Solution to --
HW assignment #4
Chapter 31: Electric Current and Resistance
Note: you may need to use the following simplified relations to solve the following HW problems
1)
dt
dQ
I
=
,
2)
E
j
σ
=
,
3)
Ed
V
=
,
4)
σ
ρ
1
=
5)
A
I
j
/
=
,
6)
A
L
R
ρ
=
---------------------------------------------------------------------------------------------------------------
Solve:
(a)
The current associated with the moving film is the rate at which the charge on the film moves past a certain point. The
tangential speed of the film is
(
)(
)
rev
1 min
2
rad
90 rpm
4.0 cm
90
1.0 cm
9.425 cm/s
min
60 s
1 rev
v
r
π
ω
=
=
=
×
×
×
=
In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is
(9.425 cm)(4.0 cm)
=
37.7 cm
2
. The amount of charge that passes to the right in 1.0 s is
Q
=
(37.7 cm
2
)(
−
2.0
×
10
−
9
C/cm
2
)
=
−
75.4
×
10
−
9
C
Since
I
Q
t
=
Δ
, we have
(
)
9
75.4
10
C
75.4 nA
1 s
I
−
−
×
=
=
The current is 75 nA.
(b)
Having found the current in part (a), we can once again use
I
Q
t
=
Δ
to obtain
Δ
t
:
6
9
10
10
C
133 s
75.4
10
A
Q
t
I
−
−
−
×
Δ =
=
=
×
----------------------------------------------------------------------------------------------------------------------------------
Solve:
(a)
The values of
I
(A) for selected values of
t
are
t
(
μ
s)
0
1
2
4
6
8
10
I
(A)
2.0
1.21
0.74
0.27
0.10
0.036
0.014
(b)
Because
I
dQ dt
=
,
(
)
(
)
(
)
(
)
2.0
s
2.0 s
2.0 s
0
0
2.0 A
4.0 A s
4.0 C
1
t
t
t
t
t
Q
I dt
e
dt
e
e
μ
μ
μ
μ
μ
−
−
−
⎡
⎤
⎡
=
=
=
−
=
−
⎤
⎣
⎦
⎣
∫
∫
⎦
where we have used the condition
Q
=
0 C at
t
=
0
μ
s.
(c)
The values of
Q
(
μ
C) for selected values of
t
are
t
(
μ
s)
0
1
2
4
6
8
10
Q
(
μ
C)
0
1.57
2.52
3.46
3.80
3.92
3.98

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