Short Version : 27. Electromagnetic Induction

27.1. Induced Currents

4 results from Faraday / Henry (1831)

1. Current induced in coil by moving magnet bar.

2. Moving the coil instead of the magnet gives the same result.

v = 0, I = 0 v > 0, I > 0

v >> 0, I >> 0 v < 0, I < 0

3. An induced current also results when a current-carrying circuit replaces the magnet.

4. A current is also induced when the current in an adjacent circuit changes.

changing B induces currents (electromagnetic induction)

27.2. Faraday’s Law

• Magnetic flux

• Flux & Induced EMF

Magnetic flux

B d B AMagnetic flux:

Reminder: 0d B A

For a uniform B on a flat surface:

cosB BA B AMove magnet right more lines thru loop

Example 27.1. Solenoid

A solenoid of circular cross section has radius R, consists of n turns per unit length, and carries current I.Find the magnetic flux through each turn of the solenoid.

BA 20 n I R

B out of planeI B

Example 27.2. Nonuniform Field

A long, straight wire carries current I.A rectangular wire loop of dimensions l by w lies in a plane containing the wire, with its closest edge a distance a from the wire, and its dimension l parallel to the wire.Find the magnetic flux through the loop.

0

2

a w

a

Il dr

r

B d B A 0

2

a w

a

I l d r

r

0 ln2

I l a w

a

Area element for integration

Flux & Induced EMF

Faraday’s law of induction:

The induced emf in a circuit is proportional to the rate of change of

magnetic flux through any surface bounded by that circuit.

Note: dB/dt can be due to

• changing B caused by

• relative motion between circuit & magnet,

• changing current in adjacent circuit,

• changing area of circuit,

• changing orientation between B & circuit.

Bd

d t

E

Cd E l

dd

d t S

B A C is CCW about S.

CC

Example 27.3. Changing B

A wire loop of radius 10 cm has resistance 2.0 .

The plane of the loop is perpendicular to a uniform B that’s increasing at 0.10 T/s.

Find the magnitude of the induced current in the loop.

2B rB BA

2Bd dB r

d t d t

33.14 10 V

2 d Br

d t

Bd

d t

E 2

0.1 0.10 /m T s

IR

E 33.14 10

2.0

V

31.57 10 A

CCW

S

CI

Example 27.4. Changing Area

Two parallel conducting rails a distance l apart are connected at one end by a resistance R.

A conducting bar completes the circuit, joining the two rails electrically but free to slide along.

The whole circuit is perpendicular to a uniform B, as show in figure.

Find the current when the bar is pulled to the right with constant speed v.

B l xB B A

Bd

d t

E

Let x = 0 be at the left end of rail.

B l v

IR

E B l v

R

CCW

S

x

C I

27.3. Induction & Energy

Direction of emf is to oppose magnet’s motion.

Lenz’s law :Direction of induced emf is such that B created by the induced current opposes the changes in that created the current.

Magnet moving right

RH rule: thumb // m.Loop ~ magnet with N to left.

RH rule: thumb // m.Loop ~ magnet with S to left.

Magnet moving left

m

I

I

m

Motional EMF & Len’s Law

Motional emf: induced emf due to motion of conductor in B.

Square loop of sides L & resistance R pulled with constant speed v out of uniform B.

Force on e: e F v B downward force upward I (CW)

Force on current carrying wire:

mag I F L B

,mag net appliedF F

B B L x

BdB L v

d t

x B L vE

P I E2

RE

IR

E B L v

R

2B L v

R

F v F v I L B v 2B L v

R

Work done is used to heat up circuit ( E conservation ).

S

C

d xv

d t

I

CW

< 0

0Bd

d t

Application. Electric Generators

World electricity generation ~ 2TW.

Sinusoidal AC output

Hand-cranked generator ~ 100W

Work required due to Lenz’s law.

Rotating loop changes & induces emf.

Rotating conducting loop

Stationary brushes

Rotating slip rings.

Electric load

Example 27.5. Designing a Generator

An electric generator consists of a 100-turn circular coil 50 cm in diameter.It’s rotated at f = 60 rev/s to produce standard 60 Hz alternating current.Find B needed for a peak output voltage of 170 V,which is the actual peak in standard 120 V household wiring.

1 costurn BA

2

cos 22

d dN B f t

d t

Bd

d t

E

2

cos 22

dB f t

2

22 sin 22

dN B f f t

2

2 0.5170 2 100 60 /

2

mV B s

222

2peak

dN B f

E

323 10B T

Loop rotation

Swiping a credit card. Patterns of magnetization on the strip induce currents in the coil.

EM induction is basis of magnetic recording ( audio, video, computer disks, …).

Modern hard disks:Giant magnetoresistance.

CoilIron

Magnetic strip

Card motion

Information stored in magnetization pattern

Eddy Currents

Application: Metal Detectors

Eddy current: current in solid conductor induced by changing .

Usage: non-frictional brakes for rotating saw blades, train wheels, …

Transmitter coil

Receiver coil

Current detector

Nothing between coils

Metal between coils

AC

Induced Current

Strong I

Weak I : alarm.

Closed & Open Circuits

Bin

B

B of induced I points out of page

RH rule gives CCW I

Setting n // Bin

C is CCW & < 0

Bin d /d t < 0

> 0

I is CCW

> 0

+_

C

27.4. Inductance

Mutual Inductance:Changing current in one circuit induces an emf in the other.

Self-Inductance:Changing current induces emf in own circuit & opposes further changes.

Inductance:

B selfL I

Large inductance: two coils are wound on same iron core.

Applications: Transformers, ignition coil, battery chargers, …

Applications: Inductors frequency generator / detector …

[ L ] = T m2 / A = Henry

Example 27.6. Solenoid

A long solenoid of cross section area A and length l has n turns per unit length.Find its self-inductance.

1 turn BA

0B n IB of solenoid:

0 n I A

1 turnn l 20 n l I A

20L n l A

I

B L I

Bd

d t

E

d IL

d t back emf

Rapid switching of inductive devices can destroy delicate electronic devices.

+direction = V along I.

d I /d t < 0

Inductor

+ (higher voltage)

0d I

d t

E

E

0d I

Ld t

E

+ (higher voltage)

(lower voltage) (lower voltage)

Example 27.7. Dangerous Inductor

A 5.0-A current is flowing in a 2.0-H inductor.

The current is then reduced steadily to zero over 1.0 ms.

Find the magnitude & direction of the inductor emf during this time.

10000V

3

5.02.0

1.0 10

AH

s

E

+ here ( > 0 ) helps keep I flowing

0d I

Ld t

E

Inductors in Circuits

Current through inductor can’t change instantaneously.

Switch open: I = 0

Switch just closed : I = 0, dI/dt 0; | L | = 0

L ~ open circuit.

Long after switch closing: I 0, dI/dt = 0; L = 0;L ~ wire.

L < 0 ; | L |

2

20

d I d IR L

d t d t

0 0LI R E E

0 0d I

I R Ld t

E

0LL

R d

L d t

EE

/0

R t LL eE E 00L t E E

0

1LI

R E E /0 1 R t Le

R

E

Inductive time constant = L / R

c.f. capacitive time constant = RC

I V = IR But rate is

+_

Example 27.8. Firing Up a Electromagnet

A large electromagnet used for lifting scrap iron has self-inductance L = 56 H.

It’s connected to a constant 440-V power source;

the total resistance of the circuit is 2.8 .

Find the time it takes for the current to reach 75% of its final value.

/0 1 R t LI eR

E /1 R t LI e

2.80.75 1 exp

56

t

H

20ln 0.25t 28 s

/0

R t LI I e

Short times: IL can’t change instantaneously.

Long times: L = 0 ; inductor wire.

Switch at A, I . Switch at B, battery’s shorted out. I exponentially.

0 0d I

I R Ld t

E

/0 1 R t LI eR

E

0d I

I R Ld t

0 0t

0 tR

E

0 0I t

0 t

Making the Connection

Verify that the current in just after the switch is reopened has the value indicated

2

0

1RI R

E

0

1 2

IR R

E

Immediately after the switch is closed:

Long time after the switch is closed:2

0RI

Immediately after 2nd switch opening :

L ~ open circuit.

L ~ short circuit. I in L ~ continues.

27.5. Magnetic Energy

Any B contains energy.

This eruption of a huge prominence from the sun’s surface releases energy stored in magnetic fields.

Magnetic Energy in an Inductor

RL circuit: 0 0LI R E E 20 0LI I R I E E

20 0

d II I R L I

d t E

Power from battery

Power dissipated

Power taken by inductor

L

d IP L I

d t

U P dtd I

L I d td t

21

2L I 0 0U I

Energy stored in inductor:

Example 27.10. MRI Disaster

Superconducting electromagnets like solenoids in MRI scanners store a lot of magnetic energy.

Loss of coolant is dangerous since current quickly decays due to resistance.

A particular MRI solenoid carries 2.4 kA and has a 0.53 H inductance.

When it loses superconductivity, its resistance goes abruptly to 31 m. Find

(a)the stored magnetic energy, and

(b) the rate of energy release at the instance the superconductivity is lost.

61.5 10 J 2310.53 2.4 10

2H A 21

2U L I

2P I R 23 32.4 10 31 10A 51.8 10 W

In practice, Cu / Ag are incorporated into the superconducting wires to reduce R.

Magnetic Energy Density

Solenoid with length l & cross-section area A : 20L n A l (Eg.

27.6)2 2

0

1

2n A l I21

2U L I 2

0

1

2B A l

0B n I

Magnetic Energy Density :2

0

1

2Bu B

c.f. electric energy density : 20

1

2Eu E

27.6. Induced Electric Fields

EMF acts to separate charges:Battery: chemical reactionMotional emf: F = v B.Stationary loop in changing B : induced E

Bd

d t

Ed E l

dd

d t B A Faraday’s law

Static E begins / ends on charge.

Induced E forms loop.

Example 27.11. Solenoid

A long solenoid has circular cross section of radius R.

The solenoid current is increasing, & as a result so is B in solenoid.

The field strength is given by B = b t, where b is a constant.

Find the induced E outside the solenoid, a distance r from the axis.

2db t R

d t

2d r E E l

Symmetry E lines are circles.

Bd

d t

2b R

2

2

b RE

r

Loop for Faraday’s law

S out C ccw

CCW

S

Conservative & Nonconservative Electric Fields

For stationary charges (electrostatics) :

0d E l E is conservative

Induced fields (electromagnetics) :

0d E l E is non-conservative

W against E

E does W

E

GOT IT? 27.7The figure shows three resistors in series surrounding an infinitely long solenoid with a changing magnetic field;

the resulting induced electric field drives a current counterclockwise, as shown.

Two identical voltmeters are shown connected to the same points A and B.

What does each read?

Explain any apparent contradiction.

Hint: this is a challenging question!

VA VB = IR

3R IE

VA VB = 2IR

Diamagnetism

Superconductor is a perfect diamagnet (Meissner effect).

B 0

B = 0: net = 0

This e slows down. This e speeds up.

net 0

Classical model of diamagnetism (not quite right)