Home Work Solutions 6
1. How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and
the electrons travel through a copper wire with cross-sectional area 0.21 cm2 and length 0.85 m. The number of
charge carriers per unit volume is 8.49 × 1028 m-3.
Sol
We use vd = J/ne = i/Ane. Thus,
14
2
28
3
19
L
L
LAne  0.85m   0.2110 m  8.47 10 / m  1.60 10 C 
t 


vd i / Ane
i
300A
 8.1102 s  13min.
2. Figure 26-28 shows wire section 1 of diameter D1 = 4.00R and wire section 2 of diameter D2 = 2.00R, connected
by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed
across any cross-sectional area through the wire's width. The electric potential change V along the length L = 2.00
m shown in section 2 is 10.0 mV. The number of charge carriers per unit volume is 8.49 × 1028 m-3. What is the
drift speed of the conduction electrons in section 1?
Figure 26-28 Problem 34.
Sol
The number density of conduction electrons in copper is n = 8.49 × 1028 /m3. The electric field
in section 2 is (10.0 V)/(1.75 m) = 5.71 V/m. Since = 1.69 × 108 ·m for copper (see Table
26-1) then Eq. 26-10 leads to a current density vector of magnitude J2 = (5.71 V/m)/(1.69 × 108
·m) = 338 A/m2 in section 2. Conservation of electric current from section 1 into section 2
implies
J1 A1  J 2 A2
 J1 (4 R2 )  J 2 ( R 2 )
(see Eq. 26-5). This leads to J1 = 84.5 A/m2. Now, for the drift speed of conduction-electrons in
section 1, Eq. 26-7 immediately yields
vd 
J1
 6.22 109 m/s
ne
3. Earth's lower atmosphere contains negative and positive ions that are produced by radioactive elements in the soil
and cosmic rays from space. In a certain region, the atmospheric electric field strength is 120 V/m and the field is
directed vertically down. This field causes singly charged positive ions, at a density of 620 cm-3, to drift
downward and singly charged negative ions, at a density of 550 cm-3, to drift upward (Fig. 26-27). The measured
conductivity of the air in that region is 2.70 × 10–14 (Ω · m)–1. Calculate (a) the magnitude of the current density
and (b) the ion drift speed, assumed to be the same for positive and negative ions.
Figure 26-27 Problem 32.
Sol
We use J =  E = (n+ + n–)evd, which combines Eq. 26-13 and Eq. 26-7.
(a) The magnitude of the current density is
J =  E = (2.70  10–14 / ·m) (120 V/m) = 3.24  10–12 A/m2.
(b) The drift velocity is
vd 
E
 n  n  e

 2.70 10
14
  m  120 V m 
 640  550  cm3  1.60 1019 C 
 1.70 cm s.
4. Swimming during a storm. Figure 26-30 shows a swimmer at distance D = 35.0 m from a lightning strike to the
water, with current I = 78 kA. The water has resistivity 30Ω · m, the width of the swimmer along a radial line
from the strike is 0.70 m, and his resistance across that width is 4.00 kΩ. Assume that the current spreads through
the water over a hemisphere centered on the strike point. What is the current through the swimmer?
Figure 26-30 Problem 36.
Sol
Since the current spreads uniformly over the hemisphere, the current density at any given
radius r from the striking point is J  I / 2 r 2 . From Eq. 26-10, the magnitude of the electric
field at a radial distance r is
 I
E  w J  w 2 ,
2 r
where w  30  m is the resistivity of water. The potential difference between a point at
radial distance D and a point at D  r is
V   
D r
D
Edr   
D r
D
w I
 I 1
 I
1
r
,
dr  w 
  w
2
2 r
2  D  r D 
2 D( D  r )
which implies that the current across the swimmer is
i
| V |  w I
r
.

R
2 R D( D  r )
Substituting the values given, we obtain
(30.0   m)(7.80 104 A)
0.70 m
i
 4.43 102 A .
3
2 (4.00 10 )
(38.0 m)(38.0 m  0.70 m)
5. In Fig. 26-29, current is set up through a truncated right circular cone of resistivity 731 Ω · m, left radius a = 2.00
mm, right radius b = 2.30 mm, and length L = 1.94 cm. Assume that the current density is uniform across any
cross section taken perpendicular to the length. What is the resistance of the cone?
Figure 26-29
Sol
(a) The current i is shown in Fig. 26-29 entering the truncated cone at the left end and leaving at
the right. This is our choice of positive x direction. We make the assumption that the current
density J at each value of x may be found by taking the ratio i/A where A = r2 is the cone’s

cross-section area at that particular value of x. The direction of J is identical to that shown in the
figure for i (our +x direction). Using Eq. 26-11, we then find an expression for the electric field at
each value of x, and next find the potential difference V by integrating the field along the x axis, in
accordance with the ideas of Chapter 25. Finally, the resistance of the cone is given by R = V/i.
Thus,
J
i
E

2
r 
where we must deduce how r depends on x in order to proceed. We note that the radius increases
linearly with x, so (with c1 and c2 to be determined later) we may write
r  c1  c2 x.
Choosing the origin at the left end of the truncated cone, the coefficient c1 is chosen so that r = a
(when x = 0); therefore, c1 = a. Also, the coefficient c2 must be chosen so that (at the right end of
the truncated cone) we have r = b (when x = L); therefore, c2  (b  a) / L . Our expression, then,
becomes
r a
b  aI
F
G
HL J
Kx.
Substituting this into our previous statement and solving for the field, we find
2
i 
ba 
E  a
x .
 
L 
Consequently, the potential difference between the faces of the cone is
L
i
0

V    E dx  


L
0
2
ba 
i L 
ba 

x  dx 
x
a 
a 
L 
 ba
L 

1 L
0
i
L  1 1  i L b  a i L



.
 b  a  a b   b  a ab  ab
The resistance is therefore
R
V L
(731  m)(1.94 102 m)


 9.81105 
3
3
i  ab  (2.00 10 m)(2.30 10 m)
Note that if b = a, then R = L/a2 = L/A, where A = a2 is the cross-sectional area of the cylinder.