chapter 22: electromagnetic induction · chapter 22: electromagnetic induction ¥ induced emf and...

Chapter 22: Electromagnetic Induction

• Induced emf and current

• Magnetic flux

• Faraday’s and Lenz’s laws

• Electric generators, back emf

• Omit 22.8, 22.9, (inductance and transformers)

1Monday, February 19, 2007

Induced emf

When the magnet moves relative

to coil, a current is induced in the

coil.

Reversing the magnet N and S

poles reverses the deflection.

Moving the coil to the magnet

produces the same deflection as

moving the magnet to the coil –

only the relative motion of coil

and magnet matters.

zero


!v

++

+

– – –

I II

I

Charges inside the moving rod experience a force

due to the magnetic field...

Conductor

The moving conductor acts as a generator.

Electromagnetic Induction

The basis of electromagnetic induction.


!B

A charge !q inside the wire moves with the coil relative to the magnetic

field. A component of field, B", is perpendicular to the velocity of the coil.

!B!

!F

x!F

I

!B!

Moving coil relative to magnet

!v

Motion of coil

toward the

magnet

The magnetic forces induce a current to flow around the coil.


!B

The charges in the coil are no longer moving as the coil is at rest, but the

induced current is the same as before...

There must be some more basic reason for the induced current.

! Changing magnetic field at the position of the coil.

!B!I

!B!

Moving magnet relative to coil

!v


Motional emf

The magnetic forces Fm

separate +

and – charges in the conductor.

Fm = !q v B

Fm

The separated + and – charges give

rise to an electric field E in the

conductor.

!E

Fq = !q E

At equilibrium, the electrostatic force:

balances the magnetic force.

Fq = !q E

That is:

Also, E = V/L

So the induced potential difference between the ends of the rod is: V = vLB

22.4

+!q

–!q

Fq = Fm

!qE = !qvB


Induced emf

The emf induced between the ends of a conductor that is moving in

a magnetic field is:

V = vLB

The induced emf is the same whether the coil moves or the

magnet moves, only the relative motion matters.

(V = vLBsin! when the angle between !B and!v is !)


Prob. 22.2/4: “Tethered Satellite Experiment”. A 20,000 m length of

wire is trailed behind the shuttle while in orbit around the earth. The

orbital speed of the shuttle is 7600 m/s.

If the earth’s magnetic field at the position of the shuttle is 5.1!10-5 T

and the wire moves perpendicular to the field, what is the induced emf

between the ends of the wire?

V = vLB = 7600 ! 20,000 ! 5.1 ! 10-5 = 7752 V

Negative at the top.

!B!!v

Wire

+

–


Prob. 22.4/2: The drawing shows a type of blood flow meter. Blood is

conductive enough that it can be treated as a moving conductor. When it

flows perpendicular to a magnetic field, electrodes can be used to measure

the small voltage that develops across the vessel.

Suppose the speed of the blood is 0.3 m/s, the diameter of the vessel is 5.6

mm and B = 0.6 T. What is the magnitude of the voltage that is measured?

!vL+ + +

– – –

!B

Blood – a moving

conductor


Prob. 22.5: Each rod of length L = 1.3 m moves at speed v = 2.7 m/s in a

magnetic field, B = 0.45 T. Find the motional emf for each.

+

–


Prob. 22.C6: Initially the rod is at rest. Describe the rod’s motion after

the switch is closed. Be sure to account for the effect of any motional emf

that develops.

V0


The rod experiences a magnetic force to the right and accelerates.

I F = ILBV0


The moving rod now generates its own emf that opposes the emf of the

battery (a “back emf”). The current therefore decreases. The rod

continues to accelerate until the current is reduced to zero (assuming no

friction).

v

+

–

IV = vLBV0 F = ILB

Speed constant when vLB = V0


V = vLBV0

I

I

Induced emf – equivalent circuit

R

Resistance of rails and bar

I = 0 when V = V0


22.7/660 W bulb, R = 240 "

Motional emf between ends of sliding rod, V = vLB

Power dissipated, W = VI = V2/R = 60 W, so (vLB)2/R = 60 W

B = 0.4 T L = 0.6 m

Therefore, v=!60R

LB=

!60" (240 !)

(0.6 m)" (0.4 T)= 500 m/s

In 0.5 s, the rod slides 250 m!

How long do the

rails have to be to

light the 60 W bulb

for 0.5 s?

!Fm

Fapplied


B = 0.4 T L = 0.6 m

Work done to light the lamp

There is an induced

current I in the bar

when the bar is moving

in the magnetic field.

Magnetic force on the

bar, Fm

= ILB, opposes

the motion of the bar.

Fapplied

In 1 s, work done by the applied force in opposing Fm

is W = Fm

v

W = Fm

v = (ILB) v = I (LBv) = I V = 60 W

That is, the power to light the bulb is supplied by doing work against the

magnetic force.

!Fm

60 W lamp


Induced emf

The emf induced between the ends of a conductor that is moving in

a magnetic field is:

V = vLB

The induced emf is the same whether the rod moves or the

magnet moves, only the relative motion matters.

(V = vLBsin! when the angle between !B and!v is !)


Fapplied

!Fm

• Sliding the bar along the rails generates an emf

• When the circuit is completed, a current flows and lights up the lamp

• A magnetic force acts on the current in the rod to oppose the motion

! of the rod (a consequence of Lenz’s law – later)

• The work done in pushing the rod is equal to the electrical energy

# dissipated in the lamp – mechanical energy is converted into electrical.

Work done to light the lamp

!B


The emf induced between the ends of the falling

rod is:

V = vLB

No current is flowing, so there is no magnetic

force on the rod.

The resistor R completes the circuit, so that

current flows and there is now a magnetic force

resisting the gravitational force that

accelerates the rod downwards.

The rod stops accelerating when the magnetic

force is equal to the weight of the rod.

Motional emf

22.9

L

L

!v

!v


Prob. 22.9: A conducting rod 1.3 m long slides down between two

frictionless vertical copper tracks at a constant speed of 4 m/s

perpendicular to a 0.5 T magnetic field.

a) What is the mass of the rod?

b) Find the change in gravitational PE in 0.2 s.

c) Find the electrical energy dissipated in the resistor in 0.2 s.

R = 0.75 "

L = 1.3 m


Induced emf

Changing the area of the loop also

induces an emf.

An emf is induced in the coil

whenever the number of field lines

passing through the coil changes.

The number of field lines is a

measure of “magnetic flux”.

! There is an induced emf

whenever there is a change of

magnetic flux passing through the

coil.


Induced emf

The emf induced between the ends of the moving rod is: V = vLB

Between time t0 and t, the rod moves a distance x – x

0 = v(t – t

0), so

V = vLB= B(x! x0)L(t! t0)

= B

!A!A0t! t0

"=

(BA)! (BA)0t! t0

A0 = x

0L

A = xL

With B perpendicular to the loop, (BA) is the “magnetic flux” passing

through the loop. The induced emf is equal to the rate of change of

magnetic flux passing through the loop – Faraday’s Law.

=!(BA)!t

LA0 A

L


Magnetic Flux

Magnetic flux, != Bcos"!A= BAcos"

Faraday’s Law: the induced emf is equal to the rate of change of magnetic flux

Unit of Magnetic Flux:

1 Weber (Wb) = 1 T.m2


Magnetic Flux and Field Lines

!= BAcos0! = BA != BAcos60

! = BA/2 != BAcos90! = 0

The magnetic flux is proportional to the number of magnetic field lines

passing through the coil.


B = 0.4 T L = 0.6 m

Fapplied

!Fm

Induced emf and rate of change of flux

As the rod is moved to the right, the area of the closed loop increases and

the magnetic flux passing through the loop increases. There is increasing

magnetic flux passing through the loop and pointing into the page.

R

!BI

!BI

!BI

The induced current, I, produces a magnetic field, BI, pointing out of the

page, that opposes the change in magnetic flux. This is Lenz’s law.

Faraday’s Law:

V =!"

!t= IR

= B!A

!t

= BLv


Faraday’s Law

The induced emf is equal to

the rate of change of

magnetic flux.

The direction of the induced current is such that

the magnetic field produced by the current

opposes the change in magnetic flux that

generated the current.

Lenz’s Law

Magnetic flux: != BAcos"


Prob. 22.C11: Use Lenz’s law to verify

that the induced current is in the direction

in the diagram.

• The flux through the loop is

into the page and is decreasing as

the area of the loop decreases.

• The induced current produces a

magnetic field that opposes the

decreasing flux.

! The magnetic field produced

by the induced current must

point into the page.

! The current flows clockwise

in the loop


Conducting ring falling through a magnetic field

The magnetic flux passing through the ring

is constant (zero), so there is no induced

emf or current.

!BI

I

I = 0

!Fm

A magnetic force is generated that opposes the motion of the ring.

The magnetic flux passing through the

ring is increasing and is directed into

the page.

The induced current produces a

magnetic field, BI, that opposes the

increase of flux.


The magnetic flux passing

through the coil is constant, so

there is no induced emf or

current and no magnetic force.

The induced emf generates a current that

opposes the change in magnetic flux, producing

a magnetic field, BI, into the page.

!BIx

I

22.70

!Fm = 0

!Fm

A magnetic force opposes the motion of the coil.

The magnetic flux passing

through the coil is decreasing and

points into the page.


Prob. 22.32/70: A bar magnet is falling through a metal ring. In part a

the ring is solid all the round around, but in part b it has been cut

through.

Explain why the motion of the magnet in part a is retarded, whereas it is

not in part b.


Faraday’s Law

The induced emf is equal to

the rate of change of

magnetic flux.

The direction of the induced current is such that

the magnetic field produced by the current

opposes the change in magnetic flux that

generated the current.

Lenz’s Law

Magnetic flux: != BAcos"


Prob. 22.33: A circular loop of wire rests on a table. A long, straight wire

lies on this loop over its centre.

The current I in the straight wire is increasing. In what direction is the

induced current, if any, in the loop?

" ""

"

• What is the total magnetic flux through the loop?

• Does it change?

!B


A wire is bent into a circular loop as shown. The radius of the circle is 2

cm. A constant magnetic field B = 0.55 T is directed perpendicular to the

plane of the loop. Someone grabs the ends of the wire and pulls it taut, so

the radius shrinks to zero in 0.25 s.

Find the magnitude of the average induced emf between the ends of the

wire.

B

I


Prob. 22.70/32: Indicate the direction of the electric field between the

plates of the capacitor if the magnetic field is decreasing in time.

++++

– – – –E

I

I

I

BI

BI

BI

Induced magnetic field


#

Prob. 22.26

A 0.5 m copper bar, AC, sweeps around a

conducting circular track at 15 rad/s.

A uniform magnetic field points into the

page, B = 0.0038 T.

Find the current in the loop ABC.

The loop forms a closed circuit of

increasing area, so the magnetic flux

passing through the loop increases and

an emf is generated.

and I = V/R = 0.00713/3

# = 2.4 mA

IBI

$ = 15 rad/s

r = 0.5 m

The flux passing through the loop is: != BA= B

!"

2#

"!#r2 = Br

2!/2

The induced emf is: V =!"

!t=Br

2

2! !#

!t=Br

2$

2

V =0.0038!0.52!15

2= 7.13!10"3 V

$

I


22.73/27: Two 0.68 m long conducting rods are rotating at the same

speed in opposite directions and both are perpendicular to a 4.7 T

magnetic field. The ends of these rods come to within 1 mm of each

other as they rotate. The fixed ends of the rods are connected by a wire,

so the fixed ends are at the same potential.

If a potential difference of 4500 V is needed to cause a 1 mm spark, what

is the angular speed of the rods when a spark jumps across the gap?

% %

L LV0 V0

V0V0

+–+–


Guitar pickup

• The strings are magnetizable

• A permanent magnet magnetizes them

• The vibration of a string changes the magnetic flux through a coil

# close to it at the frequency of vibration of the string

• An emf is induced in the coil at that frequency


Playback of tape recording

• Recorded tape is magnetized in N-S patches

• The tape passes by the playback head which channels and concentrates

# the magnetic field through an iron core

• The changing magnetic flux induces an emf in the coil


Moving coil microphone

• Sound waves cause the diaphragm of the microphone to move in/out

• A coil moves with the diaphragm relative to a permanent magnet,

# causing the magnetic flux through the coil to change in step with the

# pressure variations of the sound wave

• An emf is set up in the coil at the frequency of the sound wave


Ground fault detector

If the return current (green) is equal to the supply current (red), the magnetic

fluxes around the iron ring are equal and opposite and cancel.

If the currents differ, the fluxes do not

cancel and there is a net flux varying at

60 Hz, which induces a current in the

sensing coil.


Electric Generator

The magnetic flux passing through

the coil varies as the coil rotates –

an emf is generated.

Flux, != BAcos"= BAcos(#t)


V0

V =V0 sin(!t)

EMF from electric generator, using

rate of change of flux

Flux, != BAcos"= BAcos(#t)

V =V0 sin(!t), V0 = BA!

(diff. calculus)Then, the induced emf is V = (!)!"!t

= BA! sin(!t)

Reminder of Lenz’s Law


V = BLvsin!

V = BLvsin!

Electric GeneratorEMF generated in moving conductors

Total emf generated: Vtot = 2BLvsin!

Therefore, Vtot = BLW!sin"= BA!sin"

If the coil has N turns, then: Vtot = NBA!sin"=V0 sin"

W

Area, A = LW

v= r!=!W

2

"!, != angular frequency of rotating coil

VV+

+

–

–

2V

= V0 sin!t


Prob. 22.63/39

The cross-sectional area of

the coil is 0.02 m2 and the

coil has 150 turns. Find the

rotation frequency of the

coil and the magnetic field.

The period is T = 0.42 s

Therefore, B=V0

NA!=

28

150!0.02!14.96 = 0.624 T

!=2"

T=

2"

0.42 s= 14.96 rad/s != 2" f , so f =

14.96

2"= 2.38 Hz

Peak voltage, V0 = 28 V = NBA$.

Alternating Current Electric Generator

V0 = 28 V

f = 1/T = 2.38 Hz

V = NBA!sin!t


Time

V = V0 sin $t

Time

Root Mean Square (rms)

Power ∝ V2

V0 = NBA$V0

V2 = V02 sin2 $t

Time

V02

Mean = V02/2

Mean power ∝ Vrms2 = V0

2/2

Vrms = V0/&2 = rms voltage


Prob. 22.40: A generator uses a coil that has 100 turns and a 0.5 T

magnetic field. The frequency of the generator is 60 Hz and its emf has

an rms value of 120 V.

Assuming that each turn of the coil is a square, determine the length of

the wire from which the coil is made.

• What is the peak voltage generated?

• What is the area of the coil?

# A = a2...


Electric Generator

Electric Motor

Just like a generator, but use a current to

cause the coil to rotate.

Once the coil is rotating, it acts as a generator,

producing an emf that opposes the rotation of

the coil! – the “back emf.”

L

L As the coil rotates and a current is induced in it, a

magnetic force is generated that opposes the

rotation of the coil.

Work has to be done to rotate the coil against

this torque.


Electric motor, back emf

• The current flowing around the coil generates a torque

• The coil rotates

• The rotating coil acts as a generator

• The generated emf opposes the motion of the coil. This is the back emf.

From above

!v

!!v


Back emf

Vback

A power supply drives the motor The motor acts as a generator

Kirchhoff’s loop law: V !Vback = IR

Symbol for

AC generator


Vback

Prob. 22.38/36

A vacuum cleaner is plugged into a 120 V socket and draws 3 A of current

in normal operation (motor running at full speed, back emf at maximum

value) and the back emf is then 72 V.

Find the coil resistance of the motor.

V !Vback = IR

So, 120 – 72 = 3R

R = (120 – 72)/3 = 16 "

When the motor is first switched on, while it’s still spinning slowly, the back

emf is small and: V – 0 = IR,

Then, I = (120 V)/(16 ") = 7.5 A % the motor draws extra current while it is

speeding up. This is the time it’s most likely to trip the breaker.


Prob. 22.36/38: A 120 volt motor draws a current of 7 A when running at

normal speed. The resistance of the armature wire is 0.72 ".

a) Determine the back emf generated by the motor.

b) What is the current at the instant the motor is turned on and has not

# begun to rotate?

c) What series resistance must be added to limit the starting current to

# 15 A?


Prob. 22.74/43: The armature of an electric drill has a resistance of 15 ".

When connected to a 120 V outlet, the motor rotates at its normal speed

and develops a back emf of 108 V.

a) What is the current through the motor?

b) If the armature freezes up and can no longer rotate, what is the current

# in the stationary armature?

c) What is the current when the motor is running at half speed?


Prob. 22.25/71: A conducting coil of 1850 turns is connected to a

galvanometer. The total resistance of the circuit is 45 ". The area of

each turn is 4.7 ! 10-4 m2.

The coil is moved into a magnetic field, the normal to the coil being

kept parallel to the magnetic field. The amount of charge that is induced

to flow around the circuit is 8.87 ! 10-3 C.

Find the magnitude of the magnetic field.


Summary of Chapter 22

• EMF induced in a moving conductor: V = vLB sin#

• Mechanical work has to be performed to produce

# electrical energy – you don’t get something for

# nothing

• Magnetic flux, & = BA cos', proportional to

# number of field lines.

• Faraday’s law, V = N !&/!t,

# Lenz’s law – the magnetic field

# produced by the induced current opposes

# the changing magnetic flux.

• Electric generator, back emf


chapter 22: electromagnetic induction · chapter 22: electromagnetic induction ¥ induced emf and...

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