27. Electromagnetic Induction

49
27. Electromagnetic Induction 1. Induced Currents 2. Faraday’s Law 3. Induction & Energy 4. Inductance 5. Magnetic Energy 6. Induced Electric Fields

description

27. Electromagnetic Induction. Induced Currents Faraday’s Law Induction & Energy Inductance Magnetic Energy Induced Electric Fields. It takes fourteen 110-car trainloads of each week to fuel this power plant. - PowerPoint PPT Presentation

Transcript of 27. Electromagnetic Induction

Page 1: 27.  Electromagnetic  Induction

27. Electromagnetic Induction

1. Induced Currents

2. Faraday’s Law

3. Induction & Energy

4. Inductance

5. Magnetic Energy

6. Induced Electric Fields

Page 2: 27.  Electromagnetic  Induction

It takes fourteen 110-car trainloads of each week to fuel this power plant. What feature of the equation ε = - dΦB/dt demands this prodigious fuel consumption?

The minus sign, which denotes energy conservation in electromagnetic induction.

Page 3: 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

Page 4: 27.  Electromagnetic  Induction

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)

Page 5: 27.  Electromagnetic  Induction

27.2. Faraday’s Law

• Magnetic flux

• Flux & Induced EMF

Page 6: 27.  Electromagnetic  Induction

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

Page 7: 27.  Electromagnetic  Induction

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

Page 8: 27.  Electromagnetic  Induction

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 drr

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

Page 9: 27.  Electromagnetic  Induction

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 SB A C is CCW about S.

CC

Page 10: 27.  Electromagnetic  Induction

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 Brd 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

Page 11: 27.  Electromagnetic  Induction

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 Brd 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

Page 12: 27.  Electromagnetic  Induction

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

Page 13: 27.  Electromagnetic  Induction

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

Page 14: 27.  Electromagnetic  Induction

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

Page 15: 27.  Electromagnetic  Induction

GOT IT? 27.1

You push a bar magnet towards a loop, with the north pole toward the loop.

If you keep pushing the magnet straight through the loop,

what will be the direction of the current as you pull it out the other side?

Will you need to do work, or will work be done on you?

Reversed 逆轉

mm

II

Page 16: 27.  Electromagnetic  Induction

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

Page 17: 27.  Electromagnetic  Induction

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

Page 18: 27.  Electromagnetic  Induction

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

Page 19: 27.  Electromagnetic  Induction

GOT IT? 27.2

What will be the direction of the current when the loop first enters the field from the left side?

Current is CCW

S

0Bd

d t

0E 0Bd

d t

0B

C

Page 20: 27.  Electromagnetic  Induction

GOT IT? 27.2

What will be the direction of the current when the loop first enters the field from the left side?

Current is CCW

S

0Bd

d t

0E 0Bd

d t

0B

C

Page 21: 27.  Electromagnetic  Induction

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

Page 22: 27.  Electromagnetic  Induction

GOT IT? 27.3

If you lower the electrical resistance connected across a generator while turning the generator at a constant speed,

will the generator get easier or harder to turn?

constant speed fixed peak emflower R higher power P I V

2V

R

Page 23: 27.  Electromagnetic  Induction

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

Page 24: 27.  Electromagnetic  Induction

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

Page 25: 27.  Electromagnetic  Induction

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.

Page 26: 27.  Electromagnetic  Induction

GOT IT? 27.4

A copper penny falls on a path that takes it between the poles of a magnet.Does it hit the ground going(a) faster,(b) slower, (c) at the same speed as if the magnet weren’t present?

Eddy current dissipates KE.

Page 27: 27.  Electromagnetic  Induction

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

Page 28: 27.  Electromagnetic  Induction

GOT IT? 27.5

A long wire carries a current I as shown.

What’s the direction of the current in the circular conducting loop when I is(a) increasing and (b) decreasing?

CCW

CW

B

Setting n // B

C is CW & > 0

I B

d /d t > 0

< 0

Iind is CCW

Page 29: 27.  Electromagnetic  Induction

GOT IT? 27.5

A long wire carries a current I as shown.

What’s the direction of the current in the circular conducting loop when I is(a) increasing and(b) decreasing?

CCW

CW

B

Setting n // B

C is CCW & < 0

I B

d /d t < 0

> 0

Iind is CCW

Page 30: 27.  Electromagnetic  Induction

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

Page 31: 27.  Electromagnetic  Induction

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

Page 32: 27.  Electromagnetic  Induction

B L I

Bd

d t

E

d ILd t

back emf

Rapid switching of inductive devices can destroy delicate electronic devices.

+direction = V along I.

d I /d t < 0

Page 33: 27.  Electromagnetic  Induction

GOT IT? 27.6

Current flows left to right through the inductor shown.A voltmeter connected across the inductor gives a constant reading, and shows that the left end of the inductor is positive. Is the current in the inductor (a) increasing, (b) decreasing, or(c) steady? Why?

V along I < 0 dI/dt > 0

Page 34: 27.  Electromagnetic  Induction

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

Page 35: 27.  Electromagnetic  Induction

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.

Page 36: 27.  Electromagnetic  Induction

L < 0 ; | L |

2

20

d I d IR Ld 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

+_

Page 37: 27.  Electromagnetic  Induction

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

Page 38: 27.  Electromagnetic  Induction

/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

Page 39: 27.  Electromagnetic  Induction

Example 27.9. Short & long Times

The switch in Fig is initially open.

Find the current in R2 immediately and a long time after the switch is closed.

Long after the switch is closed, it’s reopened.

Find the current in R2 immediately after the second switch opening.

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 :

(b) L ~ open circuit.

(c) L ~ short

circuit.(d) I in L ~

continues.

(a)

Page 40: 27.  Electromagnetic  Induction

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.

Page 41: 27.  Electromagnetic  Induction

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:

Page 42: 27.  Electromagnetic  Induction

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.

Page 43: 27.  Electromagnetic  Induction

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

Page 44: 27.  Electromagnetic  Induction

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.

Page 45: 27.  Electromagnetic  Induction

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 in C cw

CCW

S

Page 46: 27.  Electromagnetic  Induction

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 in C ccw

CCW

S

Page 47: 27.  Electromagnetic  Induction

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

I

Page 48: 27.  Electromagnetic  Induction

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

Page 49: 27.  Electromagnetic  Induction

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)