CHAPTER 30 HOMEWORK SOLUTIONS
30.5.IDENTIFY and SET UP: Apply Eq.(30.5).
400 0.0320 Wb 
 1.96 H
EXECUTE: (a) M  N 2 B 2  
6.52 A
i1
(b)
M
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.
30.8.IDENTIFY:
SET UP:
 B is the average flux through one turn of the solenoid.
EXECUTE:
EVALUATE:
30.12.
Combine the two expressions for L: L  N  B / i and L  E/(di/dt ).
Solving for N we have N  Ei/ B (di/dt ) 
(12.6  10 3 V)(1.40 A)
 238 turns.
(0.00285 Wb)(0.0260 A/s)
The induced emf depends on the time rate of change of the total flux through the 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.
30.14.
IDENTIFY:
energy.
(a) SET UP:
EXECUTE:
(b) SET UP:
EXECUTE:
A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic
The magnetic field inside a toroidal solenoid is B 
B
0 (300)(5.00 A)
 2.50  10 3 T  2.50 mT
2 (0.120 m)
The self-inductance of a toroidal solenoid is L 
L
0 NI
.
2 r
0 N 2 A
.
2 r
(4  107 T  m/A)(300) 2 (4.00  104 m 2 )
 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
(d) SET UP:
B2
.
2 0
(2.50  103 T) 2
 2.49 J/m3
2(4  107 T  m/A)
energy energy
7.50  104 J
u


 2.49 J/m3
volume 2 rA 2 (0.120 m)(4.00  104 m 2 )
EXECUTE:
(e)
The energy density in a magnetic field is u 
EVALUATE:
u
An inductor stores its energy in the magnetic field inside of it.
30.19.
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
E  vR  v L  0
EXECUTE:
E  iR  L
E
di
 0 so i  1  e ( R / L )t 
dt
R
E L
(a) Initially (t = 0), i = 0 so
di
0
dt
di E 6.00 V
 
 2.40 A/s
dt L 2.50 H
(b) E  iR  L di  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 )

 0.800 A/s
Thus 
dt
L
2.50 H
6.00 V 
E
 (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  
(d) Final steady state means
i
di
 0, so E  iR  0.
dt
E 6.00 V

 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.
30.22.
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.
EXECUTE:
(b)
t   and
i  Ie  ( R/L )t
t
2U
2(0.260 J)

 2.13 A. E  IR  (2.13 A)(120 )  256 V.
L
0.115 H
(a)
U  12 LI 2
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 ,
and
I
so
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 .
30.34. IDENTIFY:
SET UP: q  Q
EXECUTE:
(a)
Apply Eq.(30.25).
i  0 . i  imax when q  0 . 1/ LC  1917 s 1 .
when
1
2
2
Limax

Q2
2C
.
Q  imax LC  (0.850 103 A) (0.0850 H)(3.20  106 F)  4.43 107 C
2
(b)
 5.00 104 A 
7
q  Q  LCi  (4.43 10 C)  
  3.58 10 C .
1
 1917s

2
7
2
2
EVALUATE: The value of q calculated in part (b) is less than the maximum value Q
calculated in part (a).
30.52.
SET UP:
EXECUTE:
(b)
This is an R-L circuit and i (t ) is given by Eq.(30.14).
IDENTIFY:
When
(a)
i  if (1  e  ( R / L )t )
t   , i  if  V / R .
V
12.0 V
 1860 .
R 
if 6.45 103 A
so
Rt
 ln(1  i / if )
L
and L 
 Rt
(1860 )(7.25 104 s)

 0.963 H .
ln(1  i / if )
ln(1  (4.86 / 6.45))
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.
30.60.
i (t ) is given by Eq.(30.14).
IDENTIFY:
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.
(b) From the graph, as t  , VR  25 V, so there is no voltage drop across the inductor, so
its internal resistance must be zero. VR  Vmax (1  et / r ) . When
From the graph,
V  0.63Vmax  16 V at t  0.5 ms .
t 
,
 1
VR  Vmax 1    0.63Vmax .
 e
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
the inductor since
di / dt  0 .
i0.
At
t 
appears across
all the battery voltage is across the resistance, since
Figure 30.60
30.67.
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 vR  i2 R2  0
2
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.
(e) After the switch has been closed a long time, di2  0 so vL  0. Then E  vR  0 and i2 R2  E
dt
so i2 
2
E 60.0 V

 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  vR  vR  0
1
2
vL  vR1  vR2
vR1  vab  96.0 V; vR2  i2 R2  (2.40 A)(25.0 )  60.0 V
Then vL  vR  vR  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
1
2
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.