GROUP - 6

ELECTRICITY AND MAGNETISM

A form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.

OR

 The set of physical phenomena associated with the presence and flow of electric charge.

What is Electricity ?

Time rate of flow of charge through any cross section of a conductor is called current

Mathematically, it can be expressed asI=Δq Δt

Where, Δq is the charge passing through any cross section of a conductor

And Δt is the time taken by the chargeThe SI unit of current is ampere

Electric Current

Conventional Current The current which passes from a point of high

potential to a point of low potential as if it is represented by the movement of positive charges.

Electric CurrentThe current which passes from a point of low potential to a point of high potential as if it is represented by a movement of negative charges

Types of Current

When two conductors at different potential are connected through a metallic wire, then the current will start to flow.TYPES OF CURRENT SOURCES

1. Cells convert the chemical energy into electrical energy

2. Electric Generators convert mechanical energy into electrical energy

3. Thermo-couples convert heat energy into electrical energy

4. Solar Cells convert sunlight into electrical energy

Sources of Current

1. Heating Effect• Current flows through the conductor due to

motion of the free electrons• The free electrons collide with the atoms of

metal• The electrons lose their kinetic energy and

transfer it to the atoms of metal• The vibrational K.E of the atoms is increased,

which produces heat in the wire.• Mathematically, the heat can be measured by

H=I²Rt• Applications: Heater, toaster,

electric iron, etc.

Effects of Current

2. Chemical Effect There are certain liquids through which

electricity can pass through due to some chemical reaction

The study of this process is called electrolysis The chemical changes produced during the

electrolysis are due to the chemical effects of the current

Depends upon the nature of the liquid and the quantity of electricity passed through liquids

Applications: Electroplating and purification of metals

3. Magnetic Effect When a current passes through the wire, a

magnetic field is produced around it The strength of the magnetic field depends

upon the magnitude of current and the distance from current carrying conductor.

Applications: Used to detect and measure current, used in the machines involving electric motors

ELECTROMAGNETISM

The branch of physics which inter-relates electricity and magnetism

In 1820, Prof. Hans Oersted described the magnetic field due to current in a wire

EXPERIMENT• Take a copper wire that passes vertically

through a horizontal piece of cardboard• Place small magnetic compass needles

on the card board along a circle with centre at the wire• All the compass needles point in the direction of the

north-south. When a heavy current passes through wire , the compass needles set themselves along the tangent to the circle.

• Reverse the direction of the current , the direction of needles is also reversed

• When the current through the wire is stopped, all the needles again point in the north-south direction

CONCLUSIONS1. A magnetic field is set up around current carrying

conductor2. The lines of forces are circular and their direction

depends upon the direction of current3. The magnetic field lasts only as long as the

current is flowing through it

DIRECTION OF MAGNETIC FIELD RIGHT HAND RULE

If the wire is grasped in the fist of the right

hand with the thumb pointing in the directionof current , then the curled fingers indicate the direction of magnetic field

MAGNETIC FLUX

The number of magnetic lines of forcing passing through certain element of area is called magnetic flux

Mathematically,

Φᵦ= B.A Φᵦ =BACosθ

The SI unit of magnetic flux is Weber or Nm/A

MAGNETIC FLUX DENSITY (Magnetic Induction)

The magnetic flux per unit area of a surface is perpendicular to the magnetic field is called magnetic flux density

Mathematically,

B= Φᵦ A

SI unit of magnetic flux density is Wbmˉ² or NAˉ¹mˉ¹ or tesla

A force in experienced by a current carrying conductor placed in a uniform magnetic field. In actual, the magnetic field exerts the force on the charged particles moving in the conductor.

Mathematically,

Force On A Moving Charge In A Magnetic Field

sin

)(

qvBF

BvqF

Consider a segment of wire of length L and area of cross section A placed in a magnetic field B

Let I be the current flowing through the wire, then magnetic fore acting on the conductor is,

Fв=I( L × B ) (1)Let,

• Area of cross section of wire = A• Length of conductor = L• Volume of the wire segment = AL• Number of charge carrier per unit volume = n• Total number of charge carriers in the wire

segment = nAL

• Velocity of each charge = v• Charge on a charge carrier = q• Total charge on nAL charge carriers = ΔQ = nALqAs current is defined as I = ΔQ

ΔtOR I = nALq [As Δt = L/v]

L/vOR I = nAqv (2)Using this value of current in Eq. (1), we get F = nAqv (L × B) (3)

As the direction of segment LA is similar to v, it can be written as

LA = v̂So, L = LLA = Lv̂

Thus Eq. 3 becomes,

OR

OR]vv̂v[ )Bv(nAqLF

)Bv̂v(nAqLF

)Bv̂L(nAqvF

L

L

L

This is the force exerted on nAL chargesNow the force experienced by a single charge

is,

(4)This is the general equation which holds for

any charge carrier moving in a magnetic field

)Bv(qF

nAL

)Bv(nAqLF

nAL

FF L

For an ElectronF = -e ( v × B )

For a Proton F = +e ( v × B )

Direction of force can be determined by right hand rule. The direction of force is given by the direction of vector ( v × B )

Special Cases• When θ=90˚, F=qvB• When θ=0˚, F=0• When θ=180˚, F=0• When charge particle v=0 , F=0

When the charge particle q is moving with velocity v in a region where there is an electric and magnetic field, then the total force F is the vector sum of electric and magnetic force

Mathematically,

Electrical force does work only while the magnetic force deflects the charged particle. It performs no work

Lorrentz Force

] EqF[ )Bv(qEqF

FFF

e

Be

Consider a current carrying rectangular coil which is capable of rotating about its axis

Suppose it is placed in uniform magnetic field B with its plane along the field.

The force acting on a conductor of length L in uniform magnetic field is,

F = I (L×B)F = ILBsinθ

Where θ is the angle b/w L and BDivide the length of rectangular coil into four

segments AB, BC, CD and DA

Torque On A Current Carrying Coil

Forces acting on the sides of the coil can be calculated as:• Force on segment AB As AB is anti-parallel to magnetic field, θ=180˚

• Force on segment BCAs BC is perpendicular to magnetic field, θ=90˚

• Force on segment CDAs CD is parallel to magnetic field, θ=0˚

• Force on segment DA

As DA is perpendicular to magnetic field, θ=90˚

From right hand rule, its direction is out of the paper

0)0(180sin ILBILBF

ILBILBILBF )1(90sin2

0)0(0sin ILBILBFCD

ILBILBILBF )1(90sin1

As F₁ and F₂ are equal and opposite, they form a couple which tends to rotate the coil about its axis

Now,Torque due to couple=(magnitude of either force) ×

(couple arm)τ = ILB × a

Where a is the perpendicular distance b/w two forces called couple arm and is equal to AB or CD and La is the area of the coil. (i.e. La=A)

So, τ = IBA If the field makes an angle α with the plane of the coil,

then the couple arm becomes acosα τ = ILB × acosα τ = IB(La) cosα τ = IBA cosα

If there are N number of turns of a coil, then τ = NIBA cosα

Special case• The torque acting on current carrying coil will

be maximum when the plane of coil is parallel to the magnetic field i.e. when α=0˚, so

τ max= N IB A cos 0˚= N IB ANUMERICAL PROBLEM

• A coil of 0.1 m × 0.1 m and of 200 turns carrying a current of 1.0 mA is placed in a uniform magnetic field of 0.1 T . Calculate the maximum torque acting on the coil

• Area of coil = A = 0.1 m × 0.1 m = 0.01 m²• Number of turns = N = 200 turns• Current = I = 1.0 × 10¯³ A• Magnetic field = B = 0.1 T• Maximum torque = τ max = ?

• Formula : τ = NIBA cosα• Answer : 0.2 × 10¯³ OR 2.0 × 10¯⁴ Nm

If the magnetic field through a circuit changes, an emf and a current are induced in the circuit

Applications:Used in power generating systemTransformers use this principle to produce

emfsATM card also uses this principleAny appliance we plug into a wall socket uses

induced emf INDUCED EMF AND INDUCED CURRENT

If a conductor moves through a magnetic field then due to change in magnetic flux, an emf is induced called induced emf. If the circuit is closed, it will cause an electric current called induced current

ELECTROMAGNETIC INDUCTION

EXPERIMENTS TO PRODUCE INDUCED EMF AND CURRENTCase 1

When the coil is stationary and the bar magnet is moved

Case 2When area of the coil is changed in a uniform magnetic field

Case 3By rotating the coil of constant area in uniform magnetic field

The phenomenon in which changing of current in one coil induces an emf in another coil

ExplanationThe coil connected with a battery through a

switch and a rheostat is called primary coilThe one connected to the galvanometer is called

secondary coilWhen variation is produced in the rheostat,

current in the primary coil changes thus magnetic flux changes.

This produces an induced emf in the secondary coil

According to Faraday’s Law, induced emf in secondary coil is

εs= -Ns Δφs (1)

Δt

MUTUAL INDUCTION

As M is constant, soεs= -MΔIp

(5) Δt

Above Eq. shows that induced emf in the secondary coil is directly proportional to rate of change of current in the primary coil

From Eq. 5,

Hence, mutual inductance can be defined asThe emf induced in the secondary coil when the rate of change of current in the primary coil is one ampere/sec

Its SI unit is VsA¯¹ known as henry (H)

t

M s

If φs is flux passing through one turn of coil then flux through whole coil is Nsφs

This magnetic flux is directly proportional to current Ip flowing through the primary, so

(2)

(3)Where M is the constant of proportionality known as

mutual inductance of two coilsFrom Eq. 1,

εs= - Δ(Ns φs) (4) Δt

Using Eq 2 in 4 ,εs= - Δ(MIp)

Δt

P

SS

PSS

PSS

I

NM

MIN

IN

The phenomenon in which a changing current in a coil induces an emf in itself

ExplanationConsider a coil connected in series

with a battery & a rheostatDue to varying the resistance of the

rheostat ,the current in the coil is changedChanging flux produces an induced emf in the coilLet φ be the flux passing through one loop of the

coil then the total flux for N no. of turns will be Nφ

Since flux (φ=B.A) is proportional to magnetic field, and magnetic field (B = µoI) is proportional to current. So,

SELF INDUCTION

Nφ INφ = LI L = Nφ (1)

IWhere L is the proportionality constant of self

inductance. According to faraday’s Law,εL= -N Δφ (2)

ΔtOr εL = -Δ (Nφ/Δt)Or εL = -Δ (LI) [ As

Nφ=LI] Δt

But L is constant, so εL = - L ΔI (3)

Δt

This equation shows that emf is directly proportional to the rate of change of current in the coil.

Thus from Eq. 3,

Hence self inductance can be defined as,The ratio of induced emf produced in a coil to

rate of change of current in the coil , in the same coil

Its SI unit is VsA¯¹ known as henry (H)

t

L L

Transformer is an electrical device which changes a given AC voltage into a larger or smaller AC voltage

PRINCIPLE• Mutual induction b/w two coils

wound on same iron coreCONSTRUCTION

• It consists of two copper coils which are magnetically linked to each other

• Primary coil : The coil to which AC power is supplied

• Secondary coil : The coil which delivers power to the output circuit

WORKING• Let no. of turns of the primary coil = Np

• No. of turns of the secondary coil = Ns

• Alternating emf applied to the primary coil = Vp

TRANSFORMER

Rate of change of flux in the primary coil = Δφ/ΔtDue to this change of flux, a back emf is also

induced in the primary which opposes the applied voltage. Thus, Self-induced emf in primary coil is

Self induced emf = - Np Δφ Δt

If the resistance of the coil is negligible, then the back emf is equal and opposite to applied voltage Vp so,

Vp = - back emf [Vp =ε + IR, If R=0; Vp = ε]

Vp = Np Δφ (1) Δt

Emf induced across the secondary coil is Vs = Ns (Δφ/Δt) (2)

Transformation Ratio Ns Δφ

Vs = Δt [Dividing Eq.1 and 2]

Vp Np Δφ Δt

OR Vs = Ns

Vp Np

Efficiency of a Transformer can be expressed by, output power input power

Efficiency of a transformer can be improved byi. The core should be made of ironii. The core should be assembled by laminated sheetsiii. Resistance of primary and secondary coil should be

minimum

100

In a ideal transformer, the output power is nearly equal to input power

Principle for Transmission

Power loss in transformers can be due to one of the following reasons:• Eddy Currents: These currents cause

energy loss in the core due to heat produced in it. It can be reduced by using laminated core with insulation b/w the lamination of the sheet

• Hysteresis Loss: Due to repeated magnetization and demagnetization of core due to flow of AC. It can be reduced by using soft iron core

p

s

s

p

sspp

I

I

V

V

IVIV

outputPower inputPower

TYPES OF TRANSFORMERS

STEP UP TRANSFORMERA transformer in which voltage across the

secondary is greater than primary voltageCondition: No. of turns in the secondary

is greater than no. of turns in primary coili.e. Ns > Np

STEP DOWN TRANSFORMERA transformer in which voltage across the

secondary is less than primary voltageCondition: No. of turns in the secondary

coil is less than no. of turns in primary coili.e. Ns < Np

NUMERICAL PROBLEM

An ideal step down transformer is connected to main supply of 240 V. It is desired to operate a 12 V, 30 W lamp. Find the current in the primary coil and the transformation ratio

Primary voltage =Vp = 240 VSecondary voltage = Vs = 12 VOutput Power = Po = 30 wattCurrent in primary = Ip = ?Transformation Ratio = ?

Formulae: Ip = Po/Vp & (Ns/Np )= (Vs/Vp)

Answers : Ip= 0.125 ANs = 1 Np 20