24.1
ELECTRIC CURRENT
24
EXERCISES
Section 24.1 Electric Current
14.
The current is the amount of charge passing a given point in the wire, per unit time, so in one second,
(1.5 A)(1s)
1.5 C.
qI
t
Δ=Δ=
=
The number of electrons in this amount of charge is1.5
19
C/1.6 10
−
×
18
C9
.
3
81
0
.
=×
t
15.
INTERPRET
This problem is about the number of charges involved in setting up an electric current.
DEVELOP
For a steady current, the amount of charge crossing a given area in time
Δ
can be found by using
Equation 24.1a,
/.
I
Qt
=Δ Δ
EVALUATE
A battery rated at 80 A h
⋅
can supply a net charge of
5
(80 C/s)(3600 s)
2.88 10 C
QI
t
=
×
ASSESS
By definition,1A
or1C
1C/s,
=
1A s.
=
⋅
So the result makes sense.
16.
The charge moving through the membrane each second is
Since singlycharged ions carry one elementary
charge (about160
this corresponds to
30 nC.
zC),
11
23
30 nC/(160 zC/ion)
(1.88 10 ion)/(6.02 10 ion/mol)
0.311
×
=
pmol.(See Example 24.2(a), Appendix B, and Table 1.1.) Chemical drugtesting instrumentation can detect
amounts of substances this low.
17.
INTERPRET
In this problem we are asked to find the current density, given the electric current and the cross section.
DEVELOP
The current density
J
, is defined as the current per unit area, or
J
IA
=
The area of the cross section is
22
/4.
AR
d
ππ
==
EVALUATE
The cross section of a wire is uniform, so the density is
62
23
2
10 A
7.65 10 A/m
/4
(1.29 10 m) /4
II
J
Ad
−
=
= ×
×
ASSESS
If the current
I
is kept fixed, the smaller the crosssectional area
A
, the greater the current density
J
.
Section 24.2 Conduction Mechanisms
18.
Aluminum obeys Ohm’s law, so
(see Table 24.1).
82
/
(0.085 V/m)/(2.65 10
m)
3.21 MA/m
JE
E
σρ
−
===
×
Ω
⋅
=
19.
INTERPRET
In this problem we are asked to calculate the electric field in a currentcarrying conductor.
DEVELOP
To find the electric field, we make use of Ohm’s law (which applies to silver),
/,
J
E
ρ
=
and the
definition of current density, which is assumed to be uniform in the wire.
EVALUATE
Using Table 24.1, we find the electric field to be
8
24
2
(1.59 10
m)(7.5 A)
0.168 V/m
/4
(9.5 10 m) /4
I
EJ
d
−
−
×Ω
⋅
=
=
×
ASSESS
The value is a lot smaller than the electric field we discussed in electrostatic situations. Since silver is
such a good conductor, a small field can drive a substantial current.
20.
Assuming a uniform current density obeying Ohm’s law, we find
2
1
4
//
,
o
r
4
/
J
EId
d
I
E
ρπ
π
=
=
1/2
2[(0.22
m)(350 mA)/ (21 V/m)]
6.83 cm
Ω⋅
=
(see Table 24.1).
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Chapter 24
21.
INTERPRET
This problem is about applying Ohm’s law to find the resistivity of a rod.
DEVELOP
If the rod has a uniform current density and obeys Ohm’s law (Equations 24.4a and 4b), then its
resistivity is
2
(/
4
)
/
EEE
d
JI
A
I
π
ρ
== =
EVALUATE
Substituting the values given in the problem, we find the resistivity of the rod to be
22
2
6
(
/4)
(1.4 V/m) (10 m) /4
2.20 10
m
50 A
Ed
I
ππ
−
−
==
=
×
Ω
⋅
ASSESS
With reference to Table 24.1, we see that the value is within the range of resistivity of a metallic conductor.
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 Fall '10
 Pheong
 Electric charge, Electrical resistance

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