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ENGR -2150 HOMEWORK-10
Exercise 30.5
30.5.
IDENTIFY and SET UP: Apply Eq. (30.5).
N 
400(00320 Wb)
EXECUTE: (a) M  2 B 2 
 196 H
i1
652 A
N1B1
Mi
(196 H)(254 A)
so B1  2 
 711  103 Wb
i2
N1
700
EVALUATE: M relates the current in one coil to the flux through the other coil. Eq. (30.5) shows that M is the
same for a pair of coils, no matter which one has the current and which one has the flux.
(b) M 
Exercise 30.8
30.8.
IDENTIFY: A changing current in an inductor induces an emf in it.
(a) SET UP: The self-inductance of a toroidal solenoid is L 
EXECUTE: L 
0 N 2 A
.
2 r
(4 107 T  m/A)(500)2 (625 104 m2 )
 781 104 H
2 (00400 m)
(b) SET UP: The magnitude of the induced emf is
EXECUTE:
  L di .
dt
A
  (781104 H)  500 A  200
  0781 V
3
 300 10 s 
(c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to
b, making b at a higher potential than a.
EVALUATE: This is a reasonable value for self-inductance, in the range of a mH.
Exercise 30.13
30.13.IDENTIFY: The inductance depends only on the geometry of the object, and the resistance of the wire depends on
its length.
SET UP: L 
0 N 2 A
.
2 r
EXECUTE: (a) N 
2 rL
(0120 m)(0100 103 H)

 100  103 turns.
0 A
(2 107 T  m/A)(0600 104 m2 )
(b) A   d 2/4 and c   d , so c  4 A  4 (0600 104 m2 )  002746 m. The total length of the wire is
(1000)(002746 m)  2746 m. Therefore R  (00760 /m)(2746 m)  209.
EVALUATE: A resistance of 2  is large enough to be significant in a circuit.
Exercise 30.19
30.19.IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy.
(a) SET UP: The magnetic field inside a solenoid is B  0nI .
EXECUTE: B 
(4 107 T  m/A)(400)(800 A)
 0161 T
0250 m
(b) SET UP: The energy density in a magnetic field is u 
EXECUTE: u 
B2
.
2 0
(0161T)
 2
 103  104 J/m3
2(4 107 T  m/A)
(c) SET UP: The total stored energy is U  uV .
EXECUTE: U  uV  u(lA)  (103 104 J/m3 )(0250 m)(0500 104 m2 )  0129 J
(d) SET UP: The energy stored in an inductor is U  12 LI 2 .
EXECUTE: Solving for L and putting in the numbers gives
2U 2(0129 J)
L 2 
 402  105 H
I
(800 A)2
EVALUATE: An inductor stores its energy in the magnetic field inside of it.
Exercise 30.27
Exercise 30.36
1
 2 f
LC
SET UP:  is the angular frequency in rad/s and f is the corresponding frequency in Hz.
1
1
 237  103 H.
EXECUTE: (a) L  2 2  2
6
2
12
4 f C 4 (16 10 Hz) (418 10
F)
(b) The maximum capacitance corresponds to the minimum frequency.
1
1
Cmax  2 2
 2
 367  1011 F  367 pF
5
4 f min L 4 (540  10 Hz)2 (237  103 H)
EVALUATE: To vary f by a factor of three (approximately the range in this problem), C must be varied by a
factor of nine.
30.36.IDENTIFY:  
Exercise 30.40
30.40.IDENTIFY: The presence of resistance in an L-R-C circuit affects the frequency of oscillation and causes the
amplitude of the oscillations to decrease over time.
(a) SET UP: The frequency of damped oscillations is  
EXECUTE:  
1
(22 103 H)(150 109 F)

1
R2
 2.
LC 4 L
(750 )2
4(22 103 H) 2
 55  104 rad/s
 550 104 rad/s

 876  103 Hz  876 kHz.
2
2
(b) SET UP: The amplitude decreases as A(t )  A0 e –( R/2 L)t .
The frequency f is f 
Execute: Solving for t and putting in the numbers gives:
t
2L ln( A/A0 ) 2(220 103 H)ln(0100)

 135  103 s  135 ms
R
750 
(c) SET UP: At critical damping, R  4L/C .
EXECUTE: R 
4(220 103 H)
150  109 F
 2420 
EVALUATE: The frequency with damping is almost the same as the resonance frequency of this circuit
(1/ LC ), which is plausible because the 75- resistance is considerably less than the 2420  required for
critical damping.
Problem 30.57
30.57.IDENTIFY and SET UP: Use UC  12 CVC2 (energy stored in a capacitor) to solve for C. Then use
Eq. (30.22) and   2 f to solve for the L that gives the desired current oscillation frequency.
EXECUTE: VC  120 V; UC  12 CVC2 so C  2UC /VC2  2(00160 J)/(120 V)2  222F
f 
1
so L 
1
2 LC
(2 f )2 C
f  3500 Hz gives L  931H
EVALUATE: f is in Hz and  is in rad/s; we must be careful not to confuse the two.
Problem 30.61
30.61.IDENTIFY: The current through an inductor doesn’t change abruptly. After a long time the current isn’t changing
and the voltage across each inductor is zero.
SET UP: First combine the inductors.
EXECUTE: (a) Just after the switch is closed there is no current in the inductors. There is no current in the
resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V.
(b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the
inductors can be replaced by short-circuits. The circuit becomes equivalent to the circuit shown in
Figure 30.61a. I  (200 V)/(750 )  0267 A. The voltage between points a and b is zero, so the voltmeter
reads zero.
(c) Combine the inductor network into its equivalent, as shown in Figure 30.61b. R  750  is the equivalent
resistance. Eq. (30.14) says i  ( /R)(1  et/ ) with   L/R  (108 mH)/(750 )  0144 ms.
  200 V,
R  750 , t  0115 ms so i  0147A. VR  iR  (0147 A)(750 )  110 V. 200 V  VR  VL  0 and
VL  200 V  VR  90 V. The ammeter reads 0.147 A and the voltmeter reads 9.0 V.
EVALUATE: The current through the battery increases from zero to a final value of 0.267 A . The voltage
across the inductor network drops from 20.0 V to zero.
Figure 30.61
Problem 30.69
30.69.IDENTIFY: Apply the loop rule to each parallel branch. The voltage across a resistor is given by iR and the voltage
across an inductor is given by L di/dt . The rate of change of current through the inductor is limited.
SET UP: With S closed the circuit is sketched in Figure 30.69a.
The rate of change of the current through
the inductor is limited by the induced emf.
Just after the switch is closed the current
in the inductor has not had time to increase
from zero, so i2  0.
Figure 30.69a
EXECUTE : (a)   vab  0, so vab  600 V
(b) The voltage drops across R, as we travel through the resistor in the direction of the current, so point a is at
higher potential.
(c) i2  0 so vR  i2 R2  0
  vR
2
2
 vL  0 so vL    600 V
(d) The voltage rises when we go from b to a through the emf, so it must drop when we go from a to b through
the inductor. Point c must be at higher potential than point d.
di
(e) After the switch has been closed a long time, 2  0 so vL  0. Then   vR2  0 and i2 R2  
dt

600 V
so i2 

 240 A.
R2 250 
SET UP: The rate of change of the current through the inductor is limited by the induced emf. Just after the
switch is opened again the current through the inductor hasn’t had time to change and is still i2  240 A. The
circuit is sketched in Figure 30.69b.
EXECUTE: The current through
R1 is i2  240 A in the direction b to a.
Thus vab  i2R1  (240 A)(400 )
vab  960 V.
Figure 30.69b
(f) Point where current enters resistor is at higher potential; point b is at higher potential.
(g) vL  vR1  vR2  0
vL  vR1  vR2
vR1  vab  960 V; vR2  i2R2  (240 A)(250 )  600 V
Then vL  vR1  vR2  960 V  600 V  156 V.
As you travel counterclockwise around the circuit in the direction of the current, the voltage drops across each
resistor, so it must rise across the inductor and point d is at higher potential than point c. The current is
decreasing, so the induced emf in the inductor is directed in the direction of the current. Thus, vcd  156 V.
(h) Point d is at higher potential.
EVALUATE: The voltage across R1 is constant once the switch is closed. In the branch containing R2 , just
after S is closed the voltage drop is all across L and after a long time it is all across R2 . Just after S is opened
the same current flows in the single loop as had been flowing through the inductor and the sum of the voltage
across the resistors equals the voltage across the inductor. This voltage dies away, as the energy stored in the
inductor is dissipated in the resistors.