31
Faraday’s Law
CHAPTER OUTLINE
31.1
31.2
31.3
31.4
31.5
31.6
31.7
Faraday’s Law of Induction
Motional emf
Lenz’s Law
Induced emf and Electric
Fields
Generators and Motors
Eddy Currents
Maxwell’s Equations
ANSWERS TO QUESTIONS
Q31.1
Magnetic flux measures the “flow” of the magnetic field
through a given area of a loop—even though the field does not
actually flow. By changing the size of the loop, or the
orientation of the loop and the field, one can change the
magnetic flux through the loop, but the magnetic field will not
change.
Q31.2
The magnetic flux is Φ B = BA cos θ . Therefore the flux is
maximum when B is perpendicular to the loop of wire and zero
when there is no component of magnetic field perpendicular to
the loop. The flux is zero when the loop is turned so that the
field lies in the plane of its area.
Q31.3
The force on positive charges in the bar is F = q v × B . If the bar
is moving to the left, positive charge will move downward and
accumulate at the bottom end of the bar, so that an electric field
will be established upward.
a
f
Q31.4
No. The magnetic force acts within the bar, but has no influence on the forward motion of the bar.
Q31.5
By the magnetic force law F = q v × B : the positive charges in the moving bar will flow downward
and therefore clockwise in the circuit. If the bar is moving to the left, the positive charge in the bar
will flow upward and therefore counterclockwise in the circuit.
Q31.6
We ignore mechanical friction between the bar and the rails. Moving the conducting bar through the
magnetic field will force charges to move around the circuit to constitute clockwise current. The
downward current in the bar feels a magnetic force to the left. Then a counterbalancing applied
force to the right is required to maintain the motion.
Q31.7
A current could be set up in the bracelet by moving the bracelet through the magnetic field, or if the
field rapidly changed.
Q31.8
Moving a magnet inside the hole of the doughnut-shaped toroid will not change the magnetic flux
through any turn of wire in the toroid, and thus not induce any current.
a
f
213
214
Faraday’s Law
Q31.9
As water falls, it gains speed and kinetic energy. It then pushes against turbine blades, transferring
its energy to the rotor coils of a large AC generator. The rotor of the generator turns within a strong
magnetic field. Because the rotor is spinning, the magnetic flux through its turns changes in time as
−NdΦ B
. This induced emf is the
Φ B = BA cos ω t . Generated in the rotor is an induced emf of ε =
dt
voltage driving the current in our electric power lines.
Q31.10
Yes. Eddy currents will be induced around the circumference of the copper tube so as to fight the
changing magnetic flux by the falling magnet. If a bar magnet is dropped with its north pole
downwards, a ring of counterclockwise current will surround its approaching bottom end and a ring
of clockwise current will surround the receding south pole at its top end. The magnetic fields created
by these loops of current will exert forces on the magnet to slow the fall of the magnet quite
significantly. Some of the original gravitational energy of the magnet will appear as internal energy
in the walls of the tube.
Q31.11
Yes. The induced eddy currents on the surface of the aluminum will slow the descent of the
aluminum. It may fall very slowly.
Q31.12
The maximum induced emf will increase, increasing the terminal voltage of the generator.
Q31.13
The increasing counterclockwise current in
the solenoid coil produces an upward
magnetic field that increases rapidly. The
increasing upward flux of this field
through the ring induces an emf to
produce clockwise current in the ring. The
magnetic field of the solenoid has a
radially outward component at each point
on the ring. This field component exerts
upward force on the current in the ring
there. The whole ring feels a total upward
force larger than its weight.
FIG. Q31.13
Q31.14
Oscillating current in the solenoid produces an always-changing magnetic field. Vertical flux
through the ring, alternately increasing and decreasing, produces current in it with a direction that
is alternately clockwise and counterclockwise. The current through the ring’s resistance produces
internal energy at the rate I 2 R .
Q31.15
(a)
The south pole of the magnet produces an upward magnetic field that increases as the
magnet approaches. The loop opposes change by making its own downward magnetic field;
it carries current clockwise, which goes to the left through the resistor.
(b)
The north pole of the magnet produces an upward magnetic field. The loop sees decreasing
upward flux as the magnet falls away, and tries to make an upward magnetic field of its
own by carrying current counterclockwise, to the right in the resistor.
Chapter 31
Q31.16
(a)
The battery makes counterclockwise current
I 1 in the primary coil, so its magnetic field
B1 is to the right and increasing just after the
switch is closed. The secondary coil will
oppose the change with a leftward field B 2 ,
which comes from an induced clockwise
current I 2 that goes to the right in the
resistor.
(b)
At steady state the primary magnetic field is
unchanging, so no emf is induced in the
secondary.
(c)
The primary’s field is to the right and
decreasing as the switch is opened. The
secondary coil opposes this decrease by
making its own field to the right, carrying
counterclockwise current to the left in the
resistor.
215
FIG. Q31.16
Q31.17
The motional emf between the wingtips cannot be used to run a light bulb. To connect the light, an
extra insulated wire would have to be run out along each wing, making contact with the wing tip.
The wings with the extra wires and the bulb constitute a single-loop circuit. As the plane flies
through a uniform magnetic field, the magnetic flux through this loop is constant and zero emf is
generated. On the other hand, if the magnetic field is not uniform, a large loop towed through it will
generate pulses of positive and negative emf. This phenomenon has been demonstrated with a cable
unreeled from the Space Shuttle.
Q31.18
No, they do not. Specifically, Gauss’s law in magnetism prohibits magnetic monopoles. If magnetic
monopoles existed, then the magnetic field lines would not have to be closed loops, but could begin
or terminate on a magnetic monopole, as they can in Gauss’s law in electrostatics.
Q31.19
(a)
A current is induced by the changing magnetic flux through the a ring of the tube, produced
by the high frequency alternating current in the coil.
(b)
The higher frequency implies a greater rate of change in the magnetic field, for a larger
induced voltage.
(c)
The resistance of one cubic centimeter in the bulk sheet metal is low, so the I 2 R rate of
production of internal energy is low. At the seam, the current starts out crowded into a small
area with high resistance, so the temperature rises rapidly, and the edges melt together.
(d)
The edges must be in contact to allow the induced emf to create an electric current around
the circumference of the tube. Additionally, (duh) the two edges must be in contact to be
welded at all, just as you can’t glue two pieces of paper together without putting them in
contact with each other.
216
Faraday’s Law
SOLUTIONS TO PROBLEMS
Section 31.1
Faraday’s Law of Induction
Section 31.3
Lenz’s Law
a
f
∆Φ B ∆ NBA
=
= 500 mV
∆t
∆t
P31.1
ε=
P31.2
2.50 T − 0.500 T 8.00 × 10 −4 m 2
∆Φ B ∆ B ⋅ A
ε =
=
=
1.00 s
∆t
∆t
a f a
ε = 1.60 mV and I loop =
P31.3
ε = −N
fe
j FG 1 N ⋅ s IJ FG 1 V ⋅ C IJ
H 1 T ⋅C ⋅ mK H 1 N ⋅ mK
ε 1.60 mV
=
= 0.800 mA
R
2.00 Ω
FG
H
IJ
K
°− cos 0° I
f FGH cos 180
JK
0.200 s
cos θ f − cos θ i
∆BA cos θ
= − NBπ r 2
= −25.0 50.0 × 10 −6 T π 0.500 m
∆t
∆t
j a
e
2
ε = +9.82 mV
P31.4
dΦ B
ABmax − t τ
dB
e
= −A
=
dt
dt
τ
(a)
ε=−
(b)
e0.160 m ja0.350 Tf e
ε=
(c)
At t = 0 ε = 28.0 mV
2
P31.5
−4.00 2.00
2.00 s
= 3.79 mV
Noting unit conversions from F = qv × B and U = qV , the induced voltage is
a f
FG
H
IJ
K
a
fe
j
FG
H
+200 1.60 T 0.200 m 2 cos 0° 1 N ⋅ s
d B⋅A
0 − Bi A cos θ
ε = −N
= −N
=
dt
∆t
1 T ⋅C ⋅ m
20.0 × 10 −3 s
ε 3 200 V
= 160 A
I= =
R 20.0 Ω
P31.6
ε = −N
∆t =
a
N BA − 0
dΦ B
=−
dt
∆t
NBA
ε
=
f
e j = 500a0.200fπ e5.00 × 10 j
−2 2
NB π r 2
ε
10.0 × 10 3
= 7.85 × 10 −5 s
IJ FG 1 V ⋅ C IJ = 3 200 V
KH N ⋅m K
Chapter 31
P31.7
ε =
217
a f
d BA
dI
= 0.500 µ 0 nA = 0.480 × 10 −3 V
dt
dt
(a)
I ring =
ε 4.80 × 10 −4
=
= 1.60 A
R 3.00 × 10 −4
(b)
Bring =
µ0I
= 20.1 µT
2rring
(c)
Coil’s field points downward, and is increasing, so
Bring points upward .
FIG. P31.7
P31.8
ε =
a f
d BA
dI
∆I
= 0.500 µ 0 nA = 0.500 µ 0 nπ r22
dt
dt
∆t
µ 0nπ r22 ∆I
ε
=
R
2 R ∆t
(a)
I ring =
(b)
B=
(c)
The coil’s field points downward, and is increasing, so
Bring points upward .
µ 02nπ r22 ∆I
µ0I
=
2r1
4r1 R ∆t
FIG. P31.8
P31.9
(a)
dΦ B = B ⋅ dA =
(b)
ε=−
ε=−
LM
N
IJ OP = − LM µ L lnFG h + w IJ OP dI
K Q N 2π H h K Q dt
T ⋅ m A ja1.00 mf F 1.00 + 10.0 I
lnG
H 1.00 JK b10.0 A sg =
2π
FG
H
dΦ B
d µ 0 IL
h+w
=−
ln
dt
dt 2π
h
e4π × 10
−7
FG
H
z
h+ w
µ0I
µ 0 IL dx
µ 0 IL
h+ w
Ldx : Φ B =
=
ln
2π x
h
2π x
2π
h
IJ
K
0
−4.80 µV
The long wire produces magnetic flux into the page through the rectangle,
shown by the first hand in the figure to the right.
As the magnetic flux increases, the rectangle produces its own magnetic
field out of the page, which it does by carrying counterclockwise current
(second hand in the figure).
FIG. P31.9
218
P31.10
Faraday’s Law
b
g
Φ B = µ 0 nI Asolenoid
ε = −N
dΦ B
dI
2
= − Nµ 0 n π rsolenoid
dt
dt
e
e
j
g b600 A sg cosa120tf
jb
je
ε = −15.0 4π × 10 −7 T ⋅ m A 1.00 × 10 3 m −1 π 0.020 0 m
a f
2
ε = −14.2 cos 120 t mV
P31.11
For a counterclockwise trip around the left-hand loop, with
B = At
a f
d
At 2 a 2 cos 0° − I 1 5 R − I PQ R = 0
dt
e j
and for the right-hand loop,
a f
d
Ata 2 + I PQ R − I 2 3 R = 0
dt
where I PQ = I 1 − I 2 is the upward current in QP.
i
Aa + I R = I a3 Rf
5
2 Aa − 6 RI − e Aa
3
2
and
PQ
2
I PQ
ε =
2
2
PQ
I PQ =
P31.12
j
+ I PQ R = 0
Aa 2
upward, and since R = 0.100 Ω m 0.650 m = 0.065 0 Ω
23 R
b
e1.00 × 10
=
−3
b
ja
23 0.065 0 Ω
FG IJ
H K
f
T s 0.650 m
g
f
ga
2
= 283 µA upward .
∆Φ B
dB
A = N 0.010 0 + 0.080 0t A
=N
dt
∆t
b
g
b
g b
fe
j
At t = 5.00 s , ε = 30.0 0.410 T s π 0.040 0 m
P31.13
FIG. P31.11
d
2 Aa 2 − 5 R I PQ + I 2 − I PQ R = 0
Thus,
a
B = µ 0 nI = µ 0 n 30.0 A 1 − e −1.60 t
a fe
jz dA
Φ = µ na30.0 A fe1 − e
jπ R
dΦ
ε = −N
= − Nµ na30.0 A fπ R a1.60fe
dt
g
2
= 61.8 mV
z
Φ B = BdA = µ 0 n 30.0 A 1 − e −1.60 t
B
−1.60 t
0
B
a fe
a68.2 mVfe
ε = − 250 4π × 10
ε=
2
0
−7
−1.60 t
2
N A
2
−1.60 t
je400 m ja30.0 Af π b0.060 0 mg
−1
counterclockwise
FIG. P31.13
2
−1 −1.60 t
1.60 s e
219
Chapter 31
*P31.14
(a)
Each coil has a pulse of voltage tending to produce
counterclockwise current as the projectile
approaches, and then a pulse of clockwise voltage
as the projectile recedes.
v=
(b)
d
1.50 m
=
= 625 m s
t 2.40 × 10 −3 s
∆V
t
0
V1
V2
FIG. P31.14
P31.15
ε=
d
NA 2 ∆B cos θ
NBA 2 cos θ =
dt
∆t
e
j
e80.0 × 10 Vja0.400 sf
a50fe600 × 10 T − 200 × 10 Tj cosa30.0°f = 1.36 m
Length = 4AN = 4a1.36 mfa50f = 272 m
−3
ε ∆t
A=
=
N∆B cos θ
*P31.16
(a)
−6
−6
Suppose, first, that the central wire is long and straight. The enclosed current of unknown
amplitude creates a circular magnetic field around it, with the magnitude of the field given
by Ampere’s Law.
z
B ⋅ ds = µ 0 I : B =
µ 0 I max sin ω t
2π R
at the location of the Rogowski coil, which we assume is centered on the wire. This field
passes perpendicularly through each turn of the toroid, producing flux
B⋅A =
µ 0 I max A sin ω t
.
2π R
The toroid has 2π Rn turns. As the magnetic field varies, the emf induced in it is
ε = −N
µ I A d
d
sin ω t = − µ 0 I max nAω cos ω t .
B ⋅ A = −2π Rn 0 max
dt
2π R dt
This is an alternating voltage with amplitude ε max = µ 0 nAω I max . Measuring the amplitude
determines the size I max of the central current. Our assumptions that the central wire is long
and straight and passes perpendicularly through the center of the Rogowski coil are all
unnecessary.
(b)
If the wire is not centered, the coil will respond to stronger magnetic fields on one side, but
to correspondingly weaker fields on the opposite side. The emf induced in the coil is
proportional to the line integral of the magnetic field around the circular axis of the toroid.
Ampere’s Law says that this line integral depends only on the amount of current the coil
encloses. It does not depend on the shape or location of the current within the coil, or on
any currents outside the coil.
220
P31.17
Faraday’s Law
In a toroid, all the flux is confined to the inside of the
toroid.
B=
µ 0 NI 500 µ 0 I
=
2π r
2π r
z
Φ B = BdA =
ΦB =
=
*P31.18
FG
H
IJ
K
500 µ 0 I max
b+R
a sin ω t ln
R
2π
ε = N′
ε=
z
500 µ 0 I max
adr
sin ω t
r
2π
FG
H
IJ
K
FG
H
IJ
K
dΦ B
500 µ 0 I max
b+R
= 20
cos ω t
ω a ln
R
dt
2π
FIG. P31.17
g FGH a
f IJ
K
3.00 + 4.00 cm
10 4
4π × 10 −7 N A 2 50.0 A 377 rad s 0.020 0 m ln
cos ω t
2π
4.00 cm
ja
e
fb
gb
a0.422 Vf cos ω t
a
f
The upper loop has area π 0.05 m
ε = −N
2
= 7.85 × 10 −3 m 2 . The induced emf in it is
b
g
d
dB
BA cos θ = −1 A cos 0°
= −7.85 × 10 −3 m 2 2 T s = −1.57 × 10 −2 V .
dt
dt
The minus sign indicates that it tends to produce counterclockwise current, to make its own
magnetic field out of the page. Similarly, the induced emf in the lower loop is
ε = − NA cos θ
a
f
dB
2
= −π 0.09 m 2 T s = −5.09 × 10 −2 V = +5.09 × 10 −2 V to produce
dt
counterclockwise current in the lower loop, which becomes clockwise current in the upper loop .
The net emf for current in this sense around the figure 8 is
5.09 × 10 −2 V − 1.57 × 10 −2 V = 3.52 × 10 −2 V .
a
f
a
f
It pushes current in this sense through series resistance 2π 0.05 m + 2π 0.09 m 3 Ω m = 2.64 Ω.
ε 3.52 × 10 −2 V
= 13.3 mA .
The current is I = =
R
2.64 Ω
Section 31.2
Motional emf
Section 31.3
Lenz’s Law
P31.19
(a)
For maximum induced emf, with positive charge at the top of the antenna,
a
f
F+ = q + v × B , so the auto must move east .
(b)
e
ja
ε = BAv = 5.00 × 10 −5 T 1.20 m
fFGH 65.30 ×60010 s m IJK cos 65.0° =
3
4.58 × 10 −4 V
Chapter 31
P31.20
I=
ε
R
=
221
BAv
R
v = 1.00 m s
FIG. P31.20
P31.21
(a)
FB = I A × B = IAB
and
ε
R
ε = BAv
we get
FB =
I=
When
a f a1.20f a2.00f = 3.00 N .
2.50
BAv
B2A2 v
AB =
=
R
R
a f
2
2
6.00
The applied force is 3.00 N to the right .
(b)
P31.22
*P31.23
P = I2R =
FIG. P31.21
B2A2 v2
= 6.00 W or P = Fv = 6.00 W
R
FB = IAB and ε = BAv
IR
ε BAv
I= =
so B =
R
R
Av
I 2 AR
and I =
Av
FB v
= 0.500 A
R
(a)
FB =
(b)
I 2 R = 2.00 W
(c)
For constant force, P = F ⋅ v = 1.00 N 2.00 m s = 2.00 W .
a
fb
g
Model the magnetic flux inside the metallic tube as constant as it shrinks form radius R to radius r:
e j
F RI
= 2.50 TG J
HrK
2.50 T πR 2 = B f π r 2
Bf
2
a f
= 2.50 T 12
2
= 360 T
222
*P31.24
Faraday’s Law
Observe that the homopolar generator has no commutator and
produces a voltage constant in time: DC with no ripple. In time dt, the
disk turns by angle dθ = ωdt . The outer brush slides over distance rdθ .
The radial line to the outer brush sweeps over area
1
1
dA = rrdθ = r 2ωdt .
2
2
d
ε = −N B⋅ A
The emf generated is
dt
dA
1
ε = − 1 B cos 0°
= − B r 2ω
dt
2
(We could think of this as following from the result of Example 31.4.)
The magnitude of the emf is
FG
H
af
IJ
K
FIG. P31.24
FG 1 r ω IJ = b0.9 N ⋅ s C ⋅ mgLM 1 a0.4 mf b3 200 rev mingOPFG 2π rad rev IJ
H2 K
N2
QH 60 s min K
ε =B
2
2
ε = 24.1 V
A free positive charge q shown, turning with the disk, feels a magnetic force qv × B
radially
outward. Thus the outer contact is positive .
*P31.25
The speed of waves on the wire is
v=
T
µ
=
267 N ⋅ m
= 298 m s .
3 × 10 −3 kg
In the simplest standing-wave vibration state,
d NN = 0.64 m =
and f =
v
λ
=
λ
2
λ = 1.28 m
298 m s
= 233 Hz .
1.28 m
(a)
The changing flux of magnetic field through the circuit containing the wire will drive
current to the left in the wire as it moves up and to the right as it moves down. The emf will
have this same frequency of 233 Hz .
(b)
The vertical coordinate of the center of the wire is described by
a
f b
g
dx
= −a1.5 cmfb 2π 233 sg sinb 2π 233 t sg .
Its velocity is v =
dt
Its maximum speed is 1.5 cma 2π f 233 s = 22.0 m s .
x = A cos ω t = 1.5 cm cos 2π 233 t s .
The induced emf is ε = −BAv , with amplitude
a
f
ε max = BAv max = 4.50 × 10 −3 T 0.02 m 22 m s = 1.98 × 10 −3 V .
Chapter 31
P31.26
ε = −N
a
FG IJ
H K
d
∆A
BA cos θ = − NB cos θ
dt
∆t
f
ε = −1 0.100 T cos 0°
I=
223
a3.00 m × 3.00 m sin 60.0°f − a3.00 mf
2
0.100 s
1.21 V
= 0.121 A
10.0 Ω
= 1. 21 V
FIG. P31.26
The flux is into the page and decreasing. The loop makes its own magnetic field into the page by
carrying clockwise current.
P31.27
b
ε=
P31.28
gb
g
ω = 2.00 rev s 2π rad rev = 4.00π rad s
(a)
1
BωA 2 = 2.83 mV
2
B ext = B ext i and Bext decreases; therefore, the
induced field is B = B i (to the right) and the
0
0
current in the resistor is directed to the right .
(b)
e j
= B e+ ij is to the right, and the current in
B ext = B ext − i increases; therefore, the induced
field B0
0
the resistor is directed to the right .
(c)
e j
B ext = B ext − k into the paper and Bext decreases;
e j
therefore, the induced field is B0 = B0 − k into the
paper, and the current in the resistor is directed
to the right .
(d)
a
f
FIG. P31.28
By the magnetic force law, FB = q v × B . Therefore, a positive charge will move to the top of
the bar if B is into the paper .
224
P31.29
Faraday’s Law
(a)
The force on the side of the coil entering the field
(consisting of N wires) is
a f a f
F = N ILB = N IwB .
The induced emf in the coil is
ε =N
so the current is I =
a f
d Bwx
dΦ B
=N
= NBwv .
dt
dt
ε NBwv
=
counterclockwise.
R
R
The force on the leading side of the coil is then:
F=N
(b)
FG NBwv IJ wB =
H R K
N 2B2w 2 v
to the left .
R
Once the coil is entirely inside the field,
Φ B = NBA = constant ,
ε =0, I =0,
so
and
F= 0 .
FIG. P31.29
As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude of the
current is the same as in part (a), but now the current is clockwise. Thus, the force exerted on
the trailing side of the coil is:
(c)
F=
N 2B2 w 2 v
to the left again .
R
P31.30
Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The
loop will create a field out of the page by carrying a counterclockwise current. Therefore, current
must flow from b to a through the resistor. Hence, Va − Vb will be negative .
P31.31
Name the currents as shown in the diagram:
Left loop:
+ Bdv 2 − I 2 R 2 − I 1 R1 = 0
Right loop:
+ Bdv 3 − I 3 R3 + I 1 R1 = 0
At the junction:
I 2 = I1 + I 3
Then,
Bdv 2 − I 1 R 2 − I 3 R 2 − I 1 R1 = 0
I3 =
Bdv 3 I 1 R1
+
.
R3
R3
FIG. P31.31
b
Bdv 3 R 2 I 1 R1 R 2
−
=0
R3
R3
FG v R − v R IJ upward
HR R +R R +R R K
L b4.00 m sga15.0 Ωf − b2.00 m sga10.0 Ωf OP =
= b0.010 0 Tga0.100 mfM
MN a5.00 Ωfa10.0 Ωf + a5.00 Ωfa15.0 Ωf + a10.0 Ωfa15.0 Ωf PQ
I 1 = Bd
I1
g
Bdv 2 − I 1 R1 + R 2 −
So,
2
1
2
3
3
1
3
2
2
3
145 µA upward.
Chapter 31
Section 31.4
P31.32
(a)
Induced emf and Electric Fields
dB
= 6.00t 2 − 8.00t
dt
dΦ B
dt
ε =
At t = 2.00 s ,
E=
b
π R 2 dB dt
2π r2
g = 8.00π b0.025 0g
2π b0.050 0g
2
F = qE = 8.00 × 10 −21 N
FIG. P31.32
(b)
P31.33
2
When 6.00t − 8.00t = 0 ,
dB
= 0.060 0t
dt
ε =
t = 1.33 s
dΦ B
dB
= π r12
= 2π r1 E
dt
dt
At t = 3.00 s ,
E=
F π r I dB = 0.020 0 m e0.060 0 T s jb3.00 sgFG 1 N ⋅ s IJ
GH 2π r JK dt
H 1 T ⋅C ⋅ mK
2
2
1
2
1
E = 1.80 × 10 −3 N C perpendicular to r1 and counterclockwise
P31.34
(a)
z
E ⋅ dA =
dΦ B
dt
e j dBdt
2π rE = π r 2
(b)
Section 31.5
P31.35
FIG. P31.33
E=
so
b9.87 mV mg cosb100π tg
The E field is always opposite to increasing B. ∴ clockwise .
Generators and Motors
b
ga
fa
fa
f
(a)
ε max = NABω = 1 000 0.100 0.200 120π = 7.54 kV
(b)
ε t = NBAω sin ω t = NBAω sin θ
af
ε is maximal when sin θ = 1
or θ = ±
π
2
so the plane of coil is parallel to B .
FIG. P31.35
225
226
P31.36
Faraday’s Law
b
ε = −N
P31.37
rad I F 1 min I
gFGH 21πrev
JK GH 60 s JK = 314 rad s
T ⋅ m j cosb314t sg = +250e 2.50 × 10
For the alternator, ω = 3 000 rev min
dΦ B
d
= −250
2.50 × 10 −4
dt
dt
e
a
2
−4
g a f
jb
T ⋅ m 2 314 s sin 314t
f a f
(a)
ε = 19.6 V sin 314t
(b)
ε max = 19.6 V
e
ja
je
f
B = µ 0 nI = 4π × 10 −7 T ⋅ m A 200 m −1 15.0 A = 3.77 × 10 −3 T
e j
For the small coil, Φ B = NB ⋅ A = NBA cos ω t = NB π r 2 cos ω t .
ε=−
Thus,
dΦ B
= NBπ r 2ω sin ω t
dt
a fe
g e4.00π s j sinb4.00π tg = a28.6 mVf sinb4.00π tg
jb
ε = 30.0 3.77 × 10 −3 T π 0.080 0 m
P31.38
As the magnet rotates, the flux
through the coil varies
sinusoidally in time with Φ B = 0
at t = 0 . Choosing the flux as
positive when the field passes
from left to right through the
area of the coil, the flux at any
time may be written as
Φ B = − Φ max sin ω t so the
induced emf is given by
ε=−
2
−1
1
0.5
I/I max
0
0
0.5
1
1.5
–1
t/T = ( ω t/2 π )
FIG. P31.38
ε ω Φ max
=
cos ω t = I max cos ω t .
R
R
850 mA
M
120 V
2
–0.5
dΦ B
= ω Φ max cos ω t .
dt
The current in the coil is then I =
*P31.39
850 mA
To analyze the actual circuit, we model it as 120 V
a
f
(a)
The loop rule gives +120 V − 0.85 A 11.8 Ω − ε back = 0
(b)
The resistor is the device changing electrical work input into internal energy:
2
P = I 2 R = 0.85 A 11.8 Ω = 8.53 W .
(c)
With no motion, the motor does not function as a generator, and ε back = 0 . Then
a
fa
c
2
ε back = 110 V .
f
I = 10.2 A
a f
P = I R = a10.2 A f a11.8 Ωf = 1.22 kW
120 V − I c 11.8 Ω = 0
2
c
.
c
11.8 Ω
ε back
.
227
Chapter 31
P31.40
(a)
FG 1 π R IJω
H2 K
π
= a1.30 Tf a0. 250 mf b 4.00π rad sg
2
ε max
ε
2
ε max = BAω = B
t
2
ε max = 1.60 V
(b)
ε=
z
2π
0
P31.41
z
2π
ε
t
sin θdθ = 0
0
(c)
The maximum and average ε would remain
unchanged.
(d)
See Figure 1 at the right.
(e)
See Figure 2 at the right.
(a)
Φ B = BA cos θ = BA cos ω t = 0.800 T 0.010 0 m 2 cos 2π 60.0 t =
(b)
ε=−
(c)
I=
(d)
P = I2R =
(e)
P = Fv = τω so τ =
Section 31.6
P31.42
BAω
ε
dθ =
2π
2π
Figure 1
a
dΦ B
=
dt
ε
=
R
fe
Figure 2
FIG. P31.40
j
a f e8.00 mT ⋅ m j cosa377tf
2
a3.02 Vf sina377tf
a3.02 Af sina377tf
a9.10 Wf sin a377tf
2
P
=
ω
a24.1 mN ⋅ mf sin a377tf
2
Eddy Currents
The current in the magnet creates an
upward magnetic field, so the N and S poles on the
solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as
the coil approaches, so a current is induced here to create a downward magnetic field. This is
clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the
upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic
field by being
counterclockwise as the picture correctly shows.
228
P31.43
Faraday’s Law
(a)
At terminal speed,
Mg = FB = IwB =
MgR
B 2ω 2
2
T
.
The emf is directly proportional to vT , but the
current is inversely proportional to R. A large R
means a small current at a given speed, so the loop
must travel faster to get FB = mg .
(b)
(c)
P31.44
2
T
vT =
or
Section 31.7
FG ε IJ wB = FG Bwv IJ wB = B w v
H RK H R K
R
FIG. P31.43
At a given speed, the current is directly proportional to the magnetic field. But the force is
proportional to the product of the current and the field. For a small B, the speed must
increase to compensate for both the small B and also the current, so vT ∝ B 2 .
Maxwell’s Equations
i
j
k
−e
E + v × B where v × B = 10.0 0
F = ma = qE + qv × B so a =
0 = −4.00 j
m
0
0 0.400
e−1.60 × 10 j 2.50 i + 5.00 j − 4.00 j = e−1.76 × 10 j 2.50 i + 1.00j
9.11 × 10
a = e −4.39 × 10 i − 1.76 × 10 jj m s
−19
a=
11
−31
11
P31.45
2
11
F = ma = qE + qv × B
j
i
k
e
a=
E + v × B where v × B = 200
0
0 = −200 0.400 j + 200 0.300 k
m
0.200 0.300 0.400
a
a
f
1.60 × 10 −19
50.0 j − 80.0 j + 60.0k = 9.58 × 10 7 −30.0 j + 60.0k
1.67 × 10 −27
a = 2.87 × 10 9 − j + 2k m s 2 = −2.87 × 10 9 j +5.75 × 10 9 k m s 2
a=
e
j
Additional Problems
P31.46
a
f e j FGH
a fLMN e
j OPQa f
ε = −a30.0fLMπ e 2.70 × 10 mj OPe3.20 × 10
N
Q
ε = −e7.22 × 10 V j cos 2π e523t s j
IJ
K
d
dB
BA cos θ = − N π r 2 cos 0°
dt
dt
2
d
ε = − 30.0 π 2.70 × 10 −3 m
1
50.0 mT + 3.20 mT sin 2π 523 t s −1
dt
ε = −N
−3
−3
2
−1
−3
a
j e
T 2π 523 s −1
f e
j
j cose2π 523t s j
−1
f
Chapter 31
P31.47
(a)
229
Doubling the number of turns.
Amplitude doubles: period unchanged
(b)
Doubling the angular velocity.
doubles the amplitude: cuts the period in half
(c)
Doubling the angular velocity while reducing the
number of turns to one half the original value.
FIG. P31.47
Amplitude unchanged: cuts the period in half
P31.48
P31.49
ε = −N
a
f
ja fFGH
IJ
K
∆
∆B
1.50 T − 5.00 T
BA cos θ = − N π r 2 cos 0°
= −1 0.005 00 m 2 1
= 0.875 V
∆t
∆t
20.0 × 10 −3 s
ε
e
0.875 V
= 43.8 A
0.020 0 Ω
(a)
I=
(b)
P = εI = 0.875 V 43.8 A = 38.3 W
R
=
e j
a
fa
f
In the loop on the left, the induced emf is
ε =
a
f b100 T sg = π V
dΦ B
dB
=A
= π 0.100 m
dt
dt
2
and it attempts to produce a counterclockwise
current in this loop.
In the loop on the right, the induced emf is
ε =
a
f b100 T sg = 2.25π V
dΦ B
= π 0.150 m
dt
2
FIG. P31.49
and it attempts to produce a clockwise current. Assume that I 1 flows down through the 6.00-Ω
resistor, I 2 flows down through the 5.00-Ω resistor, and that I 3 flows up through the 3.00-Ω
resistor.
From Kirchhoff’s junction rule:
I3 = I1 + I 2
(1)
Using the loop rule on the left loop:
6.00 I 1 + 3.00 I 3 = π
(2)
Using the loop rule on the right loop:
5.00 I 2 + 3.00 I 3 = 2.25π
(3)
Solving these three equations simultaneously,
I1 = 0.062 3 A , I 2 = 0.860 A , and I 3 = 0.923 A .
230
P31.50
Faraday’s Law
The emf induced between the ends of the moving bar is
a
fa
fb
g
ε = BAv = 2.50 T 0.350 m 8.00 m s = 7.00 V .
The left-hand loop contains decreasing flux away from you, so the induced current in it will be
clockwise, to produce its own field directed away from you. Let I 1 represent the current flowing
upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise current. Let I 3
be the upward current in the 5.00-Ω resistor.
(a)
a f
+7.00 V − I a5.00 Ωf = 0
+7.00 V − I 1 2.00 Ω = 0
Kirchhoff’s loop rule then gives:
and
(b)
3
I1 = 3.50 A
I 3 = 1.40 A .
The total power dissipated in the resistors of the circuit is
g a
b
fa
f
P = εI1 + εI 3 = ε I 1 + I 3 = 7.00 V 3.50 A + 1.40 A = 34.3 W .
(c)
Method 1: The current in the sliding conductor is downward with value
I 2 = 3.50 A + 1.40 A = 4.90 A . The magnetic field exerts a force of
a
fa
fa
f
Fm = IAB = 4.90 A 0.350 m 2.50 T = 4.29 N directed
toward the right on this
conductor. An outside agent must then exert a force of 4.29 N
to the left to keep the bar
moving.
Method 2: The agent moving the bar must supply the power according to P = F ⋅ v = Fv cos 0° .
The force required is then:
F=
P31.51
P
34.3 W
=
= 4.29 N .
v 8.00 m s
Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we
use a bar magnet to produce field 10 −3 T through the coil in one direction along its axis. Suppose
we then flip the magnet to reverse the flux in 10 −1 s . The average induced emf is then
e jFGH
IJ
K
∆ BA cos θ
∆Φ B
cos 180°− cos 0°
= −N
= − NB π r 2
∆t
∆t
∆t
−2
2
ε = − 20 10 −3 T π 0.015 0 m
~ 10 −4 V
10 −1 s
ε = −N
a fe
jb
g FGH
IJ
K
Chapter 31
P31.52
I=
ε + ε induced
R
ε induced = −
and
a f
d
BA
dt
dv
= IBd
dt
dv IBd Bd
=
=
ε + ε induced
dt
m
mR
dv Bd
=
ε − Bvd
dt mR
u = ε − Bvd
du
dv
= − Bd
dt
dt
1 du Bd
−
=
u
Bd dt mR
F=m
b
a
To solve the differential equation, let
z
u
so
u0
f
za f
or
2
t
Bd
du
dt .
=−
u
mR
0
a f
Since v = 0 when t = 0 ,
u0 = ε
and
u = ε − Bvd
2
ε − Bvd = ε e − B
v=
Therefore,
ε
Bd
e1 − e
2 2
d t mR
.
2 2
− B d t mR
j
.
Φ B = BA = Bπ r 2 .
The enclosed flux is
The particle moves according to
∑ F = ma :
qvB sin 90° =
r=
−6
je
T ⋅ m 2 30 × 10 −9 C
Bπ m 2 v 2
q2B2
.
j a0.6 Tf =
2
2.54 × 10 5 m s
(a)
v=
(b)
Energy for the particle-electric field system is conserved in the firing process:
π m2
Ui = K f :
=
e15 × 10
mv 2
r
mv
.
qB
ΦB =
Then
ΦBq 2B
FIG. P31.52
Bd
u
t
=−
ln
u0
mR
2 2
u
= e − B d t mR .
u0
Integrating from t = 0 to t = t ,
*P31.53
g
e
π 2 × 10
q∆V =
−16
kg
j
2
1
mv 2
2
e
je
2 × 10 −16 kg 2.54 × 10 5 m s
mv 2
∆V =
=
2q
2 30 × 10 −9 C
e
j
j
2
= 215 V .
231
232
*P31.54
Faraday’s Law
(a)
Consider an annulus of radius r, width dr, height b, and resistivity ρ. Around its
circumference, a voltage is induced according to
ε = −N
b
g
d
d
B ⋅ A = −1 Bmax cos ω t π r 2 = + Bmaxπ r 2ω sin ω t .
dt
dt
b g
ρA ρ 2π r
=
.
Ax
bdr
The resistance around the loop is
The eddy current in the ring is dI =
b g
dPi = ε dI =
The instantaneous power is
b
g
Bmax π r 2ω sin ω t bdr Bmax rbωdr sin ω t
ε
.
=
=
2ρ
resistance
ρ 2π r
2
π r 3 bω 2 dr sin 2 ω t
Bmax
.
2ρ
sin 2 ω t =
The time average of the function
1 1
1
1
− cos 2ω t is − 0 =
2
2 2
2
so the time-averaged power delivered to the annulus is
dP =
2
π r 3 bω 2 dr
Bmax
.
4ρ
z z
P = dP =
The power delivered to the disk is
P=
P31.55
I
JK
2
Bmax
πbω 2 3
r dr
4ρ
0
2
2
Bmax
R 4 bω 2
π bω 2 R 4
π Bmax
−0 =
.
4ρ
4
16 ρ
2
Bmax
and P get 4 times larger.
(b)
When Bmax gets two times larger,
(c)
When f and ω = 2π f double, ω 2 and P get 4 times larger.
(d)
When R doubles,
I=
(a)
R 4 and P become 2 4 = 16 times larger.
ε B A
=
R R ∆t
so q = I∆t =
P31.56
F
GH
R
I=
b15.0 µTga0.200 mf
0.500 Ω
2
= 1.20 µC
dq ε
dΦ B
=
where ε = −N
so
dt
dt R
z
dq =
N
R
z
Φ2
dΦ B
Φ1
and the charge through the circuit will be Q =
(b)
IJ OP = BAN
KQ R
RQ a 200 Ωfe5.00 × 10 C j
=
=
so B =
NA a100fe 40.0 × 10 m j
Q=
LM
N
FG
H
N
π
BA cos 0 − BA cos
R
2
−4
−4
2
b
g
N
Φ 2 − Φ1 .
R
0.250 T .
Chapter 31
P31.57
ε
= 0.900 A
R
(a)
ε = BAv = 0.360 V
(b)
FB = IAB = 0.108 N
(c)
Since the magnetic flux B ⋅ A is in effect decreasing, the
induced current flow through R is from b to a. Point b is
I=
233
FIG. P31.57
at higher potential.
No . Magnetic flux will increase through a loop to the left of ab. Here counterclockwise
(d)
current will flow to produce upward magnetic field. The current in R is still from b to a.
P31.58
ε = BAv at a distance r from wire
ε =
F µ I I Av
GH 2π r JK
0
v
FIG. P31.58
P31.59
(a)
a
fe j
a
f
At time t, the flux through the loop is Φ B = BA cos θ = a + bt π r 2 cos 0° = π a + bt r 2 .
At t = 0 , Φ B = π ar 2 .
P31.60
a
ε=−
(c)
I=
(d)
P =εI= −
ε=−
π br 2
ε
= −
R
R
F
GH
a
f
d a + bt
dΦ B
= −π r 2
= −π br 2
dt
dt
(b)
f
π br 2
R
I e−π br j =
JK
2
π 2b 2r 4
R
FG IJ
H K
d
dB
NBA = −1
π a2 = π a2K
dt
dt
(a)
Q = Cε = Cπ a 2 K
(b)
B into the paper is decreasing; therefore, current will attempt to counteract this. Positive
charge will go to upper plate .
(c)
The changing magnetic field through the enclosed area induces an electric field ,
surrounding the B-field, and this pushes on charges in the wire.
234
P31.61
Faraday’s Law
The flux through the coil is Φ B = B ⋅ A = BA cos θ = BA cos ω t . The induced emf is
ε = −N
b
g
d cos ω t
dΦ B
= − NBA
= NBAω sin ω t .
dt
dt
a
fe
jb
ε max = NBAω = 60.0 1.00 T 0.100 × 0.200 m 2 30.0 rad s = 36.0 V
(b)
dΦ B ε
dΦ B
= , thus
dt
N
dt
=
max
ε max 36.0 V
=
= 0.600 V = 0.600 Wb s
N
60.0
(c)
At t = 0.050 0 s , ω t = 1.50 rad and
ε = ε max sin 1.50 rad = 36.0 V sin 1.50 rad = 35.9 V .
(d)
The torque on the coil at any time is
a
f a
f a
f
FG ε IJ FG ε IJ sin ω t .
H ω KH RK
a36.0 V f
ε
=
=
, sin ω t = 1.00 and τ =
ω R b30.0 rad sga10.0 Ωf
a
f
max
τ = µ × B = NIA × B = NAB I sin ω t =
When ε = ε max
P31.62
g
(a)
(a)
2
2
max
∆Φ B
, with N = 1 .
∆t
Taking a = 5.00 × 10 −3 m to be the radius of the washer, and h = 0.500 m ,
We use ε = −N
b
g
∆Φ B = B2 A − B1 A = A B2 − B1 = π a 2
F µ I − µ I I = a µ I FG 1 − 1 IJ = − µ ahI .
GH 2π ah + af 2π a JK 2 H h + a a K 2ah + af
0
2
0
0
The time for the washer to drop a distance h (from rest) is:
Therefore, ε =
µ 0 ahI
µ ahI
= 0
2 h + a ∆t 2 h + a
a f
−7
P31.63
g
µ 0 aI
0
∆t =
gh
−3
2
97.4 nV .
Since the magnetic flux going through the washer (into the plane of the paper) is decreasing
in time, a current will form in the washer so as to oppose that decrease. Therefore, the
current will flow in a clockwise direction .
Find an expression for the flux through a rectangular area “swept out”
I
by the bar in time t. The magnetic field at a distance x from wire is
B=
2h
.
g
a f 2 h = 2a h + a f 2
e4π × 10 T ⋅ m Aje5.00 × 10 mja10.0 Af e9.80 m s ja0.500 mf =
ε=
2
2b0.500 m + 0.005 00 mg
and
(b)
4.32 N ⋅ m .
z
µ0I
and Φ B = BdA . Therefore,
2π x
z
µ Ivt r + A dx
where vt is the distance the bar has moved in time t.
ΦB = 0
2π r x
Then, ε =
FG
H
dΦ B
µ 0 Iv
A
=
ln 1 +
dt
r
2π
IJ
K
.
r
v
A
FIG. P31.63
vt
Chapter 31
P31.64
The magnetic field at a distance x from a long wire is B =
through the loop.
dΦ B =
µ0I
Adx so
2π x
a f
Therefore,
P31.65
FG
H
z
ε=−
dΦ B µ 0 IAv w
µ 0 IAv w
ε
=
and I = =
dt
2π r r + w
R
2π Rr r + w
f
a
e
f
.
j
We are given
Φ B = 6.00t 3 − 18.0t 2 T ⋅ m 2
and
ε=−
dΦ B
= −18.0t 2 + 36.0t .
dt
which gives
dε
= −36.0t + 36.0 = 0
dt
t = 1.00 s .
Therefore, the maximum current (at t = 1.00 s ) is
I=
For the suspended mass, M:
Mg −
zb
v
0
ε BAv
=
R
R
Mg
dv
B2A2v
a=
=
−
dt m + M R M + m
f
a
z
t
dv
= dt where
α − βv 0
g
α=
v=
Therefore, the velocity varies with time as
(a)
ε = −N
f
I=
B2A2 v
= m + M a or
R
a
a
−18.0 + 36.0 V
ε
=
= 6.00 A .
R
3.00 Ω
∑ F = Mg − T = Ma .
∑ F = T − IAB = ma , where
For the sliding bar, m:
P31.67
IJ
K
µ 0 IA r + w dx µ 0 IA
w
=
ln 1 +
r
2π r x
2π
Maximum E occurs when
P31.66
µ0I
. Find an expression for the flux
2π x
ΦB =
a
f
Mg
B2A2
.
and β =
R M+m
M+m
a
f
2 2
MgR
α
1 − e−β t =
1 − e − B A t Ra M + m f
2 2
β
B A
e
j
dΦ B
dB
d
= − NA
= − NA
µ 0nI where A = area of coil
dt
dt
dt
b
g
N = number of turns in coil
(b)
and
n = number of turns per unit length in solenoid.
Therefore,
ε = Nµ 0 An
I=
∆V
R
235
ε =
a f b g
b g
g e
e
j b
ja f b
a1.19 Vf cosb120π tg
and
P = ∆VI =
d
4 sin 120π t = Nµ 0 An 480π cos 120π t
dt
2
ε = 40 4π × 10 −7 π 0.050 0 m 2.00 × 10 3 480π cos 120π t
cos 2 θ =
From
1
the average value of cos θ is , so
2
2
a
a
a1.19 V f
2
b
cos 2 120π t
8.00 Ω
1 1
+ cos 2θ
2 2
f
2
1 1.19 V
P =
= 88.5 mW .
2 8.00 Ω
f
g
g
.
236
*P31.68
Faraday’s Law
(a)
d θ a2
Ba 2 dθ
1
1
d
2
BA cos θ = −1 B
cos 0° = −
= − Ba 2 ω = − 0.5 T 0.5 m 2 rad s
dt
dt
2
2 dt
2
2
= −0.125 V = 0.125 V clockwise
a
ε = −N
fa
f
The – sign indicates that the induced emf produces clockwise current, to make its own
magnetic field into the page.
(b)
*P31.69
a
f
At this instant θ = ω t = 2 rad s 0.25 s = 0.5 rad . The arc PQ has length
rθ == 0.5 rad 0.5 m = 0.25 m . The length of the circuit is 0.5 m + 0.5 m + 0.25 m = 1.25 m its
0.125 V
resistance is 1.25 m 5 Ω m = 6.25 Ω . The current is
= 0.020 0 A clockwise .
6.25 Ω
a
fa
b
f
g
Suppose the field is vertically down. When an electron is moving away
from you the force on it is in the direction given by
b
g
qv × B c as − away × down = −
= − left = right .
Therefore, the electrons circulate clockwise.
FIG. P31.69
(a)
As the downward field increases, an emf is induced to produce some current that in turn
produces an upward field. This current is directed
counterclockwise, carried by
negative electrons moving clockwise. Therefore the original electron motion speeds up.
(b)
At the circumference, we have
∑ Fc = ma c :
q vBc sin 90° =
mv 2
r
mv = q rBc .
The increasing magnetic field Bav in the area enclosed by the orbit produces a tangential
electric field according to
dB
d
r dBav
E 2π r = π r 2 av
E=
E ⋅ ds = − B av ⋅ A
.
dt
dt
2 dt
dv
qE=m
.
An electron feels a tangential force according to ∑ Ft = mat :
dt
r dBav
r
dv
q
q B av = mv = q rBc
=m
Then
dt
2 dt
2
Bav = 2Bc .
and
z
P31.70
b g
The induced emf is ε = BAv where B =
y f = yi −
µ0I
, v f = vi + gt = 9.80 m s 2 t , and
2π y
e
j
1 2
gt = 0.800 m − 4.90 m s 2 t 2 .
2
e
j
f a0.300 mf 9.80 m s t = e1.18 × 10 jt
e
j 0.800 − 4.90t
2π 0.800 m − e 4.90 m s jt
e1.18 × 10 ja0.300f V = 98.3 µV .
At t = 0.300 s , ε =
0.800 − 4.90a0.300 f
ε=
e4π × 10
−7
ja
−4
T ⋅ m A 200 A
2
2
2
2
−4
2
V
Chapter 31
P31.71
237
The magnetic field produced by the current in the straight wire is
perpendicular to the plane of the coil at all points within the coil. The
µ I
magnitude of the field is B = 0 . Thus, the flux linkage is
2π r
NΦ B =
FG
H
z
IJ b
K
µ 0 NIL h+ w dr µ 0 NI max L
h+ w
=
ln
sin ω t + φ .
h
r
2π
2π
h
g
Finally, the induced emf is
FIG. P31.71
FG IJ b g
H K
e4π × 10 ja100fa50.0fa0.200 mfe200π s j lnFG 1 + 5.00 cm IJ cosbω t + φ g
ε=−
H 5.00 cm K
2π
ε = −a87.1 mV f cosb 200π t + φ g
The term sinbω t + φ g in the expression for the current in the straight wire does not change
ε=−
µ 0 NI max Lω
w
ln 1 +
cos ω t + φ
h
2π
−7
−1
appreciably when ω t changes by 0.10 rad or less. Thus, the current does not change appreciably
during a time interval
∆t <
0.10
e200π s j
−1
= 1.6 × 10 −4 s .
e
je
j
We define a critical length, c∆t = 3.00 × 10 8 m s 1.6 × 10 −4 s = 4.8 × 10 4 m equal to the distance to
which field changes could be propagated during an interval of 1.6 × 10 −4 s . This length is so much
larger than any dimension of the coil or its distance from the wire that, although we consider the
straight wire to be infinitely long, we can also safely ignore the field propagation effects in the
vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the
vicinity of the coil.
If the frequency w were much larger, say, 200π × 10 5 s −1 , the corresponding critical length would be
only 48 cm. In this situation propagation effects would be important and the above expression for ε
would require modification. As a “rule of thumb” we can consider field propagation effects for
circuits of laboratory size to be negligible for frequencies, f =
P31.72
Φ B = BA cos θ
ω
2π
, that are less than about 10 6 Hz.
dΦ B
= −ωBA sin θ ;
dt
I ∝ − sin θ
θ
τ ∝ IB sin θ ∝ − sin 2 θ
FIG. P31.72
238
Faraday’s Law
ANSWERS TO EVEN PROBLEMS
P31.2
0.800 mA
P31.40
P31.4
(a) see the solution; (b) 3.79 mV ;
(c) 28.0 mV
(a) 1.60 V ; (b) 0; (c) no change;
(d) and (e) see the solution
P31.42
both are correct; see the solution
78.5 µs
P31.44
P31.6
P31.8
µ 0 nπ r22
2R
µ 02nπ r22
(b)
4r1 R
(a)
∆I
counterclockwise;
∆t
∆I
; (c) upward
∆t
a f
P31.10
−14.2 mV cos 120 t
P31.12
61.8 mV
P31.14
(a) see the solution; (b) 625 m/s
P31.16
see the solution
P31.18
13.3 mA counterclockwise in the lower
loop and clockwise in the upper loop.
P31.20
1.00 m s
P31.22
(a) 500 mA; (b) 2.00 W ; (c) 2.00 W
P31.24
24.1 V with the outer contact positive
P31.26
121 mA clockwise
P31.28
(a) to the right; (b) to the right;
(c) to the right; (d) into the paper
P31.30
negative; see the solution
P31.32
(a) 8.00 × 10 −21 N downward
perpendicular to r1 ; (b) 1.33 s
P31.34
(a) 9.87 mV m cos 100π t ; (b) clockwise
P31.36
b
g b g
(a) a19.6 V f sina314tf ; (b) 19.6 V
P31.38
see the solution
P31.46
e−4.39i − 1.76jj10 m s
−a7.22 mV f cosb 2π 523 t sg
P31.48
(a) 43.8 A ; (b) 38.3 W
P31.50
(a) 3.50 A up in 2 Ω and 1.40 A up in 5 Ω;
(b) 34.3 W ; (c) 4.29 N
P31.52
see the solution
P31.54
(a)
P31.56
(a) see the solution; (b) 0.250 T
P31.58
see the solution
P31.60
(a) Cπ a 2 K ; (b) the upper plate;
(c) see the solution
P31.62
(a) 97.4 nV ; (b) clockwise
P31.64
µ 0 I Av w
2π Rr r + w
P31.66
P31.68
11
2
2
R 4 bω 2
πBmax
; (b) 4 times larger;
16 ρ
(c) 4 times larger; (d) 16 times larger
a
MgR
2 2
B A
f
1 − e−B
2 2
a
A t R M +m
f
(a) 0.125 V to produce clockwise current;
(b) 20.0 mA clockwise
P31.70
1.18 × 10 −4
; 98.3 µV
0.800 − 4.90t 2
P31.72
see the solution