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Physics III
Homework IV
CJ
Chapter 29; 48, 49
Chapter 30; 4, 9, 22, 52, 60, 67
29.48. IDENTIFY: A changing magnetic field causes a changing flux through a coil and
therefore induces an emf in the coil.
SET UP: Faraday’s law says that the induced emf is E   d  B and the magnetic flux
dt
through a coil is defined as
 B  BA cos.
EXECUTE: In this case,  B  BA, where A is constant. So the emf is proportional to the
negative slope of the magnetic field. The result is shown in Figure 29.48.
EVALUATE: It is the rate at which the magnetic field is changing, not the field’s
magnitude, that determines the induced emf. When the field is constant, even though it
may have a large value, the induced emf is zero.
Figure 29.48
29.49. (a) IDENTIFY: (i)
E
dB
. The flux is changing because the magnitude of the magnetic
dt
field of the wire decreases with distance from the wire. Find the flux through a narrow strip
of area and integrate over the loop to find the total flux.
SET UP:
Consider a narrow strip of width dx and a
distance x from the long wire, as
shown in Figure 29.49a. The
magnetic field of the wire at the strip
is B  0 I/2 x. The flux through the
strip is d B  Bb dx  (0 Ib/2 )(dx/x)
Figure 29.49a
r a
EXECUTE: The total flux through the loop is  B   d B   0 Ib   r dx
 2 
x
  Ib   r  a 
 B   0  ln 

 2   r 

d  B d  B dr 0 Ib 
a


 
v
dt
dt dt
2  r  r  a  
0 Iabv
E
2 r  r  a 
(ii) IDENTIFY: E  Bvl for a bar of length l moving at speed v perpendicular to a
magnetic field B. Calculate the induced emf in each side of the loop, and combine the
emfs according to their polarity.
SET UP: The four segments of the loop are shown in Figure 29.49b.
EXECUTE: The emf in each side
of the loop is E1   0 I  vb,
 2 r 


0 I
E2  
 vb, E2  E4  0
 2 r (r  a ) 
Figure 29.49b
Both emfs
E1 and E2
are directed toward the top of the loop so oppose each other. The net
emf is E  E1  E2  0 Ivb  1  1   0 Iabv
2  r r  a  2 r (r  a)
This expression agrees with what was obtained in (i) using Faraday’s law.
(b) (i) IDENTIFY and SET UP: The flux of the induced current opposes the change in
flux.
EXECUTE: B is  .  B is  and decreasing, so the flux
and the current is clockwise.
ind
of the induced current is

(ii) IDENTIFY and SET UP: Use the right-hand rule to find the force on the positive
charges in each side of the loop. The forces on positive charges in segments 1 and 2 of
the loop are shown in Figure 29.49c.
Figure 29.49c
EXECUTE: B is larger at segment 1 since it is closer to the long wire, so FB is larger in
segment 1 and the induced current in the loop is clockwise. This agrees with the direction
deduced in (i) using Lenz’s law.
(c) EVALUATE: When v = 0 the induced emf should be zero; the expression in part (a)
gives this. When a  0 the flux goes to zero and the emf should approach zero; the
expression in part (a) gives this. When r   the magnetic field through the loop goes to
zero and the emf should go to zero; the expression in part (a) gives this.
30.4. IDENTIFY: Changing flux from one object induces an emf in another object.
(a) SET UP: The magnetic field due to a solenoid is
B  0nI .
EXECUTE: The above formula gives
B1 
 4 10
7
T  m/A  (300)(0.120 A)
0.250 m
=1.81104 T
The average flux through each turn of the inner solenoid is therefore
 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
(c) SET UP: The induced emf is E2  M
EXECUTE:
di1
.
dt
E2   1.18  105 H  (1750 A/s)  0.0207 V
EVALUATE: A mutual inductance around 105 H is not unreasonable.
30.9. IDENTIFY and SET UP: Apply
of the induced emf in the coil.
E  L di/dt .
Apply Lenz’s law to determine the direction
EXECUTE: (a) E  L(di/dt )  (0.260 H)(0.0180 A/s)  4.68 103 V
(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.
30.22. IDENTIFY: With S1 closed and S 2 open, i(t ) is given by Eq.(30.14). With
S 2 closed, i(t ) is given by Eq.(30.18).
SET UP:
U  12 Li 2 .
After
S1
S1 open
and
has been closed a long time, i has reached its final value of
I  E/R.
EXECUTE: (a)
U  12 LI 2
(b) i  Ie( R/L)t and
t
and
I
2U
2(0.260 J)

 2.13 A. E  IR  (2.13 A)(120 )  256 V.
L
0.115 H
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
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 .
30.52. IDENTIFY: This is an R-L circuit and i(t ) is given by Eq.(30.14).
SET UP: When t  ,
EXECUTE: (a)
(b)
R
i  if  V / R .
V
12.0 V

 1860 .
if 6.45 103 A
so Rt  ln(1  i / if ) and L 
i  if (1  e  ( R / L ) t )
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. IDENTIFY:
i(t )
is given by Eq.(30.14).
SET UP: The graph shows
large times.
V 0
at
t  0 and
V approaches the constant value of 25 V at
EXECUTE: (a) The voltage behaves the same as the current. Since
i, the scope must be across the 150  resistor.
VR
is proportional to
(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 t   ,
 1
VR  Vmax 1    0.63Vmax .
 e
L / R  0.5 ms
gives
From the graph,
V  0.63Vmax  16 V at t  0.5 ms .
L  (0.5 ms) (150 )  0.075 H .
Therefore   0.5 ms .
(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 .
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 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.
(e) After the switch has been closed a long time,
E  vR2  0 and i2 R2  E so i2 
di2
 0 so vL  0.
dt
Then
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  vR1  vR2  0
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.
1
2
(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.