13 4 electromagnetic induction

Honors/AP Physics: An Electronic TextbookChapter 13

ElectromagnetismMaking Electricity Work for You

© 2009 Stan Eisenstein

Lesson 13.4

Electromagnetic Induction

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Lesson 13.4 Table of Contents

We have learned about two aspects of magnetism.

Each aspect has its own equations and rules for

determining direction.

Introduction

v

B F+V

FI

Aspect 1: CurrentCreates B Field

Aspect 2: B Field Exerts Force on Current and Moving Charge

Author Schorschi2; Public Domain Author Stan Eisenstein

http://de.wikipedia.org/wiki/User:Schorschi2

Introduction

Aspect 1 is that current creates a B field. The B

field lines near a long straight wire form circles. The

direction of the those circles can be determined

using the Right Hand Rule:

If you point the thumb of your right hand in the direction of the current, then your fingers will curl in the direction of the magnetic field lines.

Author Schorschi2; Public Domain


Introduction

The size of the B field produced by a long straight

wire is given by the equation:

B = m0I/2pR or B = kI/R

Where ‘I’ represents the size of the current, ‘R’

represents the distance from the wire, and m0 represents a constant equal to 4p x 10-7 Tesla·meters/Ampere.

k = m0/2p = 2.0 x 10-7 T·m/A

Introduction

A coil of wire, called a solenoid, creates a strong

uniform field inside the coil and a weak, near-zero,

field outside the coil.

+-

x x x x xx

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B field

Author Zureks Public Domain

Introduction

Aspect 2: Currents and charges moving through

B fields experience forces. The direction of the force

can be determined using the Right Hand Rule:

Thumb = v, I Fingers = B

Palm = F on I, + Back = F on -

+

VF

I

Author Stan Eisenstein

v

BF

Introduction

Aspect 2: The size of the force is given by these

equations:

F = qvBsinq F = ILBsinq

+

VF

I


v

BF

Introduction

Aspect 2: Charge particles moving through a B field

move in a circular or spiral path (or a straight line, if

v is parallel to B)X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X+ V

+

V

+

VF

F

F

R

+V┴

V║

+

Introduction

Aspect 2: The forces on a current loop in a magnetic

field causes rotation. One important application is an

electric motor.

N S

A

C

View from the Side

x

N SA

B

C

D

View from Above

x●

● ●● ●● ●● ●● ●A B

0.20 m

2.0 A

0.10 m

F

Bwire B

Iwire

Introduction

It is important to be clear about which aspect of

magnetism is being asked about in problems. We

have already seen one problem in which both Aspects 1 and 2 are included. One current

created a B field (Aspect 1) and a second current in that

field experienced a force (Aspect 2).

I

B

F

Author Schorschi2; Public DomainAuthor Stan Eisenstein


Introduction

In this lesson we introduce a 3rd Aspect of magnetism: the generation of a current using magnetism, a process called “Electromagnetic Induction”.

X X X X X X X

X X X X X X X

X X X X X X X

X X X X X X X

Introduction

The process of Electromagnetic Induction is at the

heart of such applications as 1) electric generators,

2) transformers, and 3) microphones.

Author - http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG; Public Domain Author Stan EisensteinAuthor Stan Eisenstein

http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG

Consider a wire that is pushed across a magnetic

field. Consider that there are positive charges in the

wire crossing the field.

Question 1: Use the Right Hand Rule to determine

how those charges are affected.


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+

+ v

+


Answer 1: The charges are forced downward.

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vB

F

F+

+ v

+ Author Stan Eisenstein


The result of these charges being forced downward

is that the bottom of the wire is charged positive and

the top is charged negative.

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F

-

v

+


This charge separation creates an electric potential

difference (i.e. a voltage). We call this potential

difference an induced Electromotive Force (EMF).

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F

-

v

+

E


If the ends of the wire are connected by a conductor,

then a complete circuit is created and a current is

generated.

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F

-

v

+

EI


It doesn’t matter whether the wire loop is moving

across the field, or whether the field is moving across the wire loop, the result is the same –

an induced EMF generating a current.

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F

-

v

+

I E


Once the loop is entirely inside the B field, each of

the wires moving across the field experiences forces

on the charges within them. An equal EMF of same

polarity is generated in each wire. The result is that

no current is generated, though one side of the loop

is positive and the other negative.

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vF

-

+

EF

-

+

-- -

+ + +

I = 0

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In the 1830’s Michael Faraday derived an equation

that described the induced EMF in a wire loop under

various conditions: Moving a loop or coil through a

B field; changing the strength of a B field; rotating a

coil within a B field. Click to animate

Faraday’s Law

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x

A

C

Faraday’s Law states that the induced EMF is equal

to the number of loops in a coil times the rate at

which the “Magnetic Flux” changes.

E = N(DFB/Dt)

Where E is the induced EMF, N is the number of

loops in a coil, FB is the magnetic flux (defined in the

next slides), t is time.

Faraday’s Law

Before we can use Faraday’s Law, we must first

define “Magnetic Flux”. Magnetic Flux can be considered to be a measure of the amount of B field that passes through an area.

Faraday’s Law

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●

x

A B C D

In case A, a wire rectangle lies in the plane of the

screen/page. To get a sense of the meaning of Magnetic Flux, we can count the number of B

field lines passing through the loop. For case A

there are 12 lines passing through the loop.

Faraday’s Law

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●

x

A B C D

One way to change the Magnetic Flux is to change

the area of the loop. In case B, a loop with a smaller

area is in the same B field. This time there are only 6

B field lines passing through the loop. The flux is

reduced compared to case A.

Faraday’s Law

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B

Flux a Area

Another way to change the Magnetic Flux is to change the strength of the B field. In case C, a

loop with the same area is in a weaker B field (as depicted by the field line density). This time

there are only 4 B field lines passing through the

loop. The flux is reduced compared to case A.

Faraday’s Law

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A C

Flux a B Field

A third way to change the Magnetic Flux is to change orientation between the loop and the

B field. In case D, a loop with the same area is in the

same B field. The loop has been rotated, with the

top of the loop moved toward you and the bottom

moved away. There are 6 B

field lines passing through the loop. The flux is reduced compared to case A.

Faraday’s Law

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●

x

D

Flux a Orientation

If we view the loop and field from the side, we see

that the flux is a maximum when the plane of the

loop is perpendicular to the field (A). It is a minimum

(i.e. zero) when the plane of the loop is parallel to

the field (E).

Faraday’s Law

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●

x

DA D E

In order to quantify the orientation between the loop

and the field, we define a “direction” for an Area

“vector” to be perpendicular to the plane of the area.

In case A, the Area Vector points right (or left). In

case E, the Area Vector points down (or up).

Faraday’s Law

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●

x

DA D E

Area Vector

Magnetic Flux is defined by the equation:

FB = BAcosq

where FB represents magnetic flux, B represents the

strength of the field, A represents the area of the

loop, and q represents the angle between the Area

and B field vectors.

Faraday’s Law

FB a A FB a OrientationFB a B

FB = BAcosq

We see that this equation corresponds to the proportionality relationships we explored

when counting B field lines passing through the

loop. As B increases, FB increases.

Faraday’s Law

FB a B

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A C

FB = BAcosq

As A increases, FB increases.

Faraday’s Law

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B

FB a A

FB = BAcosq

As q increases from 0° to 90°, FB decreases from a

maximum to zero.

FB a Orientation

Faraday’s Law

A D E

0°

90°

FB = BAcosq

Question 2: What is the unit for magnetic flux?

Faraday’s Law

FB a A FB a OrientationFB a B●

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FB = BAcosq

Answer 2: Tesla·meter2. This unit is also called a

Weber. In fact, sometimes B field strength is given

the unit Weber/meter2, which is the same unit as a

Tesla.

Faraday’s Law

FB a A FB a OrientationFB a B●

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E = N(DFB/Dt) FB = BAcosq

Faraday’s Law states that induced EMF is directly

proportional to the number of loops in a coil and to

the rate at which the magnetic flux changes. The

faster the flux changes, the greater the EMF.

Faraday’s Law

Electromagnetic Induction (1:27) Turn Sound On; Click on Video Frame to Start

Demonstration


Question 3: Based on Faraday’s Law, what is the unit for EMF?

Faraday’s Law


Answer 3: Tesla·meter2/second. This unit is equivalent to a volt.

B = F/qv Tesla = Newton·second/coulomb·meter

Tesla·meter2/second =

(Newton·second/coulomb·meter)(meter2/second) =

Newton·meter/coulomb = Joule/coulomb = volt

Faraday’s Law


Question 4: For a given coil of wire, what are the

three different ways that you can generate an EMF?

Faraday’s Law

Author - http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG; Public Domain

http://commons.wikimedia.org/wiki/File:Hoover_Dam_turbines.JPG


Answer 4: An EMF is generated by changing FB.

This is done by a) changing B, b) changing A or

c) changing q. Let’s explore each of these scenarios.

Faraday’s Law

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●

x

A B C D


Question 5: A rectangular coil of wire with 50 loops

has a width of 10 cm, a length of 20 cm, and a resistance of 50 W. The coil is placed in a B

field that rises from 2 T to 6 T in 5 seconds. The Area vector of the coil points parallel to the field. Click to animate.

(Question continued on next slide)

Using Faraday’s Law

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10 cm

20

cm2 T6 T


Question 5:

a) What are the initial and final values of the magnetic flux?

b) What is the size of the induced EMF?c) What is the size of the

induced current?


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10 cm

20

cm

2 6 T

5 s

50 loops

50 W


Answer 5: a) 0.04 T·m2 and 0.12 T·m2; b) 0.8 v; c) 0.016 A

FBi = BiAcosq = (2 T)(0.20 m)(0.10 m)(cos 0°) = 0.04 T·m2

FBf = (6 T)(0.20 m)(0.10 m)(cos 0°) = 0.12 T·m2

E = N(DFB/Dt) = (50)(0.12 – 0.04 T·m2)/(5 s) = 0.8 v

I = E/R = (0.8 v)/(50 W) = 0.016 A



Question 6: A single wire loop, 20 cm long and

10 cm wide, is pushed into a 5 T B field at a rate of

2 cm/s. What is the size of the induced EMF in the

loop a) before it fully enters the field? b) after it is

fully in the field but still moving? Click to animate.


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5 T

10 cm

20

cm 2 cm/s


Answer 6: a) 0.02 v; b) 0.0 v

There are two methods for solving part a)

Method 1: First, determine the flux before entering

the field (0 T·m2) and after fully entering the field:

FBf = (5 T)(0.02 m2)(cos 0°)

FBf = 0.10 T·m2


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5 T

10 cm

20

cm 2 cm/s


Answer 6: a) 0.02 v; b) 0.0 v

Method 1: FBf = 0.10 T·m2 FBi = 0.00 T·m2

Next – determine the time it takes the loop to fully

enter the field.

v = d/t t = d/v

t = (10 cm)/(2 cm/s) = 5 s


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5 T

10 cm

20

cm 2 cm/s


Answer 6: a) 0.02 v; b) 0.0 v

Method 1: FBf = 0.10 T·m2 FBi = 0.00 T·m2 t = 5 s

Finally, use Faraday’s Law to determine the EMF.

E = N(DFB/Dt)

E = (1)(0.10 – 0.00 T·m2)/(5 s)

E = 0.02 v


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5 T

10 cm

20

cm 2 cm/s


Answer 6: a) 0.02 v; b) 0.0 v

For Method 2 we will derive a generic equation for

the EMF of a wire moving through a B field. ‘L’ represents the length of the wire crossing the field and ‘W’ represents the width of the loop already in the field.


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B

W

L v


Answer 6: a) 0.02 v; b) 0.0 v

Method 2: The first step is to determine an expression for FB at a given moment.

FB = BAcosq = BLw


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B

W

L v


Answer 6: a) 0.02 v; b) 0.0 v

Method 2: FB = BLw The next step is use Faraday’s Law to

determine the EMF.

E = N(DFB/Dt) = (1)( D (BLw) /Dt)

B and L are constant.

E = BL( D w/Dt) = BLv


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B

W

L v


Answer 6: a) 0.02 v; b) 0.0 v

Method 2: E = BLv

This equation holds for wires moving across B fields.

E = (5 T)(0.20 m)(0.02 m/s)

E = 0.02 v


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5 T

0.2

0 m 0.02 m/s


Answer 6: a) 0.02 v; b) 0.0 v

b) Once the loop is entirely within the field, FB is constant; it is no longer changing. This makes the induced EMF equal to zero.

E = (1)(0.10 – 0.10 T·m2)/(5 s) = 0 volts


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Answer 6: a) 0.02 v; b) 0.0 v

b) In actuality, each wire crossing the field has an EMF induced equal to BLv. But since these EMFs oppose each other, the net EMF is zero. Charge will separate to opposite ends of the loop, but no current will flow.


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B

L v

-

+

--

+ +

We have already learned one method of determining

the direction of induced current – the Right Hand

Rule. As a wire crosses a B field, we consider the

force on the + charges in the wire. This force is the

direction of the current in the wire.

Direction of Induced Current

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F v+

EI ++

vB

FAuthor Stan Eisenstein

Question 7: As the wire loop enters the field shown

here, the current is induced in a clockwise direction.

Use the Right Hand Rule to determine the direction

of the current as the loop leaves the field.


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● ● ● ● ● ● ●

● ● ● ● ● ● ●

F v+

EI ++

v

Answer 7: Counterclockwise. The force on the left-

hand wire is down (just like the right-hand wire as

the loop enters the field). This downward force

causes a CCW current.


● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

FI

vvB

FAuthor Stan Eisenstein

Unfortunately, the Right Hand Rule is not easy to

apply in all induction situations. For instance, if the

wire loop does not move, but the B field changes

size, then there is no velocity and the Right Hand

Rule becomes difficult or impossible to use.


● ● ● ●

● ● ● ●

● ● ● ●

● ● ● ●

10 cm

20

cm

2 6 T

5 s

50 loops

50 W

?B

?Author Stan Eisenstein

A second rule that may be used to determine direction of induced current is the “Left Hand

Rule”. This rule was made up for this text and is not generally recognized. However, this rule will accurately determine the direction of induced

current for all situations.


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● ● ● ●

● ● ● ●

● ● ● ●

10 cm

20

cm

2 6 T

5 s

50 loops

50 W

Author Schorschi2 at de.wikipedia; Public Domain; Adapted


http://de.wikipedia.org/

Left Hand Rule:

If Flux is increasing, point Thumb in direction of B

If Flux is decreasing, point Thumb opposite B Fingers curl in the direction of the current.


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● ● ● ●

● ● ● ●

10 cm

20

cm

2 6 T

5 s

50 loops

50 W




If FB ↑, Thumb = B; If FB ↓, Thumb opp. B; Fingers = I.

Question 8a: Use the Left Hand Rule to determine

the direction of the current in the wire loop shown

below as it enters and leaves the magnetic field.


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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

v v


Answer 8a: Clockwise; Counterclockwise

Entering: As the area ↑, FB ↑; Point Thumb in direction of B – toward you. Fingers curl CW.


● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI


B

I


Answer 8a: Leaving: As the area ↓, FB ↓; Point Thumb opp direction of B – away. Fingers curl

CCW.


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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI

B

I



Answer 8a: These directions match the directions

determined by the Right Hand Rule.


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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI

vI


Question 8b: For the loop shown below, the B field

changes from 5 T away to 5 T toward you. Determine the direction of the current while

the field away from you decreases and then again

while the field toward you increases.


X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X


Question 8b: Clockwise for both. While the field and

flux are decreasing away from you, the thumb points

opp. B (toward you). While the field and flux are increasing toward you, the thumb points with the

field, again toward you.


X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

Opp.B B


I


I


Question 8c: A coil of wire is rotating in a clockwise

direction in a field pointing to the right. Determine

the direction of the current in wires A and C for

positions 1 and 2 of the coil in its rotation.


A A

C C

1 2


Answer 8c: The currents are shown in the diagram.

At position 1, the flux is increasing. The Thumb

points Right and the fingers curl away at A and

toward you at C. At position 2, the flux is decreasing.

The Thumb points left and the fingers curl toward

you at A and away at C.


A A

C C

1 2

x

x ●

●

1 2A A

CC





Question 8d: A wire with an increasing current lies

next to a rectangular loop of wire. What is the direction of the induced current in the loop?



Answer 8d: CW.

There are two parts to this problem: Aspect 1 – a

current creates a B field; Aspect 3 – a changing flux induces a current.


Author Schorschi2 at de.wikipedia; Public Domain; Adapted Author Stan Eisenstein

I




Answer 8d: CW.

Using the Right Hand Rule, we find that the wire with the current creates a B field toward you at the position of the loop.


●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●Author Schorschi2 at de.wikipedia; Public Domain; Adapted




Answer 8d: CW.

Since the current is increasing, the strength of the

field and the flux are increasing. The Left Hand Rule shows that the induced current must be CW.


BI

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●Author Stan Eisenstein

The more traditional method of determining the

direction of induced current is called “Lenz’s Law”.

Lenz’s Law states that a current is induced in a

direction that resists the change in flux.

Lenz’s Law

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vI

vI

Let’s explore Lenz’s Law in some of the examples

we have already discussed.

For a loop being pushed into a B field, the FB is clearly increasing. Lenz’s Law states that the induced current should resist this increase in

FB. How might this occur?

Lenz’s Law

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● ● ● ● ● ● ●

vI A

We know from the Right and Left Hand Rules that a

CW current is induced in the loop. This is according

to Aspect 3 – Electromagnetic Induction.

Now consider Wire A. It conducts a current across a

B field. Consider Aspect 2 of Magnetism – a B field

exerts a force on a Current. The B field will exert a

force on this induced current.

Lenz’s Law

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vI A

Question 9a: Use the Right Hand Rule to determine

the direction of the force of the B field on the current

in Wire A.

Lenz’s Law

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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI A

Answer 9a: Left.

The fingers point toward you, in the direction of the

B field. The Thumb points down, in the direction of

the current. The palm points Left, in the direction of

the Force on the current.

Lenz’s Law

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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI FB

BF

IAuthor Stan Eisenstein

Answer 9a: Left.

Notice that as the loop tries to enter the field and

increase the flux, the field tries to push the loop back

out, resisting the increase.

Lenz’s Law

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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI FB

BF


Question 9b: Now consider as the loop leaves the

field. We have already determined that the induced

current flows CCW. Use the Right Hand Rule to determine the direction of the force that the B

field exerts on the current in wire C.

Lenz’s Law

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● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

v

I

C

Answer 9b: Left. The force points in the same direction that it did when the loop was

entering the field. Now the flux is reducing as the loop

leaves the field. The B field exerts a force that resists the reduction of flux – it tries to pull the loop back

into the field.

Lenz’s Law

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

v

I

CF

BF


Lenz’s Law relates to the conservation of energy.

The force from the field must oppose the velocity. If

the force were in the same direction as the velocity,

the speed would keep getting faster, creating Kinetic

Energy out of nothing.

Lenz’s Law

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

● ● ● ● ● ● ●

vI

vIFF

Consider now a loop in a B field that points away

that is decreasing in strength. We have already

determined that a clockwise current is induced.

Since the flux is reducing, the induced current must

somehow try it resist the change. The new current

creates its own B field within the loop (Aspect 1).

Lenz’s Law

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

B decreasing

Question 9c: Use the Right Hand Rule to determine

the direction of the B field created by the induced

current.

Lenz’s Law

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

B decreasing

Answer 9c: The induced current generates a field

away from you inside the loop.

As the decreasing B field reduces the flux, the induced current generates its own field to

reinforce the original field, resisting the decrease in flux.

Lenz’s Law

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

X X X X X X X X X X

B decreasing

x x x x

x x xx

Author Schorschi2 at de.wikipedia; Public Domain



Question 10: A conducting bar moves frictionlessly

on conducting rails in a 5 T field point away. A 10 W

resistor connects the rails at the left end (the rest of

the circuit has minimal resistance). The bar is pushed so that it moves at a constant speed

of 5 m/s. (continued on next slide)

Practice Problem

x x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x xx x x x x x x x x x x x x x x

10 W

20 cm

5 m/s5 T

1 m 10 m

Question 10: The rails are 20 cm apart. The bar

starts 1 meter from the left end of the rails and

moves 10 meters. a) What is the direction of the induced current

in the moving bar?(continued on next slide)

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Question 10: b) What is the size of the EMF induced in the

bar?c) What is the size of the current?d) What is the force needed to move the bar

at a constant speed?(continued on next slide)

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Question 10: e) How much work is done moving the bar 10

m?f) How much thermal energy is generated in

the resistor in the time it takes to move the bar

10 m?

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: a) The current in the bar is up. LH Rule: Thumb = Away (Flux increases)

Fingers = I = CCW

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: b) 5 volts. Method 1: E = BLv = (5 T)(0.20 m)(5 m/s) = 5

v

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

b) 5 volts. Method 2: FBi = BA = (5 T)(0.20 m)(1 m) = 1

T·m2

FBf = BA = (5 T)(0.20 m)(11 m) = 11 T·m2

t = d/v = (10 m)/(5 m/s) = 2 sE = /DF Dt = (11 – 1 T·m2)/2 s = 5 v

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: c) 0.5 amperes.

I = E/R = (5 v)/(10 W) = 0.5 A

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: d) 0.5 newtons.

F = ILB = (0.5 A)(0.20 m)(5 T) = 0.5 n

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: e) 5 joules.

W = Fd = (0.5 n)(10 m) = 5 joules

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

Answer 10: f) 5 joules.

E = Pt = IVt = (0.5 A)(5 v)(2 s) = 5 joules

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

W = Fd = 5 Joules E = IVt = 5 joules

Note that the work done in moving the bar is converted to electric energy and ultimately

thermal energy in the resistor. Energy is conserved.

Practice Problem


10 W

20 cm

5 m/s5 T

1 m 10 m

13 4 electromagnetic induction

Technology

x x x x x x b field

x x x xf x

animatea x c

constant equal to4 x

current aspect

current loop

b fieldon current

author stan eisenstein