction opposes any change in flux. (b) and (c) are two other situations.Verify for yourself that the direction of the induced Bcoil shown indeed opposes the change in flux and that the current direction shown is consistent with RHR-2.

Problem-Solving Strategy for Lenz’s Law

To use Lenz’s law to determine the directions of the induced magnetic fields, currents, and emfs:

1. Make a sketch of the situation for use in visualizing and recording directions.2. Determine the direction of the magnetic field B.3. Determine whether the flux is increasing or decreasing.4. Now determine the direction of the induced magnetic field B. It opposes the change in flux by adding or subtracting from the original field.5. Use RHR-2 to determine the direction of the induced current I that is responsible for the induced magnetic field B.6. The direction (or polarity) of the induced emf will now drive a current in this direction and can be represented as current emerging from the

positive terminal of the emf and returning to its negative terminal.

For practice, apply these steps to the situations shown in Figure 6.7 and to others that are part of the following text material.

Applications of Electromagnetic InductionThere are many applications of Faraday’s Law of induction, as we will explore in this chapter and others. At this juncture, let us mention several thathave to do with data storage and magnetic fields. A very important application has to do with audio and video recording tapes. A plastic tape, coatedwith iron oxide, moves past a recording head. This recording head is basically a round iron ring about which is wrapped a coil of wire—anelectromagnet (Figure 6.8). A signal in the form of a varying input current from a microphone or camera goes to the recording head. These signals(which are a function of the signal amplitude and frequency) produce varying magnetic fields at the recording head. As the tape moves past therecording head, the magnetic field orientations of the iron oxide molecules on the tape are changed thus recording the signal. In the playback mode,the magnetized tape is run past another head, similar in structure to the recording head. The different magnetic field orientations of the iron oxidemolecules on the tape induces an emf in the coil of wire in the playback head. This signal then is sent to a loudspeaker or video player.


Figure 6.8 Recording and playback heads used with audio and video magnetic tapes. (credit: Steve Jurvetson)

Similar principles apply to computer hard drives, except at a much faster rate. Here recordings are on a coated, spinning disk. Read heads historicallywere made to work on the principle of induction. However, the input information is carried in digital rather than analog form – a series of 0’s or 1’s arewritten upon the spinning hard drive. Today, most hard drive readout devices do not work on the principle of induction, but use a technique known asgiant magnetoresistance. (The discovery that weak changes in a magnetic field in a thin film of iron and chromium could bring about much largerchanges in electrical resistance was one of the first large successes of nanotechnology.) Another application of induction is found on the magneticstripe on the back of your personal credit card as used at the grocery store or the ATM machine. This works on the same principle as the audio orvideo tape mentioned in the last paragraph in which a head reads personal information from your card.

Another application of electromagnetic induction is when electrical signals need to be transmitted across a barrier. Consider the cochlear implantshown below. Sound is picked up by a microphone on the outside of the skull and is used to set up a varying magnetic field. A current is induced in areceiver secured in the bone beneath the skin and transmitted to electrodes in the inner ear. Electromagnetic induction can be used in otherinstances where electric signals need to be conveyed across various media.

Figure 6.9 Electromagnetic induction used in transmitting electric currents across mediums. The device on the baby’s head induces an electrical current in a receiver securedin the bone beneath the skin. (credit: Bjorn Knetsch)

Another contemporary area of research in which electromagnetic induction is being successfully implemented (and with substantial potential) istranscranial magnetic simulation. A host of disorders, including depression and hallucinations can be traced to irregular localized electrical activity inthe brain. In transcranial magnetic stimulation, a rapidly varying and very localized magnetic field is placed close to certain sites identified in the brain.Weak electric currents are induced in the identified sites and can result in recovery of electrical functioning in the brain tissue.

Sleep apnea (“the cessation of breath”) affects both adults and infants (especially premature babies and it may be a cause of sudden infant deaths[SID]). In such individuals, breath can stop repeatedly during their sleep. A cessation of more than 20 seconds can be very dangerous. Stroke, heartfailure, and tiredness are just some of the possible consequences for a person having sleep apnea. The concern in infants is the stopping of breathfor these longer times. One type of monitor to alert parents when a child is not breathing uses electromagnetic induction. A wire wrapped around theinfant’s chest has an alternating current running through it. The expansion and contraction of the infant’s chest as the infant breathes changes thearea through the coil. A pickup coil located nearby has an alternating current induced in it due to the changing magnetic field of the initial wire. If thechild stops breathing, there will be a change in the induced current, and so a parent can be alerted.

Making Connections: Conservation of Energy

Lenz’s law is a manifestation of the conservation of energy. The induced emf produces a current that opposes the change in flux, because achange in flux means a change in energy. Energy can enter or leave, but not instantaneously. Lenz’s law is a consequence. As the changebegins, the law says induction opposes and, thus, slows the change. In fact, if the induced emf were in the same direction as the change in flux,there would be a positive feedback that would give us free energy from no apparent source—conservation of energy would be violated.

Example 6.1 Calculating Emf: How Great Is the Induced Emf?

Calculate the magnitude of the induced emf when the magnet in Figure 6.7(a) is thrust into the coil, given the following information: the singleloop coil has a radius of 6.00 cm and the average value of B cos θ (this is given, since the bar magnet’s field is complex) increases from 0.0500T to 0.250 T in 0.100 s.

Strategy

To find the magnitude of emf, we use Faraday’s law of induction as stated by emf = −N ΔΦΔt , but without the minus sign that indicates

direction:

(6.3)emf = N ΔΦΔt .

Solution



We are given that N = 1 and Δt = 0.100 s , but we must determine the change in flux ΔΦ before we can find emf. Since the area of theloop is fixed, we see that

(6.4)ΔΦ = Δ(BA cos θ) = AΔ(B cos θ).

Now Δ(B cos θ) = 0.200 T , since it was given that B cos θ changes from 0.0500 to 0.250 T. The area of the loop is

A = πr2 = (3.14...)(0.060 m)2 = 1.13×10−2 m2 . Thus,

(6.5)ΔΦ = (1.13×10−2 m2)(0.200 T).

Entering the determined values into the expression for emf gives

(6.6)Emf = N ΔΦ

Δt = (1.13×10−2 m2)(0.200 T)0.100 s = 22.6 mV.

Discussion

While this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in the coil, a stronger magnet,and faster movement make induction the practical source of voltages that it is.

PhET Explorations: Faraday's Electromagnetic Lab

Play with a bar magnet and coils to learn about Faraday's law. Move a bar magnet near one or two coils to make a light bulb glow. View themagnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field lines or use a meter to show the directionand magnitude of the current. You can also play with electromagnets, generators and transformers!

Figure 6.10 Faraday's Electromagnetic Lab (http://cnx.org/content/m42392/1.3/faraday_en.jar)

6.3 Motional EmfAs we have seen, any change in magnetic flux induces an emf opposing that change—a process known as induction. Motion is one of the majorcauses of induction. For example, a magnet moved toward a coil induces an emf, and a coil moved toward a magnet produces a similar emf. In thissection, we concentrate on motion in a magnetic field that is stationary relative to the Earth, producing what is loosely called motional emf.

One situation where motional emf occurs is known as the Hall effect and has already been examined. Charges moving in a magnetic field experiencethe magnetic force F = qvB sin θ , which moves opposite charges in opposite directions and produces an emf = Bℓv . We saw that the Hall effect

has applications, including measurements of B and v . We will now see that the Hall effect is one aspect of the broader phenomenon of induction,and we will find that motional emf can be used as a power source.

Consider the situation shown in Figure 6.11. A rod is moved at a speed v along a pair of conducting rails separated by a distance ℓ in a uniform

magnetic field B . The rails are stationary relative to B and are connected to a stationary resistor R . The resistor could be anything from a light bulb

to a voltmeter. Consider the area enclosed by the moving rod, rails, and resistor. B is perpendicular to this area, and the area is increasing as therod moves. Thus the magnetic flux enclosed by the rails, rod, and resistor is increasing. When flux changes, an emf is induced according toFaraday’s law of induction.


http://cnx.org/content/m42392/1.3/faraday_en.jar

Figure 6.11 (a) A motional emf = Bℓv is induced between the rails when this rod moves to the right in the uniform magnetic field. The magnetic field B is into the page,perpendicular to the moving rod and rails and, hence, to the area enclosed by them. (b) Lenz’s law gives the directions of the induced field and current, and the polarity of theinduced emf. Since the flux is increasing, the induced field is in the opposite direction, or out of the page. RHR-2 gives the current direction shown, and the polarity of the rodwill drive such a current. RHR-1 also indicates the same polarity for the rod. (Note that the script E symbol used in the equivalent circuit at the bottom of part (b) representsemf.)

To find the magnitude of emf induced along the moving rod, we use Faraday’s law of induction without the sign:

(6.7)emf = N ΔΦΔt .

Here and below, “emf” implies the magnitude of the emf. In this equation, N = 1 and the flux Φ = BA cos θ . We have θ = 0º and cos θ = 1 ,

since B is perpendicular to A . Now ΔΦ = Δ(BA) = BΔA , since B is uniform. Note that the area swept out by the rod is ΔA = ℓ Δ x .

Entering these quantities into the expression for emf yields

(6.8)emf = BΔAΔt = BℓΔx

Δt .

Finally, note that Δx / Δt = v , the velocity of the rod. Entering this into the last expression shows that

(6.9)emf = Bℓv (B, ℓ, and v perpendicular)

is the motional emf. This is the same expression given for the Hall effect previously.

Making Connections: Unification of Forces

There are many connections between the electric force and the magnetic force. The fact that a moving electric field produces a magnetic fieldand, conversely, a moving magnetic field produces an electric field is part of why electric and magnetic forces are now considered to be differentmanifestations of the same force. This classic unification of electric and magnetic forces into what is called the electromagnetic force is theinspiration for contemporary efforts to unify other basic forces.



To find the direction of the induced field, the direction of the current, and the polarity of the induced emf, we apply Lenz’s law as explained inFaraday's Law of Induction: Lenz's Law. (See Figure 6.11(b).) Flux is increasing, since the area enclosed is increasing. Thus the induced fieldmust oppose the existing one and be out of the page. And so the RHR-2 requires that I be counterclockwise, which in turn means the top of the rod ispositive as shown.

Motional emf also occurs if the magnetic field moves and the rod (or other object) is stationary relative to the Earth (or some observer). We have seenan example of this in the situation where a moving magnet induces an emf in a stationary coil. It is the relative motion that is important. What isemerging in these observations is a connection between magnetic and electric fields. A moving magnetic field produces an electric field through itsinduced emf. We already have seen that a moving electric field produces a magnetic field—moving charge implies moving electric field and movingcharge produces a magnetic field.

Motional emfs in the Earth’s weak magnetic field are not ordinarily very large, or we would notice voltage along metal rods, such as a screwdriver,during ordinary motions. For example, a simple calculation of the motional emf of a 1 m rod moving at 3.0 m/s perpendicular to the Earth’s field gives

emf = Bℓv = (5.0×10−5 T)(1.0 m)(3.0 m/s) = 150 μV . This small value is consistent with experience. There is a spectacular exception,

however. In 1992 and 1996, attempts were made with the space shuttle to create large motional emfs. The Tethered Satellite was to be let out on a20 km length of wire as shown in Figure 6.12, to create a 5 kV emf by moving at orbital speed through the Earth’s field. This emf could be used toconvert some of the shuttle’s kinetic and potential energy into electrical energy if a complete circuit could be made. To complete the circuit, thestationary ionosphere was to supply a return path for the current to flow. (The ionosphere is the rarefied and partially ionized atmosphere at orbitalaltitudes. It conducts because of the ionization. The ionosphere serves the same function as the stationary rails and connecting resistor in Figure6.11, without which there would not be a complete circuit.) Drag on the current in the cable due to the magnetic force F = IℓB sin θ does the workthat reduces the shuttle’s kinetic and potential energy and allows it to be converted to electrical energy. The tests were both unsuccessful. In the first,the cable hung up and could only be extended a couple of hundred meters; in the second, the cable broke when almost fully extended. Example 6.2indicates feasibility in principle.

Example 6.2 Calculating the Large Motional Emf of an Object in Orbit

Figure 6.12 Motional emf as electrical power conversion for the space shuttle is the motivation for the Tethered Satellite experiment. A 5 kV emf was predicted to beinduced in the 20 km long tether while moving at orbital speed in the Earth’s magnetic field. The circuit is completed by a return path through the stationary ionosphere.

Calculate the motional emf induced along a 20.0 km long conductor moving at an orbital speed of 7.80 km/s perpendicular to the Earth’s

5.00×10−5 T magnetic field.

Strategy

This is a straightforward application of the expression for motional emf— emf = Bℓv .

Solution

Entering the given values into emf = Bℓv gives

(6.10)emf = Bℓv= (5.00×10−5 T)(2.0×104 m)(7.80×103 m/s)= 7.80×103 V.

Discussion

The value obtained is greater than the 5 kV measured voltage for the shuttle experiment, since the actual orbital motion of the tether is notperpendicular to the Earth’s field. The 7.80 kV value is the maximum emf obtained when θ = 90º and sin θ = 1 .


6.4 Eddy Currents and Magnetic Damping

Eddy Currents and Magnetic DampingAs discussed in Motional Emf, motional emf is induced when a conductor moves in a magnetic field or when a magnetic field moves relative to aconductor. If motional emf can cause a current loop in the conductor, we refer to that current as an eddy current. Eddy currents can producesignificant drag, called magnetic damping, on the motion involved. Consider the apparatus shown in Figure 6.13, which swings a pendulum bobbetween the poles of a strong magnet. (This is another favorite physics lab activity.) If the bob is metal, there is significant drag on the bob as it entersand leaves the field, quickly damping the motion. If, however, the bob is a slotted metal plate, as shown in Figure 6.13(b), there is a much smallereffect due to the magnet. There is no discernible effect on a bob made of an insulator. Why is there drag in both directions, and are there any uses formagnetic drag?

Figure 6.13 A common physics demonstration device for exploring eddy currents and magnetic damping. (a) The motion of a metal pendulum bob swinging between the polesof a magnet is quickly damped by the action of eddy currents. (b) There is little effect on the motion of a slotted metal bob, implying that eddy currents are made less effective.(c) There is also no magnetic damping on a nonconducting bob, since the eddy currents are extremely small.

Figure 6.14 shows what happens to the metal plate as it enters and leaves the magnetic field. In both cases, it experiences a force opposing itsmotion. As it enters from the left, flux increases, and so an eddy current is set up (Faraday’s law) in the counterclockwise direction (Lenz’s law), asshown. Only the right-hand side of the current loop is in the field, so that there is an unopposed force on it to the left (RHR-1). When the metal plate iscompletely inside the field, there is no eddy current if the field is uniform, since the flux remains constant in this region. But when the plate leaves thefield on the right, flux decreases, causing an eddy current in the clockwise direction that, again, experiences a force to the left, further slowing themotion. A similar analysis of what happens when the plate swings from the right toward the left shows that its motion is also damped when enteringand leaving the field.

Figure 6.14 A more detailed look at the conducting plate passing between the poles of a magnet. As it enters and leaves the field, the change in flux produces an eddycurrent. Magnetic force on the current loop opposes the motion. There is no current and no magnetic drag when the plate is completely inside the uniform field.

When a slotted metal plate enters the field, as shown in Figure 6.15, an emf is induced by the change in flux, but it is less effective because the slotslimit the size of the current loops. Moreover, adjacent loops have currents in opposite directions, and their effects cancel. When an insulating materialis used, the eddy current is extremely small, and so magnetic damping on insulators is negligible. If eddy currents are to be avoided in conductors,then they can be slotted or constructed of thin layers of conducting material separated by insulating sheets.



Figure 6.15 Eddy currents induced in a slotted metal plate entering a magnetic field form small loops, and the forces on them tend to cancel, thereby making magnetic dragalmost zero.

Applications of Magnetic DampingOne use of magnetic damping is found in sensitive laboratory balances. To have maximum sensitivity and accuracy, the balance must be as friction-free as possible. But if it is friction-free, then it will oscillate for a very long time. Magnetic damping is a simple and ideal solution. With magneticdamping, drag is proportional to speed and becomes zero at zero velocity. Thus the oscillations are quickly damped, after which the damping forcedisappears, allowing the balance to be very sensitive. (See Figure 6.16.) In most balances, magnetic damping is accomplished with a conductingdisc that rotates in a fixed field.

Figure 6.16 Magnetic damping of this sensitive balance slows its oscillations. Since Faraday’s law of induction gives the greatest effect for the most rapid change, damping isgreatest for large oscillations and goes to zero as the motion stops.

Since eddy currents and magnetic damping occur only in conductors, recycling centers can use magnets to separate metals from other materials.Trash is dumped in batches down a ramp, beneath which lies a powerful magnet. Conductors in the trash are slowed by magnetic damping whilenonmetals in the trash move on, separating from the metals. (See Figure 6.17.) This works for all metals, not just ferromagnetic ones. A magnet canseparate out the ferromagnetic materials alone by acting on stationary trash.


Figure 6.17 Metals can be separated from other trash by magnetic drag. Eddy currents and magnetic drag are created in the metals sent down this ramp by the powerfulmagnet beneath it. Nonmetals move on.

Other major applications of eddy currents are in metal detectors and braking systems in trains and roller coasters. Portable metal detectors (Figure6.18) consist of a primary coil carrying an alternating current and a secondary coil in which a current is induced. An eddy current will be induced in apiece of metal close to the detector which will cause a change in the induced current within the secondary coil, leading to some sort of signal like ashrill noise. Braking using eddy currents is safer because factors such as rain do not affect the braking and the braking is smoother. However, eddycurrents cannot bring the motion to a complete stop, since the force produced decreases with speed. Thus, speed can be reduced from say 20 m/s to5 m/s, but another form of braking is needed to completely stop the vehicle. Generally, powerful rare earth magnets such as neodymium magnets areused in roller coasters. Figure 6.19 shows rows of magnets in such an application. The vehicle has metal fins (normally containing copper) whichpass through the magnetic field slowing the vehicle down in much the same way as with the pendulum bob shown in Figure 6.13.

Figure 6.18 A soldier in Iraq uses a metal detector to search for explosives and weapons. (credit: U.S. Army)

Figure 6.19 The rows of rare earth magnets (protruding horizontally) are used for magnetic braking in roller coasters. (credit: Stefan Scheer, Wikimedia Commons)

Induction cooktops have electromagnets under their surface. The magnetic field is varied rapidly producing eddy currents in the base of the pot,causing the pot and its contents to increase in temperature. Induction cooktops have high efficiencies and good response times but the base of thepot needs to be ferromagnetic, iron or steel for induction to work.



6.5 Electric GeneratorsElectric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux. We will nowexplore generators in more detail. Consider the following example.

Example 6.3 Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 6.20 is rotated through one-fourth of a revolution (from θ = 0º to θ = 90º ) in 15.0 ms. The 200-turncircular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?

Figure 6.20 When this generator coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero, inducing an emf.

Strategy

We use Faraday’s law of induction to find the average emf induced over a time Δt :

(6.11)emf = −N ΔΦΔt .

We know that N = 200 and Δt = 15.0 ms , and so we must determine the change in flux ΔΦ to find emf.

Solution

Since the area of the loop and the magnetic field strength are constant, we see that

(6.12)ΔΦ = Δ(BA cos θ) = ABΔ(cos θ).

Now, Δ(cos θ) = −1.0 , since it was given that θ goes from 0º to 90º . Thus ΔΦ = −AB , and

(6.13)emf = N ABΔt .

The area of the loop is A = πr2 = (3.14...)(0.0500 m)2 = 7.85×10−3 m2 . Entering this value gives

(6.14)emf = 200(7.85×10−3 m2)(1.25 T)

15.0×10−3 s= 131 V.

Discussion

This is a practical average value, similar to the 120 V used in household power.

The emf calculated in Example 6.3 is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the anglebetween the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf ona rotating rectangular coil of width w and height ℓ in a uniform magnetic field, as illustrated in Figure 6.21.


Figure 6.21 A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that varies sinusoidally in time. Note thegenerator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around.

Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experienceforces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause acurrent. We can thus find the induced emf by considering only the side wires. Motional emf is given to be emf = Bℓv , where the velocity v is

perpendicular to the magnetic field B . Here the velocity is at an angle θ with B , so that its component perpendicular to B is v sin θ (see Figure6.21). Thus in this case the emf induced on each side is emf = Bℓv sin θ , and they are in the same direction. The total emf around the loop is then

(6.15)emf = 2Bℓv sin θ.This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constantangular velocity ω . The angle θ is related to angular velocity by θ = ωt , so that

(6.16)emf = 2Bℓv sin ωt.

Now, linear velocity v is related to angular velocity ω by v = rω . Here r = w / 2 , so that v = (w / 2)ω , and

(6.17)emf = 2Bℓw2 ω sin ωt = (ℓw)Bω sin ωt.

Noting that the area of the loop is A = ℓw , and allowing for N loops, we find that

(6.18)emf = NABω sin ωt

is the emf induced in a generator coil of N turns and area A rotating at a constant angular velocity ω in a uniform magnetic field B . This canalso be expressed as

(6.19)emf = emf0 sin ωt,

where

(6.20)emf0 = NABω

is the maximum (peak) emf. Note that the frequency of the oscillation is f = ω / 2π , and the period is T = 1 / f = 2π / ω . Figure 6.22 shows a

graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.

Figure 6.22 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time. emf0is the peak emf. The period is T = 1 / f = 2π / ω , where f is the frequency. Note that the script E stands for emf.

The fact that the peak emf, emf0 = NABω , makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the

greater the output voltage. It is interesting that the faster the generator is spun (greater ω ), the greater the emf. This is noticeable on bicyclegenerators—at least the cheaper varieties. One of the authors as a juvenile found it amusing to ride his bicycle fast enough to burn out his lights, untilhe had to ride home lightless one dark night.



Figure 6.23 shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements of multiple coils and split ringscan produce smoother DC, although electronic rather than mechanical means are usually used to make ripple-free DC.

Figure 6.23 Split rings, called commutators, produce a pulsed DC emf output in this configuration.

Example 6.4 Calculating the Maximum Emf of a Generator

Calculate the maximum emf, emf0 , of the generator that was the subject of Example 6.3.

Strategy

Once ω , the angular velocity, is determined, emf0 = NABω can be used to find emf0 . All other quantities are known.

Solution

Angular velocity is defined to be the change in angle per unit time:

(6.21)ω = ΔθΔt .

One-fourth of a revolution is π/2 radians, and the time is 0.0150 s; thus,

(6.22)ω = π / 2 rad0.0150 s

= 104.7 rad/s.

104.7 rad/s is exactly 1000 rpm. We substitute this value for ω and the information from the previous example into emf0 = NABω , yielding

(6.23)emf0 = NABω

= 200(7.85×10−3 m2)(1.25 T)(104.7 rad/s)= 206 V

.

Discussion

The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be.

In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energythat turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 6.24 shows acutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator.


Figure 6.24 Steam turbine/generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to the generator. (credit:Nabonaco, Wikimedia Commons)

Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes agenerator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Back Emf, we shall further explore the action ofa motor as a generator.

6.6 Back EmfIt has been noted that motors and generators are very similar. Generators convert mechanical energy into electrical energy, whereas motors convertelectrical energy into mechanical energy. Furthermore, motors and generators have the same construction. When the coil of a motor is turned,magnetic flux changes, and an emf (consistent with Faraday’s law of induction) is induced. The motor thus acts as a generator whenever its coilrotates. This will happen whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor isdoing work and its shaft is turning, an emf is generated. Lenz’s law tells us the emf opposes any change, so that the input emf that powers the motorwill be opposed by the motor’s self-generated emf, called the back emf of the motor. (See Figure 6.25.)

Figure 6.25 The coil of a DC motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes the one driving the motor. Backemf is zero when the motor is not turning, and it increases proportionally to the motor’s angular velocity.

Back emf is the generator output of a motor, and so it is proportional to the motor’s angular velocity ω . It is zero when the motor is first turned on,meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns fasterand faster, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. Thiseffect is noticeable in a number of situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dimbriefly due to the IR drop produced in feeder lines by the large current drawn by the motor. When a motor first comes on, it draws more current thanwhen it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motorslows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overheat it

(via resistive power in the coil, P = I 2R ), perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it will increaseits angular velocity ω until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction.

Consider, for example, the motor coils represented in Figure 6.25. The coils have a 0.400 Ω equivalent resistance and are driven by a 48.0 V

emf. Shortly after being turned on, they draw a current I = V/R = (48.0 V)/(0.400 Ω ) = 120 A and, thus, dissipate P = I 2R = 5.76 kW of

energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltageacross the coils is 8.0 V (48.0 V minus the 40.0 V back emf), and the current drawn is I = V/R = (8.0 V)/(0.400 Ω ) = 20 A . Under normal

load, then, the power dissipated is P = IV = (20 A) / (8.0 V) = 160 W . The latter will not cause a problem for this motor, whereas the former

5.76 kW would burn out the coils if sustained.

6.7 TransformersTransformers do what their name implies—they transform voltages from one value to another (The term voltage is used rather than emf, becausetransformers have internal resistance). For example, many cell phones, laptops, video games, and power tools and small appliances have atransformer built into their plug-in unit (like that in Figure 6.26) that changes 120 V or 240 V AC into whatever voltage the device uses. Transformers



are also used at several points in the power distribution systems, such as illustrated in Figure 6.27. Power is sent long distances at high voltages,because less current is required for a given amount of power, and this means less line loss, as was discussed previously. But high voltages posegreater hazards, so that transformers are employed to produce lower voltage at the user’s location.

Figure 6.26 The plug-in transformer has become increasingly familiar with the proliferation of electronic devices that operate on voltages other than common 120 V AC. Mostare in the 3 to 12 V range. (credit: Shop Xtreme)

Figure 6.27 Transformers change voltages at several points in a power distribution system. Electric power is usually generated at greater than 10 kV, and transmitted longdistances at voltages over 200 kV—sometimes as great as 700 kV—to limit energy losses. Local power distribution to neighborhoods or industries goes through a substationand is sent short distances at voltages ranging from 5 to 13 kV. This is reduced to 120, 240, or 480 V for safety at the individual user site.

The type of transformer considered in this text—see Figure 6.28—is based on Faraday’s law of induction and is very similar in construction to theapparatus Faraday used to demonstrate magnetic fields could cause currents. The two coils are called the primary and secondary coils. In normaluse, the input voltage is placed on the primary, and the secondary produces the transformed output voltage. Not only does the iron core trap themagnetic field created by the primary coil, its magnetization increases the field strength. Since the input voltage is AC, a time-varying magnetic flux issent to the secondary, inducing its AC output voltage.

Figure 6.28 A typical construction of a simple transformer has two coils wound on a ferromagnetic core that is laminated to minimize eddy currents. The magnetic field createdby the primary is mostly confined to and increased by the core, which transmits it to the secondary coil. Any change in current in the primary induces a current in thesecondary.

For the simple transformer shown in Figure 6.28, the output voltage Vs depends almost entirely on the input voltage Vp and the ratio of the

number of loops in the primary and secondary coils. Faraday’s law of induction for the secondary coil gives its induced output voltage Vs to be


(6.24)Vs = −NsΔΦΔt ,

where Ns is the number of loops in the secondary coil and ΔΦ / Δt is the rate of change of magnetic flux. Note that the output voltage equals the

induced emf ( Vs = emfs ), provided coil resistance is small (a reasonable assumption for transformers). The cross-sectional area of the coils is the

same on either side, as is the magnetic field strength, and so ΔΦ / Δt is the same on either side. The input primary voltage Vp is also related to

changing flux by

(6.25)V p = −NpΔΦΔt .

The reason for this is a little more subtle. Lenz’s law tells us that the primary coil opposes the change in flux caused by the input voltage Vp , hence

the minus sign (This is an example of self-inductance, a topic to be explored in some detail in later sections). Assuming negligible coil resistance,Kirchhoff’s loop rule tells us that the induced emf exactly equals the input voltage. Taking the ratio of these last two equations yields a usefulrelationship:

(6.26)VsVp

= NsNp

.

This is known as the transformer equation, and it simply states that the ratio of the secondary to primary voltages in a transformer equals the ratioof the number of loops in their coils.

The output voltage of a transformer can be less than, greater than, or equal to the input voltage, depending on the ratio of the number of loops in theircoils. Some transformers even provide a variable output by allowing connection to be made at different points on the secondary coil. A step-uptransformer is one that increases voltage, whereas a step-down transformer decreases voltage. Assuming, as we have, that resistance isnegligible, the electrical power output of a transformer equals its input. This is nearly true in practice—transformer efficiency often exceeds 99%.Equating the power input and output,

(6.27)Pp = IpVp = IsVs = Ps.

Rearranging terms gives

(6.28)VsVp

=IpIs

.

Combining this withVsVp

= NsNp

, we find that

(6.29)IsIp

=NpNs

is the relationship between the output and input currents of a transformer. So if voltage increases, current decreases. Conversely, if voltagedecreases, current increases.

Example 6.5 Calculating Characteristics of a Step-Up Transformer

A portable x-ray unit has a step-up transformer, the 120 V input of which is transformed to the 100 kV output needed by the x-ray tube. Theprimary has 50 loops and draws a current of 10.00 A when in use. (a) What is the number of loops in the secondary? (b) Find the current outputof the secondary.


We solveVsVp

= NsNp

for Ns , the number of loops in the secondary, and enter the known values. This gives

(6.30)Ns = NpVsVp

= (50)100,000 V120 V = 4.17×104 .

Discussion for (a)

A large number of loops in the secondary (compared with the primary) is required to produce such a large voltage. This would be true for neonsign transformers and those supplying high voltage inside TVs and CRTs.


We can similarly find the output current of the secondary by solvingIsIp

=NpNs

for Is and entering known values. This gives



(6.31)Is = Ip

NpNs

= (10.00 A) 504.17×104 = 12.0 mA.

Discussion for (b)

As expected, the current output is significantly less than the input. In certain spectacular demonstrations, very large voltages are used to producelong arcs, but they are relatively safe because the transformer output does not supply a large current. Note that the power input here isPp = IpVp = (10.00 A)(120 V) = 1.20 kW . This equals the power output Pp = IsVs = (12.0 mA)(100 kV) = 1.20 kW , as we

assumed in the derivation of the equations used.

The fact that transformers are based on Faraday’s law of induction makes it clear why we cannot use transformers to change DC voltages. If there isno change in primary voltage, there is no voltage induced in the secondary. One possibility is to connect DC to the primary coil through a switch. Asthe switch is opened and closed, the secondary produces a voltage like that in Figure 6.29. This is not really a practical alternative, and AC is incommon use wherever it is necessary to increase or decrease voltages.

Figure 6.29 Transformers do not work for pure DC voltage input, but if it is switched on and off as on the top graph, the output will look something like that on the bottomgraph. This is not the sinusoidal AC most AC appliances need.

Example 6.6 Calculating Characteristics of a Step-Down Transformer

A battery charger meant for a series connection of ten nickel-cadmium batteries (total emf of 12.5 V DC) needs to have a 15.0 V output to chargethe batteries. It uses a step-down transformer with a 200-loop primary and a 120 V input. (a) How many loops should there be in the secondarycoil? (b) If the charging current is 16.0 A, what is the input current?


You would expect the secondary to have a small number of loops. SolvingVsVp

= NsNp

for Ns and entering known values gives

(6.32)Ns = NpVsVp

= (200)15.0 V120 V = 25.


The current input can be obtained by solvingIsIp

=NpNs

for Ip and entering known values. This gives

(6.33)Ip = IsNsNp

= (16.0 A) 25200 = 2.00 A.

Discussion


The number of loops in the secondary is small, as expected for a step-down transformer. We also see that a small input current produces alarger output current in a step-down transformer. When transformers are used to operate large magnets, they sometimes have a small number ofvery heavy loops in the secondary. This allows the secondary to have low internal resistance and produce large currents. Note again that thissolution is based on the assumption of 100% efficiency—or power out equals power in ( Pp = Ps )—reasonable for good transformers. In this

case the primary and secondary power is 240 W. (Verify this for yourself as a consistency check.) Note that the Ni-Cd batteries need to becharged from a DC power source (as would a 12 V battery). So the AC output of the secondary coil needs to be converted into DC. This is doneusing something called a rectifier, which uses devices called diodes that allow only a one-way flow of current.

Transformers have many applications in electrical safety systems, which are discussed in Electrical Safety: Systems and Devices.




6.8 Electrical Safety: Systems and DevicesElectricity has two hazards. A thermal hazard occurs when there is electrical overheating. A shock hazard occurs when electric current passesthrough a person. Both hazards have already been discussed. Here we will concentrate on systems and devices that prevent electrical hazards.

Figure 6.31 shows the schematic for a simple AC circuit with no safety features. This is not how power is distributed in practice. Modern householdand industrial wiring requires the three-wire system, shown schematically in Figure 6.32, which has several safety features. First is the familiarcircuit breaker (or fuse) to prevent thermal overload. Second, there is a protective case around the appliance, such as a toaster or refrigerator. Thecase’s safety feature is that it prevents a person from touching exposed wires and coming into electrical contact with the circuit, helping preventshocks.

Figure 6.31 Schematic of a simple AC circuit with a voltage source and a single appliance represented by the resistance R . There are no safety features in this circuit.

Figure 6.32 The three-wire system connects the neutral wire to the earth at the voltage source and user location, forcing it to be at zero volts and supplying an alternativereturn path for the current through the earth. Also grounded to zero volts is the case of the appliance. A circuit breaker or fuse protects against thermal overload and is inseries on the active (live/hot) wire. Note that wire insulation colors vary with region and it is essential to check locally to determine which color codes are in use (and even ifthey were followed in the particular installation).




There are three connections to earth or ground (hereafter referred to as “earth/ground”) shown in Figure 6.32. Recall that an earth/ground connectionis a low-resistance path directly to the earth. The two earth/ground connections on the neutral wire force it to be at zero volts relative to the earth,giving the wire its name. This wire is therefore safe to touch even if its insulation, usually white, is missing. The neutral wire is the return path for thecurrent to follow to complete the circuit. Furthermore, the two earth/ground connections supply an alternative path through the earth, a goodconductor, to complete the circuit. The earth/ground connection closest to the power source could be at the generating plant, while the other is at theuser’s location. The third earth/ground is to the case of the appliance, through the green earth/ground wire, forcing the case, too, to be at zero volts.The live or hot wire (hereafter referred to as “live/hot”) supplies voltage and current to operate the appliance. Figure 6.33 shows a more pictorialversion of how the three-wire system is connected through a three-prong plug to an appliance.

Figure 6.33 The standard three-prong plug can only be inserted in one way, to assure proper function of the three-wire system.

A note on insulation color-coding: Insulating plastic is color-coded to identify live/hot, neutral and ground wires but these codes vary around the world.Live/hot wires may be brown, red, black, blue or grey. Neutral wire may be blue, black or white. Since the same color may be used for live/hot orneutral in different parts of the world, it is essential to determine the color code in your region. The only exception is the earth/ground wire which isoften green but may be yellow or just bare wire. Striped coatings are sometimes used for the benefit of those who are colorblind.

The three-wire system replaced the older two-wire system, which lacks an earth/ground wire. Under ordinary circumstances, insulation on the live/hotand neutral wires prevents the case from being directly in the circuit, so that the earth/ground wire may seem like double protection. Grounding thecase solves more than one problem, however. The simplest problem is worn insulation on the live/hot wire that allows it to contact the case, as shownin Figure 6.34. Lacking an earth/ground connection (some people cut the third prong off the plug because they only have outdated two holereceptacles), a severe shock is possible. This is particularly dangerous in the kitchen, where a good connection to earth/ground is available throughwater on the floor or a water faucet. With the earth/ground connection intact, the circuit breaker will trip, forcing repair of the appliance. Why are someappliances still sold with two-prong plugs? These have nonconducting cases, such as power tools with impact resistant plastic cases, and are calleddoubly insulated. Modern two-prong plugs can be inserted into the asymmetric standard outlet in only one way, to ensure proper connection of live/hot and neutral wires.


Figure 6.34 Worn insulation allows the live/hot wire to come into direct contact with the metal case of this appliance. (a) The earth/ground connection being broken, the personis severely shocked. The appliance may operate normally in this situation. (b) With a proper earth/ground, the circuit breaker trips, forcing repair of the appliance.

Electromagnetic induction causes a more subtle problem that is solved by grounding the case. The AC current in appliances can induce an emf onthe case. If grounded, the case voltage is kept near zero, but if the case is not grounded, a shock can occur as pictured in Figure 6.35. Currentdriven by the induced case emf is called a leakage current, although current does not necessarily pass from the resistor to the case.

Figure 6.35 AC currents can induce an emf on the case of an appliance. The voltage can be large enough to cause a shock. If the case is grounded, the induced emf is keptnear zero.



A ground fault interrupter (GFI) is a safety device found in updated kitchen and bathroom wiring that works based on electromagnetic induction. GFIscompare the currents in the live/hot and neutral wires. When live/hot and neutral currents are not equal, it is almost always because current in theneutral is less than in the live/hot wire. Then some of the current, again called a leakage current, is returning to the voltage source by a path otherthan through the neutral wire. It is assumed that this path presents a hazard, such as shown in Figure 6.36. GFIs are usually set to interrupt thecircuit if the leakage current is greater than 5 mA, the accepted maximum harmless shock. Even if the leakage current goes safely to earth/groundthrough an intact earth/ground wire, the GFI will trip, forcing repair of the leakage.

Figure 6.36 A ground fault interrupter (GFI) compares the currents in the live/hot and neutral wires and will trip if their difference exceeds a safe value. The leakage currenthere follows a hazardous path that could have been prevented by an intact earth/ground wire.

Figure 6.37 shows how a GFI works. If the currents in the live/hot and neutral wires are equal, then they induce equal and opposite emfs in the coil. Ifnot, then the circuit breaker will trip.

Figure 6.37 A GFI compares currents by using both to induce an emf in the same coil. If the currents are equal, they will induce equal but opposite emfs.

Another induction-based safety device is the isolation transformer, shown in Figure 6.38. Most isolation transformers have equal input and outputvoltages. Their function is to put a large resistance between the original voltage source and the device being operated. This prevents a completecircuit between them, even in the circumstance shown. There is a complete circuit through the appliance. But there is not a complete circuit forcurrent to flow through the person in the figure, who is touching only one of the transformer’s output wires, and neither output wire is grounded. Theappliance is isolated from the original voltage source by the high resistance of the material between the transformer coils, hence the name isolationtransformer. For current to flow through the person, it must pass through the high-resistance material between the coils, through the wire, the person,and back through the earth—a path with such a large resistance that the current is negligible.

Figure 6.38 An isolation transformer puts a large resistance between the original voltage source and the device, preventing a complete circuit between them.


The basics of electrical safety presented here help prevent many electrical hazards. Electrical safety can be pursued to greater depths. There are, forexample, problems related to different earth/ground connections for appliances in close proximity. Many other examples are found in hospitals.Microshock-sensitive patients, for instance, require special protection. For these people, currents as low as 0.1 mA may cause ventricular fibrillation.The interested reader can use the material presented here as a basis for further study.

6.9 Inductance

InductorsInduction is the process in which an emf is induced by changing magnetic flux. Many examples have been discussed so far, some more effective thanothers. Transformers, for example, are designed to be particularly effective at inducing a desired voltage and current with very little loss of energy toother forms. Is there a useful physical quantity related to how “effective” a given device is? The answer is yes, and that physical quantity is calledinductance.

Mutual inductance is the effect of Faraday’s law of induction for one device upon another, such as the primary coil in transmitting energy to thesecondary in a transformer. See Figure 6.39, where simple coils induce emfs in one another.

Figure 6.39 These coils can induce emfs in one another like an inefficient transformer. Their mutual inductance M indicates the effectiveness of the coupling between them.Here a change in current in coil 1 is seen to induce an emf in coil 2. (Note that " E2 induced" represents the induced emf in coil 2.)

In the many cases where the geometry of the devices is fixed, flux is changed by varying current. We therefore concentrate on the rate of change ofcurrent, ΔI/Δt , as the cause of induction. A change in the current I1 in one device, coil 1 in the figure, induces an emf2 in the other. We express

this in equation form as

(6.34)emf2 = −MΔI1

Δt ,

where M is defined to be the mutual inductance between the two devices. The minus sign is an expression of Lenz’s law. The larger the mutual

inductance M , the more effective the coupling. For example, the coils in Figure 6.39 have a small M compared with the transformer coils in Figure6.28. Units for M are (V ⋅ s)/A = Ω ⋅ s , which is named a henry (H), after Joseph Henry. That is, 1 H = 1 Ω ⋅ s .

Nature is symmetric here. If we change the current I2 in coil 2, we induce an emf1 in coil 1, which is given by

(6.35)emf1 = −MΔI2

Δt ,

where M is the same as for the reverse process. Transformers run backward with the same effectiveness, or mutual inductance M .

A large mutual inductance M may or may not be desirable. We want a transformer to have a large mutual inductance. But an appliance, such as anelectric clothes dryer, can induce a dangerous emf on its case if the mutual inductance between its coils and the case is large. One way to reducemutual inductance M is to counterwind coils to cancel the magnetic field produced. (See Figure 6.40.)



Figure 6.40 The heating coils of an electric clothes dryer can be counter-wound so that their magnetic fields cancel one another, greatly reducing the mutual inductance withthe case of the dryer.

Self-inductance, the effect of Faraday’s law of induction of a device on itself, also exists. When, for example, current through a coil is increased, themagnetic field and flux also increase, inducing a counter emf, as required by Lenz’s law. Conversely, if the current is decreased, an emf is inducedthat opposes the decrease. Most devices have a fixed geometry, and so the change in flux is due entirely to the change in current ΔI through thedevice. The induced emf is related to the physical geometry of the device and the rate of change of current. It is given by

(6.36)emf = −LΔIΔt ,

where L is the self-inductance of the device. A device that exhibits significant self-inductance is called an inductor, and given the symbol in Figure6.41.

Figure 6.41

The minus sign is an expression of Lenz’s law, indicating that emf opposes the change in current. Units of self-inductance are henries (H) just as formutual inductance. The larger the self-inductance L of a device, the greater its opposition to any change in current through it. For example, a large

coil with many turns and an iron core has a large L and will not allow current to change quickly. To avoid this effect, a small L must be achieved,such as by counterwinding coils as in Figure 6.40.

A 1 H inductor is a large inductor. To illustrate this, consider a device with L = 1.0 H that has a 10 A current flowing through it. What happens if we

try to shut off the current rapidly, perhaps in only 1.0 ms? An emf, given by emf = −L(ΔI / Δt) , will oppose the change. Thus an emf will be

induced given by emf = −L(ΔI / Δt) = (1.0 H)[(10 A) / (1.0 ms)] = 10,000 V . The positive sign means this large voltage is in the same

direction as the current, opposing its decrease. Such large emfs can cause arcs, damaging switching equipment, and so it may be necessary tochange current more slowly.

There are uses for such a large induced voltage. Camera flashes use a battery, two inductors that function as a transformer, and a switching systemor oscillator to induce large voltages. (Remember that we need a changing magnetic field, brought about by a changing current, to induce a voltage inanother coil.) The oscillator system will do this many times as the battery voltage is boosted to over one thousand volts. (You may hear the highpitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash. (SeeFigure 6.42.)


Figure 6.42 Through rapid switching of an inductor, 1.5 V batteries can be used to induce emfs of several thousand volts. This voltage can be used to store charge in acapacitor for later use, such as in a camera flash attachment.

It is possible to calculate L for an inductor given its geometry (size and shape) and knowing the magnetic field that it produces. This is difficult in

most cases, because of the complexity of the field created. So in this text the inductance L is usually a given quantity. One exception is the solenoid,because it has a very uniform field inside, a nearly zero field outside, and a simple shape. It is instructive to derive an equation for its inductance. Westart by noting that the induced emf is given by Faraday’s law of induction as emf = −N(ΔΦ / Δt) and, by the definition of self-inductance, as

emf = −L(ΔI / Δt) . Equating these yields

(6.37)emf = −N ΔΦΔt = −LΔI

Δt .

Solving for L gives

(6.38)L = N ΔΦΔI .

This equation for the self-inductance L of a device is always valid. It means that self-inductance L depends on how effective the current is in

creating flux; the more effective, the greater ΔΦ / ΔI is.

Let us use this last equation to find an expression for the inductance of a solenoid. Since the area A of a solenoid is fixed, the change in flux is

ΔΦ = Δ(BA) = AΔB . To find ΔB , we note that the magnetic field of a solenoid is given by B = μ0nI = μ0NIℓ . (Here n = N / ℓ , where N is

the number of coils and ℓ is the solenoid’s length.) Only the current changes, so that ΔΦ = AΔB = μ0NAΔIℓ . Substituting ΔΦ into

L = N ΔΦΔI gives

(6.39)L = N ΔΦ

ΔI = Nμ0 NAΔI

ℓΔI .

This simplifies to

(6.40)L = μ0 N 2 A

ℓ (solenoid).

This is the self-inductance of a solenoid of cross-sectional area A and length ℓ . Note that the inductance depends only on the physicalcharacteristics of the solenoid, consistent with its definition.

Example 6.7 Calculating the Self-inductance of a Moderate Size Solenoid

Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.

Strategy

This is a straightforward application of L = μ0 N 2 Aℓ , since all quantities in the equation except L are known.

Solution

Use the following expression for the self-inductance of a solenoid:

(6.41)L = μ0 N 2 A

ℓ .

The cross-sectional area in this example is A = πr2 = (3.14...)(0.0200 m)2 = 1.26×10−3 m2 , N is given to be 200, and the length ℓ is

0.100 m. We know the permeability of free space is μ0 = 4π×10−7 T ⋅ m/A . Substituting these into the expression for L gives



(6.42)L = (4π×10−7 T ⋅ m/A)(200)2(1.26×10−3 m2)

0.100 m= 0.632 mH.

Discussion

This solenoid is moderate in size. Its inductance of nearly a millihenry is also considered moderate.

One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. An electrical circuit with aninductor is placed in the road under the place a waiting car will stop over. The body of the car increases the inductance and the circuit changessending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductorin the metal detector frame acts as both a transmitter and a receiver. The pulsed signal in the transmitter coil induces a signal in the receiver. Theself-inductance of the circuit is affected by any metal object in the path. Such detectors can be adjusted for sensitivity and also can indicate theapproximate location of metal found on a person. (But they will not be able to detect any plastic explosive such as that found on the “underwearbomber.”) See Figure 6.43.

Figure 6.43 The familiar security gate at an airport can not only detect metals but also indicate their approximate height above the floor. (credit: Alexbuirds, WikimediaCommons)

Energy Stored in an InductorWe know from Lenz’s law that inductances oppose changes in current. There is an alternative way to look at this opposition that is based on energy.Energy is stored in a magnetic field. It takes time to build up energy, and it also takes time to deplete energy; hence, there is an opposition to rapidchange. In an inductor, the magnetic field is directly proportional to current and to the inductance of the device. It can be shown that the energystored in an inductor Eind is given by

(6.43)Eind = 12LI 2.

This expression is similar to that for the energy stored in a capacitor.

Example 6.8 Calculating the Energy Stored in the Field of a Solenoid

How much energy is stored in the 0.632 mH inductor of the preceding example when a 30.0 A current flows through it?

Strategy

The energy is given by the equation Eind = 12LI 2 , and all quantities except Eind are known.

Solution

Substituting the value for L found in the previous example and the given current into Eind = 12LI 2 gives

(6.44)Eind = 12LI 2

= 0.5(0.632×10−3 H)(30.0 A)2 = 0.284 J.

Discussion

This amount of energy is certainly enough to cause a spark if the current is suddenly switched off. It cannot be built up instantaneously unlessthe power input is infinite.

6.10 RL CircuitsWe know that the current through an inductor L cannot be turned on or off instantaneously. The change in current changes flux, inducing an emfopposing the change (Lenz’s law). How long does the opposition last? Current will flow and can be turned off, but how long does it take? Figure 6.44shows a switching circuit that can be used to examine current through an inductor as a function of time.


Figure 6.44 (a) An RL circuit with a switch to turn current on and off. When in position 1, the battery, resistor, and inductor are in series and a current is established. In position2, the battery is removed and the current eventually stops because of energy loss in the resistor. (b) A graph of current growth versus time when the switch is moved toposition 1. (c) A graph of current decay when the switch is moved to position 2.

When the switch is first moved to position 1 (at t = 0 ), the current is zero and it eventually rises to I0 = V/R , where R is the total resistance of

the circuit. The opposition of the inductor L is greatest at the beginning, because the amount of change is greatest. The opposition it poses is in theform of an induced emf, which decreases to zero as the current approaches its final value. The opposing emf is proportional to the amount of changeleft. This is the hallmark of an exponential behavior, and it can be shown with calculus that

(6.45)I = I0(1 − e−t / τ) (turning on),

is the current in an RL circuit when switched on (Note the similarity to the exponential behavior of the voltage on a charging capacitor). The initialcurrent is zero and approaches I0 = V/R with a characteristic time constant τ for an RL circuit, given by

(6.46)τ = LR,

where τ has units of seconds, since 1 H=1 Ω·s . In the first period of time τ , the current rises from zero to 0.632I0 , since

I = I0(1 − e−1) = I0(1 − 0.368) = 0.632I0 . The current will go 0.632 of the remainder in the next time τ . A well-known property of the

exponential is that the final value is never exactly reached, but 0.632 of the remainder to that value is achieved in every characteristic time τ . In justa few multiples of the time τ , the final value is very nearly achieved, as the graph in Figure 6.44(b) illustrates.

The characteristic time τ depends on only two factors, the inductance L and the resistance R . The greater the inductance L , the greater τ is,

which makes sense since a large inductance is very effective in opposing change. The smaller the resistance R , the greater τ is. Again this makes

sense, since a small resistance means a large final current and a greater change to get there. In both cases—large L and small R —more energyis stored in the inductor and more time is required to get it in and out.

When the switch in Figure 6.44(a) is moved to position 2 and cuts the battery out of the circuit, the current drops because of energy dissipation bythe resistor. But this is also not instantaneous, since the inductor opposes the decrease in current by inducing an emf in the same direction as the

battery that drove the current. Furthermore, there is a certain amount of energy, (1/2)LI02 , stored in the inductor, and it is dissipated at a finite rate.

As the current approaches zero, the rate of decrease slows, since the energy dissipation rate is I 2 R . Once again the behavior is exponential, and

I is found to be

(6.47)I = I0e−t / τ (turning off).

(See Figure 6.44(c).) In the first period of time τ = L / R after the switch is closed, the current falls to 0.368 of its initial value, since

I = I0e−1 = 0.368I0 . In each successive time τ , the current falls to 0.368 of the preceding value, and in a few multiples of τ , the current

becomes very close to zero, as seen in the graph in Figure 6.44(c).

Example 6.9 Calculating Characteristic Time and Current in an RL Circuit

(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 Ω resistor? (b) Find the current 5.00 ms after the switchis moved to position 2 to disconnect the battery, if it is initially 10.0 A.

Strategy for (a)

The time constant for an RL circuit is defined by τ = L / R .

Solution for (a)

Entering known values into the expression for τ given in τ = L / R yields

(6.48)τ = LR = 7.50 mH

3.00 Ω = 2.50 ms.

Discussion for (a)

This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.



Strategy for (b)

We can find the current by using I = I0e−t / τ , or by considering the decline in steps. Since the time is twice the characteristic time, we consider

the process in steps.

Solution for (b)

In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

(6.49)I = 0.368I0 = (0.368)(10.0 A)= 3.68 A at t = 2.50 ms.

After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,

(6.50)I′ = 0.368I = (0.368)(3.68 A)= 1.35 A at t = 5.00 ms.

Discussion for (b)

After another 5.00 ms has passed, the current will be 0.183 A (see Exercise 6.69); so, although it does die out, the current certainly does not goto zero instantaneously.

In summary, when the voltage applied to an inductor is changed, the current also changes, but the change in current lags the change in voltage in anRL circuit. In Reactance, Inductive and Capacitive, we explore how an RL circuit behaves when a sinusoidal AC voltage is applied.

6.11 Reactance, Inductive and CapacitiveMany circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen how capacitors and inductorsrespond to DC voltage when it is switched on and off. We will now explore how inductors and capacitors react to sinusoidal AC voltage.

Inductors and Inductive ReactanceSuppose an inductor is connected directly to an AC voltage source, as shown in Figure 6.45. It is reasonable to assume negligible resistance, sincein practice we can make the resistance of an inductor so small that it has a negligible effect on the circuit. Also shown is a graph of voltage andcurrent as functions of time.

Figure 6.45 (a) An AC voltage source in series with an inductor having negligible resistance. (b) Graph of current and voltage across the inductor as functions of time.

The graph in Figure 6.45(b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak after the voltage that drives it,just as was the case when DC voltage was switched on in the preceding section. When the voltage becomes negative at point a, the current beginsto decrease; it becomes zero at point b, where voltage is its most negative. The current then becomes negative, again following the voltage. Thevoltage becomes positive at point c and begins to make the current less negative. At point d, the current goes through zero just as the voltagereaches its positive peak to start another cycle. This behavior is summarized as follows:

AC Voltage in an Inductor

When a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a cycle, or by a 90º phase angle.

Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf V = −L(ΔI / Δt) . This is considered

to be an effective resistance of the inductor to AC. The rms current I through an inductor L is given by a version of Ohm’s law:

(6.51)I = VXL

,

where V is the rms voltage across the inductor and XL is defined to be

(6.52)XL = 2π fL,


with f the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff’s loop rule and calculus actually produces this

expression). XL is called the inductive reactance, because the inductor reacts to impede the current. XL has units of ohms ( 1 H = 1 Ω ⋅ s , so

that frequency times inductance has units of (cycles/s)( Ω ⋅ s) = Ω ), consistent with its role as an effective resistance. It makes sense that XLis proportional to L , since the greater the induction the greater its resistance to change. It is also reasonable that XL is proportional to frequency f, since greater frequency means greater change in current. That is, ΔI/Δt is large for large frequencies (large f , small Δt ). The greater the

change, the greater the opposition of an inductor.

Example 6.10 Calculating Inductive Reactance and then Current

(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current ateach frequency if the applied rms voltage is 120 V?

Strategy

The inductive reactance is found directly from the expression XL = 2π fL . Once XL has been found at each frequency, Ohm’s law as stated in

the Equation I = V / XL can be used to find the current at each frequency.

Solution for (a)

Entering the frequency and inductance into Equation XL = 2π fL gives

(6.53)XL = 2π fL = 6.28(60.0 / s)(3.00 mH) = 1.13 Ω at 60 Hz.

Similarly, at 10 kHz,

(6.54)XL = 2π fL = 6.28(1.00×104 /s)(3.00 mH) = 188 Ω at 10 kHz.

Solution for (b)

The rms current is now found using the version of Ohm’s law in Equation I = V / XL , given the applied rms voltage is 120 V. For the first

frequency, this yields

(6.55)I = VXL

= 120 V1.13 Ω = 106 A at 60 Hz.


(6.56)I = VXL

= 120 V188 Ω = 0.637 A at 10 kHz.

Discussion

The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small,consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most. Inductors can be used to filter out highfrequencies; for example, a large inductor can be put in series with a sound reproduction system or in series with your home computer to reducehigh-frequency sound output from your speakers or high-frequency power spikes into your computer.

Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes itsflow. With AC, there is no time for the current to become extremely large.

Capacitors and Capacitive ReactanceConsider the capacitor connected directly to an AC voltage source as shown in Figure 6.46. The resistance of a circuit like this can be made so smallthat it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current aregraphed as functions of time in the figure.

Figure 6.46 (a) An AC voltage source in series with a capacitor C having negligible resistance. (b) Graph of current and voltage across the capacitor as functions of time.

The graph in Figure 6.46 starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fullycharged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully



discharged ( Q = 0 on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the

capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positiveafter point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum.Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltagefollows what the current is doing by one-fourth of a cycle:

AC Voltage in a Capacitor

When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a 90º phase angle.

The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rmscurrent, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current I in the circuit

containing only a capacitor C is given by another version of Ohm’s law to be

(6.57)I = VXC

,

where V is the rms voltage and XC is defined (As with XL , this expression for XC results from an analysis of the circuit using Kirchhoff’s rules

and calculus) to be

(6.58)XC = 12π fC ,

where XC is called the capacitive reactance, because the capacitor reacts to impede the current. XC has units of ohms (verification left as an

exercise for the reader). XC is inversely proportional to the capacitance C ; the larger the capacitor, the greater the charge it can store and the

greater the current that can flow. It is also inversely proportional to the frequency f ; the greater the frequency, the less time there is to fully charge

the capacitor, and so it impedes current less.

Example 6.11 Calculating Capacitive Reactance and then Current

(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current ifthe applied rms voltage is 120 V?

Strategy

The capacitive reactance is found directly from the expression in XC = 12π fC . Once XC has been found at each frequency, Ohm’s law stated

as I = V / XC can be used to find the current at each frequency.

Solution for (a)

Entering the frequency and capacitance into XC = 12π fC gives

(6.59)XC = 12π fC

= 16.28(60.0 / s)(5.00 μF) = 531 Ω at 60 Hz.


(6.60)XC = 12π fC = 1

6.28(1.00×104 / s)(5.00 μF)= 3.18 Ω at 10 kHz

.

Solution for (b)

The rms current is now found using the version of Ohm’s law in I = V / XC , given the applied rms voltage is 120 V. For the first frequency, this

yields

(6.61)I = VXC

= 120 V531 Ω = 0.226 A at 60 Hz.


(6.62)I = VXC

= 120 V3.18 Ω = 37.7 A at 10 kHz.

Discussion

The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency,its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies


the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies.For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.

Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor. This is because thevoltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC), XC tends to infinity, and the current is

zero once the capacitor is charged. At very high frequencies, the capacitor’s reactance tends to zero—it has a negligible reactance and does notimpede the current (it acts like a simple wire). Capacitors have the opposite effect on AC circuits that inductors have.

Resistors in an AC CircuitJust as a reminder, consider Figure 6.47, which shows an AC voltage applied to a resistor and a graph of voltage and current versus time. Thevoltage and current are exactly in phase in a resistor. There is no frequency dependence to the behavior of plain resistance in a circuit:

Figure 6.47 (a) An AC voltage source in series with a resistor. (b) Graph of current and voltage across the resistor as functions of time, showing them to be exactly in phase.

AC Voltage in a Resistor

When a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current—they have a 0º phase angle.

6.12 RLC Series AC Circuits

ImpedanceWhen alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly,their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally canceleach other’s effect. Figure 6.48 shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section. The cruxof the analysis of an RLC circuit is the frequency dependence of XL and XC , and the effect they have on the phase of voltage versus current

(established in the preceding section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are thebasis of many applications, such as radio tuners.

Figure 6.48 An RLC series circuit with an AC voltage source.

The combined effect of resistance R , inductive reactance XL , and capacitive reactance XC is defined to be impedance, an AC analogue to

resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law:

(6.63)I0 = V0

Z or Irms = VrmsZ .



Here I0 is the peak current, V0 the peak source voltage, and Z is the impedance of the circuit. The units of impedance are ohms, and its effect on

the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for Z in terms of R , XL , and XC , we

will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VR , VL , and VCin Figure 6.48.

Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in R , L , and C are

equal and in phase. But we know from the preceding section that the voltage across the inductor VL leads the current by one-fourth of a cycle, the

voltage across the capacitor VC follows the current by one-fourth of a cycle, and the voltage across the resistor VR is exactly in phase with the

current. Figure 6.49 shows these relationships in one graph, as well as showing the total voltage around the circuit V = VR + VL + VC , where all

four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit V is also the voltage of the source.

You can see from Figure 6.49 that while VR is in phase with the current, VL leads by 90º , and VC follows by 90º . Thus VL and VC are

180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are

not aligned (not in phase), the peak voltage V0 of the source does not equal the sum of the peak voltages across R , L , and C . The actual

relationship is

(6.64)V0 = V0R2 +(V0L − V0C)2,

where V0R , V0L , and V0C are the peak voltages across R , L , and C , respectively. Now, using Ohm’s law and definitions from Reactance,

Inductive and Capacitive, we substitute V0 = I0Z into the above, as well as V0R = I0R , V0L = I0XL , and V0C = I0XC , yielding

(6.65)I0 Z = I02 R2 + (I0 XL − I0XC )2 = I0 R2 + (XL − XC)2.

I0 cancels to yield an expression for Z :

(6.66)Z = R2 + (XL − XC)2,

which is the impedance of an RLC series AC circuit. For circuits without a resistor, take R = 0 ; for those without an inductor, take XL = 0 ; and for

those without a capacitor, take XC = 0 .

Figure 6.49 This graph shows the relationships of the voltages in an RLC circuit to the current. The voltages across the circuit elements add to equal the voltage of the source,which is seen to be out of phase with the current.

Example 6.12 Calculating Impedance and Current

An RLC series circuit has a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0

kHz, noting that these frequencies and the values for L and C are the same as in Example 6.10 and Example 6.11. (b) If the voltage source

has Vrms = 120 V , what is Irms at each frequency?

Strategy

For each frequency, we use Z = R2 + (XL − XC)2 to find the impedance and then Ohm’s law to find current. We can take advantage of the

results of the previous two examples rather than calculate the reactances again.

Solution for (a)


At 60.0 Hz, the values of the reactances were found in Example 6.10 to be XL = 1.13 Ω and in Example 6.11 to be XC = 531 Ω .

Entering these and the given 40.0 Ω for resistance into Z = R2 + (XL − XC)2 yields

(6.67)Z = R2 + (XL − XC)2

= (40.0 Ω )2 + (1.13 Ω − 531 Ω )2

= 531 Ω at 60.0 Hz.

Similarly, at 10.0 kHz, XL = 188 Ω and XC = 3.18 Ω , so that

(6.68)Z = (40.0 Ω )2 + (188 Ω − 3.18 Ω )2

= 190 Ω at 10.0 kHz.Discussion for (a)

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clearthat XL dominates at high frequency and XC dominates at low frequency.

Solution for (b)

The current Irms can be found using the AC version of Ohm’s law in Equation Irms = Vrms / Z :

Irms = VrmsZ = 120 V

531 Ω = 0.226 A at 60.0 Hz

Finally, at 10.0 kHz, we find

Irms = VrmsZ = 120 V

190 Ω = 0.633 A at 10.0 kHz

Discussion for (a)

The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in Example 6.11. The capacitor dominates at low frequency.The current at 10.0 kHz is only slightly different from that found for the inductor alone in Example 6.10. The inductor dominates at highfrequency.

Resonance in RLC Series AC Circuits

How does an RLC circuit behave as a function of the frequency of the driving voltage source? Combining Ohm’s law, Irms = Vrms / Z , and the

expression for impedance Z from Z = R2 + (XL − XC)2 gives

(6.69)Irms = Vrms

R2 + (XL − XC)2.

The reactances vary with frequency, with XL large at high frequencies and XC large at low frequencies, as we have seen in three previous

examples. At some intermediate frequency f0 , the reactances will be equal and cancel, giving Z = R —this is a minimum value for impedance,

and a maximum value for Irms results. We can get an expression for f0 by taking

(6.70)XL = XC.

Substituting the definitions of XL and XC ,

(6.71)2πf0 L = 12πf0 C .

Solving this expression for f0 yields

(6.72)f0 = 12π LC

,

where f0 is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if not driven by

the voltage source. At f0 , the effects of the inductor and capacitor cancel, so that Z = R , and Irms is a maximum.

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in this case, forced by thevoltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit that oscillates best at its f0 . A variable capacitor is

often used to adjust f0 to receive a desired frequency and to reject others. Figure 6.50 is a graph of current as a function of frequency, illustrating a

resonant peak in Irms at f0 . The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower



and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radioreceiver, for example.

Figure 6.50 A graph of current versus frequency for two RLC series circuits differing only in the amount of resistance. Both have a resonance at f0 , but that for the higher

resistance is lower and broader. The driving AC voltage source has a fixed amplitude V0 .

Example 6.13 Calculating Resonant Frequency and Current

For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF capacitor: (a) Find the resonant frequency. (b)

Calculate Irms at resonance if Vrms is 120 V.

Strategy

The resonant frequency is found by using the expression in f0 = 12π LC

. The current at that frequency is the same as if the resistor alone

were in the circuit.

Solution for (a)

Entering the given values for L and C into the expression given for f0 in f0 = 12π LC

yields

(6.73)f0 = 12π LC

= 12π (3.00×10−3 H)(5.00×10−6 F)

= 1.30 kHz.

Discussion for (a)

We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected,since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at thisintermediate frequency.

Solution for (b)

The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone.Thus,

(6.74)Irms = VrmsZ = 120 V

40.0 Ω = 3.00 A.

Discussion for (b)

At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.

Power in RLC Series AC CircuitsIf current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average power is not simplycurrent times voltage, as it is in purely resistive circuits. As was seen in Figure 6.49, voltage and current are out of phase in an RLC circuit. There isa phase angle ϕ between the source voltage V and the current I , which can be found from

(6.75)cos ϕ = RZ .

For example, at the resonant frequency or in a purely resistive circuit Z = R , so that cos ϕ = 1 . This implies that ϕ = 0 º and that voltage and

current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage andcurrent are out of phase and because Irms is lower. The fact that source voltage and current are out of phase affects the power delivered to the

circuit. It can be shown that the average power is

(6.76)Pave = IrmsVrms cos ϕ,


Thus cos ϕ is called the power factor, which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for

example. At the resonant frequency, cos ϕ = 1 .

Example 6.14 Calculating the Power Factor and Power

For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, a 5.00 μF capacitor, and a voltage source with a Vrms of 120

V: (a) Calculate the power factor and phase angle for f = 60.0Hz . (b) What is the average power at 50.0 Hz? (c) Find the average power at

the circuit’s resonant frequency.


The power factor at 60.0 Hz is found from

(6.77)cos ϕ = RZ .

We know Z= 531 Ω from Example 6.12, so that

(6.78)cos ϕ = 40.0 Ω531 Ω = 0.0753 at 60.0 Hz.

This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is

(6.79)ϕ = cos−1 0.0753 = 85.7º at 60.0 Hz.

Discussion for (a)

The phase angle is close to 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure RC circuit has its

voltage and current 90º out of phase).


The average power at 60.0 Hz is

(6.80)Pave = IrmsVrms cos ϕ.

Irms was found to be 0.226 A in Example 6.12. Entering the known values gives

(6.81)Pave = (0.226 A)(120 V)(0.0753) = 2.04 W at 60.0 Hz.


At the resonant frequency, we know cos ϕ = 1 , and Irms was found to be 6.00 A in Example 6.13. Thus,

Pave = (3.00 A)(120 V)(1) = 360 W at resonance (1.30 kHz)

Discussion

Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.

Power delivered to an RLC series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but donot dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltagesource puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Suchradiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as isthe case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in Figure 6.51. The regularly spacedbumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance dampingand limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, andenergy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of acapacitor). The amplitude of the wheels’ motion is a maximum if the bumps in the road are hit at the resonant frequency.



Figure 6.51 The forced but damped motion of the wheel on the car spring is analogous to an RLC series AC circuit. The shock absorber damps the motion and dissipatesenergy, analogous to the resistance in an RLC circuit. The mass and spring determine the resonant frequency.

A pure LC circuit with negligible resistance oscillates at f0 , the same resonant frequency as an RLC circuit. It can serve as a frequency standard or

clock circuit—for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain theoscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time.Figure 6.52 shows the analogy between an LC circuit and a mass on a spring.

Figure 6.52 An LC circuit is analogous to a mass oscillating on a spring with no friction and no driving force. Energy moves back and forth between the inductor and capacitor,just as it moves from kinetic to potential in the mass-spring system.

PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab

Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters andammeters.


back emf:

capacitive reactance:

characteristic time constant:

eddy current:

electric generator:

electromagnetic induction:

emf induced in a generator coil:

energy stored in an inductor:

Faraday’s law of induction:

henry:

impedance:

inductance:

induction:

inductive reactance:

inductor:

Lenz’s law:

magnetic damping:

magnetic flux:

mutual inductance:

peak emf:

phase angle:

power factor:

resonant frequency:

self-inductance:

Figure 6.53 Circuit Construction Kit (AC+DC), Virtual Lab (http://cnx.org/content/m42431/1.4/circuit-construction-kit-ac-virtual-lab_en.jar)

Glossarythe emf generated by a running motor, because it consists of a coil turning in a magnetic field; it opposes the voltage powering the

motor

the opposition of a capacitor to a change in current; calculated by XC = 12π fC

denoted by τ , of a particular series RL circuit is calculated by τ = LR , where L is the inductance and R is the

resistance

a current loop in a conductor caused by motional emf

a device for converting mechanical work into electric energy; it induces an emf by rotating a coil in a magnetic field

the process of inducing an emf (voltage) with a change in magnetic flux

emf = NABω sin ωt , where A is the area of an N -turn coil rotated at a constant angular velocity ω in a

uniform magnetic field B , over a period of time t

self-explanatory; calculated by Eind = 12LI 2

the means of calculating the emf in a coil due to changing magnetic flux, given by emf = −N ΔΦΔt

the unit of inductance; 1 H = 1 Ω ⋅ s

the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and capacitive reactance in

the form Z = R2 + (XL − XC)2

a property of a device describing how efficient it is at inducing emf in another device

(magnetic induction) the creation of emfs and hence currents by magnetic fields

the opposition of an inductor to a change in current; calculated by XL = 2π fL

a device that exhibits significant self-inductance

the minus sign in Faraday’s law, signifying that the emf induced in a coil opposes the change in magnetic flux

the drag produced by eddy currents

the amount of magnetic field going through a particular area, calculated with Φ = BA cos θ where B is the magnetic field

strength over an area A at an angle θ with the perpendicular to the area

how effective a pair of devices are at inducing emfs in each other

emf0 = NABω

denoted by ϕ , the amount by which the voltage and current are out of phase with each other in a circuit

the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and currentbeing out of phase; calculated by cos ϕ

the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the circuit would oscillate

if not driven by a voltage source; calculated by f0 = 12π LC

how effective a device is at inducing emf in itself



http://cnx.org/content/m42431/1.4/circuit-construction-kit-ac-virtual-lab_en.jar

shock hazard:

step-down transformer:

step-up transformer:

thermal hazard:

three-wire system:

transformer equation:

transformer:

the term for electrical hazards due to current passing through a human

a transformer that decreases voltage

a transformer that increases voltage

the term for electrical hazards due to overheating

the wiring system used at present for safety reasons, with live, neutral, and ground wires

the equation showing that the ratio of the secondary to primary voltages in a transformer equals the ratio of the number of

loops in their coils;VsVp

= NsNp

a device that transforms voltages from one value to another using induction

Section Summary

6.1 Induced Emf and Magnetic Flux• The crucial quantity in induction is magnetic flux Φ , defined to be Φ = BA cos θ , where B is the magnetic field strength over an area A at

an angle θ with the perpendicular to the area.

• Units of magnetic flux Φ are T ⋅ m2 .

• Any change in magnetic flux Φ induces an emf—the process is defined to be electromagnetic induction.

6.2 Faraday’s Law of Induction: Lenz’s Law• Faraday’s law of induction states that the emfinduced by a change in magnetic flux is

emf = −N ΔΦΔt

when flux changes by ΔΦ in a time Δt .

• If emf is induced in a coil, N is its number of turns.

• The minus sign means that the emf creates a current I and magnetic field B that oppose the change in flux ΔΦ —this opposition is knownas Lenz’s law.

6.3 Motional Emf• An emf induced by motion relative to a magnetic field B is called a motional emf and is given by

emf = Bℓv (B, ℓ, and v perpendicular),where ℓ is the length of the object moving at speed v relative to the field.

6.4 Eddy Currents and Magnetic Damping• Current loops induced in moving conductors are called eddy currents.• They can create significant drag, called magnetic damping.

6.5 Electric Generators• An electric generator rotates a coil in a magnetic field, inducing an emfgiven as a function of time by

emf = NABω sin ωt,where A is the area of an N -turn coil rotated at a constant angular velocity ω in a uniform magnetic field B .

• The peak emf emf0 of a generator is

emf0 = NABω.

6.6 Back Emf• Any rotating coil will have an induced emf—in motors, this is called back emf, since it opposes the emf input to the motor.

6.7 Transformers• Transformers use induction to transform voltages from one value to another.• For a transformer, the voltages across the primary and secondary coils are related by

VsVp

= NsNp

,

where Vp and Vs are the voltages across primary and secondary coils having Np and Ns turns.

• The currents Ip and Is in the primary and secondary coils are related byIsIp

=NpNs

.

• A step-up transformer increases voltage and decreases current, whereas a step-down transformer decreases voltage and increases current.

6.8 Electrical Safety: Systems and Devices• Electrical safety systems and devices are employed to prevent thermal and shock hazards.


• Circuit breakers and fuses interrupt excessive currents to prevent thermal hazards.• The three-wire system guards against thermal and shock hazards, utilizing live/hot, neutral, and earth/ground wires, and grounding the neutral

wire and case of the appliance.• A ground fault interrupter (GFI) prevents shock by detecting the loss of current to unintentional paths.• An isolation transformer insulates the device being powered from the original source, also to prevent shock.• Many of these devices use induction to perform their basic function.

6.9 Inductance• Inductance is the property of a device that tells how effectively it induces an emf in another device.• Mutual inductance is the effect of two devices in inducing emfs in each other.• A change in current ΔI1 / Δt in one induces an emf emf2 in the second:

emf2 = −MΔI1Δt ,

where M is defined to be the mutual inductance between the two devices, and the minus sign is due to Lenz’s law.

• Symmetrically, a change in current ΔI2 / Δt through the second device induces an emf emf1 in the first:

emf1 = −MΔI2Δt ,

where M is the same mutual inductance as in the reverse process.• Current changes in a device induce an emf in the device itself.• Self-inductance is the effect of the device inducing emf in itself.• The device is called an inductor, and the emf induced in it by a change in current through it is

emf = −LΔIΔt ,

where L is the self-inductance of the inductor, and ΔI / Δt is the rate of change of current through it. The minus sign indicates that emfopposes the change in current, as required by Lenz’s law.

• The unit of self- and mutual inductance is the henry (H), where 1 H = 1 Ω ⋅ s .

• The self-inductance L of an inductor is proportional to how much flux changes with current. For an N -turn inductor,

L = N ΔΦΔI .

• The self-inductance of a solenoid is

L = μ0 N 2 Aℓ (solenoid),

where N is its number of turns in the solenoid, A is its cross-sectional area, ℓ is its length, and μ0 = 4π×10−7 T ⋅ m/A is the

permeability of free space.• The energy stored in an inductor Eind is

Eind = 12LI 2.

6.10 RL Circuits• When a series connection of a resistor and an inductor—an RL circuit—is connected to a voltage source, the time variation of the current is

I = I0(1 − e−t / τ) (turning on).where I0 = V / R is the final current.

• The characteristic time constant τ is τ = LR , where L is the inductance and R is the resistance.

• In the first time constant τ , the current rises from zero to 0.632I0 , and 0.632 of the remainder in every subsequent time interval τ .

• When the inductor is shorted through a resistor, current decreases as

I = I0e−t / τ (turning off).Here I0 is the initial current.

• Current falls to 0.368I0 in the first time interval τ , and 0.368 of the remainder toward zero in each subsequent time τ .

6.11 Reactance, Inductive and Capacitive• For inductors in AC circuits, we find that when a sinusoidal voltage is applied to an inductor, the voltage leads the current by one-fourth of a

cycle, or by a 90º phase angle.• The opposition of an inductor to a change in current is expressed as a type of AC resistance.• Ohm’s law for an inductor is

I = VXL

,

where V is the rms voltage across the inductor.

• XL is defined to be the inductive reactance, given by



XL = 2π fL,with f the frequency of the AC voltage source in hertz.

• Inductive reactance XL has units of ohms and is greatest at high frequencies.

• For capacitors, we find that when a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a90º phase angle.

• Since a capacitor can stop current when fully charged, it limits current and offers another form of AC resistance; Ohm’s law for a capacitor is

I = VXC

,

where V is the rms voltage across the capacitor.

• XC is defined to be the capacitive reactance, given by

XC = 12π fC .

• XC has units of ohms and is greatest at low frequencies.

6.12 RLC Series AC Circuits• The AC analogy to resistance is impedance Z , the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm’s

law:

I0 = V0Z or Irms = Vrms

Z ,

where I0 is the peak current and V0 is the peak source voltage.

• Impedance has units of ohms and is given by Z = R2 + (XL − XC)2 .

• The resonant frequency f0 , at which XL = XC , is

f0 = 12π LC

.

• In an AC circuit, there is a phase angle ϕ between source voltage V and the current I , which can be found from

cos ϕ = RZ ,

• ϕ = 0º for a purely resistive circuit or an RLC circuit at resonance.

• The average power delivered to an RLC circuit is affected by the phase angle and is given byPave = IrmsVrms cos ϕ,

cos ϕ is called the power factor, which ranges from 0 to 1.


6.1 Induced Emf and Magnetic Flux1. How do the multiple-loop coils and iron ring in the version of Faraday’s apparatus shown in Figure 6.3 enhance the observation of induced emf?

2. When a magnet is thrust into a coil as in Figure 6.4(a), what is the direction of the force exerted by the coil on the magnet? Draw a diagramshowing the direction of the current induced in the coil and the magnetic field it produces, to justify your response. How does the magnitude of theforce depend on the resistance of the galvanometer?

3. Explain how magnetic flux can be zero when the magnetic field is not zero.

4. Is an emf induced in the coil in Figure 6.54 when it is stretched? If so, state why and give the direction of the induced current.

Figure 6.54 A circular coil of wire is stretched in a magnetic field.

6.2 Faraday’s Law of Induction: Lenz’s Law5. A person who works with large magnets sometimes places her head inside a strong field. She reports feeling dizzy as she quickly turns her head.How might this be associated with induction?

6. A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrapped around the pipe coulddetect the passage of individual particles. Sketch a graph of the voltage output of the coil as a single particle passes through it.


6.3 Motional Emf7. Why must part of the circuit be moving relative to other parts, to have usable motional emf? Consider, for example, that the rails in Figure 6.11 arestationary relative to the magnetic field, while the rod moves.

8. A powerful induction cannon can be made by placing a metal cylinder inside a solenoid coil. The cylinder is forcefully expelled when solenoidcurrent is turned on rapidly. Use Faraday’s and Lenz’s laws to explain how this works. Why might the cylinder get live/hot when the cannon is fired?

9. An induction stove heats a pot with a coil carrying an alternating current located beneath the pot (and without a hot surface). Can the stove surfacebe a conductor? Why won’t a coil carrying a direct current work?

10. Explain how you could thaw out a frozen water pipe by wrapping a coil carrying an alternating current around it. Does it matter whether or not thepipe is a conductor? Explain.

6.4 Eddy Currents and Magnetic Damping11. Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation.

12. Explain how electromagnetic induction can be used to detect metals? This technique is particularly important in detecting buried landmines fordisposal, geophysical prospecting and at airports.

6.5 Electric Generators13. Using RHR-1, show that the emfs in the sides of the generator loop in Figure 6.23 are in the same sense and thus add.

14. The source of a generator’s electrical energy output is the work done to turn its coils. How is the work needed to turn the generator related toLenz’s law?

6.6 Back Emf15. Suppose you find that the belt drive connecting a powerful motor to an air conditioning unit is broken and the motor is running freely. Should yoube worried that the motor is consuming a great deal of energy for no useful purpose? Explain why or why not.

6.7 Transformers16. Explain what causes physical vibrations in transformers at twice the frequency of the AC power involved.

6.8 Electrical Safety: Systems and Devices17. Does plastic insulation on live/hot wires prevent shock hazards, thermal hazards, or both?

18. Why are ordinary circuit breakers and fuses ineffective in preventing shocks?

19. A GFI may trip just because the live/hot and neutral wires connected to it are significantly different in length. Explain why.

6.9 Inductance20. How would you place two identical flat coils in contact so that they had the greatest mutual inductance? The least?

21. How would you shape a given length of wire to give it the greatest self-inductance? The least?

22. Verify, as was concluded without proof in Example 6.7, that units of T ⋅ m2 / A = Ω ⋅ s = H .

6.11 Reactance, Inductive and Capacitive23. Presbycusis is a hearing loss due to age that progressively affects higher frequencies. A hearing aid amplifier is designed to amplify allfrequencies equally. To adjust its output for presbycusis, would you put a capacitor in series or parallel with the hearing aid’s speaker? Explain.

24. Would you use a large inductance or a large capacitance in series with a system to filter out low frequencies, such as the 100 Hz hum in a soundsystem? Explain.

25. High-frequency noise in AC power can damage computers. Does the plug-in unit designed to prevent this damage use a large inductance or alarge capacitance (in series with the computer) to filter out such high frequencies? Explain.

26. Does inductance depend on current, frequency, or both? What about inductive reactance?

27. Explain why the capacitor in Figure 6.55(a) acts as a low-frequency filter between the two circuits, whereas that in Figure 6.55(b) acts as a high-frequency filter.



Figure 6.55 Capacitors and inductors. Capacitor with high frequency and low frequency.

28. If the capacitors in Figure 6.55 are replaced by inductors, which acts as a low-frequency filter and which as a high-frequency filter?

6.12 RLC Series AC Circuits29. Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.

30. Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the power factor as a method ofimproving the motor’s output, rather than to increase the voltage input.



6.1 Induced Emf and Magnetic Flux1. What is the value of the magnetic flux at coil 2 in Figure 6.56 due tocoil 1?

Figure 6.56 (a) The planes of the two coils are perpendicular. (b) The wire isperpendicular to the plane of the coil.

2. What is the value of the magnetic flux through the coil in Figure6.56(b) due to the wire?

6.2 Faraday’s Law of Induction: Lenz’s Law3. Referring to Figure 6.57(a), what is the direction of the current inducedin coil 2: (a) If the current in coil 1 increases? (b) If the current in coil 1decreases? (c) If the current in coil 1 is constant? Explicitly show howyou follow the steps in the Problem-Solving Strategy for Lenz's Law.

Figure 6.57 (a) The coils lie in the same plane. (b) The wire is in the plane of the coil

4. Referring to Figure 6.57(b), what is the direction of the current inducedin the coil: (a) If the current in the wire increases? (b) If the current in thewire decreases? (c) If the current in the wire suddenly changes direction?Explicitly show how you follow the steps in the Problem-SolvingStrategy for Lenz’s Law.

5. Referring to Figure 6.58, what are the directions of the currents in coils1, 2, and 3 (assume that the coils are lying in the plane of the circuit): (a)When the switch is first closed? (b) When the switch has been closed fora long time? (c) Just after the switch is opened?

Figure 6.58

6. Repeat the previous problem with the battery reversed.

7. Verify that the units of ΔΦ / Δt are volts. That is, show that

1 T ⋅ m2 / s = 1 V .

8. Suppose a 50-turn coil lies in the plane of the page in a uniformmagnetic field that is directed into the page. The coil originally has an

area of 0.250 m2 . It is stretched to have no area in 0.100 s. What is thedirection and magnitude of the induced emf if the uniform magnetic fieldhas a strength of 1.50 T?

9. (a) An MRI technician moves his hand from a region of very lowmagnetic field strength into an MRI scanner’s 2.00 T field with his fingerspointing in the direction of the field. Find the average emf induced in hiswedding ring, given its diameter is 2.20 cm and assuming it takes 0.250 sto move it into the field. (b) Discuss whether this current wouldsignificantly change the temperature of the ring.


Referring to the situation in the previous problem: (a) What current isinduced in the ring if its resistance is 0.0100 Ω ? (b) What averagepower is dissipated? (c) What magnetic field is induced at the center ofthe ring? (d) What is the direction of the induced magnetic field relative tothe MRI’s field?

11. An emf is induced by rotating a 1000-turn, 20.0 cm diameter coil in

the Earth’s 5.00×10−5 T magnetic field. What average emf is induced,given the plane of the coil is originally perpendicular to the Earth’s fieldand is rotated to be parallel to the field in 10.0 ms?

12. A 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in4.17 ms, originally having its plane perpendicular to a uniform magneticfield. (This is 60 rev/s.) Find the magnetic field strength needed to inducean average emf of 10,000 V.


Approximately how does the emf induced in the loop in Figure 6.57(b)depend on the distance of the center of the loop from the wire?


(a) A lightning bolt produces a rapidly varying magnetic field. If the boltstrikes the earth vertically and acts like a current in a long straight wire, itwill induce a voltage in a loop aligned like that in Figure 6.57(b). Whatvoltage is induced in a 1.00 m diameter loop 50.0 m from a

2.00×106 A lightning strike, if the current falls to zero in 25.0 μs ? (b)

Discuss circumstances under which such a voltage would producenoticeable consequences.

6.3 Motional Emf15. Use Faraday’s law, Lenz’s law, and RHR-1 to show that the magneticforce on the current in the moving rod in Figure 6.11 is in the oppositedirection of its velocity.

16. If a current flows in the Satellite Tether shown in Figure 6.12, useFaraday’s law, Lenz’s law, and RHR-1 to show that there is a magneticforce on the tether in the direction opposite to its velocity.

17. (a) A jet airplane with a 75.0 m wingspan is flying at 280 m/s. Whatemf is induced between wing tips if the vertical component of the Earth’s

field is 3.00×10−5 T ? (b) Is an emf of this magnitude likely to have anyconsequences? Explain.

18. (a) A nonferrous screwdriver is being used in a 2.00 T magnetic field.What maximum emf can be induced along its 12.0 cm length when itmoves at 6.00 m/s? (b) Is it likely that this emf will have anyconsequences or even be noticed?

19. At what speed must the sliding rod in Figure 6.11 move to producean emf of 1.00 V in a 1.50 T field, given the rod’s length is 30.0 cm?

20. The 12.0 cm long rod in Figure 6.11 moves at 4.00 m/s. What is thestrength of the magnetic field if a 95.0 V emf is induced?

21. Prove that when B , ℓ , and v are not mutually perpendicular,

motional emf is given by emf = Bℓvsin θ . If v is perpendicular to B ,

then θ is the angle between ℓ and B . If ℓ is perpendicular to B ,

then θ is the angle between v and B .



22. In the August 1992 space shuttle flight, only 250 m of the conductingtether considered in Example 6.2 could be let out. A 40.0 V motional emf

was generated in the Earth’s 5.00×10−5 T field, while moving at

7.80×103 m/s . What was the angle between the shuttle’s velocity andthe Earth’s field, assuming the conductor was perpendicular to the field?


Derive an expression for the current in a system like that in Figure 6.11,under the following conditions. The resistance between the rails is R ,

the rails and the moving rod are identical in cross section A and havethe same resistivity ρ . The distance between the rails is l, and the rod

moves at constant speed v perpendicular to the uniform field B . At time

zero, the moving rod is next to the resistance R .


The Tethered Satellite in Figure 6.12 has a mass of 525 kg and is at theend of a 20.0 km long, 2.50 mm diameter cable with the tensile strengthof steel. (a) How much does the cable stretch if a 100 N force is exertedto pull the satellite in? (Assume the satellite and shuttle are at the samealtitude above the Earth.) (b) What is the effective force constant of thecable? (c) How much energy is stored in it when stretched by the 100 Nforce?


The Tethered Satellite discussed in this module is producing 5.00 kV, anda current of 10.0 A flows. (a) What magnetic drag force does this produceif the system is moving at 7.80 km/s? (b) How much kinetic energy isremoved from the system in 1.00 h, neglecting any change in altitude orvelocity during that time? (c) What is the change in velocity if the mass ofthe system is 100,000 kg? (d) Discuss the long term consequences (say,a week-long mission) on the space shuttle’s orbit, noting what effect adecrease in velocity has and assessing the magnitude of the effect.

6.4 Eddy Currents and Magnetic Damping26. Make a drawing similar to Figure 6.14, but with the pendulum movingin the opposite direction. Then use Faraday’s law, Lenz’s law, and RHR-1to show that magnetic force opposes motion.

27.

Figure 6.59 A coil is moved into and out of a region of uniform magnetic field. A coilis moved through a magnetic field as shown in Figure 6.59. The field isuniform inside the rectangle and zero outside. What is the direction of theinduced current and what is the direction of the magnetic force on the coilat each position shown?

6.5 Electric Generators28. Calculate the peak voltage of a generator that rotates its 200-turn,0.100 m diameter coil at 3600 rpm in a 0.800 T field.

29. At what angular velocity in rpm will the peak voltage of a generator be480 V, if its 500-turn, 8.00 cm diameter coil rotates in a 0.250 T field?

30. What is the peak emf generated by rotating a 1000-turn, 20.0 cm

diameter coil in the Earth’s 5.00×10−5 T magnetic field, given theplane of the coil is originally perpendicular to the Earth’s field and isrotated to be parallel to the field in 10.0 ms?

31. What is the peak emf generated by a 0.250 m radius, 500-turn coil isrotated one-fourth of a revolution in 4.17 ms, originally having its planeperpendicular to a uniform magnetic field. (This is 60 rev/s.)

32. (a) A bicycle generator rotates at 1875 rad/s, producing an 18.0 Vpeak emf. It has a 1.00 by 3.00 cm rectangular coil in a 0.640 T field.How many turns are in the coil? (b) Is this number of turns of wirepractical for a 1.00 by 3.00 cm coil?


This problem refers to the bicycle generator considered in the previousproblem. It is driven by a 1.60 cm diameter wheel that rolls on the outsiderim of the bicycle tire. (a) What is the velocity of the bicycle if thegenerator’s angular velocity is 1875 rad/s? (b) What is the maximum emfof the generator when the bicycle moves at 10.0 m/s, noting that it was18.0 V under the original conditions? (c) If the sophisticated generatorcan vary its own magnetic field, what field strength will it need at 5.00 m/sto produce a 9.00 V maximum emf?

34. (a) A car generator turns at 400 rpm when the engine is idling. Its300-turn, 5.00 by 8.00 cm rectangular coil rotates in an adjustablemagnetic field so that it can produce sufficient voltage even at low rpms.What is the field strength needed to produce a 24.0 V peak emf? (b)Discuss how this required field strength compares to those available inpermanent and electromagnets.

35. Show that if a coil rotates at an angular velocity ω , the period of its

AC output is 2π/ω .

36. A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00rad/s in a 1.25 T field, starting with the plane of the coil parallel to thefield. (a) What is the peak emf? (b) At what time is the peak emf firstreached? (c) At what time is the emf first at its most negative? (d) What isthe period of the AC voltage output?

37. (a) If the emf of a coil rotating in a magnetic field is zero at t = 0 ,

and increases to its first peak at t = 0.100 ms , what is the angularvelocity of the coil? (b) At what time will its next maximum occur? (c)What is the period of the output? (d) When is the output first one-fourth ofits maximum? (e) When is it next one-fourth of its maximum?


A 500-turn coil with a 0.250 m2 area is spun in the Earth’s

5.00×10−5 T field, producing a 12.0 kV maximum emf. (a) At whatangular velocity must the coil be spun? (b) What is unreasonable aboutthis result? (c) Which assumption or premise is responsible?

6.6 Back Emf39. Suppose a motor connected to a 120 V source draws 10.0 A when itfirst starts. (a) What is its resistance? (b) What current does it draw at itsnormal operating speed when it develops a 100 V back emf?

40. A motor operating on 240 V electricity has a 180 V back emf atoperating speed and draws a 12.0 A current. (a) What is its resistance?(b) What current does it draw when it is first started?

41. What is the back emf of a 120 V motor that draws 8.00 A at its normalspeed and 20.0 A when first starting?

42. The motor in a toy car operates on 6.00 V, developing a 4.50 V backemf at normal speed. If it draws 3.00 A at normal speed, what currentdoes it draw when starting?


The motor in a toy car is powered by four batteries in series, whichproduce a total emf of 6.00 V. The motor draws 3.00 A and develops a4.50 V back emf at normal speed. Each battery has a 0.100 Ω internalresistance. What is the resistance of the motor?

6.7 Transformers44. A plug-in transformer, like that in Figure 6.29, supplies 9.00 V to avideo game system. (a) How many turns are in its secondary coil, if itsinput voltage is 120 V and the primary coil has 400 turns? (b) What is itsinput current when its output is 1.30 A?

45. An American traveler in New Zealand carries a transformer to convertNew Zealand’s standard 240 V to 120 V so that she can use some smallappliances on her trip. (a) What is the ratio of turns in the primary andsecondary coils of her transformer? (b) What is the ratio of input to outputcurrent? (c) How could a New Zealander traveling in the United Statesuse this same transformer to power her 240 V appliances from 120 V?

46. A cassette recorder uses a plug-in transformer to convert 120 V to12.0 V, with a maximum current output of 200 mA. (a) What is the current


input? (b) What is the power input? (c) Is this amount of powerreasonable for a small appliance?

47. (a) What is the voltage output of a transformer used for rechargeableflashlight batteries, if its primary has 500 turns, its secondary 4 turns, andthe input voltage is 120 V? (b) What input current is required to produce a4.00 A output? (c) What is the power input?

48. (a) The plug-in transformer for a laptop computer puts out 7.50 V andcan supply a maximum current of 2.00 A. What is the maximum inputcurrent if the input voltage is 240 V? Assume 100% efficiency. (b) If theactual efficiency is less than 100%, would the input current need to begreater or smaller? Explain.

49. A multipurpose transformer has a secondary coil with several pointsat which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480V. (a) The input voltage is 240 V to a primary coil of 280 turns. What arethe numbers of turns in the parts of the secondary used to produce theoutput voltages? (b) If the maximum input current is 5.00 A, what are themaximum output currents (each used alone)?

50. A large power plant generates electricity at 12.0 kV. Its oldtransformer once converted the voltage to 335 kV. The secondary of thistransformer is being replaced so that its output can be 750 kV for moreefficient cross-country transmission on upgraded transmission lines. (a)What is the ratio of turns in the new secondary compared with the oldsecondary? (b) What is the ratio of new current output to old output (at335 kV) for the same power? (c) If the upgraded transmission lines havethe same resistance, what is the ratio of new line power loss to old?

51. If the power output in the previous problem is 1000 MW and lineresistance is 2.00 Ω , what were the old and new line losses?


The 335 kV AC electricity from a power transmission line is fed into theprimary coil of a transformer. The ratio of the number of turns in thesecondary to the number in the primary is Ns / Np = 1000 . (a) What

voltage is induced in the secondary? (b) What is unreasonable about thisresult? (c) Which assumption or premise is responsible?


Consider a double transformer to be used to create very large voltages.The device consists of two stages. The first is a transformer thatproduces a much larger output voltage than its input. The output of thefirst transformer is used as input to a second transformer that furtherincreases the voltage. Construct a problem in which you calculate theoutput voltage of the final stage based on the input voltage of the firststage and the number of turns or loops in both parts of both transformers(four coils in all). Also calculate the maximum output current of the finalstage based on the input current. Discuss the possibility of power lossesin the devices and the effect on the output current and power.

6.8 Electrical Safety: Systems and Devices54. Integrated Concepts

A short circuit to the grounded metal case of an appliance occurs asshown in Figure 6.60. The person touching the case is wet and only hasa 3.00 kΩ resistance to earth/ground. (a) What is the voltage on thecase if 5.00 mA flows through the person? (b) What is the current in theshort circuit if the resistance of the earth/ground wire is 0.200 Ω ? (c)Will this trigger the 20.0 A circuit breaker supplying the appliance?

Figure 6.60 A person can be shocked even when the case of an appliance isgrounded. The large short circuit current produces a voltage on the case of theappliance, since the resistance of the earth/ground wire is not zero.

6.9 Inductance55. Two coils are placed close together in a physics lab to demonstrateFaraday’s law of induction. A current of 5.00 A in one is switched off in1.00 ms, inducing a 9.00 V emf in the other. What is their mutualinductance?

56. If two coils placed next to one another have a mutual inductance of5.00 mH, what voltage is induced in one when the 2.00 A current in theother is switched off in 30.0 ms?

57. The 4.00 A current through a 7.50 mH inductor is switched off in 8.33ms. What is the emf induced opposing this?

58. A device is turned on and 3.00 A flows through it 0.100 ms later.What is the self-inductance of the device if an induced 150 V emfopposes this?

59. Starting with emf2 = −MΔI1Δt , show that the units of inductance

are (V ⋅ s)/A = Ω ⋅ s .

60. Camera flashes charge a capacitor to high voltage by switching thecurrent through an inductor on and off rapidly. In what time must the0.100 A current through a 2.00 mH inductor be switched on or off toinduce a 500 V emf?

61. A large research solenoid has a self-inductance of 25.0 H. (a) Whatinduced emf opposes shutting it off when 100 A of current through it isswitched off in 80.0 ms? (b) How much energy is stored in the inductor atfull current? (c) At what rate in watts must energy be dissipated to switchthe current off in 80.0 ms? (d) In view of the answer to the last part, is itsurprising that shutting it down this quickly is difficult?

62. (a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diametersolenoid having 1000 loops. (b) How much energy is stored in thisinductor when 20.0 A of current flows through it? (c) How fast can it beturned off if the induced emf cannot exceed 3.00 V?

63. A precision laboratory resistor is made of a coil of wire 1.50 cm indiameter and 4.00 cm long, and it has 500 turns. (a) What is its self-inductance? (b) What average emf is induced if the 12.0 A currentthrough it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)?(c) What is its inductance if it is shortened to half its length and counter-wound (two layers of 250 turns in opposite directions)?

64. The heating coils in a hair dryer are 0.800 cm in diameter, have acombined length of 1.00 m, and a total of 400 turns. (a) What is their totalself-inductance assuming they act like a single solenoid? (b) How muchenergy is stored in them when 6.00 A flows? (c) What average emfopposes shutting them off if this is done in 5.00 ms (one-fourth of a cyclefor 50 Hz AC)?



65. When the 20.0 A current through an inductor is turned off in 1.50 ms,an 800 V emf is induced, opposing the change. What is the value of theself-inductance?

66. How fast can the 150 A current through a 0.250 H inductor be shut offif the induced emf cannot exceed 75.0 V?


A very large, superconducting solenoid such as one used in MRI scans,stores 1.00 MJ of energy in its magnetic field when 100 A flows. (a) Findits self-inductance. (b) If the coils “go normal,” they gain resistance andstart to dissipate thermal energy. What temperature increase is producedif all the stored energy goes into heating the 1000 kg magnet, given itsaverage specific heat is 200 J/kg·ºC ?


A 25.0 H inductor has 100 A of current turned off in 1.00 ms. (a) Whatvoltage is induced to oppose this? (b) What is unreasonable about thisresult? (c) Which assumption or premise is responsible?

6.10 RL Circuits69. If you want a characteristic RL time constant of 1.00 s, and you havea 500 Ω resistor, what value of self-inductance is needed?

70. Your RL circuit has a characteristic time constant of 20.0 ns, and aresistance of 5.00 MΩ . (a) What is the inductance of the circuit? (b)What resistance would give you a 1.00 ns time constant, perhaps neededfor quick response in an oscilloscope?

71. A large superconducting magnet, used for magnetic resonanceimaging, has a 50.0 H inductance. If you want current through it to beadjustable with a 1.00 s characteristic time constant, what is the minimumresistance of system?

72. Verify that after a time of 10.0 ms, the current for the situationconsidered in Example 6.9 will be 0.183 A as stated.

73. Suppose you have a supply of inductors ranging from 1.00 nH to 10.0H, and resistors ranging from 0.100 Ω to 1.00 MΩ . What is the rangeof characteristic RL time constants you can produce by connecting asingle resistor to a single inductor?

74. (a) What is the characteristic time constant of a 25.0 mH inductor thathas a resistance of 4.00 Ω ? (b) If it is connected to a 12.0 V battery,what is the current after 12.5 ms?

75. What percentage of the final current I0 flows through an inductor Lin series with a resistor R , three time constants after the circuit iscompleted?

76. The 5.00 A current through a 1.50 H inductor is dissipated by a2.00 Ω resistor in a circuit like that in Figure 6.44 with the switch inposition 2. (a) What is the initial energy in the inductor? (b) How long willit take the current to decline to 5.00% of its initial value? (c) Calculate theaverage power dissipated, and compare it with the initial powerdissipated by the resistor.

77. (a) Use the exact exponential treatment to find how much time isrequired to bring the current through an 80.0 mH inductor in series with a15.0 Ω resistor to 99.0% of its final value, starting from zero. (b)Compare your answer to the approximate treatment using integralnumbers of τ . (c) Discuss how significant the difference is.

78. (a) Using the exact exponential treatment, find the time required forthe current through a 2.00 H inductor in series with a 0.500 Ω resistorto be reduced to 0.100% of its original value. (b) Compare your answer tothe approximate treatment using integral numbers of τ . (c) Discuss howsignificant the difference is.

6.11 Reactance, Inductive and Capacitive79. At what frequency will a 30.0 mH inductor have a reactance of100 Ω ?

80. What value of inductance should be used if a 20.0 kΩ reactance isneeded at a frequency of 500 Hz?

81. What capacitance should be used to produce a 2.00 MΩ reactanceat 60.0 Hz?

82. At what frequency will an 80.0 mF capacitor have a reactance of0.250 Ω ?

83. (a) Find the current through a 0.500 H inductor connected to a 60.0Hz, 480 V AC source. (b) What would the current be at 100 kHz?

84. (a) What current flows when a 60.0 Hz, 480 V AC source isconnected to a 0.250 μF capacitor? (b) What would the current be at

25.0 kHz?

85. A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00A current. What is the inductance?

86. A 20.0 Hz, 16.0 V source produces a 2.00 mA current whenconnected to a capacitor. What is the capacitance?

87. (a) An inductor designed to filter high-frequency noise from powersupplied to a personal computer is placed in series with the computer.What minimum inductance should it have to produce a 2.00 kΩreactance for 15.0 kHz noise? (b) What is its reactance at 60.0 Hz?

88. The capacitor in Figure 6.55(a) is designed to filter low-frequencysignals, impeding their transmission between circuits. (a) Whatcapacitance is needed to produce a 100 kΩ reactance at a frequencyof 120 Hz? (b) What would its reactance be at 1.00 MHz? (c) Discuss theimplications of your answers to (a) and (b).

89. The capacitor in Figure 6.55(b) will filter high-frequency signals byshorting them to earth/ground. (a) What capacitance is needed toproduce a reactance of 10.0 mΩ for a 5.00 kHz signal? (b) What wouldits reactance be at 3.00 Hz? (c) Discuss the implications of your answersto (a) and (b).


In a recording of voltages due to brain activity (an EEG), a 10.0 mVsignal with a 0.500 Hz frequency is applied to a capacitor, producing acurrent of 100 mA. Resistance is negligible. (a) What is the capacitance?(b) What is unreasonable about this result? (c) Which assumption orpremise is responsible?


Consider the use of an inductor in series with a computer operating on 60Hz electricity. Construct a problem in which you calculate the relativereduction in voltage of incoming high frequency noise compared to 60 Hzvoltage. Among the things to consider are the acceptable seriesreactance of the inductor for 60 Hz power and the likely frequencies ofnoise coming through the power lines.

6.12 RLC Series AC Circuits92. An RL circuit consists of a 40.0 Ω resistor and a 3.00 mH inductor.

(a) Find its impedance Z at 60.0 Hz and 10.0 kHz. (b) Compare these

values of Z with those found in Example 6.12 in which there was also acapacitor.

93. An RC circuit consists of a 40.0 Ω resistor and a 5.00 μFcapacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Comparethese values of Z with those found in Example 6.12, in which there wasalso an inductor.

94. An LC circuit consists of a 3.00 μH inductor and a 5.00 μFcapacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Comparethese values of Z with those found in Example 6.12 in which there wasalso a resistor.

95. What is the resonant frequency of a 0.500 mH inductor connected toa 40.0 μF capacitor?


96. To receive AM radio, you want an RLC circuit that can be made toresonate at any frequency between 500 and 1650 kHz. This isaccomplished with a fixed 1.00 μH inductor connected to a variable

capacitor. What range of capacitance is needed?

97. Suppose you have a supply of inductors ranging from 1.00 nH to 10.0H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range ofresonant frequencies that can be achieved from combinations of a singleinductor and a single capacitor?

98. What capacitance do you need to produce a resonant frequency of1.00 GHz, when using an 8.00 nH inductor?

99. What inductance do you need to produce a resonant frequency of60.0 Hz, when using a 2.00 μF capacitor?

100. The lowest frequency in the FM radio band is 88.0 MHz. (a) Whatinductance is needed to produce this resonant frequency if it isconnected to a 2.50 pF capacitor? (b) The capacitor is variable, to allowthe resonant frequency to be adjusted to as high as 108 MHz. What mustthe capacitance be at this frequency?

101. An RLC series circuit has a 2.50 Ω resistor, a 100 μH inductor,

and an 80.0 μF capacitor.(a) Find the circuit’s impedance at 120 Hz. (b)

Find the circuit’s impedance at 5.00 kHz. (c) If the voltage source hasVrms = 5.60 V , what is Irms at each frequency? (d) What is the

resonant frequency of the circuit? (e) What is Irms at resonance?

102. An RLC series circuit has a 1.00 kΩ resistor, a 150 μH inductor,

and a 25.0 nF capacitor. (a) Find the circuit’s impedance at 500 Hz. (b)Find the circuit’s impedance at 7.50 kHz. (c) If the voltage source hasVrms = 408 V , what is Irms at each frequency? (d) What is the

resonant frequency of the circuit? (e) What is Irms at resonance?

103. An RLC series circuit has a 2.50 Ω resistor, a 100 μH inductor,

and an 80.0 μF capacitor. (a) Find the power factor at f = 120 Hz .

(b) What is the phase angle at 120 Hz? (c) What is the average power at120 Hz? (d) Find the average power at the circuit’s resonant frequency.

104. An RLC series circuit has a 1.00 kΩ resistor, a 150 μH inductor,

and a 25.0 nF capacitor. (a) Find the power factor at f = 7.50 Hz . (b)

What is the phase angle at this frequency? (c) What is the average powerat this frequency? (d) Find the average power at the circuit’s resonantfrequency.

105. An RLC series circuit has a 200 Ω resistor and a 25.0 mH inductor.

At 8000 Hz, the phase angle is 45.0º . (a) What is the impedance? (b)

Find the circuit’s capacitance. (c) If Vrms = 408 V is applied, what is

the average power supplied?

106. Referring to Example 6.14, find the average power at 10.0 kHz.



A USEFUL INFORMATION*

This appendix is broken into several tables.

• Table A1, Important Constants• Table A2, Submicroscopic Masses• Table A3, Solar System Data• Table A4, Metric Prefixes for Powers of Ten and Their Symbols• Table A5, The Greek Alphabet• Table A6, SI units• Table A7, Selected British Units• Table A8, Other Units• Table A9, Useful Formulae

Table A1 Important Constants[1]

Symbol Meaning Best Value Approximate Value

c Speed of light invacuum 2.99792458 × 108 m / s 3.00 × 108 m / s

G Gravitational constant 6.67384(80) × 10−11 N ⋅ m2 / kg2 6.67 × 10−11 N ⋅ m2 / kg2

NA Avogadro’s number 6.02214129(27) × 1023 6.02 × 1023

k Boltzmann’s constant 1.3806488(13) × 10−23 J / K 1.38 × 10−23 J / K

R Gas constant 8.3144621(75) J / mol ⋅ K 8.31 J / mol ⋅ K = 1.99 cal / mol ⋅ K = 0.0821atm ⋅ L / mol ⋅ K

σ Stefan-Boltzmannconstant 5.670373(21) × 10−8 W / m2 ⋅ K 5.67 × 10−8 W / m2 ⋅ K

k Coulomb forceconstant 8.987551788... × 109 N ⋅ m2 / C2 8.99 × 109 N ⋅ m2 / C2

qe Charge on electron −1.602176565(35) × 10−19 C −1.60 × 10−19 C

ε0 Permittivity of freespace 8.854187817... × 10−12 C2 / N ⋅ m2 8.85 × 10−12 C2 / N ⋅ m2

μ0 Permeability of freespace 4π × 10−7 T ⋅ m / A 1.26 × 10−6 T ⋅ m / A

h Planck’s constant 6.62606957(29) × 10−34 J ⋅ s 6.63 × 10−34 J ⋅ s

Table A2 Submicroscopic Masses[2]

Symbol Meaning Best Value Approximate Value

me Electron mass 9.10938291(40)×10−31kg 9.11×10−31kg

m p Proton mass 1.672621777(74)×10−27kg 1.6726×10−27kg

mn Neutron mass 1.674927351(74)×10−27kg 1.6749×10−27kg

u Atomic mass unit 1.660538921(73)×10−27kg 1.6605×10−27kg

1. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu(http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined.

2. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty, www.physics.nist.gov/cuu(http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the last digits. Numbers without uncertainties are exact as defined.

APPENDIX A | USEFUL INFORMATION* 237

http://www.physics.nist.gov/cuu




Table A3 Solar System Data

Sun mass 1.99×1030kg

average radius 6.96×108m

Earth-sun distance (average) 1.496×1011m

Earth mass 5.9736×1024kg


orbital period 3.16×107s

Moon mass 7.35×1022kg


orbital period (average) 2.36×106s

Earth-moon distance (average) 3.84×108m

Table A4 Metric Prefixes for Powers of Ten and Their Symbols

Prefix Symbol Value Prefix Symbol Value

tera T 1012 deci d 10−1

giga G 109 centi c 10−2

mega M 106 milli m 10−3

kilo k 103 micro μ 10−6

hecto h 102 nano n 10−9

deka da 101 pico p 10−12

— — 100( = 1) femto f 10−15

Table A5 The Greek Alphabet

Alpha Α α Eta Η η Nu Ν ν Tau Τ τ

Beta Β β Theta Θ θ Xi Ξ ξ Upsilon Υ υ

Gamma Γ γ Iota Ι ι Omicron Ο ο Phi Φ ϕ

Delta Δ δ Kappa Κ κ Pi Π π Chi Χ χ

Epsilon Ε ε Lambda Λ λ Rho Ρ ρ Psi Ψ ψ

Zeta Ζ ζ Mu Μ μ Sigma Σ σ Omega Ω ω

238 APPENDIX A | USEFUL INFORMATION*


Table A6 SI Units

Entity Abbreviation Name

Fundamental units Length m meter

Mass kg kilogram

Time s second

Current A ampere

Supplementary unit Angle rad radian

Derived units Force N = kg ⋅ m / s2 newton

Energy J = kg ⋅ m2 / s2 joule

Power W = J / s watt

Pressure Pa = N / m2 pascal

Frequency Hz = 1 / s hertz

Electronic potential V = J / C volt

Capacitance F = C / V farad

Charge C = s ⋅ A coulomb

Resistance Ω = V / A ohm

Magnetic field T = N / (A ⋅ m) tesla

Nuclear decay rate Bq = 1 / s becquerel

Table A7 Selected British Units

Length 1 inch (in.) = 2.54 cm (exactly)

1 foot (ft) = 0.3048 m

1 mile (mi) = 1.609 km

Force 1 pound (lb) = 4.448 N

Energy 1 British thermal unit (Btu) = 1.055×103 J

Power 1 horsepower (hp) = 746 W

Pressure 1 lb / in2 = 6.895×103 Pa


Table A8 Other Units

Length 1 light year (ly) = 9.46×1015 m

1 astronomical unit (au) = 1.50×1011 m

1 nautical mile = 1.852 km

1 angstrom(Å) = 10−10 m

Area 1 acre (ac) = 4.05×103 m2

1 square foot (ft2) = 9.29×10−2 m2

1 barn (b) = 10−28 m2

Volume 1 liter (L) = 10−3 m3

1 U.S. gallon (gal) = 3.785×10−3 m3

Mass 1 solar mass = 1.99×1030 kg

1 metric ton = 103 kg

1 atomic mass unit (u) = 1.6605×10−27 kg

Time 1 year (y) = 3.16×107 s

1 day (d) = 86,400 s

Speed 1 mile per hour (mph) = 1.609 km / h

1 nautical mile per hour (naut) = 1.852 km / h

Angle 1 degree (°) = 1.745×10−2 rad

1 minute of arc ( ') = 1 / 60 degree

1 second of arc ( '') = 1 / 60 minute of arc

1 grad = 1.571×10−2 rad

Energy 1 kiloton TNT (kT) = 4.2×1012 J

1 kilowatt hour (kW ⋅ h) = 3.60×106 J

1 food calorie (kcal) = 4186 J

1 calorie (cal) = 4.186 J

1 electron volt (eV) = 1.60×10−19 J

Pressure 1 atmosphere (atm) = 1.013×105 Pa

1 millimeter of mercury (mm Hg) = 133.3 Pa

1 torricelli (torr) = 1 mm Hg = 133.3 Pa

Nuclear decay rate 1 curie (Ci) = 3.70×1010 Bq

240 APPENDIX A | USEFUL INFORMATION*


Table A9 Useful Formulae

Circumference of a circle with radius r or diameter d C = 2πr = πd

Area of a circle with radius r or diameter d A = πr2 = πd2 / 4

Area of a sphere with radius r A = 4πr2

Volume of a sphere with radius r V = (4 / 3)⎛⎝πr3⎞⎠

[3][4]

3. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty,www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the lastdigits. Numbers without uncertainties are exact as defined.

4. Stated values are according to the National Institute of Standards and Technology Reference on Constants, Units, and Uncertainty,www.physics.nist.gov/cuu (http://www.physics.nist.gov/cuu) (accessed May 18, 2012). Values in parentheses are the uncertainties in the lastdigits. Numbers without uncertainties are exact as defined.




Index

Symbols(peak) emf, 202RC circuit, 137

AAC current, 89, 100AC voltage, 89, 100Alternating current, 88alternating current, 100ammeter, 140ammeters, 130ampere, 74, 101Ampere’s law, 172, 177analog meter, 140Analog meters, 132

BB-field, 157, 177back emf, 204, 226bioelectricity, 95, 101Biot-Savart law, 172, 177bridge device, 140bridge devices, 136

Ccapacitance, 54, 64, 137, 141capacitive reactance, 219, 226capacitor, 53, 64, 137, 141characteristic time constant, 216, 226conductor, 11, 30Conductors, 22conservation laws, 127, 141Coulomb force, 16, 30Coulomb forces, 16Coulomb interaction, 22, 30Coulomb’s law, 15, 30Curie temperature, 154, 177current, 112, 141Current sensitivity, 132current sensitivity, 141

Ddefibrillator, 62, 64dielectric, 56, 64dielectric strength, 64dielectric strengths, 57digital meter, 141digital meters, 132dipole, 22, 30Direct current, 88direct current, 101direction of magnetic field lines, 157, 177domains, 154, 177drift velocity , 77, 101

Eeddy current, 198, 226electric charge, 7, 30electric current, 74, 101electric field, 23, 30electric field lines, 30electric field strength, 18electric fields, 18electric generator, 226Electric generators, 201electric potential, 42, 64electric power, 85, 101electrocardiogram (ECG), 99, 101electromagnet, 177

electromagnetic force, 6, 30electromagnetic induction, 192, 226Electromagnetism, 154electromagnetism, 177electromagnets, 154electromotive force, 120electromotive force (emf), 141electron, 30electron volt, 45, 64electrons, 7electrostatic equilibrium, 22, 30electrostatic force, 15, 30electrostatic precipitators, 28, 30Electrostatic repulsion, 12electrostatic repulsion, 30electrostatics, 26, 30emf, 127emf induced in a generator coil, 202, 226energy stored in an inductor , 215, 226equipotential line, 64equipotential lines, 51

FFaraday cage, 25, 30Faraday’s law of induction, 193, 226ferromagnetic, 154, 177field, 30force field, 16free charge, 30free charges, 22free electron, 30free electrons, 11full-scale deflection, 132, 141

Ggalvanometer, 132, 141gauss, 158, 177grounded, 27, 30grounding, 51, 64

HHall effect, 163, 177Hall emf, 163, 177henry, 212, 226

Iimpedance, 220, 226inductance, 212, 226induction, 12, 31, 190, 226inductive reactance, 218, 226inductor, 213, 226ink jet printer, 27ink-jet printer, 31insulator, 31insulators, 12internal resistance, 120, 141ionosphere, 24, 31

JJoule’s law, 113, 141junction rule, 127, 141

KKirchhoff’s rules, 126, 141

Llaser printer, 31Laser printers, 27law of conservation of charge, 10, 31Lenz’s law, 193, 226loop rule, 127, 141Lorentz force, 158, 177

Mmagnetic damping, 198, 226magnetic field, 157, 178magnetic field lines, 157, 177magnetic field strength (magnitude) producedby a long straight current-carrying wire, 171,178magnetic field strength at the center of a circularloop, 172, 178magnetic field strength inside a solenoid, 173,178magnetic flux, 192, 226magnetic force, 158, 178magnetic monopoles, 156, 178Magnetic resonance imaging (MRI), 176magnetic resonance imaging (MRI), 178magnetized, 154, 178magnetocardiogram (MCG), 177, 178magnetoencephalogram (MEG), 177, 178Maxwell’s equations, 172, 177Mechanical energy, 45mechanical energy, 64meter, 178Meters, 170microshock sensitive, 95, 101motor, 178Motors, 168Mutual inductance, 212mutual inductance, 226

NNerve conduction, 95nerve conduction, 101north magnetic pole, 152, 178nuclear magnetic resonance (NMR), 176, 178Null measurements, 134null measurements, 141

Oohm, 79, 101ohmic, 79, 101ohmmeter, 141ohmmeters, 135Ohm’s law, 79, 101, 112, 141

Pparallel, 114, 141parallel plate capacitor, 53, 65peak emf, 226permeability of free space, 171, 178phase angle, 223, 226photoconductor, 27, 31point charge, 17, 31polar molecule, 22, 31, 57, 65polarization, 12, 31polarized, 23, 31potential difference, 43, 120, 141potential difference (or voltage), 65potentiometer, 134, 141power factor, 224, 226proton, 31protons, 7

RRC circuit, 141resistance, 79, 101, 112, 141resistivity, 81, 101resistor, 112, 137, 141resonant frequency, 222, 226right hand rule 1, 158right hand rule 1 (RHR-1), 178right hand rule 2, 171

242 INDEX


right hand rule 2 (RHR-2), 178rms current, 89, 101rms voltage, 89, 101

Sscalar, 46, 65screening, 22, 31Self-inductance, 213self-inductance, 226semipermeable, 96, 101series, 112, 141shock hazard, 92, 101, 208, 227short circuit, 92, 101shunt resistance, 132, 141simple circuit, 79, 101solenoid, 172, 178south magnetic pole, 152, 178static electricity, 6, 31step-down transformer, 206, 227step-up transformer, 206, 227

Ttemperature coefficient of resistivity, 83, 101terminal voltage, 122, 141tesla, 158, 178test charge , 17, 31thermal hazard, 92, 101, 208, 227three-wire system, 208, 227transformer, 227transformer equation, 206, 227Transformers, 204

VVan de Graaff generator, 31Van de Graaff generators, 26vector, 31, 46, 65vector addition, 19, 31vectors, 18voltage, 43, 112, 141voltage drop, 112, 141voltmeter, 141Voltmeters, 130

WWheatstone bridge, 136, 141

Xxerography, 27, 31

INDEX 243

244 INDEX


ATTRIBUTIONSCollection: Electricity PhysicsEdited by: M AhmadURL: http://cnx.org/content/col11514/1.1/Copyright: M AhmadLicense: http://creativecommons.org/licenses/by/3.0/Based on: College Physics II <http://cnx.org/content/col11458/1.2> arranged by Erik Christensen.

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Module: Resistance and ResistivityBy: OpenStax CollegeURL: http://cnx.org/content/m42346/1.5/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

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Module: MagnetsBy: OpenStax CollegeURL: http://cnx.org/content/m42366/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Ferromagnets and ElectromagnetsBy: OpenStax CollegeURL: http://cnx.org/content/m42368/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Magnetic Fields and Magnetic Field LinesBy: OpenStax CollegeURL: http://cnx.org/content/m42370/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Magnetic Field Strength: Force on a Moving Charge in a Magnetic FieldBy: OpenStax CollegeURL: http://cnx.org/content/m42372/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Force on a Moving Charge in a Magnetic Field: Examples and ApplicationsBy: OpenStax CollegeURL: http://cnx.org/content/m42375/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: The Hall EffectBy: OpenStax CollegeURL: http://cnx.org/content/m42377/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Magnetic Force on a Current-Carrying ConductorBy: OpenStax CollegeURL: http://cnx.org/content/m42398/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Torque on a Current Loop: Motors and MetersBy: OpenStax College

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URL: http://cnx.org/content/m42380/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Magnetic Fields Produced by Currents: Ampere’s LawBy: OpenStax CollegeURL: http://cnx.org/content/m42382/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Magnetic Force between Two Parallel ConductorsBy: OpenStax CollegeURL: http://cnx.org/content/m42386/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: More Applications of MagnetismBy: OpenStax CollegeURL: http://cnx.org/content/m42388/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Introduction to Electromagnetic Induction, AC Circuits and Electrical TechnologiesBy: OpenStax CollegeURL: http://cnx.org/content/m42389/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Induced Emf and Magnetic FluxBy: OpenStax CollegeURL: http://cnx.org/content/m42390/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Faraday’s Law of Induction: Lenz’s LawBy: OpenStax CollegeURL: http://cnx.org/content/m42392/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Motional EmfBy: OpenStax CollegeURL: http://cnx.org/content/m42400/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Eddy Currents and Magnetic DampingBy: OpenStax CollegeURL: http://cnx.org/content/m42404/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Electric GeneratorsBy: OpenStax CollegeURL: http://cnx.org/content/m42408/1.5/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Back EmfBy: OpenStax CollegeURL: http://cnx.org/content/m42411/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

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Module: TransformersBy: OpenStax CollegeURL: http://cnx.org/content/m42414/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Electrical Safety: Systems and DevicesBy: OpenStax CollegeURL: http://cnx.org/content/m42416/1.2/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: InductanceBy: OpenStax CollegeURL: http://cnx.org/content/m42420/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: RL CircuitsBy: OpenStax CollegeURL: http://cnx.org/content/m42425/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Reactance, Inductive and CapacitiveBy: OpenStax CollegeURL: http://cnx.org/content/m42427/1.3/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: RLC Series AC CircuitsBy: OpenStax CollegeURL: http://cnx.org/content/m42431/1.4/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

Module: Useful InformationBy: OpenStax CollegeURL: http://cnx.org/content/m42720/1.5/Copyright: Rice UniversityLicense: http://creativecommons.org/licenses/by/3.0/

250 INDEX
















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Electro Science

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Transcript of Electro Science