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48 ••
A 300-turn solenoid has a radius equal to 2.00 cm; a length equal to 25.0 cm, and a
1000-turn solenoid has a radius equal to 5.00 cm and is also
25.0-cm long. The two solenoids are coaxial, with one nested completely inside the other. What
is their mutual inductance?
Picture the Problem We can find the mutual inductance of the two coaxial

solenoids using M 2,1  m2   0 n2 n1r12 .
I1
Substitute numerical values and evaluate M2,1:
 300   1000 
 
0.250 m 0.0200 m2  1.89 mH
M 2,1  4 10 7 N/A 2 
 0.250 m   0.250 m 


49 •• [SSM] An insulated wire that has a resistance of 18.0 /m and a length of 9.00 m
will be used to construct a resistor. First, the wire is bent in half and then the doubled wire is
wound on a cylindrical form ( Figure 28-50) to create a 25.0-cm-long helix that has a diameter
equal to 2.00 cm. Find both the resistance and the inductance of this wire-wound resistor.
Picture the Problem Note that the current in the two parts of the wire is in
opposite directions. Consequently, the total flux in the coil is zero. We can find
the resistance of the wire-wound resistor from the length of wire used and the
resistance per unit length.
Because the total flux in the coil is
zero:
L 0
Express the total resistance of the
wire:
Ω

R  18.0  L
m

Substitute numerical values and
evaluate R:
Ω

R  18.0 9.00 m   162 Ω
m

50
••
You are given a length  of wire that has radius a and are told to wind it into an
inductor in the shape of a helix that has a circular cross section of radius r. The windings are to
be as close together as possible without overlapping. Show that the self-inductance of this
inductor is L  14  0 r a .
Picture the Problem The wire of length  and radius a is shown in the diagram,
as is the inductor constructed with this wire and whose inductance L is to be
found. We can use the equation for the self-inductance of a cylindrical inductor to
derive an expression for L.
The self-inductance of an inductor
with length d, cross-sectional area A,
and number of turns per unit length n
is:
L   0 n 2 Ad
The number of turns N is given by:
N
Assuming that a << r, the length of
the wire  is related to N and r:
r
 d 
  N 2 r    2 r 
d
a
 2a 
Solving for d yields:
d
Substitute for d, A, and n in equation
(1) to obtain:
 a 
 1 
 
L   0    r 2 
 2a 
 r 
(1)
d
N
1
n  
2a
d 2a
a
r
2
1
4
 0 r a
51 •
Using the result of Problem 50, calculate the self-inductance of an inductor wound
from 10 cm of wire that has a diameter of 1.0 mm into a coil that has a radius of 0.25 cm.
Picture the Problem We can substitute numerical values in the expression
derived in Problem 50to find the self-inductance of the inductor.
From Problem 50 we have:
L
Substitute numerical values and evaluate L:
 0 rd
4a
L
4 10
7

N/A 2 0.25 cm 10 cm 
 0.16 H
40.50 mm 
61 •• [SSM] In the circuit shown in Figure 28-54, let 0 = 12.0 V,
R = 3.00 , and L = 0.600 H. The switch, which was initially open, is closed at time t = 0. At
time t = 0.500 s, find (a) the rate at which the battery supplies energy, (b) the rate of Joule
heating in the resistor, and (c) the rate at which energy is being stored in the inductor.


Picture the Problem We can find the current using I  I f 1  e t  , where
If = 0/R,and  = L/R, and its rate of change by differentiating this expression with
respect to time.
Express the dependence of the
current on If and  :
Evaluating If and  yields:

I t   I f 1  e t 
If 
0
R


12.0 V
 4.00 A
3.00 Ω
and

L 0.600 H

 0.200 s
R 3.00 Ω


Substitute for If and  to obtain:
I t   4.00 A  1  e t 0.200s
Express dI/dt:
dI
 4.00 A  e t 0.200 s  5.00 s 1 
dt
 20.0 A/s e t 0.200 s
(a) The rate at which the battery
supplies energy is given by:
Substituting for I and 0 yields:
The rate at which the battery
supplies energy at t = 0.500 s is:
P  I 0

 48.0 W 1  e


Pt   4.00 A  1  e t 0.200s 12.0 V 
t 0.200s

P0.500 s   48.0 W  1  e 0.500 0.200s
 44.1 W
(b) The rate of Joule heating is:
PJ  I 2 R
Substitute for I and R and simplify to
obtain:
PJ  4.00 A  1  e t 0.200s


 48.0 W 1  e
 3.00 Ω

2
t 0.200s 2

The rate of Joule heating at
t = 0.500 s is:
(c) Use the expression for the
magnetic energy stored in an
inductor to express the rate at which
energy is being stored:
PJ 0.500 s   48.0 W 1  e 0.500 s 0.200s 
2
 40.4 W
dU L d

dt
dt

1
2

LI 2  LI
dI
dt
Substitute for L, I, and dI/dt to obtain:
dU L
 0.600 H 4.00 A 1  e t 0.200s 20.0 A/s e t 0.200s
dt
 48.0 W 1  e t 0.200s e t 0.200s
Evaluate this expression for t = 0.500 s:


dU L 
 48.0 W  1  e 0.500 s 0.200s e 0.500 s 0.200s  3.62 W

dt  t 0.500 s
Remarks: Note that, to a good approximation, dUL/dt = P  PJ.
3
•
[SSM] If the frequency in the circuit shown in Figure 29-27 is doubled, the
inductance of the inductor will (a) double, (b) not change, (c) halve, (d) quadruple.
Determine the Concept The inductance of an inductor is determined by the
details of its construction and is independent of the frequency of the circuit. The
inductive reactance, on the other hand, is frequency dependent. (b) is correct.
5
•
If the frequency in the circuit in Figure 29-28 is doubled, the capacitive reactance of
the circuit will (a) double, (b) not change, (c) halve,
(d) quadruple.
Determine the Concept The capacitive reactance of an capacitor varies with the
frequency according to X C  1 C . Hence, doubling  will halve XC. (c) is
correct.
34 ••
A circuit consists of a resistor, an ideal 1.4-H inductor and an ideal
60-Hz generator, all connected in series. The rms voltage across the resistor is
30 V and the rms voltage across the inductor is 40 V. (a) What is the resistance of the resistor?
(b) What is the peak emf of the generator?
Picture the Problem We can express the ratio of VR to VL and solve this
expression for the resistance R of the circuit. In (b) we can use the fact that, in an
LR circuit, VL leads VR by 90 to find the ac input voltage.
(a) Express the potential
differences across R and L in terms
of the common current through
these components:
VL  IX L  IL
and
VR  IR
Divide the second of these
equations by the first to obtain:
V 
VR
IR
R
 R   R L


VL IL L
 VL 
Substitute numerical values and
evaluate R:
 30 V 
2 60 s 1 1.4 H   0.40 kΩ
R  
 40 V 
(b) Because VR leads VL by 90 in
an LR circuit:
Vpeak  2Vrms  2 VR2  VL2
Substitute numerical values and
evaluate Vpeak :
Vpeak  2


30 V 2  40 V 2
 71 V
35 ••
[SSM] A coil that has a resistance of 80.0  has an impedance of 200  when
driven at a frequency of 1.00 kHz. What is the inductance of the coil?
Picture the Problem We can solve the expression for the impedance in an LR
circuit for the inductive reactance and then use the definition of XL to find L.
Express the impedance of the coil in
terms of its resistance and inductive
reactance:
Z  R 2  X L2
Solve for XL to obtain:
X L  Z 2  R2
Express XL in terms of L:
X L  2fL
Equate these two expressions to
obtain:
2fL  Z 2  R 2  L 
Substitute numerical values and
evaluate L:
L
Z 2  R2
2f
200 Ω 2  80.0 Ω 2
2 1.00 kHz 
 29.2 mH
41 ••
[SSM] Figure 29-33 shows a load resistor that has a resistance of
RL = 20.0  connected to a high-pass filter consisting of an inductor that has inductance L =
3.20-mH and a resistor that has resistance R = 4.00-. The output of the ideal ac generator is
given by  = (100 V) cos(2ft). Find the rms currents in all three branches of the circuit if the
driving frequency is (a) 500 Hz and
(b) 2000 Hz. Find the fraction of the total average power supplied by the ac generator that is
delivered to the load resistor if the frequency is (c) 500 Hz and (d) 2000 Hz.
Picture the Problem   V1  V2 , where V1 is the voltage drop across R and V2 is
  
the voltage drop across the parallel combination of L and RL. ε  V1  V2 is the
 

relation for the phasors. For the parallel combination I  I RL  I L . Also, V1 is in
phase with I and V2 is in phase with I RL . First draw the phasor diagram for the
currents in the parallel combination, then add the phasors for the voltages to the
diagram.
The phasor diagram for the currents
in the circuit is:
Adding the voltage phasors to the
diagram gives:
The maximum current in the
inductor, I2, peak, is given by:
I 2, peak 
V2, peak
(1)
Z2
where Z 22  RL2  X L2
tan  is given by:
tan  

I L , peak
I R , peak

V2, peak X L
V2, peak RL
RL
R
R
 L  L
X L L 2fL
(2)
Solve for  to obtain:
 RL 

 2fL 
  tan 1 
(3)
Apply the law of cosines to the triangle formed by the voltage phasors to obtain:
2
 peak
 V1,2peak  V22, peak  2V1, peakV2, peak cos 
or
2
2
2
I peak
Z 2  I peak
R2  I peak
Z22  2I peak RI peak Z2 cos 
Dividing out the current squared
yields:
Z 2  R 2  Z 22  2RZ 2 cos 
Solving for Z yields:
Z  R 2  Z 22  2 RZ 2 cos 
The maximum current I peak in the
I peak 
circuit is given by:
Irms is related to I peak according
I rms 
to:
(a) Substitute numerical values in
equation (3) and evaluate  :
 peak
(4)
(5)
Z
1
I peak
2
(6)


20.0 Ω

 2 500 Hz 3.20 mH  
  tan 1 
 20.0 Ω 
  63.31
 tan 1 
 10.053 Ω 
Solving equation (2) for Z2 yields:
Substitute numerical values and
evaluate Z2:
Z2 
Z2 
1
2
L
R  X L2
1
20.0 Ω 2  10.053 Ω2
 8.982 Ω
Substitute numerical values and evaluate Z:
Z
4.00 Ω2  8.982 Ω2  24.00 Ω8.982 Ωcos 63.31  11.36 Ω
Substitute numerical values in
equation (5) and evaluate I peak :
I peak 
100 V
 8.806 A
11.36 
Substitute for I peak in equation
(6) and evaluate I rms :
The maximum and rms values of
V2 are given by:
1
8.806 A   6.23 A
2
I rms 
V2, peak  I peak Z 2
 8.806 A 8.982 Ω   79.095 V
and
1
V2, peak
2
1
79.095 V   55.929 V

2
V2,rms 
The rms values of I RL ,rms and
I L ,rms are:
I RL ,rms 
V2,rms
RL

55.929 V
 2.80 A
20.0 Ω
and
I L ,rms 
V2,rms
XL

55.929 V
 5.53 A
10.053 Ω
X L  40.2  ,   26.4 , Z 2  17.9  ,
(b) Proceed as in (a) with
f = 2000 Hz to obtain:
Z  21.6  , I peak  4.64 A , and
I rms  3.28 A ,
V2,max  83.0V , V2, rms  58.7 V ,
I RL ,rms  2.94 A , and I L,rms  1.46 A
(c) The power delivered by the ac source equals the sum of the power dissipated in
the two resistors. The fraction of the total power delivered by the source that is
dissipated in load resistor is given by:

P
 1  R

PRL  PR  PRL
PRL
1
2



  1  I rms R 

 I R2 ,rms RL 
L



1
Substitute numerical values for f = 500 Hz to obtain:
1
PRL
PRL  PR
f 500 Hz
 6.23 A 2 4.00   
  0.502  50.2%
 1 
2

 2.80 A  20.0   
(d) Substitute numerical values for f = 2000 Hz to obtain:
1
PRL
PRL  PR
f  2000Hz
 3.28 A 2 4.00   
  0.800  80.0%
 1 
2





2
.
94
A
20
.
0


