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Chapter 26
(# 1, 8, 9, 17, 28, 35, 40, 43, 53, 64)
1. During the 4.0 min a 5.0 A current is set up in a wire, how many (a) coulombs and
(b) electrons pass through any cross section across the wire’s width?
1. (a) The charge that passes through any cross section is the product of the current and
time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The number of electrons N is given by q = Ne, where e is the magnitude of the charge
on an electron. Thus,
N = q/e = (1200 C)/(1.60  10–19 C) = 7.5  1021.
8. Near Earth, the density of protons in the solar wind (a stream of particles from
the Sun) is 8.70 cm-3, and their speed is 47 km/s. (a) Find the current density of these
protons. (b) If Earth’s magnetic field did not deflect the protons, what total current
would Earth receive?
8. (a) Since 1 cm3 = 10–6 m3, the magnitude of the current density vector is
J  nev 
F
8.70 I
.  10 Ch
c470  10 m / sh 6.54  10
G
H10 m J
Kc160
19
6
3
3
7
A / m2 .
(b) Although the total surface area of Earth is 4 RE2 (that of a sphere), the area to be used
in a computation of how many protons in an approximately unidirectional beam (the solar
wind) will be captured by Earth is its projected area. In other words, for the beam, the
encounter is with a “target” of circular area RE2 . The rate of charge transport implied by
the influx of protons is
c
hc6.54  10
i  AJ  RE2 J   6.37  106 m
2
7
h
A / m2  8.34  107 A.
9. 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 carries per
unit volume is 8.49 × 1028 m-3.
9. We use vd = J/ne = i/Ane. Thus,
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1070
t
CHAPTER 26
14
2
28
3
19
L
L
LAne  0.85m   0.2110 m   8.47 10 / m  1.60 10 C 



vd i / Ane
i
300A
 8.1102 s  13min .
17. A human being can be electrocuted if a current as small as 50 mA passes near
the heart. An electrician working with sweaty hands makes good contact with the
two conductors he is holding, one in each hand. If his resistance is 2000 Ω, what
might the fatal voltage be?
17. Since the potential difference V and current i are related by V = iR, where R is the
resistance of the electrician, the fatal voltage is V = (50  10–3 A)(2000 ) = 100 V.
28. 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 in 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 × 1014 (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.
28. We use J =  E = (n++n–)evd, which combines Eq. 26-13 and Eq. 26-7.
(a) 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 
 620  550  cm3  1.60 1019 C 
 1.73 cm s.
35. A 120 V potential difference is applied to a space heater whose resistance is 14 Ω
when hot. (a) At what rate is electrical energy transferred to thermal energy? (b)
What is the cost for 5.0 h at US$0.05/kW · h?
35. (a) Electrical energy is converted to heat at a rate given by
V2
P
,
R
where V is the potential difference across the heater and R is the resistance of the heater.
Thus,
1071
P
(120 V) 2
 10
.  103 W  10
. kW.
14 
(b) The cost is given by (1.0kW)(5.0h)(5.0cents/kW  h)  US$0.25.
40. A 120 V potential difference is applied to a space heater that dissipates 500 W
during operation. (a) What is its resistance during operation? (b) At what rate do
electrons flow through any cross section of the heater element?
40. (a) From P = V 2/R we find R = V 2/P = (120 V)2/500 W = 28.8 .
(b) Since i = P/V, the rate of electron transport is
i
P
500 W


 2.60  1019 / s.
e eV (1.60  1019 C)(120 V)
43. A 100 W light bulb is plugged into a standard 120 V outlet. (a) How much does it
cost per 31-day month to leave the light turned on continuously? Assume electrical
energy costs US$0.06/kW · h. (b) What is the resistance of the bulb? (c) What is the
current in the bulb?
43. (a) The monthly cost is (100 W) (24 h/day) (31 day/month) (6 cents/kW  h) 
446 cents  US$4.46, assuming a 31-day month.
(b) R = V 2/P = (120 V)2/100 W = 144 .
(c) i = P/V = 100 W/120 V = 0.833 A.
53. In Fig. 26-34, a battery of potential difference V = 12 V is connected to a resistive
strip of resistance R = 6.0 Ω. When an electron moves through the strip from one
end to the other, (a) in which direction in the figure does the electron move, (b) how
much work is done on the electron by the electric field in the strip, and (c) how
much energy is transferred to the thermal energy of the strip by the electron?
53. (a) Referring to Fig. 26-34, the electric field would point down (towards the bottom
of the page) in the strip, which means the current density vector would point down, too
(by Eq. 26-11). This implies (since electrons are negatively charged) that the conductionelectrons would be “drifting” upward in the strip.
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CHAPTER 26
(b) Eq. 24-6 immediately gives 12 eV, or (using e = 1.60  1019 C) 1.9  1018 J for the
work done by the field (which equals, in magnitude, the potential energy change of the
electron).
(c) Since the electrons don’t (on average) gain kinetic energy as a result of this work done,
it is generally dissipated as heat. The answer is as in part (b): 12 eV or 1.9  1018 J.
64. The headlights of a moving car require about 10 A from the 12 V alternator,
which is driven by the engine. Assume the alternator is 80% efficient (its output
electrical power is 80% of its input mechanical power), and calculate the
horsepower the engine must supply to run the lights.
64. The horsepower required is
P
| J |
iV
(10A)(12 V)

 0.20 hp.
0.80 (0.80)(746 W/hp)
i
9.20 104 A

 1.08 104 A/m2 .
A  m)2