Physics II
Homework XI
CJ
Chapter 28: 4, 16, 22, 27, 34, 42, 51
28.4. Solve: Equation 28.2 is Ne  nAvd t . Using Table 28.1 for the electron density, we get
A
D
 D2
4



Ne
nvd t
4 1.0  1016
4 Ne

 9.26  10 4 m  0.926 mm
 nvd t
 5.8  1028 m 3 8.0  10 4 m/s 320  10 6 s




28.16. Visualize:
The current density J in a wire, as given by Equation 28.13, does not depend on the thickness of the wire.
Solve: (a) The current in the wire is




2
Iwire  Jwire Awire  450,000 A/m2   21 1.5  103 m   0.795 A


Because current is continuous, Iwire  Ifilament. Thus, Ifilament  0.795 A.
(b) The current density in the filament is
J filament 
I filament
0.795 A

 7.03  10 7 A/m2
Afilament   1 0.12  10 3 m  2
2



28.22. Model: We will use the model of conduction to relate the mean time between collisions to conductivity.
Solve:
From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for silver is
 silver 




9.11  10 31 kg 6.2  10 7 1m 1
m silver

 3.80  10 14 s
2
nsilver e2
5.8  1028 m 3 1.60  10 19 C


Similarly, the mean time between collisions for gold is  gold  2.47  1014 s .
Assess: Mean free times in metals are of the order of  1014 s.
28.27. Solve: (a) Since J   E and J  I A , the electric field is
E
I
A

I
 r 2

0.020 A
  0.25  10 m   6.2  10  m
3
2
7
1
1

 1.64  10 3 N/C
(b) Since the current density is related to vd by J  I A  nevd , the drift speed is
vd 
I
 r 2 ne

0.020 A
  0.25  10 m   5.8  10 m
2
3
28
3
1.60  10
19
C

 1.10  105 m/s
Assess: The values of n and  for silver have been taken from Table 28.1 and Table 28.2. The drift velocity is typical of
metals.
28.34. Visualize: Please refer to Figure P28.34.
Solve: (a) The current associated with the moving film is the rate at which the charge on the film moves past a
certain point. The tangential speed of the film is
rev 1 min 2 rad


 1.0 cm  9.425 cm/s
min 60 s
1 rev
In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is (9.425
cm)(4.0 cm)  37.7 cm2. The amount of charge that passes to the right in 1.0 s is
v  r   90 rpm  4.0 cm  90
Q  (37.7 cm2)(2.0  10 9 C/cm2)  75.4  10 9 C
Since I  Q t , we have
I

 75.4  10 9 C

 75.4 nA
1s
(b) Having found the current in part (a), we can once again use I  Q t to obtain t:
t 
6
Q 10  10 C

 133 s
I
75.4  10 9 A
28.42. Visualize:
Solve: (a) Consider a cylindrical surface inside the metal at a radial distance r from the center. The current is flowing
through the walls of this cylinder, which have surface area A   2 r  L. Thus
I  JA   E  2 r  L
Thus the electric field strength at radius r is
E
(b) For iron, with   1.0  107 1m1,
I
2 Lr

25 A
Einner  
 2  0.10 m  1.0  10 7 1m 1



25 A
Eouter  
 2  0.25 m  1.0  10 7 1m 1



 1 

 3.98  10 5 N/C
  0.01 m 



1


 1.59  10 5 N/C
  0.025 m 

28.51. Model: Because current is conserved, the currents in the two segments of the wire are the same.
Visualize: Please refer to Figure P28.51.
Solve: The currents in the two segments of the wire are related by I1  I2  I. But I  AJ, so we have A1J1  A2J2.
The wire’s diameter is constant, so J1  J2 and 1E1  2E2. The ratio of the electric fields is
E2  1
1
1



E1  2  2  1 2