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Chapter 22 Homework Solutions
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3. REASONING AND SOLUTION The motional emf  generated by a conductor moving perpendicular to
a magnetic field is given by Equation 22.1 as  = vBL, where v and L are the speed and length,
respectively, of the conductor, and B is the magnitude of the magnetic field. The emf would have been




  vBL  7.6  103 m/s 5.1  105 T 2.0  104 m  7800 V
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6. REASONING The situation in the drawing given with the problem statement is analogous to that in
Figure 22.4b in the text. The blood flowing at a speed v corresponds to the moving rod, and the diameter of
the blood vessel corresponds to the length L of the rod in the figure. The magnitude of the magnetic field is
B, and the measured voltage is the emf  induced by the motion. Thus, we can apply   BvL
(Equation 22.1).
SOLUTION Using Equation 22.1, we find that


  BvL   0.60 T  0.30 m/s  5.6 103 m  1.0 103 V
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11. REASONING According to Equation 22.2, the magnetic flux  is the product of the magnitude B of the
magnetic field, the area A of the surface, and the cosine of the angle between the direction of the
magnetic field and the normal to the surface. The area of a circular surface is A   r 2 , where r is the
radius.
SOLUTION The magnetic flux  through the surface is
  BA cos   B  r 2  cos    0.078 T    0.10 m  cos 25  2.2 103 Wb
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20. REASONING An emf is induced in the body because the magnetic flux is changing in time. According to

Faraday’s law, as given by Equation 22.3, the magnitude of the emf is    N
. This expression can
t
be used to determine the time interval t during which the magnetic field goes from its initial value to zero.
The magnetic flux  is obtained from Equation 22.2 as  = BA cos , where  = 0 in this problem.
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SOLUTION The magnitude  of the induced emf is
 BA cos   B0 A cos  

t

  
 = N 
  N 
 t 

Solving this relation for t gives
t  
 NA cos   B  B0 


 4.8 s
0.010 V
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

 1 0.032 m2 cos 0°  0  1.5 T 
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22. REASONING According to Equation 22.3, Faraday’s law specifies the emf induced in a coil of N loops as

  N
t
where  / t is the rate of change of the magnetic flux in a single loop. Recognizing that  / t is the
same for each of the coils, we will apply Faraday’s law to each coil to obtain our solution.
SOLUTION Applying Faraday’s law to each coil gives
1   N1

t
and
2   N2

t
Dividing these equations and remembering that the rate of change of the flux is the same for each coil, we
find that

 2  N 2 t N 2


1  N  N1
1
or
N 2  N1
2
 4.23 V 
 184 
  276
1
 2.82 V 
t
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34. REASONING The magnetic field produced by I extends throughout the space surrounding the loop.
Using RHR-2, it can be shown that the magnetic field is parallel to the normal to the loop. Thus, the
magnetic field penetrates the loop and generates a magnetic flux.
According to Faraday’s law of electromagnetic induction, an emf is induced when the magnetic flux
through the loop is changing in time. If the current I is constant, the magnetic flux is constant, and no emf
is induced in the loop. However, if the current is decreasing in time, the magnetic flux is decreasing and an
induced current exists in the loop.
Lenz’s law states that the induced magnetic field opposes the change in the magnetic field produced by the
current I. The induced magnetic field does not necessarily oppose the magnetic field itself. Thus, the
induced magnetic field does not always have a direction that is opposite to the direction of the field
produced by I.
SOLUTION At the location of the loop, the magnetic field produced by the current I is directed into the
page (this can be verified by using RHR-2). The current is decreasing, so the magnetic field is decreasing.
Therefore, the magnetic flux that penetrates the loop is decreasing. According to Lenz’s law, the induced
emf has a polarity that leads to an induced current whose direction is such that the induced magnetic field
opposes this flux change. The induced magnetic field will oppose this decrease in flux by pointing into the
page, in the same direction as the field produced by I. According to RHR-2, the induced current must flow
clockwise around the loop in order to produce such an induced field. The current then flows from
left-to-right through the resistor.
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35.
SSM REASONING In solving this problem, we apply Lenz's law, which essentially says that the
change in magnetic flux must be opposed by the induced magnetic field.
SOLUTION
a. The magnetic field due to the wire in the vicinity of position 1 is directed out of the paper. The coil is
moving closer to the wire into a region of higher magnetic field, so the flux through the coil is increasing.
Lenz’s law demands that the induced field counteract this increase. The direction of the induced field,
therefore, must be into the paper. The current in the coil must be clockwise .
b. At position 2 the magnetic field is directed into the paper and is decreasing as the coil moves away from
the wire. The induced magnetic field, therefore, must be directed into the paper, so the current in the coil
must be clockwise .
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41.
SSM WWW REASONING AND SOLUTION
a. From the drawing, we see that the period of the generator (the time for one full cycle) is 0.42 s;
therefore, the frequency of the generator is
f 
1
1

= 2.4 Hz
T 0.42 s
b. The angular speed of the generator is related to its frequency by   2 f , so the angular speed is
  2 (2.4 Hz) = 15 rad/s
c. The maximum emf  0 induced in a rotating planar coil is given by 0  NAB (see Equation 22.4).
The magnitude of the magnetic field can be found by solving this expression for B and noting from the
drawing that  0 = 28 V:
B
0
28 V

 0.62 T
NA
(150)(0.020 m2 )(15 rad/s)
44. REASONING AND SOLUTION Using Equation 22.5 to take the back emf into account, we find
V –  120.0 V  –  72.0 V 
R

 16 
I
3.0 A
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53. REASONING AND SOLUTION The induced emf in the secondary coil is proportional to the mutual
inductance. If the primary coil is assumed to be unaffected by the metal, that is I1/t is the same for both
cases, then
New emf = 3(0.46 V) = 1.4 V
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54. REASONING According to s   M
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I p
(Equation 22.7), a change ΔIp in the current in the primary coil
t
induces an emf s in the secondary, where M is the mutual inductance of the two coils and Δt is the time
interval of the current change. We are interested only in the magnitude of the current change ΔIp, so we
will omit the minus sign in Equation 22.7. The induced emf s in the secondary coil will drive a current Is,
as we see from Ohm’s law: s  Is R (Equation 20.2), where R is the resistance of the circuit that includes
the secondary coil.
I
SOLUTION Omitting the minus sign in s   M p (Equation 22.7) and solving for ΔIp yields
t
 t
I p  s
(1)
M
Substituting s  Is R (Equation 20.2) into Equation (1), we obtain
I p 
s t
Is Rt
 6.0 103 A  12    72 103 s 


 1.6 A
M
M
3.2 103 H
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60. REASONING AND SOLUTION According to the transformer equation (Equation 22.12), we have
V 
 4320 V 
Ns   s  N p  
 (21)  756
 Vp 
 120.0 V 
 
62. REASONING Since the secondary voltage (the voltage to charge the batteries) is less than the primary
voltage (the voltage at the wall socket), the transformer is a step-down transformer.
In a step-down transformer, the voltage across the secondary coil is less than the voltage across the primary
coil. However, the current in the secondary coil is greater than the current in the primary coil. Thus, the
current that goes through the batteries is greater than the current from the wall socket.
If the transformer has negligible resistance, the power delivered to the batteries is equal to the power
coming from the wall socket.
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SOLUTION
a. The turns ratio Ns / N p is equal to the ratio of secondary voltage to the primary voltage:
Ns Vs 9.0 V


 1:13
N p Vp 120 V
(22.12)
b. The current from the wall socket is given by Equation 22.13:
N 
1
3
2
I p  I s  s   225  10 A    1.7  10 A
 Np 
 13 




(22.13)
c. The average power delivered by the wall socket is the product of the primary current and voltage:


Pp  I p Vp  17  103 A 120 V   2.0 W
(20.15a)
The average power delivered to the batteries is the same as that coming from the wall socket, so
Ps  2.0 W .
65. REASONING AND SOLUTION The resistance of the primary is (see Equation 20.3)
Rp 
 Lp
A
(1)
The resistance of the secondary is
Rs 
 Ls
A
(2)
In writing Equations (1) and (2) we have used the fact that both coils are made of the same wire, so that the
resistivity  and the cross-sectional area A is the same for each. Division of the equations gives
Rs Ls 14  1



Rp Lp 56  4
Since the diameters of the coils are the same, the lengths of the wires are proportional to the number of
turns. Therefore,
Ns Ls
1


N p Lp
4
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