SOLUTIONS, ASSIGNMENT 14
25.1.
I
DENTIFY
:
/
I
Q t
=
.
S
ET
U
P
:
1.0 h
3600 s
=
E
XECUTE
:
4
(3.6 A)(3.0)(3600 s)
3.89
10 C.
Q
It
=
=
=
×
E
VALUATE
:
Compared to typical charges of objects in electrostatics, this is a huge amount of charge.
25.4.
(a)
I
DENTIFY
:
By definition,
J = I/A
and radius is onehalf the diameter.
S
ET
U
P
:
Solve for the current:
I = JA = J
π(
D
/2)
2
E
XECUTE
:
I
= (1.50
×
10
6
A/m
2
)(
π
)[(0.00102 m)/2]
2
= 1.23 A
E
VALUATE
:
This is a realistic current.
(b)
I
DENTIFY
:
The current density is
J = nqv
d
S
ET
U
P
:
Solve for the drift velocity:
v
d
= J/nq
E
XECUTE
:
Since most laboratory wire is copper, we use the value of
n
for copper, giving
6
2
d
(1.50
10
A/m )
v
=
×
/[(8.5
×
10
28
el/m
3
)(1.60
×
19
10

C) = 1.1
×
4
10

m/s = 0.11 mm/s
25.7.
I
DENTIFY
and
S
ET
U
P
:
Apply Eq. (25.1) to find the charge
dQ
in time
dt
. Integrate to find the total
charge in the whole time interval.
E
XECUTE
:
(a)
dQ
I dt
=
(
29
(
29
(
29
(
29
8.0 s
8.0 s
2
2
2
3
0
0
55 A
0.65 A/s
55 A
0.217 A/s
Q
t
dt
t
t
=

=

∫
(
29
(
29
(
29
(
29
3
2
55 A
8.0 s
0.217 A/s
8.0 s
330 C
Q
=

=
(b)
330 C
= 41 A
8.0 s
Q
I
t
=
=
E
VALUATE
:
The current decreases from 55 A to 13.4 A during the interval. The decrease is not linear
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 Fall '10
 dfs
 Charge, Electrostatics, Electric charge, EXECUTE, good metallic conductors, 1.25 V/m = 2.84 × 10−8 Ω ⋅

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