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Physics III
Homework IV
CJ
Chapter 33; 60, 72, 73, 74 and Chapter 23; 4, 9, 14, 16
33.60. Model:
Assume the magnetic field is uniform over the loop.
Visualize: The motion of the eye will change the orientation of the loop relative to the
fixed field direction resulting in an induced emf.
Solve: We are just interested in the emf due to the motion of the eye so we can ignore the
details of the time dependence of the change. From Faraday’s law,
E coil  N


N r 2 B cos f  cos i
d

 cos
N
 NAB

dt
t
t
t
20 3  10 3 m
 1.0 T  cos85  cos90
2
0.20 s
 2.46  10 4 V
Assess: This is a reasonable emf to measure, although you might need some amplification.
33.72. Model:
Assume negligible resistance in the circuit.
Visualize: Energy is conserved. The maximum voltage on the right capacitor will occur
when all of the energy from the left capacitor is transferred to the right one.
Solve: (a) The voltage is calculated as follows:
1
2
2
C1VC12  21 C2 VC2
 VC2 
C1
300
VC1 
100 V   50 V
C2
1200
(b) Closing S1 causes the charge and current of the left LC circuit to oscillate with
period TL. After one-quarter of a period, the 300 F capacitor is completely discharged and
the current through the inductor is maximum. At that instant we’ll open S1 and close S2.
Then the right LC circuit will start to oscillate with period TR and the inductor current will
charge the 1200 F capacitor. The capacitor will be fully charged after one-quarter of a
period, so we will open S2 at that time to keep the charge on the 1200 F capacitor. The
periods are
2
 2 LC1  2 (5.3 H)(300  10 6 F)  0.25 s  14 TL  0.0625 s
L
2
TR 
 2 LC2  2 (5.3 H)(1200  10 6 F)  0.50 s  14 TR  0.1250 s
R
TL 
So the procedure is to close S1 at t  0 s, open S1 and close S2 at t  0.0625 s, then open S2 at t 
0/1250 s.
33.73. Model:
Assume an ideal inductor and an ideal (resistanceless) battery.
Visualize: Please refer to figure P33.73. 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 VL  –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 10  resistor, which sees
the full battery potential of the battery. Thus Ibat  Ileft  Vbat/R  (10 V)/(10 )  1.0 A.
(b) The current through the inductor increases as time passes. Once the current Iright
reaches a steady value and is no longer changing, the potential difference across the inductor
is VL  –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 10  resistors in
parallel. The equivalent resistance is 5  and the battery current is Ibat  (10 V)/(5 )  2.0
A.
33.74. Model:
Assume an ideal inductor and an ideal (resistanceless) battery.
Visualize: Please refer to figure P33.74.
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 VL  –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. Consequently, the inductor has no effect on the
circuit. It is simply a 10  resistor and 20  resistor in series with the battery. The equivalent
resistance is 30 , so the current through the circuit (including through the 20  resistor) is I 
Vbat/Req  (30 V)/(30 )  1.0 A.
(b) After a long time, the currents in the circuit will reach steady values and no
longer change. With steady currents, the potential difference across the inductor is VL  –
L(dI/dt)  0 V. An ideal inductor has no resistance (R  0 ), so the inductor simply acts
like a wire. In this case, the inductor “shorts out” the 20  resistor. All current from the 10 
resistor flows through the resistanceless inductor, so the current through the 20  resistor is 0 A.
(c) When the switch has been closed a long time, and the inductor is shorting out the
20  resistor, the current passing through the 10  resistor and through the inductor is I 
(30 V)/(10 )  3.0 A. Because the current through an inductor cannot change
instantaneously, the current must remain 3.0 A immediately after the switch reopens. This
current must go somewhere (conservation of current), but now the open switch prevents the
current from going back to the battery. Instead, it must flow upward through the 20 
resistor. That is, the current flows around the LR circuit consisting of the 20  resistor and
the inductor. This current will decay with time, with time constant   L/R, but immediately
after the switch reopens the current is 3.0 A.
23.4. Model:
sources.
Light rays travel in straight lines. Also, the red and green light bulbs are point
Visualize
:
Solve: The width of the aperture is w  1 m. From the geometry of the figure for red light,
w2
x
 x  2w  2 1.0 m  2.0 m

1m 3m  1m
The red light illuminates the wall from x = 0.50 m to x = 4.50 m. For the green light,
x1
w4
 x1  1.0 m

1m 3m  1m
x2
3w 4
 x2  3.0 m

1m
3m  1m
Because the back wall exists only for 2.75 m to the left of the green light source, the green light has a
range from x  0 m to x  3.75 m.
23.9. Model:
Light rays travel in straight lines and follow the law of reflection.
Visualize
:
To determine the angle  , we must know the point P on the mirror where the ray is incident. P is a
distance x2 from the far wall and a horizontal distance x1 from the laser source. The ray from the
source
must
strike
P
so
that
the
angle of incidence i is equal to the angle of reflection r.
Solve: From the geometry of the diagram,
tan  

1.5 m 3 m

x2
x1
1.5 m
3m
10

 1.5 m  x1  15 m 2   3 m  x1  x1 
m
5 m  x1
x1
3
 tan 
23.14. Model:
x1  x2  5 m
3m
9

 0.90    42
x1
10
Use the ray model of light. The sun is a point source of light.
Visualize
:
A ray that arrives at the diver 50 above horizontal refracted into the water at water  40.
Solve: Using Snell’s law at the water-air boundary
nair sinair  nwater sin water  sin air 
 air
nwater
 1.33 
sin water  
 sin40
nair
 1.0 
 58.7
Thus the height above the horizon is  = 90  air = 31.3. Because the sun is far away from the
fisherman (and the diver), the fisherman will see the sun at the same angle of 31.3 above the
horizon.
23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical
angle, the ray of light undergoes total internal reflection.
Visualize
:
Solve: The critical angle of incidence is given by Equation 23.9:
 ncladding 
1  1.48 
  sin 
  67.7
n
 1.60 
 core 
 c  sin 1 
Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is
90  67.7  22.3.
Assess: We can have total internal reflection because ncore  ncladding.