1P22-

Workshop: Using Visualization in Teaching Introductory E&M

AAPT National Summer Meeting, Edmonton, Alberta, Canada.

Organizers: John Belcher, Peter Dourmashkin, Carolann Koleci, Sahana Murthy

2P22-

Faraday’s Law Presentation Materials

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MIT Class: Faraday’s Law

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Faraday’s Law

Fourth (Final) Maxwell’s Equation

Underpinning of Much Technology

5P22-

Demonstration:Falling Magnet

P22-

Magnet Falling Through a Ring

Falling magnet slows as it approaches a copper ring which has been immersed in liquid nitrogen.

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Demonstration: Jumping Rings

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Jumping Ring

An aluminum ring jumps into the air when the solenoid beneath it is energized

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What is Going On?

It looks as though the conducting loops have current in them (they behave like magnetic dipoles) even though they aren’t hooked up

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Faraday’s Law Applets Discovery

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Faraday’s Law Applets Discovery Activity

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Demonstration: Induction

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Electromagnetic Induction

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Faraday’s Law of Induction

Bd

dt

A changing magnetic flux induces an EMF

P22-

What is EMF?

d E s

Looks like potential. It’s a “driving force” for current

P22-

Faraday’s Law of Induction

Bdd

dt

E s

A changing magnetic flux induces

an EMF, a curling E field

P22-

Magnetic Flux Thru Wire Loop

c o sB B A BA B A

B

S

d Φ B A

Analogous to Electric Flux (Gauss’ Law)

(1) Uniform B

(2) Non-Uniform B

P22-

Minus Sign? Lenz’s Law

Induced EMF is in direction that opposes the change in flux that caused it

P22-19

Faraday’s Law of Induction

Bd

dt

Changing magnetic flux induces an EMF

Lenz: Induction opposes change

P22-

Ways to Induce EMF

• Quantities which can vary with time:

• Magnitude of B• Area A enclosed by the loop• Angle between B and loop normal

cosdBA

dt

P22-

Group Discussion:Magnet Falling Through a Ring

Falling magnet slows as it approaches a copper ring which has been immersed in liquid nitrogen.

P22-22

Magnet Falling Through a Ring

Falling magnet slows as it approaches a copper ring which has been immersed in liquid nitrogen.

P22-23

Example: Magnitude of B Magnet Falling Through a Ring

Falling magnet approaches a copper ringor Copper Ring approaches Magnet

P22-24

Moving Towards Dipole

As ring approaches, what happens to flux?Flux up increases

Move ring

down

P22-25

PRS Question:Faraday’s Law

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PRS: Faraday’s Law: Loop

A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil:

0%

0%

0%

0%

:00

1. Current clockwise; force up

2. Current counterclockwise; force up

3. Current clockwise; force down

4. Current counterclockwise; force down

P22-27

PRS Answer: Faraday’s Law: Loop

The I dl x B force on the coil is a force which is trying to keep the flux through the coil from increasing by slowing it down (Lenz’s Law again).

Answer: 3. Current is clockwise; force is down

The clockwise current creates a self-field downward, trying to offset the increase of magnetic flux through the coil as it moves upward into stronger fields (Lenz’s Law).

P22-28

PRS Question:Loop in Uniform Field

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PRS: Loop in Uniform Field

A rectangular wire loop is pulled thru a uniform B field penetrating its top half, as shown. The induced current and the force and torque on the loop are:

v

Bout

0%

0%

0%

0%

0% 1. Current CW, Force Left, No Torque

2. Current CW, No Force, Torque Rotates CCW

3. Current CCW, Force Left, No Torque

4. Current CCW, No Force, Torque Rotates CCW

5. No current, force or torque

0

P22-30

PRS Answer: Loop in Uniform Field

The motion does not change the magnetic flux, so Faraday’s Law says there is no induced EMF, or current, or force, or torque.Of course, if we were pulling at all up or down there would be a force to oppose that motion.

Answer: 5. No current, force or torque

v

Bout

P22-

Group Problem: Changing AreaConducting rod pulled along two conducting rails in a uniform magnetic field B at constant velocity v

1. Direction of induced current?

2. Direction of resultant force?

3. Magnitude of EMF?

4. Magnitude of current?

5. Power externally supplied to move at constant v?

P22-

Changing Angle

B BA B A

0B B A

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The last of the Maxwell’s Equations (Kind of)

34P22-

Maxwell’s Equations

0

0

(Gauss's Law)

(Faraday's Law)

0 (Magnetic Gauss's Law)

(Ampere's Law)

in

S

B

C

S

enc

C

Qd

dd

dt

d

d I

Creating Electric Fields

E A

E s

Creating Magnetic FieldsB A

B s

P22-35

Experiment 5: Faraday’s Law of Induction

P22-36

Part 1: Current & Flux

Current?

Flux?

I > 0BLACK

RED

0

( ) ' 't

t R I t dt

P22-37

PRS Predictions:Flux & Current

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PRS: Flux Measurement

Moving from above to below and back, you will measure a flux of:

t

t

t

t

(A)

(C)

(

B)

(D)

0% 0% 0% 0%0%0%0%0%

1. A then A 5. B then B

2. C then C 6. D then D

3. A then C 7. B then D

4. C then A 8. D then B5. 5

6. 6

7. 7

8. 8

0

P22-39

PRS Answer: Flux Measurement

Answer: 6. D then D

The direction of motion doesn’t matter – the field and hence flux is always upwards (positive) and it increases then decreases when moving towards and away from the magnet respectively.

t

(D)

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PRS: Current Measurement

Moving from above to below and back, you will measure a current of:

t

t

t

t

(A)

(C)

(

B)

(D)

NOTE: CCW is positive!

0% 0% 0% 0%0%0%0%0%

1. A then A 5. B then B

2. C then C 6. D then D

3. A then C 7. B then D

4. C then A 8. D then B5. 56. 67. 78. 8

0

P22-41

PRS Answer: Current Measurement

Answer: 2. C then C

The direction of motion doesn’t matter – the upward flux increases then decreases so the induced current will be clockwise to make a downward flux then counterclockwise to make an upward one.

t

(C)NOTE: CCW is positive!

P22-42

PRS: Flux Behavior

Moving from below to above, you would measure a flux best represented by which plot above (taking upward flux as positive)?

t

t

t

t

(1)

(3)

(

2)

(4)

NOTE: Magnet “Upside Down”

0% 0%0%0%

1. 1

2. 2

3. 3

4. 4

:0

P22-43

t

PRS Answer: Flux Behavior

Answer: 2.

The field is downward so the flux is negative. It will increase then decrease as you move over the magnet.

(2)

P22-44

PRS: Current Behavior

Moving from above to below, you would measure a current best represented by which plot above (taking counterclockwise current as positive)?

t

t

t

t

(1)

(3)

(

2)

(4)

NOTE: Magnet “Upside Down”

0% 0%0%0%

1. 1

2. 2

3. 3

4. 4

0

P22-45

t

PRS Answer: Current Behavior

Answer: 1.

The field is downward so the current will first oppose it (CCW to make an upward flux) then try to reinforce it (CW to make a downward flux)

(1)

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PRS Confirming Predictions?Flux & Current

P22-47

Part 2: Force Direction

Force when

Move Down?

Move Up?

Test with aluminum

sleeve

P22-48

PRS Question:Wrap-Up

Faraday’s Law

P22-49

PRS: Circuit

A circuit in the form of a rectangular piece of wire is pulled away from a long wire carrying current I in the direction shown in the sketch. The induced current in the rectangular circuit is

0%

0%

0% 1. Clockwise2. Counterclockwise 3. Neither, the current is zero

0

P22-50

PRS Answer: Circuit

•B due to I is into page; the flux through the circuit due to that field decreases as the circuit moves away. So the induced current is clockwise (to make a B into the page)

Answer: 1. Induced current is clockwise

Note: Iind dl x B force is left on the left segment and right on the right, but the force on the left is bigger. So the net force on the rectangular circuit is to the left, again trying to keep the flux from decreasing by slowing the circuit’s motion

P22-51

Faraday’s LawProblem Solving Session

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Technology

Many Applications of Faraday’s Law

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Metal Detector

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Induction Stovetops

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Ground Fault Interrupters (GFI)

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Electric Guitar

Pickups

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Electric Guitar

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Demonstration:Electric Guitar

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PRS Question:Generator

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PRS: Generator

A square coil rotates in a magnetic field directed to the right. At the time shown, the current in the square, when looking down from the top of the square loop, will be

0%

0%

0%

0% 1. Clockwise2. Counterclockwise 3. Neither, the current is zero4. I don’t know

:00

P22-61

PRS Answer: Generator

•Flux through loop decreases as normal rotates away from B. To try to keep flux from decreasing, induced current will be CCW, trying to keep the magnetic flux from decreasing (Lenz’s Law)

Answer: 1. Induced current is counterclockwise

Note: Iind dl x B force on the sides of the square loop will be such as to produce a torque that tries to stop it from rotating (Lenz’s Law).

P22-62

Group Problem: GeneratorSquare loop (side L) spins with angular frequency in a field of strength B. It is hooked to a load R.1) Write an expression for current I(t) assuming the

loop is vertical at time t = 0.2) How much work from generator per revolution? 3) To make it twice as hard to turn, what do you

do to R?

P22-63

Demonstration:Levitating Magnet

P22-64

Brakes

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Magnet Falling Through a Ring

What happened to kinetic energy of magnet?

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Demonstration:Eddy Current Braking

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Eddy Current Braking

What happened to kinetic energy of disk?(link to movie)

P22-68

Eddy Current BrakingThe magnet induces currents in the metal that

dissipate the energy through Joule heating:

XXXX

1. Current is induced counter-clockwise (out from center)

2. Force is opposing motion (creates slowing torque)

P22-69

Eddy Current BrakingThe magnet induces currents in the metal that

dissipate the energy through Joule heating:

XXXX

1. Current is induced clockwise (out from center)

2. Force is opposing motion (creates slowing torque)

3. EMF proportional to 4. . 2

RF

P22-70

Faraday’s Law of Induction

Bd

dt

Changing magnetic flux induces an EMF

Lenz: Induction opposes change

P22-71

Today:Using Inductance

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First:Mutual Inductance

P22-73

Demonstration: Remote Speaker

P22-74

Mutual Inductance

12 12 2M I

212 12

dI

dtM

12 21M M M

Current I2 in coil 2, induces magnetic flux 12 in coil 1. “Mutual inductance” M12:

Change current in coil 2?Induce EMF in coil 1:

P22-75

TransformerStep-up transformer

;p p s s

d dN N

dt dt

s s

p p

N

N

Ns > Np: step-up transformerNs < Np: step-down transformer

Flux through each turn same:

P22-76

Demonstrations:

One Turn Secondary:Nail

Many Turn Secondary:Jacob’s Ladder

P22-77

Transmission of Electric Power

Power loss can be greatly reduced if transmitted at high voltage

P22-78

Example: Transmission lines An average of 120 kW of electric power is sent from

a power plant. The transmission lines have a total resistance of 0.40 . Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V.

(a) 5

2

1.2 10500

2.4 10

P WI AV V

2 2(5.0 ) (0.40 ) 10LP I R A W

(b)

83% loss!!

0.0083% loss

2 2(500 ) (0.40 ) 100LP I R A kW

5

4

1.2 105.0

2.4 10

P WI AV V

P22-79

Group Discussion: Transmission lines

We just calculated that I2R is smaller

for bigger voltages.

What about V2/R? Isn’t that bigger?

Why doesn’t that matter?

P22-80

Self Inductance

P22-81

Self Inductance

11 11 1 M I LI

dILdt

What if we forget about coil 2 and ask about putting current into coil 1?There is “self flux”:

Faraday’s Law

P22-82

Calculating Self Inductance

Total,selfLI

1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)

V s1 H = 1

A

Unit: Henry

P22-83

Group Problem: Solenoid

Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R)

REMEMBER1. Assume a current I is flowing in your device2. Calculate the B field due to that I3. Calculate the flux due to that B field4. Calculate the self inductance (divide out I)

Self, totalL I

P22-84

Group Problem: Torus

Calculate the inductance of the above torus (square cross-section of length a, radius R, N total turns)

1) For assumed current I, what is B(r)?

2) Calculate flux, divide out I

P22-85

Review: Inductor Behavior

I

Inductor with constant current does nothing

dILdt

L

P22-86

I

Back EMF

0, 0L

dI

dt 0, 0L

dI

dt

I

dILdt

P22-87

Demos: Breaking Circuits

Big InductorMarconi Coil

The Question: What happens if big I, small t

P22-88

Internal Combustion Engine

P22-89

Ignition Overview

P22-90

The Workhorse: The Coil

Primary Coil:

~200 turns heavy Cu

DC (12 V) in to GND

Secondary Coil:

~20,000 turns fine Cu

Usually no voltage…

When primary breaks

up to ~45,000 V

P22-91

Energy in Inductors

P22-92

Inductor Behavior

I

Inductor with constant current does nothing

dILdt

L

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1. Start with “uncharged” inductor

2. Gradually increase current. Must work:

3. Integrate up to find total work done:

Energy To “Charge” Inductor

dIdW Pdt I dt L I dt LI dI

dt

212

0

I

I

W dW LI dI L I

P22-94

Energy Stored in Inductor

212LU L I

But where is energy stored?

P22-95

Example: Solenoid

2 2 2 21 12 2B oU LI n R l I

Ideal solenoid, length l, radius R, n turns/length, current I:

0B nI 2 2oL n R l

22

2Bo

BU R l

Energy

Density

Volume

P22-96

Energy Density

: Magnetic Energy Density

2

2o

E

Eu

: Electric Energy Density

Energy is stored in the magnetic field!

2

2Bo

Bu

P22-97

Group Problem: Coaxial Cable

1. How much energy is stored per unit length? 2. What is inductance per unit length?

HINTS: This does require an integralThe EASIEST way to do (2) is to use (1)

Inner wire: r = a

Outer wire: r = bXI I

P22-98

PRS Questions:Inductor in a Circuit

Stopping a Motor

P22-99

PRS: Stopping a Motor

Consider a motor (a loop of wire rotating in a B field) which is driven at a constant rate by a battery through a resistor.Now grab the motor and prevent it from rotating. What happens to the current in the circuit?

0%

0%

0%

0% 1. Increases2. Decreases3. Remains the Same4. I don’t know :20

P22-100

PRS Answer: Stopping a Motor

Answer: 1. Increases

When the motor is rotating in a magnetic field an EMF is generated which opposes the motion, that is, it reduces the current. When the motor is stopped that back EMF disappears and the full voltage of the battery is now dropped across the resistor – the current increases. For some motors this increase is very significant, and a stalled motor can lead to huge currents that burn out the windings (e.g. your blender).

P22-101

Think Harder about Faraday

P22-102

PRS Question:Faraday in Circuit

P22-103

PRS: Faraday CircuitA magnetic field B penetrates this circuit outwards, and is increasing at a rate such that a current of 1 A is induced in the circuit (which direction?).

R=100

R=10A

B

The potential difference VA-VB is:

0%

0%

0%

0%

0%

0%

0%

0%

0% 1. +10 V

2. -10 V

3. +100 V

4. -100 V

5. +110 V

6. -110 V

7. +90 V

8. -90 V

9. None of the above

0

P22-104

PRS Answer: Faraday CircuitAnswer: 9. None of the above

The question is meaningless. There is no such thing as potential difference when a changing magnetic flux is present.

R=100

R=10A

B

By Faraday’s law, a non-conservative E is induced (that is, its integral around a closed loop is non-zero). Non-conservative fields can’t have potentials associated with them.

P22-105

Non-Conservative Fields

R=100R=10

Bdd

d t

E s

E is no longer a conservative field – Potential now meaningless

I=1A

P22-106

Kirchhoff’s Modified 2nd Rule

ii

V d E s

0Bi

i

dV

d t

If all inductance is ‘localized’ in inductors then our problems go away – we just have:

0ii

d IV L

d t

Bd

d t

P22-107

Inductors in CircuitsInductor: Circuit element with self-inductance

Ideally it has zero resistance

Symbol:

P22-108

• BUT, EMF generated by an inductor is not a voltage drop across the inductor!

Ideal Inductor

d ILd t

i n d u c t o rV d E s

Because resistance is 0, E must be 0!

0

P22-109

Circuits:Applying Modified Kirchhoff’s(Really Just Faraday’s Law)

P22-110

LR Circuit

0ii

dIV L

dtIR

P22-111

LR Circuit

10

dI dIL Idt dt L R R

IR

P22-112

Need Some Math:Exponential Decay

P22-113

Exponential Decay1dAA

dt Consider function A where:

A decays exponentially:

0tA A e

P22-114

0 1 2 3 4 5 60.0A

f

0.5Af

1.0Af

A

Time t

Exponential Behavior

1f

dAA A

dt Slightly modify diff. eq.:

A “decays” to Af:

1 tfA A e

P22-115

This is one of two differential equations we expect you to

know how to solve (know the answer to).

The other is simple harmonic motion (more on that next week)

P22-116

LR Circuit

Solution to this equation when switch is closed at t = 0:

1dII

dt L R R

: time constantL

R

/( ) 1 tI t eR

(units: seconds)

P22-117

LR Circuit

t=0+: Current is trying to change. Inductor works as hard as it needs to to stop it

t=∞: Current is steady. Inductor does nothing.

P22-118

PRS Question:Voltage Across Inductor

P22-

PRS: Voltage Across Inductor

119

0%

0%

0%

0%

In the circuit at right the switch is closed at t = 0. A voltmeter hooked across the inductor will read:

/tLV e

/(1 )tLV e

0LV

1.

2.

3.

4. I don’t know 0

P22-120

PRS Answer: V Across Inductor

The inductor “works hard” at first, preventing current flow, then “relaxes” as the current becomes constant in time.

Answer: 1. tLV e

Although “voltage differences” between two points isn’t completely meaningful now, we certainly can hook a voltmeter across an inductor and measure the EMF it generates.

P22-121

LR Circuit

t=0+: Current is trying to change. Inductor works as hard as it needs to to stop it

t=∞: Current is steady. Inductor does nothing.

Readings on VoltmeterInductor (a to b)Resistor (c to a)

c

P22-122

Group Problem: Circuits

For the above circuit sketch the currents through the two bottom branches as a function of time (switch closes at t = 0, opens at t = T). State values at t = 0+, T-, T+