Homework Solutions
Physics 8B – Spring 2012
Chpt. 34 - 2,5,6,9,11,17,21,22,40,41,52,78,82
34.2. Model: Assume the magnetic field is uniform.
Visualize: Please refer to Figure Ex34.2. Since a motional emf was developed the field must be perpendicular

to v . The positive charges experienced a magnetic force to the left. By the right-hand rule the field must be out
 
of the page so that v × B is to the left.
Solve: This is a straightforward use of Equation 34.3. We have
B
=

=
vl
0.050 V
= 0.10 T
( 5.0 m/s )( 0.10 m )
Assess: This is reasonable. Laboratory fields are typically up to a few teslas in magnitude.
34.5. Model: Consider the solenoid to be long so the field is constant inside and zero outside.
Visualize: Please refer to Figure Ex34.5. The field of a solenoid is along the axis. The flux through the loop is
only nonzero inside the solenoid. Since the loop completely surrounds the solenoid, the total flux through the
loop will be the same in both the perpendicular and tilted cases.

Solve: The field is constant inside the solenoid so we will use Equation 34.10. Take A to be in the same
direction as the field. The magnetic flux is

 

2
Φ= Aloop ⋅ Bloop= Asol ⋅ Bsol= π rsol2 Bsol cosθ= π ( 0.010 m ) ( 0.20 T=
) 6.3 × 10−5 Wb


When the loop is tilted the component of B in the direction of A is less, but the effective area of the loop
surface through which the magnetic field lines cross is increased by the same factor.
34.6. Model: Assume the field is uniform inside the rectangular region and zero outside.
Visualize: The flux measures how much of the field penetrates the chosen surface. We will break the surface
up because the magnetic field has different strengths over different parts of the surface of the loop.
Solve: For convenience we choose the normal to the loop to be into the page so it is in the same direction as the
magnetic field. The total flux is
 
 
 
Φ = ∫ B ⋅ dA =
∫ B ⋅ dA + ∫ B ⋅ dA= ∫ BdA cosθ = BAinside = Bab
loop
inside
rectangle
outside
rectangle
inside
rectangle
rectangle
Assess: Only the region with nonzero magnetic field inside the loop contributes to the flux.
34.9. Visualize: Please refer to Figure Ex34.9. The changing current in the solenoid produces a changing
flux in the loop. By Lenz’s law there will be an induced current and field to oppose the change in flux.
Solve: The current shown produces a field to the right inside the solenoid. So there is flux to the right through
the surrounding loop. As the current in the solenoid increases there is more field and more flux to the right
through the loop. There is an induced current in the loop that will oppose the change by creating an induced field
and flux to the left. This requires a counterclockwise current.
34.11. Model: Assume the field is uniform.
Visualize: Please refer to Figure Ex34.11. If the changing field produces a changing flux in the loop there will
be a corresponding induced emf and current.
Solve: (a) The induced emf is 
/R. The field B is changing, but the
= d Φ/dt and the induced current is I = 


area A is not. Take A to be out of the page and parallel to B , so Φ = AB. Thus,
dB
dB
2
=
A
=
π r2
=
π ( 0.050 m ) ( 0.50 T/s ) =
3.9 × 10−3 V =
3.9 mV
dt
dt
1
 3.93 × 10−3 V
2.0 10−2 A =
20 mA
I ==
=×
R
0.2 Ω
The field is increasing out of the page. To prevent the increase, the induced field needs to point into the page.
Thus, the induced current must flow clockwise.
(b) As in part
(a), A=
=
(dB/dt ) 3.9 mV and I = 20 mA. Here the field is into the page and decreasing. To
prevent the decrease, the induced field needs to point into the page. Thus the induced current must flow
clockwise.

 

(c) Now A (left or right) is perpendicular to B and so A ⋅ B =
0 Wb. That is, the field does not penetrate the
plane of the loop. If Φ = 0 Wb, then  = |dΦ/dt| = 0 V/m and I = 0 A. There is no induced current.
Assess: Note that the induced field opposes the change.
34.17. Model: Assume the magnetic field inside the circular region is uniform. The region looks exactly like
a cross section of a solenoid of radius 2.0 cm. The proton accelerates due to the force of the induced electric
field.
Visualize: Please refer to Figure Ex34.17.
Solve: Equation 34.26 gives the strength of the induced electric field inside a solenoid with changing magnetic
r dB
, where r indicates the radial distance of each point to the
field. The magnitude at each point is thus E =
2 dt
center of the circular region.
The direction of E can be determined with Lenz’s Law. As B decreases, a clockwise electric field is induced that
would create a current that opposes the decrease if a conducting loop were present.
Using Newton’s second law,
F = eE = ma ⇒=
a
eE er dB
=
m 2m dt
The sign indicates the acceleration is in the same direction as the induced electric field. At point a the
acceleration is
a=
(
)(
− 1.6 × 10−19 C 1.0 × 10−2 m
(
2 1.6 × 10−27 kg

At point b, a = 0 m/s2. At point c,=
a
( 4.8 × 10
)
4
) ( 0.10 T/s ) ⇒ a=
( 4.8 ×10

m/s2 , down . At point d,=
a
)
4
m/s 2 , up
( 9.6 × 10
4
)
)
m/s2 , down .
34.21. Visualize: The solenoid has inductance and when a current flows there is energy stored in the
magnetic field.
Solve: The inductance and energy of the solenoid are
=
Lsol
H/m ) ( 200 ) π ( 0.015 m )
= 2.96 × 10−4 H
l
0.12 m
1
1
UL =
LI 2 =
(2.96 × 10−4 H)(0.80 A) 2 =
9.5 × 10−5 J
2
2
µ0 N 2 A
=
( 4π ×10
−7
2
2
34.22. Visualize: A capacitor in combination with an inductor will have a resonant frequency determined
by the inductance and capacitance.
Solve: We know the frequency and the capacitance so we can find the inductance. We have
ω=
1
1
⇒L=
=
LC
ω 2C
1
( 2π ×100 ×10 Hz ) (10 ×10−12 F)
6
2
= 2.5 × 10−7 H = 0.25 µ H
34.40. Model: Since the solenoid is fairly long compared to its diameter and the coil is located near the
center, assume the solenoid field is uniform inside and zero outside.
Visualize: Please refer to Figure P34.40. The solenoid’s magnetic field is parallel to the coil’s axis and creates
a flux through the coil. This flux changes as the solenoid’s current changes, causing an induced emf and
corresponding induced current in the coil.
2
Solve:
The induced current from the induced emf is given by Faraday’s law. We have
I=
coil
2
2
coil 1
d Φ coil N coilπ rcoil
d  µ0 NI 
N coilπ rcoil
µ0 N dI
=
=
=
N coil


R
R
dt
R
dt  l 
Rl
dt
2
We used the fact that the coil flux is confined to the area Acoil = π rcoil
of the coil, not the larger area of the
solenoid. The current is changing uniformly over the interval 0 s ≤ t ≤ 0.02 s at the rate |dI/dt| = 50 A/s so the
induced current during this interval is
5π ( 0.0050 m ) ( 4π × 10−7 T m/A ) (120 )
I coil =
3.7 × 10−4 A =
0.37 mA
( 50 A/s ) =
0.10
Ω
0.080
m
(
)(
)
2
The current in the solenoid is changing steadily so the induced current in the coil is constant. The current is
initially positive so the field is initially to the right and decreasing. The induced current will oppose this change
and will therefore produce an induced field to the right. This requires an induced current in the coil coming out of
the page at the top, so it is also positive. That is, it is clockwise when seen from the left.
34.41. Model: Assume that the magnetic field of coil 1 passes through coil 2 and that we can use the
magnetic field of a solenoid for coil 1.
Visualize: Please refer to Figure P34.41. The field of coil 1 produces flux in coil 2. The changing current in
coil 1 gives a changing flux in coil 2 and a corresponding induced emf and current in coil 2.
Solve: (a) From 0 s to 0.1 s and 0.3 s to 0.4 s the current in coil 1 is constant so the current in coil 2 is zero.
Thus I (0.05 s) = 0 A.
(b) From 0.1 s to 0.3 s, the induced current from the induced emf is given by Faraday’s law. The current in coil 2
is
I2
=
N2π r22 µ 0 N1 dI1
2 1
dΦ 2
dB1
1
1
d µ NI 
N2
N2 A2 =
N2π r22  0 1 1 =
=
=
R R
dt
R
dt
R
dt  l1 
Rl1
dt
(
)
20π ( 0.010 m ) 4π × 10 −7 T m/A ( 20 )
=
20 A/s =7.95 × 10 −5 A =79 µ A
( 2Ω )( 0.020 m )
2
We used the facts that the field of coil 1 is constant inside the loops of coil 2 and the flux is confined to the area
l1 N=
20 (1.0 mm
=
A2 = π r22 of coil 2. Also, we used=
) 0.020 m and |dI/dt| = 20 A/s. From 0.1 s to 0.2 s the
1d
current in coil 1 is initially negative so the field is initially to the right and the flux is decreasing. The induced
current will oppose this change and will therefore produce a field to the right. This requires an induced current in
coil 2 that comes out of the page at the top of the loops so it is negative. From 0.2 s to 0.3 s the current in coil 1 is
positive so the field is to the left and the flux is increasing. The induced current will oppose this change and will
therefore produce a field to the right. Again, this is a negative current. Hence I(0.25 s) = 79 µA right to left
through the resistor.
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34.52. Model: Assume the magnetic field is uniform over the region where the bar is sliding and that
friction between the bar and the rails is zero.
Visualize: Please refer to Figure P34.52. The battery will produce a current in the rails and bar and the bar will
experience a force. With the battery connected as shown in the figure, the current in the bar will be down and by
the right-hand rule the force on the bar will be to the right. The motion of the bar will change the flux through the
loop and there will be an induced emf that opposes the change.
Solve: (a) As the bar speeds up the induced emf will get larger until finally it equals the battery emf. At that
point, the current will go to zero and the bar will continue to move at a constant velocity. We have
 = Blvterm =  bat ⇒ vterm =
 bat
Bl
(b) The terminal speed is
=
vterm
1.0 V
= 33 m/s
( 0.50 T )( 0.060 m )
Assess: This is pretty fast, about 70 mph.
34.78. Model: Assume an ideal inductor and an ideal (resistanceless) battery.
Visualize: Please refer to Figure P34.78. The current through the battery is the sum of the currents through the
left and right branches of the circuit.
Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is
closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is
because the potential difference ∆V L = –L(dI/dt) would become infinite for an instantaneous change of current, and
that is not physically possible. Because the current through the inductor was zero before the switch was closed, it
must still be zero (or very close to it) immediately after the switch is closed. Conservation of current requires the
current through the entire right branch to be the same as that through the inductor, so it is also zero immediately
after the switch is closed. The only current is through the left 20 Ω resistor, which sees the full battery potential of
the battery. Thus I bat = I left = ∆V bat /R = (10 V)/(20 Ω) = 0.50 A.
(b) The current through the inductor increases as time passes. Once the current I right reaches a steady value and is
no longer changing, the potential difference across the inductor is ∆V L = –L(dI/dt) = 0 V. An ideal inductor has
no resistance, so the inductor simply acts like a wire and has no effect on the circuit. The circuit is that of two 20
Ω resistors in parallel. The equivalent resistance is 10 Ω and the battery current is I bat = (10 V)/(10 Ω) = 1.0 A.
34.82. Model: Assume the RC circuit has wires with zero resistance. Assume that only the bottom
horizontal wire in the RC circuit closest to the metal loop contributes to the magnetic flux through the loop.
Assume that the wire appears infinite in length compared to the metal loop.
Visualize: See Figure CP34.82. The current in the RC circuit decreases as the capacitor is discharged. There is
a changing magnetic flux through the metal loop due to the changing current through the RC circuit, inducing an
emf through the loop.
4
Solve: (a) Current I(t) = I 0 e-t/RC flows from right to left through the wire at the bottom of the RC circuit,
causing a magnetic field out of the paper through the metal loop. Since the RC circuit current is decreasing, the
induced current through the loop is counterclockwise in order to increase the magnetic flux.
(b) Using the result of Example 34.5, the flux through the metal loop due to the wire is
Φ m=
µ0 I (t )(2.0 cm)  0.5 cm + 1.0 cm 
ln 
=
2π
0.5 cm


( 4.4 ×10 ) I e
−9
−t
RC
0
The magnitude of the induced emf in the metal loop is
=

dΦ
=
dt
( 4.4 ×10 ) I e =
−t
−9
0
RC
RC
( 4.4 ×10 ) I e
−4
−t
RC
0
The initial current through the RC circuit I 0 = 20 V/2.0 Ω = 10 A. At t = 5.0 µs,  = 0.267 mV. The current
through the loop I=

=
/R 0.267 mV/0.050 Ω = 5.3 mA.
loop
Assess: As the current through the RC circuit decreases to a constant value of zero at large times, the induced
emf also approaches zero.
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