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
2
0.85m 0.21 10 m 8.47 10 / m 1.60 10 C
/ 300A
8.1 10 s 13min .
d
L L LAne
t v i Ane i
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 × 108 ·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 × 108
·m) = 338 A/m2 in section 2. Conservation of electric current from section 1 into section 2 implies
2 2
1 1 2 2 1(4 ) 2( )
J A J A J R J R
(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
1 6.22 109 m/s
d
v J 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
14
3 19
2.70 10 m 120 V m
1.70 cm s.
640 550 cm 1.60 10 C
d
v E
n n e
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/ 2r2. From Eq. 26-10, the magnitude of the electric field at a radial distance r is
2 2 w w
E J I r
,
where w 30mis the resistivity of water. The potential difference between a point at
radial distance D and a point at D ris
2
1 1
2 2 2 ( )
D r D r
w w w
D D
I I I r
V Edr dr
r D r D D D r
,which implies that the current across the swimmer is
| |
2 ( )
wI
V r
i R R D D r
.
Substituting the values given, we obtain
4
2 3
(30.0 m)(7.80 10 A) 0.70 m
4.43 10 A 2 (4.00 10 ) (38.0 m)(38.0 m 0.70 m)
i
.
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,
2
i E
Jr
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 c x2 .
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 a L x
F
H G IK J
.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
2 1
0 0
0
1 1
.
L
L i L b a i L b a
V E dx a x dx a x
L b a L
i L i L b a i L
b a a b b a ab ab
The resistance is therefore
2
5
3 3
(731 m)(1.94 10 m)
9.81 10 (2.00 10 m)(2.30 10 m)
V L
R i ab
Note that if b = a, then R = L/a2 = L/A, where A = a2 is the cross-sectional area of the cylinder.
