chapter 22: electromagnetic induction · chapter 22: electromagnetic induction ¥ induced emf and...
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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
2Monday, February 19, 2007
!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.
3Monday, February 19, 2007
!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.
4Monday, February 19, 2007
!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
5Monday, February 19, 2007
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
6Monday, February 19, 2007
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 !)
7Monday, February 19, 2007
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
+
–
8Monday, February 19, 2007
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
9Monday, February 19, 2007
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.
+
–
10Monday, February 19, 2007
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
11Monday, February 19, 2007
The rod experiences a magnetic force to the right and accelerates.
I F = ILBV0
12Monday, February 19, 2007
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
13Monday, February 19, 2007
V = vLBV0
I
I
Induced emf – equivalent circuit
R
Resistance of rails and bar
I = 0 when V = V0
14Monday, February 19, 2007
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
15Monday, February 19, 2007
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
16Monday, February 19, 2007
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 !)
17Monday, February 19, 2007
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
18Monday, February 19, 2007
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
19Monday, February 19, 2007
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
20Monday, February 19, 2007
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.
21Monday, February 19, 2007
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
22Monday, February 19, 2007
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
23Monday, February 19, 2007
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.
24Monday, February 19, 2007
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
25Monday, February 19, 2007
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"
26Monday, February 19, 2007
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
27Monday, February 19, 2007
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.
28Monday, February 19, 2007
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.
29Monday, February 19, 2007
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.
30Monday, February 19, 2007
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"
31Monday, February 19, 2007
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
32Monday, February 19, 2007
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
33Monday, February 19, 2007
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
34Monday, February 19, 2007
#
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
35Monday, February 19, 2007
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
+–+–
36Monday, February 19, 2007
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
37Monday, February 19, 2007
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
38Monday, February 19, 2007
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
39Monday, February 19, 2007
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.
40Monday, February 19, 2007
Electric Generator
The magnetic flux passing through
the coil varies as the coil rotates –
an emf is generated.
Flux, != BAcos"= BAcos(#t)
41Monday, February 19, 2007
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
42Monday, February 19, 2007
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
43Monday, February 19, 2007
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
44Monday, February 19, 2007
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
45Monday, February 19, 2007
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...
46Monday, February 19, 2007
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.
47Monday, February 19, 2007
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
48Monday, February 19, 2007
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
49Monday, February 19, 2007
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.
50Monday, February 19, 2007
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?
51Monday, February 19, 2007
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?
52Monday, February 19, 2007
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.
53Monday, February 19, 2007
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
54Monday, February 19, 2007