Mar 24, 2014 PHYS 1442-004

PHYS 1442 – Section 004Lecture #17, Review for test 2

Monday March 242014Dr. Justin Griffiths for Dr. Brandt

Loose ends and review for test 2

3/17/2014

Inductor• An electrical circuit always contains some inductance but it is often

negligible– If a circuit contains a coil of many turns, it could have a large inductance

• A coil that has significant inductance, L, is called an inductor and is express with the symbol– Precision resistors are normally wire wound

• Would have both resistance and inductance• The inductance can be minimized by winding the wire back on itself in opposite

direction to cancel magnetic flux• This is called a “non-inductive winding”

• For an AC current, the greater the inductance the less the AC current– An inductor thus acts like a resistor to impede the flow of alternating current (not

to DC, though. Why?)– The quality of an inductor is indicated by the term reactance or impedance

2PHYS 1442-004, Dr. Andrew Brandt

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

3

Energy Stored in a Magnetic Field• When an inductor of inductance L is carrying current I

which is changing at a rate I/ t, energy is supplied to the inductor at a rate–

• What is the work needed to increase the current in an inductor from 0 to I?– The work, dW, done in time dt is– Thus the total work needed to bring the current from 0 to I

in an inductor is

P

W

W

I IL

I

t

Pt LII

W

LII

I 21

2LI

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

4

Energy Stored in a Magnetic Field• The work done to the system is the same as the

energy stored in the inductor when it is carrying current I–

– This is compared to the energy stored in a capacitor, C, when the potential difference across it is V

– Just like the energy stored in a capacitor is considered to reside in the electric field between its plates

– The energy in an inductor can be considered to be stored in its magnetic field

21

2U LI

U

Energy Stored in a magnetic field inside an inductor

21

2CV

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

5

Stored Energy in terms of B• So how is the stored energy written in terms of magnetic field

B?– Inductance of an ideal solenoid without a fringe effect

– The magnetic field in a solenoid is– Thus the energy stored in an inductor is

– Thus the energy density is

– This formula is valid to any region of space– If a ferromagnetic material is present, m0 becomes m.

L B

U

u

2

0

1

2

BU Al

2

0

1

2

Bu

20 N A l

0 NI l

21

2LI

201

2

N A

l

2

0

1

2

BAl

2

0

Bl

N

Volume V

What is this?U

V

U

Al

2

0

1

2

B

What volume does Al represent? The volume inside a solenoid!!

E density

E

3/17/2014

Example Energy stored in a coaxial cable. (a) How much energy is being stored per unit length in a coaxial cable whose conductors have radii r1 and r2 and which carry a current I? (b) Where is the energy density highest?

(a) The inductance per unit length for a coaxial cable is

(b) Since the magnetic field is

L

l

Thus the energy stored per unit length is

U

l

B

And the energy density is u The energy density is highest where B is highest. B is highest close to r=r1, near the surface of the inner conductor.

0 2

1

ln2

r

r

21

2

LI

l

20 2

1

ln4

I r

r

0

2

I

r

2

0

1

2

B

6PHYS 1442-004, Dr. Andrew Brandt

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

7

Why do we care about AC circuits?• The circuits we’ve learned so far contain resistors, capacitors and

inductors and have been connected to a DC source or a fully charged capacitor– What? This does not make sense. – The inductor does not work as an impedance unless the current is changing. So

an inductor in a circuit with DC source does not make sense.– Well, actually it does. When does it impede?

• Immediately after the circuit is connected to the source so the current is still changing. So?

– It causes the change of magnetic flux.– Now does it make sense?

• Anyhow, learning the responses of resistors, capacitors and inductors in a circuit connected to an AC emf source is important. Why is this?– Since most the generators produce sinusoidal current– Any voltage that varies over time can be expressed in the superposition of sine and

cosine functions

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

8

AC Circuits – the preamble• Do you remember how the rms and peak values for

current and voltage are related?

• The symbol for an AC power source is

• We assume that the voltage gives rise to current

– where

rmsV rmsI

I 2 f

0V

20I

2

0 sin 2I ft 0 sinI t

3/17/2014 PHYS 1442-004, Dr. Andrew Brandt

9

AC Circuit w/ Resistance only• What do you think will happen when an ac source

is connected to a resistor?• From Kirchhoff’s loop rule, we obtain

• Thus

– where• What does this mean?

– Current is 0 when voltage is 0 and current is in its peak when voltage is in its peak.

– Current and voltage are “in phase”• Energy is lost via the transformation into heat at

an average rate•

0V IR

V 0 0V I R

P

0 sinI R t 0 sinV t

I V 2rmsI R 2

rmsV R

Review ch 20Magnets, magnetic fields

F Il B Force on current carrying wire due to external field

F qv B Force on moving charge due to external field

sinNIAB Torque on a current loop

Mar 24, 2014 PHYS 1442-004

ExampleElectron’s path in a uniform magnetic field. An electron travels at a speed of 2.0x107m/s in a plane perpendicular to a 0.010-T magnetic field. Describe its path.

What is the formula for the centripetal force?

Since the magnetic field is perpendicular to the motion of the electron, the magnitude of the magnetic force is

Since the magnetic force provides the centripetal force, we can establish an equation with the two forces

F

31 7

2

19

9.1 10 2.0 101.1 10

1.6 10 0.010

kg m sm

C T

F

F

r Solving for r

ma m2v

r

evB

evB 2v

mr

mv

eB

Mar 24, 2014 PHYS 1442-004

Some electronic devices and experiments need a beam of charged particles all moving at nearly the same velocity. This can be achieved using both a uniform electric field and a uniform magnetic field, arranged so they are at right angles to each other. Particles of charge q pass through slit S1 If the particles enter with different velocities, show how this device “selects” a particular velocity, and determine what this velocity is.

Figure 27-21: A velocity selector: if v = E/B, the particles passing through S1 make it through S2. Solution: Only the particles whose velocities are such that the magnetic and electric forces exactly cancel will pass through both slits. We want qE = qvB, so v = E/B.

Conceptual Example: Velocity selector

COULD I ADD GRAVITY TO THIS PROBLEM?Mar 24, 2014 PHYS 1442-004

Torque on a Current Loop

• Fa=IaB• The moment arm of the coil is b/2

– So the total torque is the sum of the torques by each of the forces

• Where A=ab is the area of the coil– What is the total net torque if the coil consists of N loops of wire?

– If the coil makes an angle q w/ the field

• So what would be the magnitude of this torque?– What is the magnitude of the force on the

section of the wire with length a?

NIAB sinNIAB

2

bIaB

2

bIaB IabB IAB

Mar 24, 2014 PHYS 1442-004

Review Chapter 200

2

IB

r

Magnetic field from long straight wire

0 1 2

2

I IF

l d

Magnetic force for two parallel wires

Ampére’s Law

0B nI solenoid

0 enclB dl I

Mar 24, 2014 PHYS 1442-004

Ampère’s LawExample : Field inside and outside a wire.

A long straight cylindrical wire conductor of radius R carries a current I of uniform current density in the conductor. Determine the magnetic field due to this current at (a) points outside the conductor (r > R) and (b) points inside the conductor (r < R). Assume that r, the radial distance from the axis, is much less than the length of the wire. (c) If R = 2.0 mm and I = 60 A, what is B at r = 1.0 mm, r = 2.0 mm, and r = 3.0 mm?

Mar 24, 2014 PHYS 1442-004

Solution: We choose a circular path around the wire; if the wire is very long the field will be tangent to the path.a. The enclosed current is the total current; this is the same as a thin wire. B = μ0I/2πr.b. Now only a fraction of the current is enclosed within the path; if the current density is uniform the fraction of the current enclosed is the fraction of area enclosed: Iencl = Ir2/R2. Substituting and integrating gives B = μ0Ir/2πR2.c. 1 mm is inside the wire and 3 mm is outside; 2 mm is at the surface (so the two results should be the same). Substitution gives B = 3.0 x 10-3 T at 1.0 mm, 6.0 x 10-3 T at 2.0 mm, and 4.0 x 10-3 T at 3.0 mm.Mar 24, 2014 PHYS 1442-004

ExampleSuspending a wire with current. A horizontal wire carries a current I1=80A DC. A second parallel wire 20cm below it must carry how much current I2 so that it doesn’t fall due to the gravity? The lower has a mass of 0.12g per meter of length.

Which direction is the gravitational force?

This force must be balanced by the magnetic force exerted on the wire by the first wire.

Downward

gF

l

2I Solving for

I2

2 3

7

2 9.8 0.12 10 0.2015

4 10 80

m s kg mA

T m A A

mg

l MF

l 0 1 2

2

I I

d

0 1

2mg d

l I

Mar 24, 2014 PHYS 1442-004

Mar 24, 2014

Solenoid Magnetic Field• Now let’s use Ampere’s law to determine the magnetic field

inside a very long, densely packed solenoid

• Let’s choose the path abcd, far away from the ends– We can consider four segments of the loop for integral– – The field outside the solenoid is negligible, and the internal field is

perpendicular to the end paths, so these terms also are 0– Thus Ampere’s law gives us

PHYS 1442-004

0 enclB l I ab bc cd daB l B l B l B l

Bl 0B nI

0 NI

Mar 24, 2014 PHYS 1442-004

ExampleCoaxial cable. A coaxial cable is a single wire surrounded by a cylindrical metallic braid, as shown in the figure. The two conductors are separated by an insulator. The central wire carries current to the other end of the cable, and the outer braid carries the return current and is usually considered ground. Describe the magnetic field (a) in the space between the conductors and (b) outside the cable.

(a) The magnetic field between the conductors is the same as the long, straight wire case since the current in the outer conductor does not impact the enclosed current. (b) Outside the cable, we can draw a similar circular path, since we expect the field to have a circular symmetry. What is the sum of the total current inside the closed path?So there is no magnetic field outside a coaxial cable. In other words, the coaxial cable is self-shielding. The outer conductor also shields against external electric fields, which could cause noise.

enclI

B

0.I I

0

2

I

r

Mar 24, 2014 PHYS 1442-004

Faraday’s Law of Induction• In terms of magnetic flux

– The emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit

Faraday’s Law of InductionBd

Ndt

• For a single loop of wire N=1, for closely wrapped loops, N is the number of loops

• The negative sign has to do with the direction of the induced emf (Lenz’s Law)

cosB B A BA B A

Mar 24, 2014 PHYS 1442-004

Lenz’s Law• It is experimentally found that

– An induced emf gives rise to a current whose magnetic field opposes the original change in flux This is known as Lenz’s Law

– We can use Lenz’s law to explain the following cases :

• When the magnet is moving into the coil– Since the external flux increases, the field inside the

coil takes the opposite direction to minimize the change and causes the current to flow clockwise

• When the magnet is moving out– Since the external flux decreases, the field inside the

coil takes the opposite direction to compensate the loss, causing the current to flow counter-clockwise

Mar 24, 2014 PHYS 1442-004

How does a transformer work?• When an AC voltage is applied to the primary, the

changing B it produces will induce voltage of the same frequency in the secondary wire

• So how would we make the voltage different?– By varying the number of loops in each coil– From Faraday’s law, the induced emf in the secondary is –

SV

PV

S S

P P

V N

V N Transformer

Equation

BS

dN

dt

BP

dN

dt

–The input primary voltage is–

–Since dFB/dt is the same, we obtain–