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1. Current Continuity and Relaxation Time
P4.4: At t = 0 seconds, 60.0 C is evenly distributed throughout a 2.00 cm diameter
pure silicon sphere. (a) Find the initial charge density. (b) How long does it take to
drop to 10% of its initial value? (c) What will be the final surface charge density?
4
Q
C
3
(a) Volume v    0.01m   4.19 x106 m 3 ;  vo   14.3 3
3
v
m
 t /
(b)  v  0.10  vo   vo e
 11.8 8.854 x10
For pure silicon,   

 4.4 x104 
ln 0.10 
t

12
  237.4 x10
9
sec
 2.303; t   2.303  547ns
Fig. P4.8 (at end of animation)
Q
60C
; area=4 r 2  1.257 x103 m 2 ;  s 
 47.8 mC 2
(c)  s 
m
area
1.257 x103 m 2
3. Faraday’s Law and Transformer EMF
P4.9: The magnetic flux density increases at the rate of 10 (Wb/m 2)/sec in the z
direction. A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y
plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch)
and magnitude of the induced current in the conducting loop.
dB
Wb
 10 2 a z
dt
ms
dB
Wb
Vemf   
dS    10 2 a z dxdya z
dt
ms
Wb Vs
Vemf  0.1
 0.1V
s Wb
0.1V
 10mA
I= I 
10
Fig. P4.9
I=10mA clockwise (when viewed
From +z)
P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It
is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t
Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and
indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center
of the loop resulting from the induced current, and compare this with the original
magnetic flux density that generated the induced current at t = 1.0 sec.
We find the distributed resistance of the loop and work the problem assuming this
resistance is lumped in one spot as shown in the figure.
(a) The induced current is Vemf divided by the distributed resistance of the wire loop.
4  0.04m) 
1 l
1m
R

 3.5m
7
 A 5.8 x10   0.0005m 2
dB
Wb
dB
Wb
 1 2 a z ; Vemf   
dS  1 2
dt
ms
dt
ms
I ind 
0.04
0.04
0
0
 dx  dy  1.6mV
1.6mV
 0.46 A (note that this answer has no time dependence)
3.5m
Fig. P4.13
(b) The field at the center of the loop from a single arm of the loop is found from Eqn.
(3.7):
a

I 
z
I
H
 2 2
  a z  
2
4   z   
2
a
So B  4o H  13
Wb
m2
 0.46 
1
1
A
 -a z  
 a z   2.59a z ;
2
m
2a
2  0.02 
az .
P4.15: A triangular wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0,
4), with dimensions in meters. A time-varying magnetic field is given by B = 4t ay
Wb/m2 (t in seconds). If the wire has a total distributed resistance of 2 , calculate
the induced current and indicate its direction in a carefully drawn sketch.
B
dS,
t
B
Wb
 4a y 2
t
ms
1
MN
S  MN
2
MN
Vemf   
M  2a x  3a y , N  2a x  4a z
S  6a x  4a y  3a z m2
Vemf = -(4)(4)=-16V
Vemf 16V
I

 8A
R
2
Fig. P4.15
4. Faraday’s Law and Motional EMF
P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a
diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The
outer ends of the bar make conductive contact with a ring to make one end of the
electrical contact to R; the other contact is made to the center of the bar. Given Bo =
100. mWb/m2, a = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0
revolution per second.
d
a   a , dL  d  a 
dt
Figure P4.22 indicates one of the paths for the circulation integral.
a
a
 Bo a 2


Vemf    a  Boa z  d  a     Bo   d  
2
0
0
Vemf 
I
  u  B  dL;
Vemf

u
 Bo a 2
R
I  22.6  A
1  rev  rad 
2  1  Vs A
3 Wb 
 1
0.06m  
 2
100 x10

2 
2R
2  s 
rev 
m 
 50  Wb V
Fig. P4.22
P4.24: Consider a sliding rail problem where the conductive rails expand as they
progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance
between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y
direction, and uy = 2.0 m/sec, find the Vemf across a 100.  resistor at the instant when
y = 10. cm if the field is Bo = 100. mT.
First we modify the figure so that the top rail is horizontal and all the spreading occurs
via the bottom rail. As before, our approach will be to find  and then d /dt. We
have:
   B dS   Boa z dxdya z
Now, notice that x and y are not independent and are in fact related: x=y+w
So we have
y yw
y
1

  Bo   dxdy  Bo   y  w dy  Bo  y 2  wy 
2

y 0 x 0
0
d
dy
Wb 

 m
  Bo  y  w    Bo  y  w  u y   0.100 2   0.1m  0.1m   2 
dt
dt
m 

 s
 40mV
Vemf  
Vemf
Alternate Method:
Vemf    u  B  dL    u y a y  Boa z  dxa x
Vemf  u y Bo
1
 y
2

dx  u y Bo  w  y 
1
w y
2
Fig. P4.24
5. Displacement Current
P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer
conductor diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage
v(t) = 10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is
grounded. Neglecting fringing fields at the ends of the coax, find the displacement
current between the inner and outer conductors.
C
Q
, so Q  Cv(t )
V
for coax (from chapter 2): C=
 D dS   D a 
2 lVo cos t
2 l
so Q 
ln  b a 
ln  b a 
 d dza   2 l D  Q
2 l Vo cos t Vo cos t
V cos t

and D  o
a .
2 l ln  b a   ln  b a 
 ln  b a  
D Vo sin t

a  Jd
t
 ln  b a  
so D 
2
l
Vo sin t 1
2 l Vo sin t
id   J d dS 
  d  dz 
ln  b a   0
ln  b a 
0
Now evaluate id with the given parameters:
2  6 x106 rad s  10.2   8.854 x1012 F m  1m  0.010V sin t  C As
id 
ln  0.006 0.002 
FV C
id  97 sin  6 x106 t   A