30
INDUCTANCE
30.1.
IDENTIFY and SET UP: Apply Eq.(30.4).
di
EXECUTE: (a) E2  M 1  (3.25 104 H)(830 A/s)  0.270 V; yes, it is constant.
dt
di2
; M is a property of the pair of coils so is the same as in part (a). Thus E1  0.270 V.
dt
EVALUATE: The induced emf is the same in either case. A constant di/dt produces a constant emf.
(b) E1  M
30.2.
E1  M
IDENTIFY:
N
i2
i
and E2  M 1 . M  2 B 2 , where  B 2 is the flux through one turn of the second
i1
t
t
coil.
SET UP: M is the same whether we consider an emf induced in coil 1 or in coil 2.
E2
1.65 103 V

 6.82 103 H  6.82 mH
EXECUTE: (a) M 
i1 / t
0.242 A/s
(b)  B 2 
Mi1 (6.82 103 H)(1.20 A)

 3.27 104 Wb
N2
25
i2
 (6.82 103 H)(0.360 A/s)  2.46 10 3 V  2.46 mV
t
EVALUATE: We can express M either in terms of the total flux through one coil produced by a current in the
other coil, or in terms of the emf induced in one coil by a changing current in the other coil.
IDENTIFY: Replace units of Wb, A and  by their equivalents.
SET UP: 1 Wb  1 T  m2. 1 T  1 N/(A  m). 1 N  m  1 J. 1 A  1 C/s. 1 V  1 J/C. 1 V/A  1 .
(c) E1  M
30.3.
30.4.
EXECUTE:
EVALUATE:
IDENTIFY:
(a) SET UP:
EXECUTE:
1 H  1 Wb/A  1 T  m 2 /A  1 N  m/A 2  1 J/A 2  1 (J/[A  C])s  1 (V/A)s  1 Ω  s.
We may use whichever equivalent unit is the most convenient in a particular problem.
Changing flux from one object induces an emf in another object.
The magnetic field due to a solenoid is B  0nI .
The above formula gives
B1 
 4 10
7
T  m/A  (300)(0.120 A)
0.250 m
The average flux through each turn of the inner solenoid is therefore
=1.81104 T
 B  B1 A  1.81 104 T  (0.0100 m) 2 = 5.68  108 Wb
(b) SET UP: The flux is the same through each turn of both solenoids due to the geometry, so
M
EXECUTE:
M
(25)  5.68 108 Wb 
0.120 A
N 2 B,2 N 2 B,1

i1
i1
 1.18 105 H
di1
.
dt
E2   1.18  105 H  (1750 A/s)  0.0207 V
(c) SET UP: The induced emf is E2  M
EXECUTE:
EVALUATE: A mutual inductance around 105 H is not unreasonable.
30-1
30-2
30.5.
Chapter 30
IDENTIFY and SET UP: Apply Eq.(30.5).
400  0.0320 Wb 
N
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.
IDENTIFY: A changing current in an inductor induces an emf in it.
 N2A
(a) SET UP: The self-inductance of a toroidal solenoid is L  0
.
2 r
(4 107 T  m/A)(500) 2 (6.25 104 m 2 )
EXECUTE: L 
 7.81104 H
2 (0.0400 m)
(b) M 
30.6.
(b) SET UP: The magnitude of the induced emf is E  L
di
.
dt
 5.00 A  2.00 A 
E   7.81104 H  
  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.
i
NB
IDENTIFY: E  L
and L 
.
i
t
EXECUTE:
30.7.
i
 0.0640 A/s
t
E
0.0160 V
EXECUTE: (a) L 

 0.250 H
i / t 0.0640 A/s
SET UP:
Li (0.250 H)(0.720 A)

 4.50 104 Wb.
N
400
EVALUATE: The self-induced emf depends on the rate of change of flux and therefore on the rate of change of
the current, not on the value of the current.
IDENTIFY: Combine the two expressions for L: L  N  B / i and L  E/(di/dt ).
(b) The average flux through each turn is  B 
30.8.
SET UP:
 B is the average flux through one turn of the solenoid.
(12.6 103 V)(1.40 A)
 238 turns.
(0.00285 Wb)(0.0260 A/s)
EVALUATE: The induced emf depends on the time rate of change of the total flux through the solenoid.
IDENTIFY and SET UP: Apply E  L di/dt . Apply Lenz’s law to determine the direction of the induced emf in
EXECUTE: Solving for N we have N  Ei/ B (di/dt ) 
30.9.
the coil.
EXECUTE: (a) E  L(di/dt )  (0.260 H)(0.0180 A/s)  4.68 103 V
30.10.
30.11.
(b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a
battery it would have the  terminal at a.
EVALUATE: The induced emf is directed so as to oppose the decrease in the current.
di
IDENTIFY: Apply E   L .
dt
SET UP: The induced emf points from low potential to high potential across the inductor.
EXECUTE: (a) The induced emf points from b to a, in the direction of the current. Therefore, the current is
decreasing and the induced emf is directed to oppose this decrease.
(b) E  L i / t , so i/t  Vab /L  (1.04 V) /(0.260 H)  4.00 A/s. In 2.00 s the decrease in i is 8.00 A and the
current at 2.00 s is 12.0 A  8.0 A  4.0 A.
EVALUATE: When the current is decreasing the end of the inductor where the current enters is at the lower
potential. This agrees with our result and with Figure 30.6d in the textbook.
IDENTIFY and SET UP: Use Eq.(30.6) to relate L to the flux through each turn of the solenoid. Use Eq.(28.23) for
the magnetic field through the solenoid.
NInductance
NB
. If the magnetic field is uniform inside the solenoid  B  BA. From Eq.(28.23),
i
 NiA
N   NiA  0 N 2 A
N
B   0 ni   0   i so  B  0
. Then L   0
.

l
i  l 
l
 l 
EVALUATE: Our result is the same as L for a torodial solenoid calculated in Example 30.3, except that the
average circumference 2 r of the toroid is replaced by the length l of the straight solenoid.
IDENTIFY and SET UP: The stored energy is U  12 LI 2 . The rate at which thermal energy is developed is P  I 2 R.
EXECUTE: (a) U  12 LI 2  12 (12.0 H)(0.300 A) 2  0.540 J
(b) P  I 2 R  (0.300 A) 2 (180 )  16.2 W  16.2 J/s
EVALUATE: (c) No. If I is constant then the stored energy U is constant. The energy being consumed by the
resistance of the inductor comes from the emf source that maintains the current; it does not come from the energy
stored in the inductor.
IDENTIFY and SET UP: Use Eq.(30.9) to relate the energy stored to the inductance. Example 30.3 gives the
 N2A
inductance of a toroidal solenoid to be L  0
, so once we know L we can solve for N.
2 r
2U 2(0.390 J)
 5.417  103 H
EXECUTE: U  12 LI 2 so L  2 
I
(12.0 A) 2
EXECUTE:
30.12.
30.13.
30-3
L
2 rL
2 (0.150 m)(5.417 103 H)

 2850.
0 A
(4 107 T  m/A)(5.00 104 m2 )
EVALUATE: L and hence U increase according to the square of N.
IDENTIFY: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy.
 NI
(a) SET UP: The magnetic field inside a toroidal solenoid is B  0 .
2 r
0 (300)(5.00 A)
3
 2.50  10 T  2.50 mT
EXECUTE: B 
2 (0.120 m)
 N2A
.
(b) SET UP: The self-inductance of a toroidal solenoid is L  0
2 r
(4 107 T  m/A)(300) 2 (4.00 104 m 2 )
EXECUTE: L 
 6.00 105 H
2 (0.0120 m)
(c) SET UP: The energy stored in an inductor is U L  12 LI 2 .
EXECUTE: U L  12 (6.00  105 H)(5.00 A) 2  7.50  104 J
N
30.14.
(d) SET UP: The energy density in a magnetic field is u 
(2.50 103 T)2
 2.49 J/m3
2(4 107 T  m/A)
energy energy
7.50 104 J
(e) u 


 2.49 J/m3
volume 2 rA 2 (0.120 m)(4.00 104 m2 )
EVALUATE: An inductor stores its energy in the magnetic field inside of it.
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 .
(4  107 T  m/A)(400)(80.0 A)
= 0.161 T
EXECUTE: B 
0.250 m
B2
(b) SET UP: The energy density in a magnetic field is u 
.
2 0
(0.161 T)2
EXECUTE: u 
 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 m 2 )  0.129 J
EXECUTE:
30.15.
B2
.
2 0
u
(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.
30-4
Chapter 30
30.16.
IDENTIFY: Energy  Pt . U  12 LI 2 .
SET UP: P  200 W  200 J/s
EXECUTE: (a) Energy  (200 W)(24 h)(3600 s/h)  1.73 107 J
2U 2(1.73 107 J)

 5.41103 H
I2
(80.0 A)2
EVALUATE: A large value of L and a large current would be required, just for one light bulb. Also, the resistance
of the inductor would have to be very small, to avoid a large P  I 2 R rate of electrical energy loss.
IDENTIFY and SET UP: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the
magnetic permeability  is used in place of  0 .
(b) L 
30.17.
EXECUTE: Using L 
30.18.
EVALUATE: For a given value of B, the energy density is less when  is larger than 0 .
IDENTIFY and SET UP: The energy density (energy per unit volume) in a magnetic field (in vacuum) is given by
U B2
(Eq.30.10).
u 
V 2 0
EXECUTE: (a) V 
(b) u 
B
30.19.
B2
N2A
 NI
and B 
gives u 
.
2 r
2 r
2
20U 2(4 107 T  m/A)(3.60 106 J)

 25.1 m3.
B2
(0.600 T)2
U B2

V 2 0
2 0U
2(4 107 T  m/A)(3.60 106 J)

 11.9 T
3
V
 0.400 m 
EVALUATE: Large-scale energy storage in a magnetic field is not practical. The volume in part (a) is quite large
and the field in part (b) would be very difficult to achieve.
IDENTIFY: Apply Kirchhoff’s loop rule to the circuit. i(t) is given by Eq.(30.14).
SET UP: The circuit is sketched in Figure 30.19.
di
is positive as the current
dt
increases from its initial value of zero.
Figure 30.19
EXECUTE:
E  vR  vL  0
di
E
E  iR  L  0 so i  1  e( R / L )t 
dt
R
di
(a) Initially (t = 0), i = 0 so E  L  0
dt
di E 6.00 V
 
 2.40 A/s
dt L 2.50 H
di
(b) E  iR  L  0 (Use this equation rather than Eq.(30.15) since i rather than t is given.)
dt
di E  iR 6.00 V  (0.500 A)(8.00 )
Thus


 0.800 A/s
dt
L
2.50 H
E
 6.00 V 
 (8.00 / 2.50 H)(0.250 s)
(c) i  1  e( R/L )t   
  0.750 A(1  e0.800 )  0.413 A
 1  e
R
 8.00  
di
(d) Final steady state means t   and
 0, so E  iR  0.
dt
E 6.00 V
i 
 0.750 A
R 8.00 
EVALUATE: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its
final value of E/R. The slope of the current in the figure, which is di/dt, decreases with t.
NInductance
30.20.
30-5
IDENTIFY: The current decays exponentially.
SET UP: After opening the switch, the current is i  I 0etR/L , and the time constant is  = L /R.
EXECUTE: (a) The initial current is I 0 = (6.30 V)/(15.0 Ω) = 0.420 A. Now solve for L and put in the numbers.
L
30.21.
tR
(2.00 ms)(15.0 )

 43.3 mH
ln(i / I 0 )
 0.210 A 
ln 

 0.420 A 
(b)  = L/R = (43.3 mH)/(15.0 Ω) = 2.89 ms
(c) Solve i  I 0e t/ for t, giving t   ln(i/I 0 )  (2.89 ms)ln(0.0100)  13.3 ms.
EVALUATE: In less than 5 time constants, the current is only 1% of its initial value.
IDENTIFY: i  E/R (1  e  t/ ), with   L/R. The energy stored in the inductor is U  12 Li 2 .
SET UP: The maximum current occurs after a long time and is equal to E/R.
EXECUTE: (a) imax  E/R so i  imax /2 when (1  e  t/τ )  12 and et/τ  12 . t/τ  ln  12 .
t
L ln 2 (ln 2)(1.25 103 H)

 17.3 μs
R
50.0 
(b) U  12 Umax when i  imax
2. 1  et/τ  1
2 , so et/τ  1  1
2  0.2929. t  L ln(0.2929)/R  30.7 μs.
EVALUATE:   L/R  2.50 10 s  25.0 s. The time in part (a) is 0.692 and the time in part (b) is 1.23 .
IDENTIFY: With S1 closed and S 2 open, i(t ) is given by Eq.(30.14). With S1 open and S 2 closed, i(t ) is given
by Eq.(30.18).
SET UP: U  12 Li 2 . After S1 has been closed a long time, i has reached its final value of I  E/R.
5
30.22.
EXECUTE: (a) U  12 LI 2 and I 
2U
2(0.260 J)

 2.13 A. E  IR  (2.13 A)(120 )  256 V.
L
0.115 H
(b) i  Ie( R/L)t and U  12 Li 2  12 LI 2e 2( R/L )t  12 U 0  12  12 LI 2  . e 2( R/L ) t  12 , so
t
30.23.
30.24.
30.25.
L
0.115 H
ln  12   
ln  12   3.32  104 s.
2R
2(120 )
EVALUATE:   L/R  9.58 104 s. The time in part (b) is  ln(2)/ 2  0.347 .
IDENTIFY: L has units of H and R has units of  .
SET UP: 1 H  1  s
EXECUTE: Units of L/R  H/   ( s)/   s  units of time.
EVALUATE: Rt/L  t/ is dimensionless.
IDENTIFY: Apply the loop rule.
SET UP: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the
inductor is Ldi / dt.
i di
R t
R
di
R
EXECUTE:  Ldi/dt  iR  0.
   dt and ln(i/I 0 )   t. i  I 0e  t ( R/L ) .
  i gives 
I
0
0 i
dt
L
L
L
EVALUATE: di / dt is negative, so there is a potential rise across the inductor; point c is at higher potential than
point b. There is a potential drop across the resistor.
IDENTIFY: Apply the concepts of current decay in an R-L circuit. Apply the loop rule to the circuit. i(t) is given
by Eq.(30.18). The voltage across the resistor depends on i and the voltage across the inductor depends on di/dt.
SET UP: The circuit with S1 closed and S 2 open is sketched in Figure 30.25a.
E  iR  L
Figure 30.25a
Constant current established means
EXECUTE:
i
di
 0.
dt
E 60.0 V

 0.250 A
R 240 
di
0
dt
30-6
Chapter 30
(a) SET UP: The circuit with S 2 closed and S1 open is shown in Figure 30.25b.
i  I 0e  ( R / L ) t
At t  0, i  I 0  0.250 A
Figure 30.25b
The inductor prevents an instantaneous change in the current; the current in the inductor just after S 2 is closed and
S1 is opened equals the current in the inductor just before this is done.
(b) EXECUTE: i  I0e( R / L)t  (0.250 A)e(240  / 0.160 H)(4.0010
(c) SET UP: See Figure 30.25c.
4
s)
 (0.250 A)e0.600  0.137 A
Figure 30.25c
EXECUTE: If we trace around the loop in the direction of the current the potential falls as we travel through the
resistor so it must rise as we pass through the inductor: vab  0 and vbc  0. So point c is at higher potential than
point b.
vab  vbc  0 and vbc  vab
Or, vcb  vab  iR  (0.137 A)(240 )  32.9 V
(d) i  I 0e ( R/L ) t
i  12 I 0 says 12 I 0  I 0e  ( R/L )t and
1
2
 e  ( R/L )t
Taking natural logs of both sides of this equation gives ln( 12 )   Rt / L
 0.160 H 
4
t 
 ln 2  4.62 10 s
 240  
30.26.
EVALUATE: The current decays, as shown in Fig. 30.13 in the textbook. The time constant is   L/R  6.67 104 s.
The values of t in the problem are less than one time constant. At any instant the potential drop across the resistor
(in the direction of the current) equals the potential rise across the inductor.
IDENTIFY: Apply Eq.(30.14).
di
SET UP: vab  iR. vbc  L . The current is increasing, so di / dt is positive.
dt
EXECUTE: (a) At t  0, i  0. vab  0 and vbc  60 V.
(b) As t  , i  E/R and di / dt  0. vab  60 V and vbc  0.
(c) When i  0.150 A, vab  iR  36.0 V and vbc  60.0 V  36.0 V  24.0 V.
30.27.
EVALUATE: At all times, E  vab  vbc , as required by the loop rule.
IDENTIFY: i(t ) is given by Eq.(30.14).
SET UP: The power input from the battery is Ei. The rate of dissipation of energy in the resistance is i 2 R. The
voltage across the inductor has magnitude Ldi/dt , so the rate at which energy is being stored in the inductor is
iLdi/dt.
E2
(6.00 V)2
(1  e (8.00  / 2.50 H)t ).
EXECUTE: (a) P  Ei  EI 0 (1  e ( R/L )t )  (1  e ( R/L )t ) 
R
8.00 
P  (4.50 W)(1  e(3.20 s
1
)t
).
1
E
(6.00 V) 2
(1  e( R/L ) t )2 
(1  e(8.00 / 2.50 H)t )2  (4.50 W)(1  e(3.20 s )t ) 2
R
8.00 
2
di E
E
 E
(c) PL  iL  (1  e( R/L )t ) L  e( R/L)t   (e( R/L) t  e2( R/L) t )
dt R
L
 R
(b) PR  i 2 R 
2
1
1
PL  (4.50 W)(e(3.20 s )t  e(6.40 s )t ).
EVALUATE: (d) Note that if we expand the square in part (b), then parts (b) and (c) add to give part (a), and the
total power delivered is dissipated in the resistor and inductor. Conservation of energy requires that this be so.
NInductance
30.28.
30-7
IDENTIFY: An L-C circuit oscillates, with the energy going back and forth between the inductor and capacitor.
1
1

(a) SET UP: The frequency is f 
and  
, giving f 
.
2
LC
2 LC
1
EXECUTE: f 
 2.13 103 Hz = 2.13 kHz
3
2  0.280 10 H  20.0 106 F 
(b) SET UP: The energy stored in a capacitor is U  12 CV 2 .
EXECUTE:
30.29.
U  12 (20.0 106 F)(150.0 V) 2  0.225 J
(c) SET UP: The current in the circuit is i  Q sin t , and the energy stored in the inductor is U  12 Li 2 .
EXECUTE: First find  and Q.  = 2π f = 1.336  104 rad/s.
Q = CV = (20.0  10–6 F)(150.0 V) = 3.00  10–3 C
Now calculate the current:
i =  (1.336  104 rad/s)(3.00  10–3 C) sin[(1.336  104 rad/s)(1.30  10–3 s)]
Notice that the argument of the sine is in radians, so convert it to degrees if necessary. The result is i =  39.92 A
Now find the energy in the inductor: U  12 Li 2  12 (0.280 103 H)(39.92 A)2  0.223 J
EVALUATE: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the capacitor.
IDENTIFY: The energy moves back and forth between the inductor and capacitor.
1
1
2

 2 LC .
(a) SET UP: The period is T  
f  / 2 
EXECUTE: Solving for L gives
L
T2
(8.60 105 s)2

 2.50 102 H = 25.0 mH
4 2C 4 2 (7.50 109 C)
(b) SET UP: The charge on a capacitor is Q = CV.
EXECUTE: Q = CV = (7.50  10–9 F)(12.0 V) = 9.00  10–8 C
(c) SET UP: The stored energy is U = Q2/2C.
9.00 10 C   5.40 10
U
2  7.50 10 F 
8
EXECUTE:
9
2
7
J
(d) SET UP: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial
energy. U L  U C  U Total . 12 LI 2  0  U Total .
EXECUTE: Solve for the current:
I
30.30.
2  5.40  107 J 
2U Total

 6.58  103 A = 6.58 mA
L
2.50  102 H
EVALUATE: The energy oscillates back and forth forever. However if there is any resistance in the circuit, no
matter how small, all this energy will eventually be dissipated as heat in the resistor.
IDENTIFY: The circuit is described in Figure 30.14 of the textbook.
SET UP: The energy stored in the inductor is U L  12 Li 2 and the energy stored in the capacitor is U C  q 2 /2C. Initially,
U C  12 CV 2, with V  12.0 V. The period of oscillation is T  2 LC  2 (12.0 103 H)(18.0 106 F)  2.92 ms .
EXECUTE: (a) Energy conservation says U L (max) = UC (max), and
imax  V C/L  (22.5 V)
1
2
2
Limax
 12 CV 2 .
18  106 F
 0.871 A. The charge on the capacitor is zero because all the energy is in
12  103 H
the inductor.
(b) From Figure 30.14 in the textbook, q  0 at t  T/4  0.730 ms and at t  3T/4  2.19 ms.
(c) q0  CV  (18  F) (22.5 V)  405  C is the maximum charge on the plates. The graphs are sketched in
Figure 30.30. q refers to the charge on one plate and the sign of i indicates the direction of the current.
EVALUATE: If the capacitor is fully charged at t  0 it is fully charged again at t  T/2, but with the opposite polarity.
Figure 30.30
30-8
Chapter 30
30.31.
IDENTIFY and SET UP: The angular frequency is given by Eq.(30.22). q(t) and i(t) are given by Eqs.(30.21) and
(30.23). The energy stored in the capacitor is U C  12 CV 2  q 2 /2C. The energy stored in the inductor is U L  12 Li 2 .
EXECUTE: (a)  
2
 0.0596 s
105.4 rad/s
(b) The circuit containing the battery and capacitor is sketched in Figure 30.31.
Q
E  0
C
Q  EC  (12.0 V)(6.00 105 F)  7.20 104 C
given by T 
2
1
1

 105.4 rad/s, which rounds to 105 rad/s. The period is
LC
(1.50 H)(6.00 105 F)


Figure 30.31
(c) U  CV  12 (6.00 105 F)(12.0 V)2  4.32 103 J
(d) q  Q cos(t   ) (Eq.30.21)
q  Q at t  0 so   0
q  Q cos  t  (7.20 104 C)cos([105.4 rad/s][0.0230 s])  5.42 104 C
The minus sign means that the capacitor has discharged fully and then partially charged again by the current
maintained by the inductor; the plate that initially had positive charge now has negative charge and the plate that
initially had negative charge now has positive charge.
(e) i  Qsin(t   ) (Eq.30.23)
1
2
2
i  (105 rad/s)(7.20 104 C)sin([105.4 rad/s][0.0230 s])  0.050 A
The negative sign means the current is counterclockwise in Figure 30.15 in the textbook.
or
q 2 Q2
1
2
1

gives i  
Q2  q 2 (Eq.30.26)
2 Li 
2C 2C
LC
i  (105 rad/s) (7.20 104 C)2  (5.42 104 C)2  0.050 A, which checks.
q 2 (5.42 104 C) 2

 2.45 103 J
2C 2(6.00 10 5 F)
U L  12 Li 2  12 (1.50 H)(0.050 A)2  1.87 103 J
(f ) U C 
EVALUATE: Note that U C  U L  2.45 103 J  1.87 103 J  4.32 103 J.
Q2 (7.20 104 C)2

 4.32 103 J.
2C 2(6.00 105 F)
Energy is conserved. At some times there is energy stored in both the capacitor and the inductor. When i = 0 all the
energy is stored in the capacitor and when q = 0 all the energy is stored in the inductor. But at all times the total
energy stored is the same.
1
IDENTIFY: ω 
 2πf
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
4 f C 4 (1.6 106 Hz) 2 (4.18 1012 F)
(b) The maximum capacitance corresponds to the minimum frequency.
1
1
Cmax 

 3.67 1011 F  36.7 pF
2 2
2
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.
IDENTIFY: Apply energy conservation and Eqs. (30.22) and (30.23).
Q2
SET UP: If I is the maximum current, 12 LI 2 
. For the inductor, U L  12 Li 2.
2C
Q2
EXECUTE: (a) 12 LI 2 
gives Q  i LC  (0.750 A) (0.0800 H)(1.25 109 F)  7.50 106 C .
2C
1
1


 1.00 105 rad/s . f 
(b)  
 1.59 104 Hz .
9
2

LC
(0.0800 H)(1.25 10 F)
This agrees with the total energy initially stored in the capacitor, U 
30.32.
30.33.
NInductance
(c) q  Q at t  0 means   0 . i  Q sin(t ) , so
i  (1.00 105 rad/s)(7.50 106 C)sin([1.00 105 rad/s][2.50 103 s])  0.7279 A .
U L  12 Li 2  12 (0.0800 H)(  0.7279 A) 2  0.0212 J .
30.34.
EVALUATE: The total energy of the system is 12 LI 2  0.0225 J . At t  2.50 ms , the current is close to its
maximum value and most of the system’s energy is stored in the inductor.
IDENTIFY: Apply Eq.(30.25).
SET UP: q  Q when i  0 . i  imax when q  0 . 1/ LC  1917 s1 .
EXECUTE: (a)
1
2
2
Limax

Q2
. Q  imax LC  (0.850 103 A) (0.0850 H)(3.20 106 F)  4.43 107 C
2C
2
30.35.
 5.00 104 A 
7
(b) q  Q 2  LCi 2  (4.43 107 C) 2  
  3.58 10 C .
1
1917s


EVALUATE: The value of q calculated in part (b) is less than the maximum value Q calculated in part (a).
IDENTIFY: q  Q cos(t   ) and i  Qsin(   )
q2
. U L  12 Li 2 .
2C
q 2 1 Q 2 cos2 (ωt   )

.
EXECUTE: (a) U C  12
C 2
C
Q 2 sin 2 (ωt   )
1
U L  12 Li 2  12 Lω2Q 2 sin 2 (ωt   )  12
, since ω2 
.
C
LC
1 Q2
1
cos2 (ωt   )  Lω2Q 2 sin 2 (ωt   )
(b) U Total  U C  U L 
2 C
2
2
2
Q
1
Q2

 2 2
1 Q
U total  12
cos2 (ωt   )  12 L 
(cos 2 (ωt   )  sin 2 (ωt   ))  12
 Q sin (ωt   )  2
C
C
C
 LC 
U Total is a constant.
EVALUATE: Eqs.(30.21) and (30.23) are consistent with conservation of energy in the L-C circuit.
d 2q
IDENTIFY: Evaluate
and insert into Eq.(20.20).
dt 2
d 2q 1

q  0.
SET UP: Equation (30.20) is
dt 2 LC
dq
d 2q
 ωQ sin(ωt   )  2  ω2Q cos(ωt   ).
EXECUTE: q  Q cos(ωt   ) 
dt
dt
d 2q 1
Q
1
1

q  ω2Q cos(ωt   ) 
cos(ωt   )  0   2 
ω
.
2
dt
LC
LC
LC
LC
EVALUATE: The value of  depends on the initial conditions, the value of q at t  0 .
IDENTIFY: The unit of L is H and the unit of C is F.
SET UP: C  q / Vab says 1 F  1 C/V . 1 H  1 V  s/A  1 V  s2 /C .
SET UP:
30.36.
30.37.
EXECUTE:
30.38.
UC 
1 H  F  (1 V  s 2 /C)(1 C/V)  1 s 2 . Therefore, LC has units of s 2 and
LC has units of s.
EVALUATE: Our result shows that t is dimensionless, since   1/ LC .
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.
1
R2
(a) SET UP: The frequency of damped oscillations is   
 2 .
LC 4 L
EXECUTE:
 
1
(75.0 ) 2

 5.5 104 rad/s
 22 103 H 15.0 109 F 4  22 103 H 2
The frequency f is f 
 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/2L)t.
EXECUTE: Solving for t and putting in the numbers gives:
2 L ln( A / A0 ) 2  22.0 10 H  ln(0.100)

 1.35 103 s = 1.35 ms
R
75.0 
3
t
30-9
30-10
Chapter 30
(c) SET UP: At critical damping, R  4L / C .
EXECUTE:
30.39.
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-Ω resistance is considerably less than the 2420 Ω required for critical damping.
IDENTIFY: Follow the procedure specified in the problem.
1
SET UP: Make the substitutions x  q, m  L, b  R, k  .
C
d 2 x b dx kx
d 2 q R dq q
EXECUTE: (a) Eq. (13.41):
.
This
becomes



0


 0 , which is Eq.(30.27).
dt 2 m dt m
dt 2 L dt LC
k
b2
1
R2
. This becomes ω 
 2 , which is Eq.(30.29).

2
LC 4 L
m 4m
(b / 2 m)t
cos(ωt   ) . This becomes q  Ae  ( R / 2 L )t cos(ωt   ) , which is Eq.(30.28).
(c) Eq. (13.42): x  Ae
EVALUATE: Equations for the L-R-C circuit and for a damped harmonic oscillator have the same form.
IDENTIFY: For part (a), evaluate the derivatives as specified in the problem. For part (b) set q  Q in Eq.(30.28)
and set dq / dt  0 in the expression for dq / dt .
(b) Eq. (13.43): ω 
30.40.
SET UP: In terms of   , Eq.(30.28) is q(t )  Ae ( R / 2 L ) t cos( t   ) .
EXECUTE: (a) q  Ae  ( R / 2 L )t cos(ωt   ) .
dq
R
  A e( R / 2 L )t cos(ωt   )  ωAe ( R / 2 L )t sin(ωt   ).
dt
2L
2
d 2q
R
 R 
 A   e ( R / 2 L )t cos(ωt   )  2ωA e  ( R / 2 L )t sin(ωt   )  ω2 Ae  ( R / 2 L )t cos(ωt   )
2
dt
2L
 2L 
  R 2
1
R2
d 2 q R dq q
R2
1 
 2.


 q      2  2 
 0 , so ω2 

2
  2L 
LC 4 L
dt
L dt LC
2L LC 

Q
dq
dq
R
(b) At t  0, q  Q, i 
and
 0 , so q  Acos  Q and

Acos  ωAsin   0 . This gives A 
cos 
dt
dt
2L
R
R
.

2Lω
2L 1/ LC  R2 / 4L2
EVALUATE: If R  0 , then A  Q and   0 .
IDENTIFY: Evaluate Eq.(30.29).
SET UP: The angular frequency of the circuit is   .
1
1

 298 rad s.
EXECUTE: (a) When R  0, ω0 
LC
(0.450 H) (2.50 105 F)
tan   
30.41.
(b) We want
R
30.42.
ω
(1 LC  R 2 4L2 )
R 2C
 0.95 , so
 1
 (0.95)2 . This gives
ω0
1 LC
4L
4L
4(0.450 H)(0.0975)
(1  (0.95) 2 ) 
 83.8 .
C
(2.50 105 F)
EVALUATE: When R increases, the angular frequency decreases and approaches zero as R  2 L / C .
IDENTIFY: L has units of H and C has units of F.
SET UP: 1 H  1  s . C  q / V says 1 F  1 C/V. V  IR says 1 V/A  1  .
EXECUTE: The units of L / C are
H  s   V


  2 . Therefore, the unit of
F CV
A
L / C is .
R2
1
and
must have the same units, so R and
4 L2
LC
same units, and we have shown that this is indeed the case.
EVALUATE:
30.43.
For Eq.(30.28) to be valid,
L / C must have the
IDENTIFY: The emf E2 in solenoid 2 produced by changing current i1 in solenoid 1 is given by E2  M
i1
. The
t
mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem
 AN N
M  0 1 2 , where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid.
l
NInductance
SET UP:
30-11
 0  4 107 T  m/A . Let the outer solenoid be solenoid 1.
EXECUTE: (a) M 
(4 107 T  m/A) (6.00 104 m) 2 (6750)(15)
 2.88 107 H  0.288  H
0.500 m
i1
 (2.88 107 H)(37.5 A/s)  1.08 105 V
t
EVALUATE: If current in the inner solenoid changed at 37.5 A/s, the emf induced in the outer solenoid would be
1.08 105 V.
di
IDENTIFY: Apply E   L
and Li  N  B .
dt
SET UP:  B is the flux through one turn.
(b) E2 
30.44.
di
d
 (3.50 103 H) ((0.680 A)cos( t /[0.0250 s])).
dt
dt

 (3.50 103 H)(0.680 A)
sin( t /[0.0250 s]) . Therefore,
0.0250 s
EXECUTE: (a) E   L

Emax  (3.50 103 H)(0.680 A)

0.0250 s
 0.299 V.
Limax (3.50 103 H)(0.680 A)

 5.95 106 Wb.
N
400
di
(c) E(t )   L   (3.50 103 H)(0.680 A)( / 0.0250 s)sin( t / 0.0250 s).
dt
E(t )   (0.299 V)sin((125.6 s 1 )t ) .Therefore, at t  0.0180 s ,
(b)  B max 
E(0.0180 s)   (0.299 V)sin((125.6 s 1 )(0.0180 s))  0.230 V . The magnitude of the induced emf is 0.230 V.
EVALUATE: The maximum emf is when i  0 and at this instant  B  0.
30.45.
di
.
dt
SET UP: During an interval in which the graph of i versus t is a straight line, di / dt is constant and equal to the
slope of that line.
EXECUTE: (a) The pattern on the oscilloscope is sketched in Figure 30.45.
EVALUATE: (b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed
proportional to the derivative of the current graph.
IDENTIFY:
E  L
Figure 30.45
30.46.
IDENTIFY: Apply E   L
di
.
dt
d
cos(t )   sin(t )
dt
di
d
EXECUTE: (a) E   L   L ((0.124 A)cos[(240 π s)t ].
dt
dt
SET UP:
E  (0.250 H) (0.124 A) (240  /s)sin((240 s)t )  (23.4 V) sin ((240 π s)t ).
The graphs are given in Figure 30.46.
(b) Emax  23.4 V; i  0, since the emf and current are 90 out of phase.
(c) imax  0.124 A; E  0, since the emf and current are 90 out of phase.
30-12
Chapter 30
EVALUATE: The induced emf depends on the rate at which the current is changing.
Figure 30.46
30.47.
di
to the series and parallel combinations.
dt
SET UP: In series, i1  i2 and the voltages add. In parallel the voltages are the same and the currents add.
di
di
di
di di di
EXECUTE: (a) Series: L1 1  L2 2  Leq , but i1  i2  i for series components so 1  2  and
dt
dt
dt
dt dt dt
L1  L2  Leq .
IDENTIFY: Apply E   L
(b) Parallel: Now L1
di1
di
di
di di1 di2
di L di
. But 1  eq
and
 L2 2  Leq , where i  i1  i2 . Therefore,


dt
dt
dt
dt dt dt
dt
L1 dt
1
30.48.
1
1
di2 Leq di di Leq di Leq di
.
and Leq     .



dt L2 dt dt L1 dt L2 dt
 L1 L2 
EVALUATE: Inductors in series and parallel combine in the same way as resistors.
IDENTIFY: Follow the steps outlined in the problem.
SET UP: The energy stored is U  12 Li 2 .
μi
EXECUTE: (a) AB  dl  μ0 I encl  B2πr  μ0i  B  0 .
2πr
μi
(b) d  B  BdA  0 ldr.
2πr
b
b
μ il dr μ0il

ln ( b a ).
(c)  B   d  B  0 
2π a r
2π
a
NB
μ
 l 0 ln( b a).
i
2π
1 2 1 μ0
μ li 2
ln( b a)i 2  0 ln( b a).
(e) U  Li  l
2
2 2π
4π
EVALUATE: The magnetic field between the conductors is due only to the current in the inner conductor.
(a) IDENTIFY and SET UP: An end view is shown in Figure 30.49.
(d) L 
30.49.
Apply Ampere’s law to a circular
path of radius r.
AB  dl = 0 Iencl
Figure 30.49
EXECUTE: AB  dl = B(2 r )
I encl  i, the current in the inner conductor
i
Thus B(2 r )  0i and B  0 .
2 r
NInductance
30-13
(b) IDENTIFY and SET UP: Follow the procedure specified in the problem.
B2
EXECUTE: u 
2 0
dU  udV , where dV  2 rl dr
1  0i 
0i 2l
(2

rl
)
dr

dr


20  2 r 
4 r
2
dU 
0i 2l b dr 0i 2l

[ln r ]ba
4  a r
4
 i 2l
 i 2l  b 
U  0 (ln b  ln a)  0 ln  
4
4
a
(c) U   dU 
(d) Eq.(30.9): U  12 Li 2
Part (c): U 
1
2
Li 2 
0i 2l  b 
ln  
4
a
 0i 2l  b 
ln  
4
a
 0l  b 
ln   .
2  a 
EVALUATE: The value of L we obtain from these energy considerations agrees with L calculated in part (d) of
Problem 30.48 by considering flux and Eq.(30.6)
N
NB
IDENTIFY: Apply L 
to each solenoid, as in Example 30.3. Use M  2 B 2 to calculate the mutual
i1
i
inductance M.
SET UP: The magnetic field produced by solenoid 1 is confined to the space within its windings and is equal to
 Ni
B1  0 1 1 .
2 r
N1 B N1 A  μ0 N1i1  μ0 N12 A
N 2 B
N A μ N i  μ N 2A

, L2 
 2  0 22 0 2 .
EXECUTE: (a) L1 


i1
i1  2π r 
2π r
i2
i2  2π r 
2π r
L
30.50.
1
2
2
 μ N N A
N 2 AB1  0 N1 N 2 A
μ N 2 A μ0 N 2 2 A

. M2  0 1 2   0 1
 L1L2 .
i1
2 r
2πr
2πr
 2πr 
EVALUATE: If the two solenoids are identical, so that N1  N 2 , then M  L .
(b) M 
30.51.
U  12 LI 2 . The self-inductance of a solenoid is found in Exercise 30.11 to be L 
IDENTIFY:
 0 AN 2
l
.
SET UP: The length l of the solenoid is the number of turns divided by the turns per unit length.
2U 2(10.0 J)
 8.89 H
EXECUTE: (a) L  2 
I
(1.50 A) 2
(b) L 
 0 AN 2
l
. If  is the number of turns per unit length, then N   l and L   0 A 2l . For this coil
  10 coils/mm  10 103 coils/m. l 
L
8.89 H

 56.3 m .
 0 A 2 (4 107 T  m/A) (0.0200 m) 2 (10 103 coils/m) 2
This is not a practical length for laboratory use.
EVALUATE: The number of turns is N  (56.3 m)(10 103 coils/m)  5.63 105 turns . The length of wire in the
solenoid is the circumference C of one turn times the number of turns. C   d   (4.00 102 m)  0.126 m . The
length of wire is (0.126 m)(5.63 105 )  7.1104 m  71 km. This length of wire will have a large resistance and
30.52.
I 2 R electrical energy loses will be very large.
IDENTIFY: This is an R-L circuit and i(t ) is given by Eq.(30.14).
SET UP: When t  , i  if  V / R .
V
12.0 V
 1860 .
EXECUTE: (a) R  
if 6.45 103 A
(b) i  if (1  e  ( R / L ) t ) so
Rt
 Rt
(1860 )(7.25 104 s)
 ln(1  i / if ) and L 

 0.963 H .
L
ln(1  i / if )
ln(1  (4.86/ 6.45))
30-14
Chapter 30
30.53.
EVALUATE: The current after a long time depends only on R and is independent of L. The value of
R / L determines how rapidly the final value of i is reached.
IDENTIFY and SET UP: Follow the procedure specified in the problem. L  2.50 H, R  8.00 ,
E  6.00 V. i  (E / R)(1  e  t /  ),   L / R
EXECUTE: (a) Eq.(30.9): U L  12 Li 2
t   so i  (E / R)(1  e1 )  (6.00 V/8.00 )(1  e1 )  0.474 A
Then U L  12 Li 2  12 (2.50 H)(0.474 A) 2  0.281 J
dU L
di
Exercise 30.27 (c): PL 
 Li
dt
dt
di  E   ( R / L ) t E  t /
 E
i    (1  e  t / );
  e
 e
R
dt
L
 
 L
2
E
 E
 E
PL  L  (1  et /  )  et /    (et /   e2 /  )
R
 L
 R

2

E  t /
E2 
 2t / 
t /
U L   PL dt 
(e  e2t / ) dt 


e

e


0
R 0
R 
2
0
2
2

E
E
t /

2
t
/


1

2
   1  12  e  12 e 
U L    e  12 e
0
R
R
2
2
 E  L 
E
1
2
1
2
1
UL  
(1

2
e

e
)

 
 L(1  2e  e )
2
2
R
R
R






2
 6.00 V 
U L  12 
 (2.50 H)(0.3996)  0.281 J, which checks.
 8.00  
E
 E
(b) Exercise 30.27(a): The rate at which the battery supplies energy is PE  Ei  E (1  et / )   (1  et / )
R
 R
2

U E   PEdt 
0

 E2 
E2 
E2
t   e t /     (   e 1   )
(1  e t / )dt 

0
R 0
R
 R
 E2 
 E2   L 
 E
U E    e 1     e 1    Le 1
R
 R
 R  R 
2
2
 6.00 V 
UE  
 (2.50 H)(0.3679)  0.517 J
 8.00  
 E2 
E2
(c) PR  i 2 R    (1  e  t / ) 2  (1  2e  t /  e 2t /  )
R
 R

U R   PR dt 
0
UR 

E2 
E2 


t /
2 t / 
(1

2
e

e
)
dt

t  2 et /  e2t / 
R 0
R 
2
0
E2 

  E2  
 
  2 e1  e2  2      2 e1  e2 

R
2
2 R  2
2 
 E2   L 
1
2
UR  
   1  4e  e 
 2R   R 
2
30.54.
2
 E
 6.00 V  1
U R    ( 12 L)  1  4e1  e2   
 2 (2.50 H)(0.3362)  0.236 J
R
 
 8.00  
(d) EVALUATE: U E  U R  U L . (0.517 J  0.236 J  0.281 J)
The energy supplied by the battery equals the sum of the energy stored in the magnetic field of the inductor and the
energy dissipated in the resistance of the inductor.
IDENTIFY: This is a decaying R-L circuit with I 0  E / R . i (t )  I 0e  ( R / L )t .
SET UP: E  60.0 V , R  240  and L  0.160 H . The rate at which energy stored in the inductor is decreasing
is iLdi / dt .
2
 60 V 
1
1  E 1
3
EXECUTE: (a) U  LI 0 2  L    (0.160 H) 
  5.00 10 J.
2
2  R 2
240



2
NInductance
(b) i 
30-15
E ( R / L)t
di
R
dU L
di
E2
dU L
(60 V)2 2(240/ 0.160)(4.00104 )
e
  i
 iL   Ri 2  e2( R / L ) t .

e
 4.52 W.
R
dt
L
dt
dt
R
dt
240 
(c) In the resistor, PR 
dU R 2
E2
(60 V)2 2(240/ 0.160)(4.00104 )
 i R  e2( R / L )t 
e
 4.52 W .
dt
R
240 

(d) PR (t )  i 2 R 
30.55.
E2 2( R / L )t E2 L (60 V) 2 (0.160 H)
E2 2( R / L )t
e


 5.00  103 J, which is the same as
e
. UR 
R 0
R 2R
2(240 )2
R
part (a).
EVALUATE: During the decay of the current all the electrical energy originally stored in the inductor is dissipated
in the resistor.
IDENTIFY and SET UP: Follow the procedure specified in the problem. 12 Li 2 is the energy stored in the inductor
di q
  0.
dt C
d
d
d
di
di qi
 di 
U L   12 Li 2   12 L  i 2   12 L  2i   Li , the
EXECUTE: Multiplying by –i gives i 2 R  Li   0.
dt
dt
dt
dt
dt C
 dt 
and q 2 / 2C is the energy stored in the capacitor. The equation is iR  L
d
d  q2  1 d 2
1
dq qi
UC  
(q ) 
(2q)
 , the third term. i 2 R  PR , the rate at which

dt
dt  2C  2C dt
2C
dt C
d
electrical energy is dissipated in the resistance. U L  PL , the rate at which the amount of energy stored in the
dt
d
inductor is changing. U C  PC , the rate at which the amount of energy stored in the capacitor is changing.
dt
EVALUATE: The equation says that PR  PL  PC  0; the net rate of change of energy in the circuit is zero. Note
second term.
that at any given time one of PC or PL is negative. If the current and U L are increasing the charge on the capacitor
and U C are decreasing, and vice versa.
30.56.
IDENTIFY: The energy stored in a capacitor is U C  12 Cv 2 . The energy stored in an inductor is U L  12 Li 2 .
Energy conservation requires that the total stored energy be constant.
SET UP: The current is a maximum when the charge on the capacitor is zero and the energy stored in the
capacitor is zero.
EXECUTE: (a) Initially v  16.0 V and i  0 . U L  0 and U C  12 Cv 2  12 (5.00 106 F)(16.0 V)2  6.40 104 J .
The total energy stored is 0.640 mJ .
(b) The current is maximum when q  0 and U C  0 . U C  U L  6.40  104 J so U L  6.40 104 J .
2(6.40 104 J)
 0.584 A .
3.75 103 H
EVALUATE: The maximum charge on the capacitor is Q  CV  80.0 C .
1
2
30.57.
2
Limax
 6.40 104 J and imax 
IDENTIFY and SET UP: Use U C  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:
1
(2

f )2 C
2 LC
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.
IDENTIFY: Apply energy conservation to the circuit.
SET UP: For a capacitor V  q / C and U  q 2 / 2C . For an inductor U  12 Li 2
f 
30.58.
1
VC  12.0 V; U C  12 CVC2 so C  2U C / VC2  2(0.0160 J)/(12.0 V) 2  222  F
so L 
Q 6.00 106 C

 0.0240 V.
C 2.50 104 F
Q2
Q
6.00 106 C

, so imax 

 1.55 103 A
2C
LC
(0.0600 H)(2.50 104 F)
EXECUTE: (a) Vmax 
(b)
1 2
Limax
2
1 2
1
(c) U max  Limax
 (0.0600 H)(1.55 103 A)2  7.21108 J.
2
2
30-16
Chapter 30
30.59.
3
1
1
(d) If i  imax then U L  U max  1.80 108 J and U C  U max 
4
4
2
3
q
Q  5.20 106 C.
4
1
1 q2
EVALUATE: U max  Li 2 
for all times.
2
2C
IDENTIFY: Set U B  K , where K  12 mv 2 .

(3/ 4)Q
2C

2

q2
. This gives
2C
SET UP: The energy density in the magnetic field is uB  B 2 / 2 0 . Consider volume V  1 m3 of sunspot
material.
EXECUTE: The energy density in the sunspot is uB  B 2 /2 μ0  6.366 10 4 J /m 3. The total energy stored in
volume V of the sunspot is U B  uBV . The mass of the material in volume V of the sunspot is m  ρV .
Vv 2  uBV . The volume divides out, and v  2uB /   2 104 m/s .
EVALUATE: The speed we calculated is about 30 times smaller than the escape speed.
IDENTIFY: i(t ) is given by Eq.(30.14).
SET UP: The graph shows V  0 at t  0 and V approaches the constant value of 25 V at large times.
EXECUTE: (a) The voltage behaves the same as the current. Since VR is proportional to i, the scope must be
across the 150  resistor.
K  U B so
30.60.
1
2
mv 2  U B .
1
2
(b) From the graph, as t  , VR  25 V, so there is no voltage drop across the inductor, so its internal resistance
 1
must be zero. VR  Vmax (1  et / r ) . When t   , VR  Vmax  1    0.63Vmax . From the graph, V  0.63Vmax  16 V at
 e
t  0.5 ms . Therefore   0.5 ms . L / R  0.5 ms gives L  (0.5 ms) (150 )  0.075 H.
(c) The graph if the scope is across the inductor is sketched in Figure 30.60.
EVALUATE: At all times VR  VL  25.0 V . At t  0 all the battery voltage appears across the inductor since
i  0 . At t   all the battery voltage is across the resistance, since di / dt  0 .
30.61.
Figure 30.60
IDENTIFY and SET UP: The current grows in the circuit as given by Eq.(30.14). In an R-L circuit the full emf
initially is across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is
represented as a resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a
resistance is given by Ohm’s law.
EXECUTE: (a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage.
The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the
voltage drops, so the oscilloscope is across the solenoid.
(b) At t   the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because
the solenoid has some resistance RL . The final voltage across the solenoid is IRL , where I is the final current in
the circuit.
(c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current
is zero and there is no voltage drop across any of the resistance in the circuit.
(d) As t  , E  IR  IRL  0
E  50 V and from the graph IRL  15 V (the final voltage across the inductor), so
IR  35 V and I  (35 V)/R  3.5 A
(e) IRL  15 V, so RL  (15 V)/(3.5 A)  4.3 
E  VL  iR  0, where VL includes the voltage across the resistance of the solenoid.
E
VL  E  iR, i 
1  et / , so VL  E1  RR (1  et / )
Rtot
tot


E  50 V, R  10 , Rtot  14.3 , so when t   , VL  27.9 V. From the graph, VL has this value when t  3.0 ms
(read approximately from the graph), so   L / Rtot  3.0 ms. Then L  (3.0 ms)(14.3 )  43 mH.
EVALUATE: At t = 0 there is no current and the 50 V measured by the oscilloscope is the induced emf due to the
inductance of the solenoid. As the current grows, there are voltage drops across the two resistances in the circuit.
NInductance
30-17
We derived an equation for VL , the voltage across the solenoid. At t = 0 it gives VL  E and at t   it gives
30.62.
30.63.
VL  ER / Rtot  iR.
IDENTIFY: At t  0 , i  0 through each inductor. At t   , the voltage is zero across each inductor.
SET UP: In each case redraw the circuit. At t  0 replace each inductor by a break in the circuit and at t  
replace each inductor by a wire.
EXECUTE: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is shown in
E 50 V
 0.333 A . V1  (100 )(0.333 A)  33.3 V. V4  (50 )(0.333 A)  16.7 V. V3  0
Figure 30.62a. i  
R 150 
since no current flows through it. V2  V4  16.7 V , since the inductor is in parallel with the 50  resistor.
A1  A3  0.333 A, A2  0.
(b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. The simplified
50 V
 0.385 A . V1  (100 )(0.385 A)  38.5 V ; V2  0;
circuit is sketched in Figure 30.62b. i  E/R 
130 
11.5 V
11.5 V
 0.153 A; i3 
 0.230 A .
V3  V4  50 V  38.5 V  11.5 V . i1  0.385 A; i2 
75 
50 
EVALUATE: Just after the switch is closed the current through the battery is 0.333 A. After a long time the
current through the battery is 0.385 A. After a long time there is an additional current path, the equivalent
resistance of the circuit is decreased and the current has increased.
Figure 30.62
IDENTIFY and SET UP: Just after the switch is closed, the current in each branch containing an inductor is zero
and the voltage across any capacitor is zero. The inductors can be treated as breaks in the circuit and the capacitors
can be replaced by wires. After a long time there is no voltage across each inductor and no current in any branch
containing a capacitor. The inductors can be replaced by wires and the capacitors by breaks in the circuit.
EXECUTE: (a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current
through the inductor, so V3  0. V2  V3  V4  V5 , and since V5  0 and V3  0, V4 and V2 are also zero.
V4  0 means V3 reads zero. V1 then must equal 40.0 V, and this means the current read by A1 is
(40.0 V)/(50.0 )  0.800 A. A2  A3  A4  A1, but A2  A3  0 so A4  A1  0.800 A. A1  A4  0.800 A; all other
ammeters read zero. V1  40.0 V and all other voltmeters read zero.
(b) After a long time the capacitor is fully charged so A4  0. The current through the inductor isn’t changing, so
V2  0. The currents can be calculated from the equivalent circuit that replaces the inductor by a short circuit, as
shown in Figure 30.63a.
Figure 30.63a
I  (40.0 V)/(83.33 )  0.480 A; A1 reads 0.480 A
V1  I (50.0 )  24.0 V
The voltage across each parallel branch is 40.0 V – 24.0 V = 16.0 V
V2  0, V3  V4  V5  16.0 V
V3  16.0 V means A2 reads 0.160 A. V4  16.0 V means A3 reads 0.320 A. A4 reads zero. Note that A2  A3  A1.
(c) V5  16.0 V so Q  CV  (12.0  F)(16.0 V)  192 C
30-18
Chapter 30
(d) At t = 0 and t  , V2  0. As the current in this branch increases from zero to 0.160 A the voltage V2 reflects
the rate of change of the current. The graph is sketched in Figure 30.63b.
Figure 30.63b
30.64.
30.65.
EVALUATE: This reduction of the circuit to resistor networks only apply at t = 0 and t . At intermediate
times the analysis is complicated.
IDENTIFY: At all times v1  v2  25.0 V . The voltage across the resistor depends on the current through it and the
voltage across the inductor depends on the rate at which the current through it is changing.
SET UP: Immediately after closing the switch the current thorough the inductor is zero. After a long time the
current is no longer changing.
EXECUTE: (a) i  0 so v1  0 and v2  25.0 V . The ammeter reading is A  0 .
v 25.0 V
(b) After a long time, v2  0 and v1  25.0 V. v1  iR and i  1 
 1.67 A. The ammeter reading is
R 15.0 
A  1.67A.
(c) None of the answers in (a) and (b) depend on L so none of them would change.
EVALUATE: The inductance L of the circuit affects the rate at which current reaches its final value. But after a
long time the inductor doesn’t affect the circuit and the final current does not depend on L.
IDENTIFY: At t  0 , i  0 through each inductor. At t   , the voltage is zero across each inductor.
SET UP: In each case redraw the circuit. At t  0 replace each inductor by a break in the circuit and at t  
replace each inductor by a wire.
EXECUTE: (a) Just after the switch is closed there is no current through either inductor and they act like breaks in
the circuit. The current is the same through the 40.0  and15.0  resistors and is equal to
(25.0 V) (40.0   15.0 )  0.455 A. A1  A4  0.455 A; A2  A3  0.
(b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be
treated as a short-circuit . The circuit is equivalent to the circuit sketched in Figure 30.65.
I  (25.0 V) (42.73 )  0.585 A . A1 reads 0.585 A. The voltage across each parallel branch is
25.0 V  (0.585 A)(40.0 )  1.60 V . A2 reads (1.60 V)/(5.0 )  0.320 A . A3 reads
(1.60 V) (10.0 )  0.160 A . A4 reads (1.60 V) (15.0 )  0.107 A.
EVALUATE: Just after the switch is closed the current through the battery is 0.455 A. After a long time the
current through the battery is 0.585 A. After a long time there are additional current paths, the equivalent resistance
of the circuit is decreased and the current has increased.
Figure 30.65
30.66.
IDENTIFY: Closing S 2 and simultaneously opening S1 produces an L-C circuit with initial current through the
inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is
2
a maximum the current is zero. Conservation of energy says that the maximum energy 12 Limax
stored in the inductor
2
qmax
stored in the capacitor.
C
 3.50 A , the current in the inductor just after the switch is closed.
equals the maximum energy
SET UP:
imax
1
2
NInductance
EXECUTE: (a)
30.67.
1
2
2
Limax
 12
30-19
2
qmax
.
C
qmax  ( LC )imax  (2.0 103 H)(5.0  106 F)(3.50 A)  3.50 104 C  0.350 mC .
(b) When q is maximum, i  0 .
EVALUATE: In the final circuit the current will oscillate.
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 30.67a.
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 30.67a
EXECUTE: (a) E  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 vR2  i2 R2  0
E  vR2  vL  0 so vL  E  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 E  vR2  0 and i2 R2  E
dt
E 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 30.67b.
EXECUTE: The current through
R1 is i2  2.40 A, in the direction b to a. Thus
vab  i2 R1  (2.40 A)(40.0 ) vab  96.0 V
Figure 30.67b
(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  i2 R2  (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
30.68.
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.
IDENTIFY: Apply the loop rule to the two loops. The current through the inductor doesn't change abruptly.
30-20
Chapter 30
di
and E is directed to oppose the change in current.
dt
EXECUTE: (a) Switch is closed, then at some later time
di
di
 50.0 A/s  vcd  L  (0.300 H) (50.0 A/s)  15.0 V.
dt
dt
60.0 V
 1.50 A.
The top circuit loop: 60.0 V  i1R1  i1 
40.0 
45.0 V
 1.80 A.
The bottom loop: 60 V  i2 R2  15.0 V  0  i2 
25.0 
60.0 V
 2.40 A, and immediately when the switch is opened, the inductor maintains
(b) After a long time: i2 
25.0 
this current, so i1  i2  2.40 A.
SET UP:
For the inductor E  L
EVALUATE: The current through R1 changes abruptly when the switch is closed.
30.69.
IDENTIFY and SET UP: The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after S1 is closed, i = 0.
After a long time i has reached its final value and di/dt = 0. The voltage across a resistor depends on i and the
voltage across an inductor depends on di/dt.
Figure 30.69a
EXECUTE: (a) At time t  0, i0  0 so vac  i0 R0  0. By the loop rule E  vac  vcb  0, so vcb  E  vac  E  36.0 V.
( i0 R  0 so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the
changing current.)
di
di
(b) After a long time 0  0 so the potential  L 0 across the inductor becomes zero. The loop rule gives
dt
dt
E  i0 ( R0  R)  0.
E
36.0 V
i0 

 0.180 A
R0  R 50.0   150 
vac  i0 R0  (0.180 A)(50.0 )  9.0 V
di
Thus vcb  i0 R  L 0  (0.180 A)(150 )  0  27.0 V (Note that vac  vcb  E. )
dt
(c) E  vac  vcb  0
di
E  iR0  iR  L  0
dt
 L  di
di
E
L  E  i ( R0  R) and 
  i 
dt
R

R
dt
R

R0
0 

di
 R  R0 

 dt
i  E /( R  R0 )  L 
Integrate from t = 0, when i = 0, to t, when i  i0 :

i0
0
i0

di
R  R0 t
E 
 R  R0 

dt   ln  i 
 
 t,

0
i  E /( R  R0 )
L
R  R0  0  L 


 E 
E 
 R  R0 
so ln  i0 
  ln 
  
t
R  R0 
 L 

 R  R0 
 i  E /( R  R0 ) 
 R  R0 
ln  0
  
t
E
/(
R

R
)
 L 
0


i0  E /( R  R0 )
E
 e  ( R  R0 )t / L and i0 
1  e( R R0 )t / L 
E /( R  R0 )
R  R0
36.0 V
Substituting in the numerical values gives i0 
1  e(200  / 4.00 H)t   (0.180 A) 1  et / 0.020 s 
50   150 
Taking exponentials of both sides gives
NInductance
30-21
At t  0, i0  (0.180 A)(1 1) 0 (agrees with part (a)). At t  , i0  (0.180 A)(1  0)  0.180 A (agrees with part (b)).
ER0
vac  i0 R0 
1  e  ( R  R0 )t / L  9.0 V 1  e t / 0.020 s 
R  R0


vcb  E  vac  36.0 V  9.0 V(1  et / 0.020 s )  9.0 V(3.00  et / 0.020 s )
At t  0, vac  0, vcb  36.0 V (agrees with part (a)). At t  , vac  9.0 V, vcb  27.0 V (agrees with part (b)).
The graphs are given in Figure 30.69b.
30.70.
Figure 30.69b
EVALUATE: The expression for i(t) we derived becomes Eq.(30.14) if the two resistors R0 and R in series are
replaced by a single equivalent resistance R0  R.
IDENTIFY: Apply the loop rule. The current through the inductor doesn't change abruptly.
SET UP: With S 2 closed, vcb must be zero.
EXECUTE: (a) Immediately after S 2 is closed, the inductor maintains the current i  0.180 A through R. The
loop rule around the outside of the circuit yields
36 V
 0.720 A .
E  EL  iR  i0 R0  36.0 V  (0.18 A)(150 )  (0.18 A)(150 )  i0 (50 )  0 . i0 
50 
vac  (0.72 A)(50 V)  36.0 V and vcb  0 .
E 36.0 V

 0.720 A , iR  0 and is 2  0.720 A .
(b) After a long time, vac  36.0 V, and vcb  0. Thus i0 
R0
50 
1
E  ( R L )t
e
(c) i0  0.720 A , iR (t ) 
and iR (t )  (0.180 A)e(12.5 s )t .
Rtotal
is 2 (t )  (0.720 A)  (0.180 A)e(12.5 s
1
)t

 (0.180 A) 4  e(12.5 s
1
)t
. The graphs of the currents are given in Figure 30.70.
EVALUATE: R0 is in a loop that contains just E and R0 , so the current through R0 is constant. After a long time
the current through the inductor isn't changing and the voltage across the inductor is zero. Since vcb is zero, the
voltage across R must be zero and iR becomes zero.
Figure 30.70
30-22
Chapter 30
30.71.
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: Problem 30.47 shows how to find the equivalent inductance of inductors in series and parallel.
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 30.71a.
I  (20.0 V) (75.0 )  0.267 A . The voltage between points a and b is zero, so the voltmeter reads zero.
(c) Use the results of Problem 30.49 to combine the inductor network into its equivalent, as shown in Figure 30.71b.
R  75.0  is the equivalent resistance. Eq.(30.14) says i  (E R )(1  e  t  ) with
  L / R  (10.8 mH) (75.0 )  0.144 ms . E  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 30.71
30.72.
IDENTIFY: At steady state with the switch in position 1, no current flows to the capacitors and the inductors can
be replaced by wires. Apply conservation of energy to the circuit with the switch in position 2.
SET UP: Replace the series combinations of inductors and capacitors by their equivalents. For the inductors use
the results of Problem 30.47.
E 75.0 V
 0.600 A .
EXECUTE: (a) At steady state i  
R 125 
(b) The equivalent circuit capacitance of the two capacitors is given by
1
1
1


and Cs  14.6  F .
Cs 25  F 35  F
Ls  15.0 mH  5.0 mH  20.0 mH . The equivalent circuit is sketched in Figure 30.72a.
q2 1 2
 Li0 . q  i0 LC  (0.600 A) (20  103 H)(14.6 106 F)  3.24 104 C . As shown
2C 2
in Figure 30.72b, the capacitors have their maximum charge at t  T / 4 .
Energy conservation:
t  14 T  14 (2 LC ) 


LC 
(20 103 H)(14.6 106 F)  8.49 104 s
2
2
EVALUATE: With the switch closed the battery stores energy in the inductors. This then is the energy in the L-C
circuit when the switch is in position 2.
Figure 30.72
30.73.
IDENTIFY:
SET UP:
Follow the steps specified in the problem.
Find the flux through a ring of height h, radius r and thickness dr. Example 28.19 shows that B 
inside the toroid.
0 Ni
2 r
NInductance
30-23
 Nih dr 0 Nih
  Ni 

ln(b a).
EXECUTE: (a)  B   B(h dr )    0  (h dr )  0
2

r
2 a r
2

a
a 
b
(b) L 
b
b
N  B 0 N 2h

ln(b a).
i
2
b  a (b  a)2
0 N 2h  b  a 




L


.
a
2a 2
2  a 
EVALUATE: h(b  a) is the cross-sectional area A of the toroid and a is approximately the radius r, so this result
is approximately the same as the result derived in Example 30.3.
IDENTIFY: The direction of the current induced in circuit A is given by Lenz’s law.
SET UP: When the switch is closed current flows counterclockwise in the circuit on the left, from the positive
plate of the capacitor. The current decreases as a function of time, as the charge and voltage of the capacitor
decrease.
EXECUTE: At loop A the magnetic field from the wire of the other circuit adjacent to A is into the page. The
magnetic field of this current is decreasing, as the current decreases. Therefore, the magnetic field of the induced
current in A is directed into the page inside A and to produce a magnetic field in this direction the induced current
is clockwise.
EVALUATE: The magnitude of the emf induced in circuit A decreases with time after the switch is closed,
because the rate of change of the current in the other circuit decreases.
(a) IDENTIFY and SET UP: With switch S closed the circuit is shown in Figure 30.75a.
(c) ln(b / a)  ln(1  (b  a) / a) 
30.74.
30.75.
Apply the loop rule to loops 1 and 2.
EXECUTE:
loop 1
E  i1R1  0
i1 
E
(independent of t)
R1
Figure 30.75a
loop (2)
E  i2 R2  L
di2
0
dt
E
1  e R2t / L 
R2
E
E
di
The expressions derived in part (a) give that as t  , i1 
and i2  . Since 2  0 at
R1
R2
dt
This is in the form of equation (30.12), so the solution is analogous to Eq.(30.14): i2 
(b) EVALUATE:
steady-state, the inductance then has no effect on the circuit. The current in R1 is constant; the current in R2 starts
at zero and rises to E / R2 .
(c) IDENTIFY and SET UP: The circuit now is as shown in Figure 30.75b.
Let t = 0 now be when S is opened.
E
At t = 0, i  .
R2
Figure 30.75b
Apply the loop rule to the single current loop.
di
di
EXECUTE: i( R1  R2 )  L  0. (Now
is negative.)
dt
dt
di
di
 R  R2 
L  i ( R1  R2 ) gives
  1
 dt
dt
i
 L 
Integrate from t = 0, when i  I 0  E / R2 , to t.

i
I0
 i 
di
 R1  R2 
 R  R2  t
  1
t
  0 dt and ln     
I
i
L
 L 


 0
Taking exponentials of both sides of this equation gives i  I 0e  ( R1  R2 )t / L 
E  ( R1  R2 )t / L
e
R2
30-24
Chapter 30
(d) IDENTIFY and SET UP: Use the equation derived in part (c) and solve for R2 and E.
EXECUTE: L  22.0 H
V2
V 2 (120 V) 2
RR1 
 40.0 W gives R1 

 360 .
R1
PR1
40.0 W
We are asked to find R2 and E. Use the expression derived in part (c).
I 0  0.600 A so E / R2  0.600 A
i  0.150 A when t  0.080 s, so i 
1
4
E  ( R1  R2 )t / L
e
gives 0.150 A  (0.600 A)e ( R1  R2 )t / L
R2
 e  ( R1  R2 ) t / L so ln 4  ( R1  R2 )t / L
L ln 4
(22.0 H)ln 4
 R1 
 360   381.2   360   21.2 
t
0.080 s
Then E  (0.600 A)R2  (0.600 A)(21.2 )  12.7 V.
(e) IDENTIFY and SET UP: Use the expressions derived in part (a).
R2 
E 12.7 V

 0.0353 A
R1 360 
EVALUATE: When the switch is opened the current through the light bulb jumps from 0.0353 A to 0.600 A. Since
the electrical power dissipated in the bulb (brightness) depend on i 2 , the bulb suddenly becomes much brighter.
IDENTIFY: Follow the steps specified in the problem.
SET UP: The current in an inductor does not change abruptly.
EXECUTE: (a) Using Kirchhoff’s loop rule on the left and right branches:
di
di
Left: E  (i1  i2 ) R  L 1  0  R(i1  i2 )  L 1  E.
dt
dt
q2
q2
Right: E  (i1  i2 ) R   0  R(i1  i2 )   E.
C
C
E
(b) Initially, with the switch just closed, i1  0, i2  and q2  0.
R
(c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat
tedious exercise but straightforward exercise. We will show that the initial conditions are satisfied:
E t
E
At t  0, q2 
e sin(t ) 
sin(0)  0.
R
R
E
E
i1 (t )  (1  e  t [(2 RC )1 sin(t )  cos(t )]  i1 (0)  (1  [cos(0)])  0.
R
R
1
1

 625 rad/s .
(d) When does i2 first equal zero?  
LC (2 RC ) 2
EXECUTE: The current through the light bulb before the switch is opened is i1 
30.76.
E t
e [(2 RC) 1 sin(t )  cos(t )]   (2 RC) 1 tan(t )  1  0 and
R
tan( t )  2 RC  2(625 rad/s)(400 )(2.00 106 F)  1.00.
0.785
 t  arctan(  1.00)  0.785  t 
 1.256 103 s.
625 rad/s
EVALUATE: As t  , i1  E / R , q2  0and i2  0 .
NB
IDENTIFY: Apply L 
to calculate L.
i
 Ni
 Ni
SET UP: In the air the magnetic field is BAir  0 . In the liquid, BL 
W
W
0 Ni
K 0 Ni
EXECUTE: (a)  B  BA  BL AL  BAir AAir 
(( D  d )W ) 
(dW )  0 Ni[( D  d )  Kd ] .
W
W
NB
d
d
 L  L0 
L
  0 N 2 [( D  d )  Kd ]  L0  L0  Lf  L0   f
d .
i
D
D
 D 
 L  L0 
2
2
d 
 D, where L0   0 N D, and L f  K  0 N D.
L

L
0 
 f
i2 (t )  0 
30.77.
NInductance
30-25
d

(b) Using K   m  1 we can find the inductance for any height L  L0  1   m  .
D

_______________________________________________________________________________
Height of Fluid
Inductance of Liquid Oxygen
Inductance of Mercury
0.63024 H
0.63000 H
dD 4
dD 2
30.78.
30.79.
0.63048 H
0.62999 H
0.63072 H
0.62999 H
d  3D 4
0.63096 H
0.62998 H
d D
________________________________________________________________________________
The values  m (O2 )  1.52 103 and  m (Hg)  2.9 105 have been used.
EVALUATE: (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an
easily detectable spread of values for the liquid oxygen, but not for the mercury.
IDENTIFY: The induced emf across the two coils is due to both the self-inductance of each and the mutual
inductance of the pair of coils.
di
SET UP: The equivalent inductance is defined by E  Leq , where E and i are the total emf and current across
dt
the combination.
di
di
di
di
di
EXECUTE: Series: L1 1  L2 2  M 21 1  M12 2  Leq .
dt
dt
dt
dt
dt
di di di
di
di
But i  i1  i2   1  2 and M12  M 21  M , so ( L1  L2  2M )  Leq and Leq  L1  L2  2M .
dt dt dt
dt
dt
di
di
di
di
di
di
di di di
Parallel: We have L1 1  M12 2  Leq
and L2 2  M 21 1  Leq , with 1  2 
and M12  M 21  M .
dt
dt
dt
dt dt dt
dt
dt
dt
di
di
di
To simplify the algebra let A  1 , B  2 , and C  . So L1 A  MB  LeqC , L2 B  MA  LeqC , A  B  C. Now
dt
dt
dt
solve for A and B in terms of C. ( L1  M ) A  (M  L2 ) B  0 using A  C  B . ( L1  M )(C  B)  (M  L2 ) B  0.
( M  L1 )
C.
( L1  M )C  ( L1  M ) B  ( M  L2 ) B  0. (2M  L1  L2 ) B  (M  L1 )C and B 
(2M  L1  L2 )
( M  L1 )C
(2M  L1  L2 )  M  L1
M  L2

C , or A 
C . Substitute A in B back
But A  C  B  C 
(2M  L1  L2 )
(2M  L1  L2 )
2 M  L1  L2
into original equation:
L1 ( M  L2 )C
M ( M  L1 )
M 2  L1L2
LL M2

C  LeqC and
.
C  LeqC. Finally, Leq  1 2
2 M  L1  L2 (2 M  L1  L2 )
2M  L1  L2
L1  L2  2M
EVALUATE: If the flux of one coil doesn't pass through the other coil, so M  0 , then the results reduce to those
of problem 30.47.
IDENTIFY: Apply Kirchhoff’s loop rule to the top and bottom branches of the circuit.
SET UP: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is
zero.
di
E
q
di
i
E
EXECUTE: E  i1R1  L 1  0  i1  (1  e  ( R1 L )t ). E  i2 R2  2  0   2 R2  2  0  i2  e (1 R2C ) t ) .
dt
R1
C
dt
C
R2
t
t
q2   i2dt  
0
E
R2Ce (1/ R2C )t   EC (1  e (1/ R2C )t ).
R2
0
E
E
48.0 V
(1  e0 )  0, i2  e0 
 9.60 103 A.
R1
R2
5000 
E
E 48.0 V
E
 1.92 A, i2  e   0. A good definition of a “long time” is
(c) As t   : i1 ()  (1  e )  
R1
R1 25.0 
R2
many time constants later.
(b) i1 (0) 
30-26
Chapter 30
(d) i1  i2 
E
(1  e  ( R1
R1
ex  1  x 
x 2 x3


2 3!
L)t
)
E  (1 R2C ) t
e
 (1  e  ( R1
R2
L)t
)
2
, we find:
R1 1  R1  2
t   t 
L
2 L 

R1  (1 R2C ) t
e
. Expanding the exponentials like
R2
R1 
t
t2
 2 2
1 
R2  RC 2 R C

 and

R
R 
R
t  1  21   O(t 2 )    1 , if we have assumed that t  1. Therefore:
L
R
C
R
2
2


  LR2C   (8.0 H)(5000 )(2.0 10 5 F) 
1 
1
3
t 


  1.6 10 s.
R2  (1 L)  (1 R2 2C )   L  R2 2C   8.0 H  (5000 ) 2 (2.0 10 5 F) 
(e) At t  1.57  103 s : i1 
E
(1  e  ( R1
R1
L )t
)
48 V
(1  e (25 8) t )  9.4 103 A.
25 
(f ) We want to know when the current is half its final value. We note that the current i2 is very small to begin with,
and just gets smaller, so we ignore it and find:
E
L
8.0 H
i1 2  0.960 A  i1  (1  e  ( R1 L ) t )  (1.92 A)(1  e  ( R1 L ) t ). e  ( R1 L )t  0.500  t  ln(0.5) 
ln(0.5)  0.22 s .
R1
R1
25 
EVALUATE:
i1 is initially zero and rises to a final value of 1.92 A. i2 is initially 9.60 mA and falls to zero, q2 is
initially zero and rises to q2  EC  960  C .