Problem 29.53

A flexible circular loop 6.50 cm in

diameter lies in a magnetic field with

magnitude 0.950 T, directed into the

plane of the page, as shown. The loop

is pulled at the points indicated by the

arrows, forming a loop of zero area in

0.250 s.

(a) Find the average induced emf in the

circuit.

(b) What is the direction of the current in

Fig. 29.46

PHYS 153 08W 1PHYS 153 08W 11PHYS 153 08W

(b) What is the direction of the current in

R, from a to b, or from b to a? Explain your

reasoning.

A dc generator and back emf in a motor

Fig. 29.10

PHYS 153 08W 2PHYS 153 08W 2PHYS 153 08W 2

In this example, steps are taken to make the emf which is generated always

positive. Due to the commutator, the brushes switch sides when the induced

emf changes direction, and hence the current produced will always be in the

same directions. The ‘bumpiness’ can be smoothed out by using a large

number of coils and commutator segments.

When using the slip rings (ac alternator), the emf had both positive and negative

values, and . With the commutator, the emf is never negative, and the

average value is positive.

0=aveε

This motor’s back emf is just the emf induced by the changing magnetic flux

through the rotating coil. For N coils in the loop,

aveave tBAN |sin| ωωε =

ωπ

ωω ∫

=

2/

0sin

|sin|

T

ave

tdtt where ω

π=2

T

πω

π

πω

ωπ

ωω

ωπ

ω ωπω

π

2]0cos)[cos1(|cos)1(sin0

0 =−−

=−

=∫ ttdt

PHYS 153 08W 3PHYS 153 08W 3PHYS 153 08W 3

ωωω

πω

εBAN

ave

2=∴

Slidewire generator (motional emf)

Fig. 29.11

This is a simple version of a dc generator,

and although it is not very useful practically,

it is useful pedagogically.

The cause of the induced emf is the increase in area.

Choose to point into the page, parallel to . →

A→

B

BABAABB ==•=Φ→→

φcos

BLvdt

LvdtB

dt

dAB

dt

d B −=−=−=Φ

−=∴ε

Note, to oppose the change that is causing the induced emf, the current

must flow counterclockwise. (Check this with the R.H. rule).

If v is constant, and the rod stays on the rails, the emf generated will be constant.

PHYS 153 08W 4PHYS 153 08W 4PHYS 153 08W 4

If v is constant, and the rod stays on the rails, the emf generated will be constant.

This is a d.c. generator.

But we now have a current in a conductor

which is moving in a magnetic field. Hence

There will be a force on it pointing opposite

to . Thus, work must be done to move

the rod to the right.

Fig. 29.12

v

For a resistance in the circuit, energy

will be dissipated from the losses.

RRI 2

How does the work done compare to the energy losses?

To keep the rod moving, a force , equal to the resisting force, must be applied in

the direction of .

→

Fv

The force is given byR

vLBLB

R

BLvLB

RILBF

22||====

ε

Hence, the rate at which work is done is R

vLBFv

222

=

The energy dissipated is R

vLBR

R

BLvRI

22222 )( ==

PHYS 153 08W 5PHYS 153 08W 5PHYS 153 08W 5

RR

The work done is equal to the energy dissipated.

The situation of the sliding rod in the slidewire generator can be analyzed in a

different way. As the rod in Fig. 29.11 moves to the right, charged particles

in it experience a force

→→→

= BxvqF which creates an excess of positive charge at the top of the rod.

(Assume that the rails on which the rod slides are insulators for the moment).

An electric field will then be created which points from top to bottom on the rod.

In equilibrium, the downward electrostatic force will equal the upward magnetic

force.

qvBqE =The difference in potential between top and bottom,

vBLELVVV bottomtop ==−=∆

The moving rod has become like a battery, with a potential difference between

the ends, and hence a source of emf. When the rest of the circuit (here, just a

PHYS 153 08W 6PHYS 153 08W 6

the ends, and hence a source of emf. When the rest of the circuit (here, just a

rectangle) is a conductor, this emf drives current around it, in the counter-

clockwise direction shown.

vBLV ==∆ ε (motional emf)

The emf (magnitude and direction) has been determined now in two

ways, Faraday’s law / Lenz’s law, and by analysis of the motion of the rod

(motional emf). Both are the same.

ε

We can generalize this.

Suppose the circuit is not a simple rectangle, and that the magnetic field is not

perpendicular to the plane of the area, and not uniform.

From above,

→→→

= BxvE

→→→→→

•=•= ldBxvldEd )(ε→→→

•=∴ ∫ ldBxv )(ε (motional emf)

PHYS 153 08W 7PHYS 153 08W 7

∫This is equivalent to the original statement of Faraday’s law,

dt

d BΦ−=ε →

B

I

Consider again a conducting loop which

encloses a magnetic field which can vary

with time.

If the strength of the field shown is increasing in time, then an emf will be

induced which drives a current in the direction shown.

What makes the charges move so that a current flows?

The conductor is not moving in a magnetic field, so there are no

magnetic forces. The conductor may not even be in a magnetic field

(but just encloses it).

What is the only thing that makes electric charges move? It is an electric

field. So the changing magnetic field must induce an electric field. →

B→

E→

EElectric fields are not only produced by charges,

but by time varying magnetic fields.

PHYS 153 08W 8PHYS 153 08W 8

→

E

Ebut by time varying magnetic fields.

dt

dldE BΦ−==•→→

∫ ε

Integrating around the loop, we get

If the loop is a conductor, and a charge q moves once

around the loop, then there is work done.

εqldEq =•→→

∫

This electric field is very different from the electrostatic field that we first

studied.

Recall, for an electrostatic field, 0=•→→

∫ ldE

Such a field is conservative, and for conservative fields we could define

a potential.

Our new electric field, produced by a changing magnetic field, is

nonconservative, and is a nonelectrostatic field.

However, the fundamental force law, , applies to both

fields., conservative and nonconservative.

→→

= EqF

PHYS 153 08W 9PHYS 153 08W 9

fields., conservative and nonconservative.

Consider this. Suppose the loop in the previous slide is nonconducting. In

fact, suppose the loop is just imaginary. Is an electric field still produced?

Applications of magnetically induced electric fields.

•Power generating stations

•Alternators in cars

•Computer disk drives

•Magnetic tapes

Problem 29.7

The current in the long straight wire AB

is upward and is increasing steadily at a

rate

(a) At an instant when the current is ,

what are the magnitude and direction of

the field at a distance r to the right of

the wire?

(b) What is the flux through the narrow

shaded strip?

idt

di

→

B

BdΦ

PHYS 153 08W 10PHYS 153 08W 10

Fig. 29.27

(c) What is the total flux through the loop?

(d) What is the induced emf in the loop?

What is the direction?

(e) If a=12.0 cm, b=36.0 cm, L=24.0 cm,

and = 9.60 A/s, calculate the value of

the induced emf. dt

di

Problem 29.67

The magnetic field , at all points within

a circular region of radius R, is uniform in

space and directed into the plane of the page.

(the region could be a cross section inside the

windings of a long, straight solenoid). If the

magnetic field is increasing at a rate ,

what are the magnitude and direction

of the force on a stationary positive point charge

q located at points a, b, and c?

→

B

dtBd→

Fig. 29.54

PHYS 153 08W 11

Fig. 29.54

Eddy Currents

v1

v2

F

x x x x x

x x x x x

inB

Eddy currents are currents set up in a

piece of bulk metal (as opposed to a clearly

defined circuit, for example), by a changing

magnetic flux.

In the figure shown, a piece of metal

sheet swings into and out of a magnetic

field (it could be between the poles of

an electromagnet). As the sheet enters

or leaves the field, induced emfs are

PHYS 153 08W 12

Fx x x x x

A metal piece swings in a

magnetic field. The metal

piece is slowed considerably

by the magnetic force on the

induced eddy currents as the

metal piece enters of leaves the

field.

or leaves the field, induced emfs are

generated to oppose the change. These

set up currents as shown by the big

arrows.

If the moving piece of metal is big, then

the section of metal which enters the field

first has an induced current, and hence a

braking force applied to it. The return current

is generally outside the field and that section

of metal experiences no force.

If the piece of metal has slots cut into it, the

large eddy current loops are broken up, and

the braking action greatly reduced.

Eddy currents can have undesirable effects.

Eddy currents in cores of transformers (which

we will study later) waste energy through

RI 2 heating losses.

Advantages:

• damp unwanted oscillations, for

PHYS 153 08W 13

If the metal sheet has of slots cut

into it, the eddy currents are

confined to narrow slots and

are greatly reduced. The braking

on the ‘pendulum’ is greatly

reduced.

• damp unwanted oscillations, for

example in sensitive mechanical

balances.

• magnetic braking in some rapid

transit cars. Electromagnets positioned

in the car over the rails can induce eddy

currents in the rails.

One particularly interesting example of the

effects if eddy currents occurs in Io, the

large moon of Jupiter.

Io is the 4th largest moon in the solar system

and the most volcanic of any body. Io moves

rapidly through Jupiter’s magnetic field and

this produces eddy currents in Io’s interior.

These currents dissipate energy at the rate

of 1012 W. This energy helps in a minor way

to keep Io’s interior hot, and is an indirect

cause of volcanic eruptions.

[Jupiter’s varying gravitational pull produces

friction in Io’s interior which causes tidal

heating. This is an even bigger cause of

PHYS 153 08W 14

heating. This is an even bigger cause of

heating and volcanic activity].

Displacement current

We are slowly acquiring the 4 equations known as Maxwell’s equations, and

which are needed for a complete understanding of electromagnetism and the

fundamental nature of light. So far we have:

0εenclQ

AdE =•→→

∫ (Gauss’s law)

0=•→→

∫ AdB (Gauss’s law for magnetism)

enclildB 0µ=•→→

∫ (Ampere’s law)

It turns out that the 3rd equation (Ampere’s law) is incomplete and we must

→→

•=Φ

−= ∫ ldEdt

d Bε (Faraday’s law)

PHYS 153 08W 15

generalize Ampere’s law by completing it.

Fig. 29.22

Imagine a circuit with a charging

capacitor. Current flows into

the left hand side, and out the right. The

electric field increases in the space

between the capacitor plates.

Ci

→

E

Apply Ampere’s law to the circular path shown.

→→

• ldB around the path shown equals (where is the conduction

current flowing through the area bounded by that path). In this case, the

conduction current flows through the flat area bounded by the path.

Ci0µ Ci

CiBut the bulging surface to the right is also an area bounded by the same

path. The current through that surface is zero. So, Ampere’s law seems to

be fine for the flat surface but not for the bulging surface.

Ampere’s law is incomplete. There is an equivalent current between the

PHYS 153 08W 16

Ampere’s law is incomplete. There is an equivalent current between the

capacitor plates and it is related to the changing field.

For a parallel plate capacitor, with plates of area separated by d, and

filled with material of permittivity

→

E

The charge on the capacitor at any time is

ε

d

AC

ε=

A

EEAEdd

AtCVtq Φ==== εε

ε)()()(

The current then between the capacitor plates which represents the continuation

of the conduction current is

dt

d

dt

dq EΦ= ε

This current was called the displacement current by Maxwell.

PHYS 153 08W 17

dt

di ED

Φ= ε (displacement current)

Ampere’s law then becomes

enclDC iildB )(0 +=•→→

∫ µThe displacement current density is .

Dj

dt

dE

Adt

dj ED εε =

Φ=

Is the displacement current really there? Can we demonstrate its existence?

The answer is yes.

If there is a ‘current’ between the charging

capacitor plates, then there will have to be an

associated magnetic field.

Fig. 29.23 shows a capacitor charging. Apply

PHYS 153 08W 18

Fig. 29.23

Fig. 29.23 shows a capacitor charging. Apply

Ampere’s law to a circle of radius r<R. The

current enclosed by this circle is

)( 2

2r

R

iAj D

D ππ

=

CD iR

ri

R

rrBldB

2

2

02

2

02 µµπ ===•∴→→

∫ CD ii =( for the charging

capacitor)

CiR

rB

2

0

2πµ

=∴ (r<R)

Cir

Bπµ2

0= (r>R)

These results are confirmed by experiment.

Maxwell’s equations of electromagnetism

Maxwell’s equations in integral form for free space are:

→

PHYS 153 08W 19

0εenclQ

AdE =•→→

∫ (Gauss’s law for )

→

E

0=•→→

∫ AdB (Gauss’s law for )

→

B

enclE

Cdt

dildB )( 00

Φ+=•

→→

∫ εµ (Ampere’s law)

dt

dldE BΦ−==•→→

∫ ε (Faraday’s law)

---(1)

---(2)

---(3)

---(4)

Note that the in (1) is the conservative one , and that the in (4) is the

nonconservative on .

→

E→

EcE

nE

But plays no part in Gauss’s law, and plays no part in Faraday’s law.

So we can take the total electric field to be the electric field

used in Maxwell’s equations.

nE cE→→→

+= nc EEE

Note the symmetry in Maxwell’s equations.

In free space, and equations (1) and (2) on the previous slide are

Identical in form.

0=enclQ

PHYS 153 08W 20

Similarly, conduction current and equations (3) and (4) become: 0=Ci

→→→→

•=Φ

=• ∫∫ AdEdt

d

dt

dldB E

0000 µεµε

→→→→

•−=Φ

−=• ∫∫ AdBdt

d

dt

dldE B

More on this later.