CHAPTER 20
20
ELECTROMAGNETIC INDUCTION
689
Electromagnetic Induction
Answers to Discussion Questions
•20.1
He expected to induce a sustained current and so could stroll into the room to observe it. Of
course, the induced current was a pulse, long gone by the time he got there.
•20.2
Solar activity produces a time varying magnetic field at the Earth that is accompanied by an
electric field and therefore by an emf along the length of the pipe.
•20.3
The handle of the device slips into a coil hidden in the base of the holder. Another coil is inside
the handle and the two are in close proximity. Connecting the coil in the well to AC generates a
time-varying B -field that passes through the plastic skin of the handle and produces a rapidly
varying (60 Hz) induced emf and an induced current. That current is converted to DC and
used to charge a battery in the handle, which powers the motor in the device. No charge and
so no “electricity” passes from the base to the handle, although electrical energy certainly is
transferred. The same scheme will work across human skin to power a mechanical heart.
•20.4
(a) On becoming superconducting, the ring expels the field from the volume occupied by its
material via the Meissner Effect. Flux is then essentially captured in the region of the “hole.”
(b) If the ring is then pulled from the field region it will effectively carry away with it the flux in
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the hole which cannot cross out of the ring. In other words, surface supercurrents will maintain
the flux in the hole after the ring is removed from the field. (c) The work done on the ring
(via Lenz’s Law) in removing it from the field region goes into setting up the surface currents
which support the field in the hole.
•20.5
As the magnet approaches, its B -field attempts to penetrate the ring. A supercurrent is thereby
induced opposing the buildup of flux and the motion of the magnet. If the ring is free to move, it
will lift off its support and hover (with its induced north pole down) above the north pole of the
magnet. Finite, persistent supercurrents on the surface of the superconductor circulate in such
a way as to shield the interior from the field. No flux enters the body of the superconductor;
there is no change in flux, no E -field induced, and no bulk current.
•20.6
There are two similar techniques that have been
used to date. In one, an air-core primary coil
(with its axis vertical) located at the center of
the toroid in the hole, and carrying a time varying current, produces a changing flux and induces
a circular E -field and thence a toroidal current
in the plasma. The time-varying coil current is
provided in bursts from a bank of capacitors and
much of the energy transferred to the plasma ends
up as joule heat within it. Another approach links
the toroid with an iron core (see the figure), but
the idea is the same.
iron core
plasma
toroid
coil
•20.7
To crank the generator without a load and therefore without a current (in the steady state),
one need only overcome friction. By contrast, lighting the 100-W bulb requires twice the power
used in lighting the 50-W bulb and that power must be supplied by the person turning the
crank. Most people will find it difficult to keep the 100-W bulb lit very long.
•20.8
The battery voltage appears across the branch of lamp 2 and R, and current almost immediately
(the resistor and lamp have some small inductance) circulates through the lamp which glows
brightly. The inductor inhibits the rise of current with a back-emf so lamp 1 initially glows
faintly increasing to full brightness gradually.
•20.9
Current would be induced in the loop and power transferred from the line to it — the timevarying B -field would induce an E -field. If a wire loop was in that region, the E -field would do
CHAPTER 20
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691
work on the free electrons, transferring energy to them. The power company could detect a loss
in energy arriving at the end of the line; there would be a slight drop in delivered current since
the voltage is fixed by the generator. If you separate the two leads from a telephone and lay
the pick-up loop (attached to a good set of headphones) next to one of the wires, you should
be able to listen in on the time-varying B -field of the conversation.
•20.10
In series the individual emfs must add together to give a net emf across the equivalent inductor.
Moreover, the current in each, and therefore the rate of change of the current, must be the same
for each. E = E1 + E2 + E3 = (∆I/∆t)(L1 + L2 + L3 ) hence it follows that L = L1 + L2 + L3 in
series. In parallel, the voltage across all will be the same as the current divides, hence the total
∆I/∆t equals the sum of the individual rates of change and so E/L = E1 /L1 + E2 /L2 + E3 /L3
and 1/L = 1/L1 + 1/L2 + 1/L3 .
•20.11
The energy stored in an inductor varies linearly with µ. Hence, inserting an iron core can
increase the energy by a factor of several thousand. In such a case the power supply must
provide an increased amount of energy over the air-core situation where the maximum current
is generated relatively quickly. With the iron core, L is large as is the time constant and it takes
a much longer time and more work to establish the same maximum current V /R. The faster
the maximum current is reached the less energy the power supply must provide to establish the
field — the lower L the less the “inertia.”
•20.12
Sound waves cause the diaphragm to move and that vibrates the rod in the gap. When the top
of the rod is closer to the left face of the C , magnetic field lines rising out of the north pole
and up the rod jump the gap to the left and return to the south pole mostly through the left
coil. By contrast, when the rod is not displaced the field lines are equally distributed between
the two coils. So incoming sound causes a change in the flux in the two coils and that induces
a voltage across each of them. These two voltages are opposite since one is produced by a flux
increase and the other by a flux decrease. Therefore the coils are wound oppositely so their
voltages have the same sense and add to produce a stronger signal inasmuch as the coils are in
series.
•20.13
Yes, a coil in a motor turning through a magnetic field will experience an induced emf and an
induced current that will send energy back to the source, which is driving the motor. With
no load, the motor produces (and returns to the power company) almost as much current as it
draws, and so costs very little to run. A free-turning motor must have its speed limited by the
back-current it generates with no losses (in the bearings, etc.) the motor will speed up until
the back-current equals the driving current and it can no longer accelerate. With a load, the
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motor does work and draws energy in excess of what it returns via the back-current. In real
life, a motor also produces a good deal of thermal energy via friction and if it is to operate
continuously for any long period of time, it must be cooled (usually by forced air). A jammed
motor will not turn and not generate a back-current. The driving current (which depends on
the motor’s resistance) now undiminished by any back-current is too great. The wiring will
heat up and the insulation will begin to burn off. If the process is not stopped soon, the motor
will be destroyed.
•20.14
Since the record head matches the read head each magnetized region of the tape fits nicely
under the soft-iron C-shaped pick-up. Field lines pass through the high-permeability core
thereby changing the flux through the coil and inducing a voltage across it.
•20.15
Recall that the B -field outside a very long tight solenoid approaches zero. Each meter reads the
potential drop across the resistor adjacent to it, the resistor with which it forms a closed circuit
excluding the solenoid. The induced emf equals the difference between the two meter readings,
or alternatively, the sum of the potential differences across the resistors in the central circuit
(1-9-10-2-7-8-1). The seemingly strange thing here is that the “voltage” measured between 1
and 2 actually depends on how you hook up the meter. The varying flux provides energy to the
induced current (energy coming from the power source driving current through the solenoid).
In part (b) both meters would read Ii R2 .
•20.16
The time varying flux through the orbit generates a circular electric field which accelerates the
electron. The faster the particle moves the more centripetal force is needed so the field must
be increased accordingly.
•20.17
electromagnetic induction —– The property of an electric circuit by which an electromotive force is induced in it as the result of a changing magnetic flux.
induced emf —– Electromotive force induced as a result of changing magnetic flux.
magnetic flux —– A measure of the quantity of magnetism, being the total number of
magnetic lines of force passing through a specified area in a magnetic field.
flux density —– The amount of flux through a unit area taken perpendicular to the direction
of the flux.
Faraday’s Induction Law —– The induced emf in a loop is proportional to the time rate
of change of the magnetic flux through the loop. The direction of the induced emf is determined
by Lenz’ Law. More specifically, E = −N (∆ΦM /∆t), as in Eq. (20.3).
CHAPTER 20
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693
Lenz’s Law —– The induced emf will produce a current that always acts to oppose the
change that originally caused it.
motional emf —– The emf generated on an object as it cuts through magnetic field lines.
search coil —– A small coaxial coil which detects the presence of an independent E -field
associated with a time-varying B -field.
dynamo —– A generator, especially one for producing direct current.
ac generator —– One that generates, especially a machine that converts mechanical energy
into electrical energy (by outputting an alternating current).
dc generator —– One that generates, especially a machine that converts mechanical energy
into electrical energy (by outputting a direct current).
commutator —– A cylindrical arrangement of insulated metal bars connected to the coils of a
direct-current electric motor or generator, providing a unidirectional current from the generator
or a reversal of current into the coils of the motor.
eddy current —– If an extended conductor translates with respect to a B -field, which is not
uniform over the entire conductor, or if different portions of the conductor move at different
velocities (i.e., it rotates) with respect to the field, currents will be introduced within it that
circulate in closed paths. These are known as eddy currents.
back-emf —– The induced emf opposes the cause of itself in accordance with Lenz’s Law, so
it is also called back-emf.
self-induction —– The generation by a changing current of an electromotive force in the
same circuit.
inductance —– The magnetic flux generated by one unit of current.
henry —– The SI unit of inductance.
inductor —– A circuit element, typically a conducting coil, in which electromotive force is
generated by electromagnetic induction.
choke —– A circuit element used to suppress or limit the flow of alternating current without
affecting the flow of direct current.
R-L circuit —– A circuit consisting of a inductor and a resistor.
time constant —– A constant which influences the temporal behavior of an R-L circuit. It is
equal to L/R. The greater the time constant, the longer it takes for the current in the circuit
to reach a steady value.
energy density —– Energy of a field per unit volume.
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Answers to Multiple Choice Questions
1.
8.
15.
22.
e
b
c
c
2. c
9. a
16. b
23. b
3. c
10. c
17. b
24. c
4. b
11. a
18. b
25. b
5. c
12. a
19. d
6. c
13. c
20. a
7. a
14. a
21. d
Solutions to Problems
20.1
The magnetic flux is given by Eq. (20.1): ΦM = B⊥ A. In this case B⊥ = 1.2 mT, and
A = 25 cm2 , so
ΦM = B⊥ A = (1.2 × 10−3 T)(25 × 10−4 m2 ) = 3.0 × 10−6 T · m2 = 3.0 µWb .
20.2
From Eq. (20.1)
ΦM = BA cos θ = (100 × 10−3 T)(0.020 m2 )(cos 30◦ ) = 1.7 × 10−3 T · m2 = 1.7 mWb .
20.3
The flux density B is, by definition, the magnetic flux ΦM per unit cross-sectional area: B =
ΦM /A, where A is the total cross-sectional area perpendicular to the B -field [see Eq. (20.1)].
Plug in ΦM = 6.0 mWb = 6.0 × 10−3 Wb and A = 50 cm2 = 50 × 10−4 m2 to obtain
B=
ΦM
6.0 × 10−3 Wb
= 1.2 Wb/m2 = 1.2 T .
=
A
50 × 10−4 m2
20.4
The magnetic field B is related to the magnetic flux via Eq. (20.1): ΦM = BA⊥ . In this case
ΦM = 0.016 Wb and A⊥ = 8.0 cm2 = 8.0 × 10−4 m2 , so
B=
ΦM
0.016 Wb
= 20 Wb/m2 = 20 T .
=
−4
2
A
8.0 × 10 m
CHAPTER 20
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ELECTROMAGNETIC INDUCTION
20.5
and B
are due south so the change in magnetic field is
Both B
i
f
=B
−B
= (0.6 T) -south − (0.5 T) -south = (0.1 T) -south .
∆B
i
f
20.6
Similar to the previous problem, the change in magnetic field is
=B
−B
= (0.5 T) -up − (0.6 T) -down = (0.5 T) -up + (0.6 T) -up = (1.1 T) -up ,
∆B
i
f
where we noted that (0.6 T) -down = −(0.6 T) -up.
20.7
The change in the magnetic field is
=B
−B
= 0 − (0.01 T)-east = −(0.01 T)-east = (0.01 T)-west ,
∆B
i
f
has a magnitude of 0.01 T and is due west.
i.e., ∆B
20.8
The initial magnetic flux ΦMi through the loop is ΦMi = Bi Ai = (100 mT)(0.020 m2 ) =
2.0 × 10−3 T · m2 , while the final flux is ΦMf = 0 as A⊥ = 0. Thus
∆ΦM
Φ − ΦMi
0 − 2.0 × 10−3 T · m2
= −1.0 Wb/s = −1.0 V .
= Mf
=
∆t
∆t
0.002 0 s
20.9
Before the loop is yanked from the B -field the magnetic flux through the loop is ΦM i = BA⊥ ,
where A⊥ is the encompassed area of the loop perpendicular to the B -field. In a time interval
∆t the B -field through the loop is gone so ΦM f = 0. The induced emf, which is equal in
magnitude to the rate of change of the magnetic flux in the loop, is then [see Eq. (20.3)]
∆ΦM
E = −N
= −N
∆t
Φ M f − ΦM i
∆t
= −N
0 − BA⊥
∆t
=
N BA⊥
.
∆t
Plug in N = 1, B = 0.40 T, A⊥ = 0.25 m2 , and ∆t = 200 ms = 0.200 s to obtain E = +0.50 V.
(Here the plus sign in E means that the induced emf would generate a current which in turn
produces a B -field whose flux is positive through the loop, as is ΦM i .)
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20.10
Before the loop is removed from the B -field the magnetic flux through the loop is ΦM i = BA⊥ ,
). In a time
where A⊥ is the area of the loop (note that the area then is perpendicular to B
interval ∆t the coil is removed from the B -field so ΦM f = 0. The resulting induced emf is then
(see the result of the previous problem) E = N BA⊥ /∆t. Plug in N = 100, A⊥ = 4.0 cm2 =
4.0 × 10−4 m2 , ∆t = 20 ms = 0.020 s, and E = 1.0 V and solve for B :
B=
E∆t
(1.0 V)(0.020 s)
=
= 0.50 T .
N A⊥
(100)(4.0 × 10−4 m2 )
20.11
The induced emf due to a flux change in a time interval ∆t is given by Eq. (20.3): E =
−N ∆ΦM /∆t. Take the absolute values to obtain its magnitude to be |E| = N |∆ΦM |/∆t.
Plug in E = 60 V, ∆ΦM = 30 mWb, and N = 100, and solve for ∆t:
∆t =
N |∆ΦM |
(150)(30 mWb)
= 75 ms .
=
|E|
60 V
20.12
From Eq. (20.3) the magnitude of the induced emf is
(100)(0.050 Wb)
∆Φ
M
=
|E| = −N
= 2.5 × 102 V = 0.25 kV .
∆t 0.020 s
20.13
The change in flux is ∆ΦM = ΦM f − ΦM i = 0.070 Wb − 0.010 Wb = 0.060 Wb, which took
place in ∆t = 0.020 s. Thus from Eq. (20.3) the magnitude of the induced emf is
(200)(0.060 Wb)
∆Φ
M
=
|E| = −N
= 8.0 V .
∆t 1.5 s
20.14
From t = 0 to t = 0.30 s B changes constantly from 0 to 0.30 T so ∆B = 0.30 T. The
corresponding change in magnetic flux is ∆ΦM = A∆B , where A = 0.25 m2 . Thus from
Eq. (20.3) the induced emf is
E = −N
∆ΦM
N A∆B
(10)(0.25 m2 )(0.30 T)
=−
=−
= −2.5 V
∆t
∆t
0.30 s − 0
(0 < t < 0.30 s) .
From t = 0.30 s to t = 0.40 s there is no change in the magnetic field so ∆ΦM = 0, which gives
E = 0.
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ELECTROMAGNETIC INDUCTION
From t = 0.40 s to t = 0.50 s B changes constantly from 0.30 T back to 0 so ∆B = −0.30 T.
The corresponding change in magnetic flux is ∆ΦM = A∆B , where again A = 0.25 m2 . Thus
∆ΦM
N A∆B
(10)(0.25 m2 )(−0.30 T)
E = −N
= +7.5 V
=−
=−
∆t
∆t
0.50 s − 0.40 s
(0.40 s < t < 0.50 s) .
20.15
The cross-sectional area of the coil is A = 0.110 m × 0.199 m = 2.189 × 10−2 m2 , over which a
uniform magnetic field B = 82.0 mT is present before the coil is yanked out of the field. The
initial magnetic flux through the coil is then ΦMi = BA = (82.0 × 10−3 T)(2.189 × 10−2 m2 ) =
1.79 × 10−3 T · m2 . After the coil is yanked out the flux through it is reduced to zero, so
∆ΦM = ΦMf − ΦMi = 0 − ΦMi = −1.79 × 10−3 T · m2 . This process takes place during a time
interval ∆t = 45.0 ms, so the average induced emf is
E = −N
∆ΦM
(−1.79 × 10−3 T · m2 )
= 4.0 V .
= −(100)
∆t
45 × 10−3 s
(b) The induced emf provides the voltage difference V that drives a current I through the coil:
E = V = IR, so
E
4.0 V
I=
=
= 2.0 A .
R
2.0 Ω
20.16
(a) The induced emf is proportional to the time rate of change of the magnetic flux: E =
−N ∆ΦM /∆t.
(b) Since ΦM = AB , ∆ΦM = ∆(AB) = A∆B ; and so E = −N ∆ΦM /∆t = −N A∆B/∆t.
(c) A = (10.0 cm2 )(10−2 m/cm)2 = 1.00 × 10−3 m2 .
(d) According to the problem statement ∆B/∆t = −0.10 T/s, where the negative sign indicates that B is decreasing with time.
(e)
∆B = 10(1.00 × 10−3 m2 ) |−0.10 T/s| = 1.0 mV .
|E| = N A ∆t (f) The induced emf provides the voltage difference that drives the current. So E = V = IR,
and
E
1.0 mV
I=
=
= 1.0 mA .
R
1.0 Ω
(g) As the downward magnetic field decreases with time the magnetic flux through the loop
also decreases. According to Lenz’s Law the induced current in the loop runs from B to A, as
it tends to maintain the original flux.
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(h) Think of the loop effectively as a battery, with terminals A and B . Since the current flows
from B to A through the loop, if one hooks up a load across the two terminals a current will
flow from A to B through the load, indicating that terminal A is at a higher potential. (Note
that, inside a battery, the current flows from the negative to the positive terminal.)
20.17
The B -field in the solenoid varies as a function of the current I it carries: B = µ0 nI . The
magnetic flux of this field through the coil of cross-sectional area A is ΦM = BA = µ0 nAI . As
the current in the solenoid changes by ∆I during a time interval ∆t the corresponding change
in ΦM is ∆ΦM = µ0 nA∆I , which causes an emf in the N -turn coil:
∆ΦM
N µ0 nA∆I
=−
∆t
∆t
−7
(240)(4π × 10 T · m/A)(10 × 102 /m)(2.0 × 10−4 m2 )(4.9 A)
=−
5.00 × 10−3 s
= −5.9 × 10−2 V = −59 mV .
E = −N
Note that n = 10/cm = 10 × 102 /m.
20.18
With the original flux chosen as positive, the change in flux is ∆ΦM = ΦMf −ΦMi = −8.0 mWb−
8.0 mWb = −16 mWb, which took place in ∆t = 200 ms. The resulting emf in the N -turn coil
follows from Eq. (20.3) to be
E = −N
∆ΦM
(100)(−16 mWb)
= 8.0 V .
=−
∆t
200 ms
20.19
The magnetic flux provided by the B -field of the electromagnet through the loop of area A
and the normal
is ΦM = BA cos θ [see Eq. (20.1)], where θ (= 30◦ ) is the angle between B
direction of the plane containing the loop. As B changes from Bi to Bf the corresponding change
in ΦM is ∆ΦM = Bf A cos θ − Bi A cos θ = (Bf − Bi )A cos θ, which causes an induced emf given
by Eq. (20.3): E = −N ∆ΦM /∆t. Plug in N = 1, Bi = 0, Bf = 0.500 T, A = 4.0 cm × 4.0 cm,
θ = 30◦ , and ∆t = 200 ms to obtain
∆ΦM
(Bf − Bi )A cos θ
= −N
∆t
∆t
(1)(0.500 T − 0)(4.0 × 4.0 × 10−4 m2 )(cos 30◦ )
=−
0.200 s
−3
= −3.5 × 10 V = −3.5 mV .
E = −N
CHAPTER 20
ELECTROMAGNETIC INDUCTION
699
Here the minus sign indicates that the induced emf tends to run a current whose magnetic field
produces a flux opposite in direction to the original flux, which is downward. Thus the induced
current must be counterclockwise viewed from the top so as to produce an upward flux.
20.20
The magnetic flux provided by the B -field through the coil of area A is ΦM = BA cos θ [see
Eq. (20.1)]. As θ changes from θi = 45◦ to θf = 90◦ due to the rotation of the coil the
corresponding change in ΦM is ∆ΦM = BA cos θf − BA cos θi = BA(cos θf − cos θi ), which
causes an induced emf given by Eq. (20.3): E = −N ∆ΦM /∆t. Plug in N = 10, B = 0.20 T,
A = π(5.0 cm)2 , and ∆t = 0.10 s to obtain
∆ΦM
BA(cos θf − cos θi )
= −N
∆t
∆t
(10)(0.20 T) π(0.050 m)2 (cos 90◦ − cos 45◦ )
=−
0.10 s
= 0.11 V .
E = −N
The flux decreases in the process so by Lenz’s Law the flux due to the induced emf is in the
same direction as the original flux of the external B -field. Thus terminal B must be in a higher
potential than terminal A.
When the coil stops moving the magnetic flux no linger changes: ∆ΦM /∆t = 0, so the induced
emf becomes zero.
20.21
First, compute the induced emf in the coil from Eq. (20.3): E = −N ∆ΦM /∆t, where according
to Eq. (20.1) ΦM = BA⊥ . In this case A⊥ is a constant while B changes at the rate of ∆B/∆t,
so ∆ΦM /∆t = ∆(BA⊥ )/∆t = A⊥ (∆B/∆t); and so the magnitude of the induced emf is
|E| = N A⊥ |∆B/∆t|.
When a steady state is reached in the coil there is no more charging or discharging going on
across the capacitor so |E| must be balanced by the voltage difference VC across the capacitor:
|E| = VC = Q/C , where Q is the steady-state charge on the capacitor (of capacitance C ).
Combine this equality with the expression for |E| obtained above to yield
∆B = VC = Q .
|E| = N A⊥ ∆t C
Substitute N = 220, |∆B/∆t| = 20 mT/s, A⊥ = π(10 cm)2 , and C = 30 µF into this equation
and solve for Q:
∆B = (30 µF)(220) π(0.10 m)2 (20 × 10−3 T/s) = 4.1 µC .
Q = CN A⊥ ∆t
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CHAPTER 20
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20.22
The magnitude of the induced emf as a result of a change in flux over a time interval ∆t is
given by Eq. (20.3) to be |E| = N |∆ΦM /∆t|, which causes an induced current I = |E|/R in
the circuit of resistance R. This current exists for the duration of the change in flux, which
takes time ∆t. The total amount of charge ∆Q passing through the circuit in the mean time
is then
N |∆ΦM /∆t| ∆t
N ∆ΦM
|E|∆t
=
=
.
∆Q = I∆t =
R
R
R
20.23
The magnetic flux before the coil is yanked out is ΦM i = BA, while afterwards it reduces to
ΦM f = 0. The magnitude of the change in ΦM is then |∆ΦM | = |ΦM f − ΦM i | = BA. Plug this
expression, along with ∆Q = Kθ, into the formula for ∆Q obtained in the previous problem
to yield
N |∆ΦM |
N BA
∆Q = Kθ =
=
.
R
R
Solve for B :
B=
RKθ
.
NA
20.24
The magnetic flux provided by field B through the coil of area A is ΦM = BA cos θ [see
Eq. (20.1)]. As θ changes from θi = 0◦ to θf = 50◦ due to the rotation of the coil the
corresponding change in ΦM is ∆ΦM = BA cos θf − BA cos θi = BA(cos θf − cos θi ), which
causes an induced emf given by Eq. (20.3): E = −N ∆ΦM /∆t. Plug in N = 20, B = 10 mT,
A = 5.0 × 10−2 m2 , and ∆t = 150 ms to obtain
∆ΦM
BA(cos θf − cos θi )
= −N
∆t
∆t
−2 2
(20)(10 mT)(5.0 × 10 m )(cos 50◦ − cos 0◦ )
=−
150 ms
= 0.024 V = 24 mV .
E = −N
20.25
In Problem(20.22) we proved the formula for ∆Q, the charge that flows through a galvanometer
in response to a magnetic flux change: ∆Q = N |∆ΦM |/R. We now find the expression for
∆ΦM in this case. The initial magnetic flux due to the B -field inside the solenoid through
the coil of area A before the current is reversed is ΦM i = BA, while afterwards it changes to
ΦM f = −BA. (Here we have arbitrarily chosen the initial value of the flux to be positive so the
final value is negative.) The magnitude of the change in ΦM is then |∆ΦM | = |ΦM f − ΦM i | =
CHAPTER 20
ELECTROMAGNETIC INDUCTION
701
| − BA − BA| = 2BA. Plug this expression for |∆ΦM | into the formula for ∆Q to yield
∆Q = N |∆ΦM |/R = N (2BA)/R, which we solve for B :
B=
R∆Q
(0.50 Ω)(2.0 × 10−6 C)
=
= 8.3 × 10−4 T = 0.83 mT .
−4
2
2N A
2(12)(0.50 × 10 m )
20.26
Choose the original flux as positive to find the net change in flux to be ∆ΦM = −10 mWb −
10 mWb = −20 mWb, which took place in ∆t = 200 ms. The resulting emf in the N -turn
solenoid follows from Eq. (20.3) to be
E = −N
∆ΦM
(210)(−20 mWb)
= 21 V .
=−
∆t
200 ms
20.27
The axle moves across the B -field, cutting the field lines, whereby generating a motional emf,
E = vBl sin θ. Here v = 20.0 m/s is the speed of the axle, B = 0.40 × 10−4 T is the Earth’s
magnetic field, and l = 1.8 m is the length of the axle, and θ = 90◦ is the angle between v and
; and so
B
E = vBl = (20.0 m/s)(0.40 × 10−4 T)(1.8 m)(sin 90◦ ) = 1.4 mV .
20.28
(a) Use the right-hand-rule. The pole closer to the viewer (the one on the right), where the
current flows in, is the magnetic north pole.
(b) It points from the magnetic north to the magnetic south, i.e., from the pole closer to the
viewer to the further one (i.e., to the left).
(c) Imagine a vertical loop whose upper border is the rod. As the rod moves upward the
magnetic flux through the loop increases, causing an induced emf in the loop which would drive
a current from B to A in order to maintain the flux through the loop. Thus the electrons in
the rod, moving against the direction of the current, move from A to B . This result can also
be reached by means of analyzing the Lorentz force exerted on the electrons. As the rod moves
upward so do the electrons it carries, and the resulting Lorentz force on the negatively charged
electrons points from A to B .
(d) From A to B —– see analysis in part (c) above.
(e) The length of the portion of the rod that’s cutting the magnetic field lines perpendicularly
is l = 5.0 cm. So
E = Blv = (500 × 10−4 T)(5.0 × 10−2 m)(5.00 m/s) = 12.5 mV .
702
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.29
Use Eq.(20.4) to find the motional emf caused by the wings, considered as a single long conductor
of total length l, sweeping across the Earth’s B -field with speed v :
E = vBl = (200 m/s)(0.050 × 10−3 T)(60 m) = 0.60 V .
20.30
Use Eq. (20.4), with v = 0.50 m/s, B =
0.140 T, and l = 10.0 cm:
N
B
E = vBl
rod
= (0.50 m/s)(0.140 T)(0.100 m)
= 7.0 × 10−3 V = 7.0 mV .
horizontal plane
E
W
The rod lies south-north and moves downward. According to Lenz’s Law the north
end of the rod is at the higher voltage
(positive).
S
v
down
20.31
The wire moves at v = 0.60 m/s relative to the magnetic field lines. Thus Eq. (20.4) gives
E = vBl = (0.60 m/s)(0.200 T)(1.00 m) = 0.12 V ,
with the east end of the wire at the higher voltage (positive).
20.32
From Eq. (20.4) the motional emf in the copper wire is
E = vBl = (0.50 m/s)(1.5 T)(0.30 m) = 0.23 V .
20.33
The bulb would not light, nor would a voltmeter connected across the wing tips be able to pick
up a reading indicating the voltage difference calculated in Problem (20.29). In fact the leads
of the meter also cut through the Earth’s B -field at the same rate as the wings so the same
amount of motional emf would appear in the voltmeter as well as the leads; and once they are
attached to the wings the closed loop consisting of the wings, the leads and the voltmeter would
have two oppositely directed emf of the same strength, so no current can circulate in the circuit.
CHAPTER 20
ELECTROMAGNETIC INDUCTION
703
Alternatively, note that the flux through the closed loop of meter-leads-wings does not change
if the B -field remains constant so no net emf appears in the circuit to drive a current: E ∝
∆ΦM = 0. You could, however, read a voltage difference if the meter stayed on the ground and
still managed to remain connected to the plane.
20.34
Solve for the magnetic field B from Eq. (20.4), E = vBl:
B=
E
0.45 V
= 0.38 T .
=
−2
vl
(600 × 10 m/s)(0.20 m)
20.35
The rod falls straight down with a velocity v while the direction of the B -field is at an angle
is v =
θ = 50◦ below the horizontal. Thus the component of v that is perpendicular to B
⊥
v cos θ, and the induced emf in the rod of length l is [see Eq. (20.4)]
E = v⊥ Bl = vBl cos θ = (2.8 m/s)(0.5 × 10−4 T)(1.00 m)(cos 50◦ ) = 9.0 × 10−5 V ,
or 90 µV. Note that in Eq. (20.4) v is the component of the velocity of the wire perpendicular
to the B -field.
20.36
is
The component of the velocity v of the tube that is perpendicular to the magnetic field B
v⊥ = v sin θ, where θ is the angle between v and B. Thus the induced voltage in the tube of
length l is
E = v⊥ Bl = (v sin θ)Bl = (0.250 m/s)(sin 60.0◦ )(0.045 T)(0.200 m) = 1.9 × 10−3 V .
20.37
As soon as the leading edge of the loop enters the rectangular region of the B -field and starts
cutting field lines, there is an emf [= Blv , see Eq. (20.4)], which produces a counterclockwise
current I = Blv/R. The emf is constant (for a time interval l/v ). When the trailing edge of
the loop enters the field, it starts to produce an oppositely directed emf and the net emf in the
loop goes to zero (which is consistent with the fact that the magnetic flux in the loop no longer
changes). It remains zero until the leading edge emerges from the field at a time 3l/v after it
entered, whereupon the emf jumps to Blv in the opposite direction (due to the now-unbalanced
emf in the trailing edge). A clockwise current I = Blv/R appears and remains constant for a
time interval l/v , until the entire loop exits from the field region.
704
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.38
As the wire of length l cuts the field lines at a speed v it produces an emf in accordance with
Eq. (20.4): E = vBl, which in turn generates a current I in the closed circuit of total resistance
R: I = E/R = vBl/R. This current-carrying wire, while moving in the B -field, is subject to
a magnetic force to the left, with a magnitude of
vBl
B 2 l2 v
FM = IlB =
lB =
.
R
R
Since the wire is undergoing no acceleration the net force exerted on it must vanish, i.e., F −
FM = 0; and so
B 2 l2 v
F = FM =
.
R
20.39
The power Pext supplied by an external agency which provides the force F is
22 B l v
(Blv)2
v=
,
Pext = F v =
R
R
where we used the result of the previous problem. The power dissipated by the lamp, meanwhile,
is
2
vBl
(Blv)2
2
,
R=
Pdiss = I R =
R
R
which is identical to Pext , as expected from the Conservation of Energy.
20.40
First find the speed v of the rod traversing in the magnetic field from Eq. (20.4):
v=
E
0.50 V
=
= 4.55 m/s .
Bl
(0.11 T)(1.00 m)
At this speed the time it takes for the wire to traverse 0.50 m is t = 0.50 m/(4.55 m/s) = 0.11 s.
20.41
The power P needed to properly operate the lamp is given by P = IV , and so V , the voltage
difference across the lamp, is V = P/I . In this case V is provided by the motional emf so
V = E = vBl, where v is the speed of the rod cutting through the magnetic field lines and l is
its length. So P/I = vBl, which gives
l=
which is quite long!
P
5.0 W
=
= 31 m ,
−3
IvB
(200 × 10 A)(2.00 m/s)(0.400 T)
CHAPTER 20
705
ELECTROMAGNETIC INDUCTION
20.42
The power needed to operate the motor is P = V 2 /R, where V is the voltage difference across
the motor and R is its resistance. Similar to the previous problem, let V = E = vBl to find v ,
the desired speed of the rod of length l moving in the magnetic field B : P = V 2 /R = (vBl)2 /R,
which we solve for v :
√
(2.0 W)(10 Ω)
PR
v=
=
= 45 m/s .
Bl
(0.20 T)(0.50 m)
From the Conservation of Energy the power needed to operate the motor is equal to that
delivered by the external force F : P = F v , so
F =
2.0 W
P
=
= 4.5 × 10−2 N .
v
44.7 m/s
20.43
With the previous problem in mind, let P = V 2 /R = E 2 /R = (vBl)2 /R, and solve for B :
(10.0 W)(5.00 Ω)
PR
=
= 11.8 T .
vl
(1.20 m/s)(0.500 Ω)
√
B=
20.44
Refer to Fig. P44 in the text. As the wire descends it cuts through the magnetic field lines,
generating an induced emf in the amount of E = vBl. A counterclockwise current I is then
established in the closed circuit, flowing from left to right in the moving wire: I = E/R =
vBl/R. The current-carrying wire is in turn subject to an upward magnetic force FM =
to its weight,
IlB = (vBl/R)lB = vB 2 l2 /R. Meanwhile the wire (of mass m) is also subject
FW = mg , downward. The net downward force on the wire is then + ↓ F = FW − FM =
mg − vB 2 l2 /R, and the equation of motion for the wire is
+
↓
vB 2 l2
F = FW − FM = mg −
= ma .
R
At first v is small so the motion of the wire is nearly a free fall dominated by the weight
of the wire, which accelerates downward
at the rate of a ≈ g . As the speed v of the wire
F = FW − FM decreases as a result. The wire continues
picks up, so does FM , and + ↓
its downward acceleration, albeit with a gradually diminishing
value of a, until v reaches its
F = FW − FM = ma = 0, i.e.,
terminal (and maximum) value, vm , at which time + ↓
FM = vm B 2 l2 /R = FW = mg , which gives
vm =
mgR
.
B 2 l2
706
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.45
The motional emf is given by Eq. (20.4) to be
E = vBl = (4.0 m/s)(0.80 T)(0.50 m) = 1.6 V .
20.46
The angular frequency of the output emf is ω = 2πf , where f is its frequency. The output emf
varies as a function of time t as E = N ABω sin ωt = 2πN ABf sin(2πf t) [Eq. (20.7)], so the
maximum output is Em = 2πN ABf (which is attained when | sin(2πf t)| = 1). Solve for B :
B=
Em
20.0 V
=
= 0.026 5 T = 26.5 mT .
2πN Af
2π(150)(8.00 × 20.0 × 10−4 m2 )(50.0 Hz)
The angular speed of the coil should be the same as the angular frequency ω of the output emf,
at ω = 2πf = 2π(50.0 Hz) = 314 rad/s.
20.47
Apply Eq. (20.4) to find the emf due to each length:
E = vBl = (22 m/s)(0.35 T)(0.10 m) = 0.77 V .
Since there are 2 lengths per turn and a total of 25 turns the total emf is 2×25×0.77 V = 39 V.
20.48
The maximum emf provided by an ac generator is given by Eq.(20.7): Em = N ABω [sin ωt]m =
N ABω . Plug in Em = 170 V, N = 100, A = 0.50 m2 , ω = 2πf with f = 60 Hz, and solve for
B , the required magnetic field:
B=
Em
170 V
=
= 9.0 × 10−3 T = 9.0 mT .
2
N Aω
(100)(0.50 m ) [(2π)(60 Hz)]
20.49
The standard form of the emf produced by an ac generator as a function of times is given by
Eq.(20.7): E = N ABω sin ωt, where N is the number of turns of the coil, A is its area, B is the
magnetic field, and ω = 2πf is the angular speed of rotation of the coil in the B -field. Here f is
the frequency of the output emf. Note that the maximum emf Em occurs when | sin ωt| reaches
its peak value of 1, whereupon E = Em = N ABω . Thus we may also write E = Em sin ωt.
Comparing this standard form with the expression given: E = 100 sin 376.99t, we get, in this
case, Em = 100 V and ωt = 2πf t = 376.99t, which yields f = 376.99 Hz/2π = 60.000 Hz.
CHAPTER 20
707
ELECTROMAGNETIC INDUCTION
20.50
The maximum emf of an ac generator is given by Eq. (20.7): Em = N ABω [sin ωt]m = N ABω .
Plug in N = 100, A = 1.00 m2 , B = 650 × 10−7 T, and ω = 2πf = 2π(100 Hz) to find
Em = N ABω = (100)(1.00 m2 )(650 × 10−7 T)(2π)(100 Hz) = 4.08 V .
20.51
Following the hint given in the problem statement,
ω
we first find ∆A/∆t, the rate at which a radial strip
of conductor sweeps out an area as a result of its
rotation. Consider a time interval ∆t, during which
∆Α
r
the disk rotates through an angle ∆θ, which satisfies
∆θ
∆θ = ω∆t, with ω the angular speed of rotation
r
for the disk. During that time interval the leads
axis
connected to the resistor between the axis and the
rim also sweeps through the same angle, covering an
area ∆A as shown to the right. Note that ∆A is a
fraction of the total area A of the disk of radius r:
∆A/A = ∆θ/2π , and so
1
∆θ
1
∆θ
2
= πr
= r2 ∆θ = r2 ω∆t .
∆A = A
2π
2π
2
2
rim
The rate of change of magnetic flux due to the rotation of the resistor is then
E=
∆ΦM
B∆A
r2 Bω∆t
1
=
=
= r2 ωB ,
∆t
∆t
2∆t
2
which is equal in magnitude to the induced emf E across the resistor. The induced current
I in the resistor follows as I = E/R = 12 r2 ωB/R. Substitute I = 1.25 mA, r = 25 cm,
ω = 360 rpm = (360 × 2π)/60 s = 37.7 rad/s, and R = 20 Ω into this equation and solve for
B:
B=
2IR
2(1.25 × 10−3 A)(20 Ω)
=
= 2.1 × 10−2 T = 21 mT .
2
2
ωr
(37.7 rad/s)(0.25 m)
20.52
The area of the coil is A = 2 × 10 cm × 15 cm = 3.0 × 102 cm2 = 3.0 × 10−2 m2 (note that
each of the two segments of the coil cutting the field has a length of 15 cm to make a total of
30 cm). As it turns at the rate of ω = 10 rev/s = 10 × 2π/s = 62.83 rad/s in a magnetic field B
(= 0.60 T) an emf of E = N ABω sin ωt is generated in the coil [see Eq. (20.7)]. At the moment
the coil moves perpendicularly to the field it cuts through the field at the greatest rate, so
E = Em = N ABω = (1)(3.0 × 10−2 m2 )(0.60 T)(62.83 rad/s) = 1.1 V .
708
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.53
The emf in a rotating rod of length r cutting through the field lines in a magnetic field B
1 2
was
at an angular speed ω was computed in Problem (20.51) to be E = 2 r ωB , where B
makes an angle
perpendicular to the plane in which the rod rotated. In the present case B
θ with the normal to the rotational plane of the rotor blades so we need to replace B with
B⊥ = B cos θ in the expression for E : E = 12 r2 ωB⊥ = 12 r2 ωB cos θ. Plug in r = 5.0 m,
ω = 240 rpm = (240 × 2π)/60 s = 25.13 rad/s, B = 0.050 mT, and θ = 40◦ to obtain
E=
1 2
1
r ωB cos θ = (5.0 m)2 (25.13 rad/s)(0.050 mT)(cos 40◦ ) = 12 mV .
2
2
20.54
Each turn of the coil has two segments, each of length l moving at speed v , cutting through
the B -field lines to produce an emf of E = v⊥ Bl = vBl sin θ, where θ is the angle between B
and v. The total emf generated in the N -turn coil is the sum of that in all the field-cutting
segments (of which there are 2N ):
Etotal = 2N E = 2N (vBl sin θ) = 2(25)(22 m/s)(0.35 T)(0.10 m)(sin 30◦ ) = 19 V .
20.55
This problem is similar to Problem(20.52). Here the area of the coil is A = 2×5.0 cm ×20 cm =
2.0 × 102 cm2 = 2.0 × 10−2 m2 . As it turns at the rate of ω = 6000 rpm = (6000 × 2π)/60 s =
(200π) rad/s = 628.3 rad/s in a magnetic field B (= 0.50 T) it generates an emf in accordance
with Eq. (20.7):
E = N ABω sin ωt
= (200)(2.0 × 10−2 m2 )(0.50 T)(628.3 rad/s) sin(200πt)
= (1.3 × 103 V) sin(200πt) ,
where t is measured in s.
20.56
Consider a time interval ∆t, during which the coil is turned through an angle ∆θ under the
influence of an external torque τ , which satisfies τ = N IAB sin φ (see Chapter 19). Here
ω = ∆θ/∆t is the angular speed of rotation of the coil and φ = ωt. The work done by the
torque during this time interval is then ∆W = τ ∆θ = (N IAB sin ωt)∆θ, and the power
delivered by the torque is
Pτ =
∆W
(N IAB sin ωt)∆θ
=
= N IABω sin ωt
∆t
∆t
CHAPTER 20
ELECTROMAGNETIC INDUCTION
709
which, by the Conservation of Energy, is equal to the power generated by the induced emf:
PE = IE . Equate the expressions for Pτ and PE above to yield N IABω sin ωt = IE , which
gives
E = N ABω sin ωt .
20.57
(a) According to Eq. (19.1) µ = 2πBr/I , where B is in T, r is in m, and I is in A. So the SI
unit of µ is T · m/A.
(b) The magnetic flux has the unit of magnetic field (B , in T) times area (A, in m2 ), i.e.,
T · m2 .
(c) Solve for L from Eq. (20.8): L = N ΦM /I . The unit of ΦM is T · m2 , that of I is A, while
N is unitless. Thus H, the unit of L, is equivalent to T · m2 /A.
(d) Solve for µ0 from Eq. (20.9): µ0 = Ll/N 2 A. Here L is in H, l is in m, A is in m2 , while
N is unitless. Thus the unit of µ0 is H · m/m2 = H/m. [Note that this is consistent with the
expression T · m/A found in part (a) above, given that 1 H = 1 T · m2 /A.]
20.58
According to Eq. (20.9) L ≈ µN 2 A/l. Here we replaced µ0 with µ to accommodate the
possibility that the core may be filled with materials other than vacuum.
(b) Its diameter is d = 0.50 cm = 5.0 × 10−3 m, so its cross-sectional area is A = πd2 /4 =
π(5.0 × 10−3 m)2 /4 = 2.0 × 10−5 m2 .
(c)
L≈
µN 2 A
(0.30 × 10−3 H/m)(349)2 (1.96 × 10−5 m2 )
= 48 mH .
=
l
1.50 × 10−2 m
20.59
The self-inductance L is defined in Eq. (20.8): N ΦM = LI , where the magnetic flux ΦM in the
coil is caused by the current I running in the N -turn coil itself. Thus
N ΦM
(500)(2.0 × 10−3 Wb)
L=
= 0.26 H .
=
I
3.8 A
20.60
The flux linkage in an N -turn coil with a magnetic flux ΦM is defined as N ΦM . Thus from
Eq. (20.8), N ΦM = LI , we get
L=
6.0 mWb
N ΦM
=
= 3.0 mH .
I
2.0 A
710
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.61
Solve for ΦM from Eq. (20.8), N ΦM = LI :
ΦM =
(4.00 × 10−3 H)(1.00 A)
LI
=
= 8.00 × 10−5 Wb = 80.0 µWb .
N
50
20.62
The inductance is proportional to the relative permeability, which in this case increases from
1 (in the air) to 2000 (in the iron core). The inductance increases correspondingly to L =
2000(3.0 mH) = 6.0 × 103 mH = 6.0 H.
20.63
Similar to Problem (20.61), we solve for ΦM from Eq. (20.8), N ΦM = LI :
ΦM =
LI
(200 × 10−3 H)(2.0 A)
=
= 8.0 × 10−4 Wb = 0.80 mWb .
N
500
20.64
Recall that the magnetic field inside a solenoid is B = µ0 nI , and so the SI unit of µ0 is
Wb/m2
Wb
[B]
[µ0 ] =
.
=
=
−1
[n][I]
(m )(A)
m·A
On the other hand from ΦM = LI/N we get [ΦM ] = Wb = [L][I]/[N ] = H · A. Thus
[µ0 ] =
Wb
H·A
=
= H/m .
m·A
m·A
20.65
Use Eq. (20.8), N ΦM = LI , to solve for N :
N=
LI
(2.5 H)(1.80 A)
=
= 2.5 × 103 .
ΦM
1.80 × 10−3 Wb
20.66
The cross-sectional area of the solenoid of diameter d is A = 14 πd2 . From Eq. (20.9) we then
obtain the self-inductance of the solenoid:
µ0 N 2 A
l
(4π × 10−7 T · m/A)(200)2 [π(4.0 × 10−2 m)2 /4]
=
0.50 m
−4
= 1.3 × 10 H = 0.13 mH .
L≈
CHAPTER 20
ELECTROMAGNETIC INDUCTION
711
20.67
The self-inductance of a solenoid is given by Eq. (20.9): L ≈ µ0 N 2 A/l, which we solve for N :
Ll
(1.0 × 10−3 H)(0.40 m)
N≈
=
= 8.4 × 102 .
µ0 A
(4π × 10−7 T · m/A)(4.5 × 10−4 m2 )
20.68
From Eq. (20.10) the emf in the coil is
E = −L
∆I
= −(10 H)(+4.0 A/s) = −40 V .
∆t
20.69
Solve for L from Eq. (20.10):
E
= 10 V = 5.0 H .
L = −
∆I/∆t 2.0 A/s
Note that we took the absolute value here since we know that L > 0. (In fact the signs of E
and ∆I/∆t are always opposite to each other so as to keep L positive.)
20.70
From Eq. (20.10) the induced emf is
∆I
0 − 2.5 A
If − Ii
E = −L
= 1.3 × 103 V = 1.3 kV .
= −L
= −(5.0 H)
∆t
∆t
10 × 10−3 s
20.71
Solve for L from Eq. (20.10):
L=−
E
−500 V
= 2.5 H .
=−
∆I/∆t
4.0 A/0.020 s
Note that here E < 0 since the current is increasing.
20.72
The self-inductance of a solenoid is given by Eq. (20.9): L ≈ µN 2 A/l, where we replaced µ0
with µ, the permeability of iron. Here A = 14 πd2 , with d the diameter of the solenoid. Solve
this for N , the number of turns needed:
Ll
(300 × 10−3 H)(0.27 m)
N≈
=
= 3.7 × 102 .
µA
[1.5 × 103 (4π × 10−7 T · m/A)][π(2.0 × 10−2 m)2 /4]
712
CHAPTER 20
ELECTROMAGNETIC INDUCTION
Note that Eq. (20.9) becomes exact only for an infinitely long solenoid, which is hardly the case
here with only a few hundred turns. So we must stress that the result is only approximate.
20.73
The change in current is ∆I = −5.0 A − (+5.0 A) = −10 A, which took place in a time interval
∆t, generating an emf of E = −L(∆I/∆t) [see Eq. (20.10)]. Plug in L = 1.5 H, E = 100 V,
and solve for ∆t:
(1.5 H)(−10 A)
L∆I
=−
∆t = −
= 0.15 s .
E
100 V
20.74
The change in current is ∆I = If − Ii = +1.0 A − (−2.0 A) = +3.0 A, where we arbitrarily
choose counterclockwise to be the positive direction. The resulting emf is given by Eq. (20.10):
∆I
3.0 A
−3
E = −L
= −12 V ,
= −(2.0 × 10 H)
∆t
500 × 10−6 s
where the negative sign indicates that the induced emf is against the change in current.
20.75
The change in current in the coil is ∆I = I −(−250 mA) = I +250 mA, where we’ve chosen the
new direction of the current to be positive. From Eq. (20.10), E = −L(∆I/∆t), we may then
solve for ∆I : ∆I = −E∆t/L, or I + 250 mA = −E∆t/L. Plug in E = −5.0 V, ∆t = 2.5 ms,
and L = 2.0 mH to obtain
I=−
E∆t
(−5.0 V)(2.5 ms)
− 250 mA = 6.0 A .
− 250 mA = −
L
2.0 mH
Note the negative sign for E , which is always against the direction for ∆I , which in turn is
chosen to be positive as it is in the same direction as I , the new current.
20.76
The B -field inside an N -turn toroid of mean radius b was computed in Chapter 19 to be
B = µ0 N I/2πb, where I is the current in the toroid. The magnetic flux through its crosssectional area is then ΦM = BA = (µ0 N I/2πb)A, and according to Eq.(20.8) its self-inductance
L is given by
N ΦM
N (µ0 N I/2πb)A
µ0 N 2 A
L=
=
=
.
I
I
2πb
20.77
Before the shaft is inserted the inductance L0 of the air-filled tube is that of an N -turn solenoid
of length l and cross-sectional area A. So according to Eq.(20.9) L0 ≈ µ0 N 2 A/l = (An2 )(µ0 l),
where n = N/l.
CHAPTER 20
713
ELECTROMAGNETIC INDUCTION
After the shaft is inserted a portion of the length d of the tube, whose total length is l, is
displaced by materials with permeability µ. Note that An2 does not change in the expression
for the inductance but now we must now replace µ0 l with µ0 (l − d) + µd, where the first term
comes from the remaining air-filled segment with permeability µ0 and length l − d while the
second term corresponds to the newly inserted shaft of length d and permeability µ. Thus the
new inductance is L ≈ An2 [µ0 (l − d) + µd], and so the change in inductance is
∆L = L − L0 ≈ An2 [µ0 (l − d) + µd] − An2 µ0 l = An2 (µ − µ0 )d =
d(µ − µ0 )N 2 A
,
l2
where in the last step we used n = N/l.
20.78
(a) Due to the presence of the inductor the current in the R-L circuit cannot change immediately
(or there would be an infinitely large value of induced emf on the inductor). So the moment the
switch is closed the reading of the milliammeter is zero, as it cannot jump abruptly to a finite
value.
(b) When a steady-state current I is attained the current is no longer changing, so the induced
emf on the inductor is now zero. The circuit effectively consists of just the voltage source and
the resistor. So V = IR and
I=
V
20 V
=
= 10 mA .
R
2.0 × 103 Ω
20.79
(a) from the problem statement 5τ ≈ 2.5 s, so τ ≈ 0.50 s.
(b) For an R-L circuit τ = L/R.
(c)
R=
1.0 H
L
=
= 2.0 Ω .
τ
0.50 s
(d) The current varies as a function of time as I = Im (1 − e−t/τ ), where Im = V /R is the
steady-state current when t τ . Plug in V = 12 V, R = 2.0 Ω, and t/τ = 10 to obtain
V 12 V −t/τ
I=
1−e
1 − e−10 = 6.0 A .
=
R
2.0 Ω
Note that the factor e−10 is negligible in comparison with 1, meaning that the current has
practically reached its steady-state value at t = 10τ .
(e) Zero, as the current through the inductor has reached a constant value: VL ∝ ∆I/∆t = 0.
714
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.80
When a steady-state current I is attained ∆I/∆t = 0, so Eq. (20.11) reads V = IR, which we
solve for I , the steady-state current:
I=
12 V
V
=
= 2.0 A .
R
6.0 Ω
20.81
Plug in e ≈ 2.718 3 to obtain 1 − 1/e ≈ 1 − 1/2.718 3 ≈ 1 − 0.368 = 0.632.
20.82
Immediately after the switch is closed the current I is zero. In fact if I jumps abruptly to
any non-zero value then ∆I/∆t at that moment would have to be infinite and so would be
the induced emf on the inductor, which is impossible. Several hours later the current is no
longer changing so ∆I/∆t → 0, and Eq. (20.11) gives V = L(∆I/∆t) + IR → IR. Plug in
V = 12.0 V and R = 8.0 Ω to obtain I = V /R = 12.0 V/8.0 Ω = 15 A.
20.83
Initially there is no current: I = 0. Thus Eq. (20.11) reads V = L(∆I/∆t), which gives the
voltage of the battery to be
∆I
V =L
= |E| = 20.0 V ,
∆t
where we used Eq. (20.10). Solve for ∆I/∆t:
∆I
V
|E|
20.0 V
= 2.00 × 103 A/s .
=
=
=
∆t
L
L
10.0 × 10−3 H
20.84
Immediately after the switch is closed there is no current (I = 0), and the magnitude |E| of
the induced emf is the same as the voltage V of the power supply: V = L(∆I/∆t) + IR =
L(∆I/∆t) = |E|. Thus
L=
|E|
V
20.0 V
= 0.20 H = 200 mH .
=
=
∆I/∆t
∆I/∆t
100 A/s
20.85
Similar to the previous problem, immediately after the switch is closed |E| = V , and
L=
60 V
|E|
=
= 0.30 H .
∆I/∆t
200 A/s
CHAPTER 20
ELECTROMAGNETIC INDUCTION
715
20.86
Apply Eq. (20.11) to the circuit: V = L(∆I/∆t) + IR. Plug in V = 100.0 V, R = 25 Ω,
∆I/∆t = 200 A/s and L = 0.400 H and solve for I :
∆I
1
1
V −L
=
[100.0 V − (0.400 H)(200 A/s)] = 0.80 A .
I=
R
∆t
25 Ω
20.87
(a) In the steady state no voltage drop is present across the inductor so the steady-state current
Im in the circuit follows Ohm’s Law to be
Im =
150 V
V
=
= 6.0 A .
R
25 Ω
(b) The time constant is given by
300 mH
L
=
= 12 ms .
R
25 Ω
(c) The current varies as a function of time as I = Im 1 − e−tR/L . At t = L/R we have
e−tR/L = e−1 = 0.367 88, whereupon
I = Im 1 − e−tR/L = (6.0 A)(1 − 0.367 88) = 3.8 A .
20.88
(a) The milliammeter is in series with an inductor so the current I is the same for both. Before
S1 is closed the current I through the inductor is of course zero. An instant after it is closed I
should remain zero, as it cannot change abruptly in an inductor.
(b) L = 10 mH = 10 × 10−3 H, R = 1.0 kΩ = 1.0 × 103 Ω.
(c) The time constant is τ = L/R = 10 × 10−3 H/1.0 × 103 Ω = 1.0 × 10−5 s = 10 µs.
(d) When the current reaches Im , its steady-state value, ∆I/∆t = 0. So the voltage difference
across the inductor, being proportional to ∆I/∆t, drops to zero. The load of the circuit is just
the 1.0-kΩ resistor, so now I = Im = E/R = 10 V/1.0 kΩ = 10 mA.
(e) Since no voltage difference exists across the inductor when I reaches its steady-state value,
the emf of the battery is applied entirely to the resistor; so V = E = 10 V across the resistor.
(f) The current varies as a function of time as I = Im (1 − e−t/τ ), where τ = L/R is the time
constant. We wish to solve for t/τ . First rewrite it as 1 − I/Im = e−t/τ , then take the natural
logarithm of both sides: ln(1 − I/Im ) = ln e−t/τ = −t/τ . Thus
716
CHAPTER 20
I
t = −τ ln 1 −
Im
9.9 mA
= −τ ln 1 −
10 mA
ELECTROMAGNETIC INDUCTION
= 4.6 τ .
(g) Since τ = 10 µs, t = 4.6 τ = 4.6(10 µs) = 46 µs.
20.89
(a) Similar to the previous problem, the milliammeter is in series with an inductor so the
current I is the same for both. Before S1 is closed the current I through the inductor is of
course zero. An instant after it is closed I should remain zero, as it cannot change abruptly in
an inductor.
(b) After a long time the current I must have reached Im , its steady-state value, whereupon
∆I/∆t = 0. So the voltage difference across the inductor, being proportional to ∆I/∆t, drops
to zero. The load of the circuit is just the 10-Ω resistor, so now I = Im = E/R = 10 V/10 Ω =
1.0 A.
(c)
τ=
L
2.0 mH
=
= 0.20 ms .
R
10 Ω
(d) Before S1 is open and S2 is closed the current reading is 1.0 A, as discussed in part (b). Just
after that, the current should remain 1.0 A, in the same direction (positive), since no abrupt
change in I is possible in an inductor.
(e) The current varies as a function of time as I = Im e−t/τ , where τ = R/L is the time
constant. We wish to solve for solve for t/τ . First rewrite it as I/Im = e−t/τ , then take the
natural logarithm of both sides: ln(I/Im ) = ln e−t/τ = −t/τ . Thus
t = −τ ln
I
0.140 A
= −τ ln
= 1.97 τ ≈ 2.0 τ .
Im
1.0 A
(f) Since τ = 0.20 ms, t = 2.0 τ = 2.0(0.20 ms) = 0.40 ms.
20.90
The voltage VL across the inductor is related to the time rate-of-change of current ∆I/∆t via
VL = −L(∆I/∆t).
From t = 0 to t = 200 ms the current increases linearly from 0 to 2.5 A so ∆I/∆t = ∆I/∆t =
2.5 A/0.200 s = 12.5 A/s, and
VL = −L
∆I
= (2.00 H)(12.5 A/s) = −25 V .
∆t
From t = 200 ms to t = 400 ms the current stays at a constant value of 2.5 A so ∆I/∆t = 0,
and VL = −L(∆I/∆t) = 0.
CHAPTER 20
717
ELECTROMAGNETIC INDUCTION
From t = 400 ms to t = 500 ms the current decreases linearly from 2.5 A back to 0 so ∆I/∆t =
∆I/∆t = −2.5 A/(0.500 s − 0.400 s) = −25 A/s, and
∆I
= −(2.00 H)(−25 A/s) = +50 V .
∆t
The voltage VL as a function of t is plotted below.
VL = −L
V (V)
50
0
100
200
300
400
500
t (ms)
−25
20.91
(a) In an R-L circuit the voltage output V of the power source is related to the current I in
the circuit via Eq. (20.11): V = L(∆I/∆t) + IR. In the steady state there is no change in
current, i.e., ∆I/∆t = 0, and so the steady-state current Im in the circuit satisfies V = Im R,
or
V
12 V
Im =
=
= 2.0 A .
R
6.0 Ω
(b) The time constant is given by
L
24 H
=
= 4.0 s .
R
6.0 Ω
(c) The current in the R-L circuit varies as a function of time as I = Im 1 − e−tR/L ,
which we rewrite as e−tR/L = 1 − I/Im . To solve for the time t for I to reach within 1%
of Im (i.e., when I/Im = 1 − 1% = 0.99), take the natural logarithm of both sides to obtain
ln e−tR/L = −tR/L = ln (1 − I/Im ), or
L
I
t = − ln 1 −
= −(4.0 s) ln (1 − 0.99) = 18 s ≈ 0.2 × 102 s .
R
Im
718
CHAPTER 20
ELECTROMAGNETIC INDUCTION
Note that we made use of the identity ln ex = x.
20.92
Plug V = 12 V, R = 6.0 Ω, L = 24 H, and t = 2.0 s into the formula for I given in the problem
statement to obtain
V
−t
12 V
−2.0 s
I=
1 − exp
=
1 − exp
= 0.79 A .
R
L/R
6.0 Ω
24 H/6.0 Ω
20.93
Start from Eq. (20.11), V = L(∆I/∆t) + IR, and solve for ∆I/∆t, the instantaneous rate of
change of current:
V − IR
∆I
=
,
∆t
L
which corresponds to the slope of the curve in Fig. (20.22) at any given instant.
At t = 0, I = 0 and ∆I/∆t = (V − IR)/L = V /L = 120 V/50 mH = 2.4 kA/s. At t = L/R
(one time constant), I ≈ 0.63V /R and ∆I/∆t = (V − IR)/L ≈ (120 V)(1 − 0.63)/50 mH =
0.89 kA/s. The current rises from zero to a maximum value of V /R = 120 V/200 Ω = 0.60 A
(whereupon ∆I/∆t = 0), so it will never reach 1.0 A.
20.94
Write down the equation for the R-L circuit: V = L(∆I/∆t) + IR. From the problem
statement we know that the steady-state current Im is 2.0 A, which occurs when ∆I/∆t = 0;
so V = Im R and
V
12 V
R=
=
= 6.0 Ω .
Im
2.0 A
Now, the time rate-of-change of the current is found in the previous problem to be ∆I/∆t =
(V − IR)/L. Plug in ∆I/∆t = 12 A/s, V = 12 V, I = 1.0 A, R = 6.0 Ω, and solve for L:
L=
V − IR
12 V − (1.0 A)(6.0 Ω)
= 0.50 H .
=
∆I/∆t
12 A/s
20.95
Use Eq. (20.14) for the energy stored in an inductor:
PEM =
1 2
1
LI = (40.0 mH)(2.0 A)2 = 80 mJ .
2
2
CHAPTER 20
719
ELECTROMAGNETIC INDUCTION
20.96
Apply Eq. (20.14) for the magnetic energy stored in an inductor:
PEM =
1 2
1
LI = (200 × 10−3 H)(10 A)2 = 10 J .
2
2
20.97
Solve for I from Eq. (20.14):
I=
2P E M
=
L
2(10 J)
= 6.3 A .
500 × 10−3 H
20.98
The steady-state current in the circuit is given by I = V /R, where V = 80.0 V and R = 10.0 Ω.
Thus from Eq. (20.14)
2
2
1 2
1
1
80.0 V
V
−3
= (40.0 × 10 H)
= 1.28 J .
PEM = LI = L
2
2
R
2
10.0 Ω
20.99
Find L from Eq. (20.14), PEM = 12 LI 2 :
L=
2PEM
2(15 J)
=
= 1.2 H .
I2
(5.0 A)2
20.100
Combine the electric energy density, uE = 12 ε0 E 2 , with the magnetic energy density uM , given
in Eq. (20.15), to obtain the total energy density of the electromagnetic field in the region to be
uEM = uE + uM =
=
1
1 B2
ε0 E 2 +
2
2 µ0
1
(0.50 × 10−4 T)2
(8.854 2 × 10−12 C2/N · m2 )(100 V/m)2 +
2
2(4π × 10−7 T · m/A)
= 4.43 × 10−8 J/m3 + 9.95 × 10−4 J/m3
= 1.0 × 10−4 J/m3 .
Note that the magnetic energy density uM is about four orders of magnitude greater than uE ,
the electric energy density.
720
CHAPTER 20
ELECTROMAGNETIC INDUCTION
20.101
The magnetic field B inside a long solenoid of cross-sectional area A with n turns per unit
length is B = µ0 nI as a current I flows through it. The magnetic energy density inside the
solenoid then follows from Eq. (20.15) to be uM = 12 B 2 /µ0 = 12 (µ0 nI)2 /µ0 = 12 µ0 n2 I 2 . Now
consider a segment of the solenoid with a cross-sectional area A and length l. The volume of
the segment is V = Al, and the total magnetic energy stored in the segment is
1
2 2
PEM = uM V =
µ n I Al .
2 0
Plug in n = 2.0/mm = 2.0 × 103 /m, I = 5.0 A, and A = πr2 = π(2.0 × 10−2 m)2 =
1.257 × 10−3 m2 (where r = 2.0 m is the radius of the solenoid), to obtain PEM /l, the magnetic
energy stored in the solenoid per unit length:
PEM
=
l
1
2 2
2 µ0 n I Al
l
=
1
µ n2 I 2 A
2 0
1
(4π × 10−7 T · m/A)(2.0 × 103 /m)2 (5.0 A)2 (1.257 × 10−3 m2 )
2
= 7.9 × 10−2 J/m = 79 mJ/m .
=
20.102
From the result of the previous problem, the magnetic energy PEM stored in the solenoid is
found to be
1 2 2
µN 2 I 2 A
µn I Al =
,
2
2l
where we used n = N/l and replaced µ0 with µ for the ferromagnetic core. Plug in µ = 500µ0 ,
N = 200, I = V /R = 12 V/1.20 Ω = 10 A, A = π(0.25 × 10−2 m)2 = 3.125 × 10−6 m2 , and
l = 0.095 m into the expression above to obtain PEM = 26 J.
PEM =
20.103
Since the radius of the Earth R⊕ is far greater than H , the thickness of the region in considera2
H . The mean magnetic energy
tion (= 200 km), the volume of the region is roughly V ≈ 4πR⊕
1 2
−4
density in the region is uM = 2 B /µ0 , where B = 0.4 × 10 T. The total magnetic energy
stored in the region is then
1 B2
PEM = uM V ≈
(4πR⊕ H)
2 µ0
(0.4 × 10−4 T)2 (4π)(6.371 × 106 m)2 (200 × 103 m)
=
2(4π × 10−7 T · m/A)
= 7 × 1016 J ,
CHAPTER 20
ELECTROMAGNETIC INDUCTION
721
which is equivalent to the energy stored in 7×1016 J/(108 J/gallon) = 7×108 gallons of gasoline.
20.104
The flux linkage is defined in Eq. (20.8):
N ΦM = LI = (200 × 10−3 H)(5.0 A) = 1.0 Wb .
The energy stored in the circuit is
PEM =
1 2
1
LI = (200 × 10−3 H)(5.0 A)2 = 2.5 J .
2
2