PHYSICS 172
WQ 2010
Solutions to Homework
#8
1. Giancoli Chapter 29, Problem 8
(a)
As the resistance is increased, the current in the outer loop will decrease. Thus
the flux through the inner loop, which is out of the page, will decrease. To oppose this
decrease, the induced current in the inner loop will produce a flux out of the page, so the
direction of the induced current will be counterclockwise.
(b) If the small loop is placed to the left, the flux through the small loop will be into the page
and will decrease. To oppose this decrease, the induced current in the inner loop will
produce a flux into the page, so the direction of the induced current will be clockwise.
2. Giancoli Chapter 29, Problem 12
As the loop is pulled from the field, the flux through the loop decreases, causing an
induced EMF whose magnitude is given by Eq. 29-3, e  Bl v. Because the inward
flux is decreasing, the induced flux will be into the page, so the induced current is
clockwise, given by I  e R . Because this current in the left-hand side of the loop is
in a downward magnetic field, there will be a magnetic force to the left. To keep the
rod moving, there must be an equal external force to the right, given by
F  I l B.
F  IlB 
e
R
lB 
Bl v
R
lB 
B 2 l 2v
R

 0.650 T 2  0.350 m 2  3.40 m s  
0.280 
0.628 N
3. Giancoli Chapter 29, Problem 16
The sinusoidal varying current in the power line creates a sinusoidal varying magnetic field
encircling the power line, given by Eq. 28-1. Using Eq. 29-1b we integrate this field over
the area of the rectangle to determine the flux through it. Differentiating the flux as in Eq.
29-2b gives the emf around the rectangle. Finally, by setting the maximum emf equal to 170
V we can solve for the necessary length of the rectangle.
I
B(t )  0 0 cos  2 ft  ;
2 r
7.0 m  I
7.0 m dr
I
I
0 0
 B (t )   BdA  
cos  2 ft  l dr  0 0 l cos  2 ft  
 0 0 ln 1.4  l cos  2 ft 
5.0 m 2 r
5.0
m
2
r
2
e  N
dB
N 0 I 0

ln 1.4  l
dt
2
e0  N 0 I 0 f ln 1.4  l 
l 
d

 dt cos  2 ft   N 0 I 0 f ln 1.4  l sin  2 ft  ;
e0
170 V

 12 m
7
N 0 I 0 f ln 1.4  10  4  10 T m A   55,000 A  60 Hz  ln 1.4 
This is unethical because the current in the rectangle creates a back emf in the initial wire.
This results in a power loss to the electric company, just as if the wire had been physically
connected to the line.
4. Giancoli Chapter 29, Problem 20
The induced emf is given by Eq. 29-2a. Since the field is uniform and is perpendicular to
the area, the flux is simply the field times the area.
dB
dA
e
 B
   0.28T  3.50  102 m 2 s  9.8 mV
dt
dt
Since the area changes at a constant rate, and the area has not shrunk to 0 at t = 2.00 s, the
emf is the same for both times.


5. Giancoli Chapter 29, Problem 30
The emf is given by Eq. 29-3 as e  Bl v. The resistance of the conductor is given by Eq.
25-3. The length in Eq. 25-3 is the length of resistive material. Since the movable rod starts
at the bottom of the U at time t = 0, in a time t it will have moved a distance vt.
I
e
R

Bl v
Bl v
B l vA


 L   2vt  l 
  2vt  l 
A
A
6. Giancoli Chapter 29, Problem 33
(a)
As the rod moves through the magnetic field an emf will be
built up across the rod, but no current can flow. Without the current,
v
B
there is no force to oppose the motion of the rod, so yes, the rod
travels at constant speed.
(b) We set the force on the moving rod, obtained in Example 29-8,
equal to the mass times the acceleration of the rod. We then write
the acceleration as the derivative of the velocity, and by separation
of variables we integrate the velocity to obtain an equation for the velocity as a function
of time.
dv
B2 l 2
dv
B2l 2
F  ma  m  
v

dt
dt
R
v
mR
B l

t
dv
B2 l 2 t
v
B2 l 2
mR



dt

ln


t

v
(
t
)

v
e
0
v0 v

0
mR
v0
mR
The magnetic force is proportional to the velocity of the rod and opposes the motion.
This results in an exponentially decreasing velocity.
2 2
v
7. Giancoli Chapter 29, Problem 38
From Eq. 29-4, the peak voltage is epeak  NB A . Solve this for the rotation speed.
epeak  NB A   
f 
epeak
NBA

120 V
480  0.550 T  0.220 m 

9.39 rad s

 1.49 rev s
2 2 rad rev
2
 9.39 rad s
¬
8. Giancoli Chapter 29, Problem 48
(a)
VS
VP
Use Eqs. 29-5 and 29-6 to relate the voltage and current ratios.

NS
NP
;
IS
IP

NP
NS
VS

VP

IP
IP
 VS  VP
IS
IS
 120 V 
0.35 A
7.5 A
 5.6 V
Because VS  VP , this is a step-down transformer.
(b)
9. Giancoli Chapter 29, Problem 50
(a)
The current in the transmission lines can be found from Eq. 25-10a, and then the
emf at the end of the lines can be calculated from Kirchhoff’s loop rule.
Ptown 65  106 W
Ptown  Vrms I rms  I rms 

 1444 A
Vrms
45  103 V
E  IR  Voutput  0 
E  IR  Voutput 
Ptown
Vrms
R  Vrms 
65  106 W
45  10 V
3
 3.0    45  103 V  49333V 
49 kV  rms 
2
R.
The power loss in the lines is given by Ploss  I rms
(b)
Fraction wasted 
Ploss
Ptotal

Ploss
Ptown  Ploss
2
1444A   3.0  



2
2
Ptown  I rms
R 65  106 W  1444 A   3.0  
2
I rms
R
 0.088  8.8%
10. Giancoli Chapter 29, Problem 53
Without the transformers, we find the delivered current, which is the current in the
transmission lines, from the delivered power, and the power lost in the transmission lines.
P
85000 W
Pout  Vout I line  I line  out 
 708.33A
Vout
120 V
2
Plost  I line
Rline   708.33A  2  0.100    100346 W
Thus there must be 85000 W  100346 W  185346W  185kW of power generated at the
start of the process.
2
With the transformers, to deliver the same power at 120 V, the delivered current from the
step-down transformer must still be 708.33 A. Using the step-down transformer efficiency,
we calculate the current in the transmission lines, and the loss in the transmission lines.
Pout  0.99 Pline  Vout I out  0.99Vline I line  I line 
end
Vout I out
0.99Vline

120 V  708.33A   71.548 A
 0.99 1200 V 
2
Plost  I line
Rline   71.548 A  2  0.100    1024 W
The power to be delivered is 85000 W. The power that must be delivered to the step-down
2
85000 W
 85859 W . The power that must be present at the start of the
0.99
transmission must be 85859 W  1024 W  86883W to compensate for the transmission line
loss. The power that must enter the transmission lines from the 99% efficient step-up
86883W
transformer is
 87761  88 kW . So the power saved is
0.99
transformer is
185346 W  87761W  97585 W  98 kW .
11. Giancoli Chapter 29, Problem 54
We choose a circular path centered at the origin with radius 10 cm. By symmetry the
electric field is uniform along this path and is parallel to the path. We then use Eq. 29-8 to
calculate the electric field at each point on this path. From the electric field we calculate the
force on the charged particle.
d
2 dB
 E d l  E  2 r    dt B    r  dt
r dB
0.10 m
F  QE  Q
  1.0  106 C 
 0.10 T/s   5.0 nN
2 dt
2
Since the magnetic field points into the page and is decreasing, Lenz’s law tells us that an
induced circular current centered at the origin would flow in the clockwise direction.
Therefore, the force on a positive charge along the positive x-axis would be down, or in the
 ˆj direction.