malsam (wgm329) – hw 12 – turner – (56705)
1
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have
16
questions.
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before answering.
001 (part 1 of 2) 10.0 points
The current in a conductor varies over time as
shown in the figure below.
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
Current(A)
Time(s)
a) How much charge passes through a cross
section of the conductor in the time interval
t
= 0 s to
t
= 3 s?
Correct answer: 12 C.
Explanation:
Let :
Δ
t
= 3
.
0 s
.
I
=
Δ
Q
Δ
t
Δ
Q
=
I
Δ
t
= (2 A)(0
.
5 s) + (4 A)(1 s)
+ (6 A)(0
.
5 s) + (4 A)(1 s)
=
12 C
.
002 (part 2 of 2) 10.0 points
b) What constant current would transport the
same total charge during the 3 s interval as
does the actual current?
Correct answer: 4 A.
Explanation:
I
=
Δ
Q
Δ
t
=
12 C
3 s
=
4 A
.
003
10.0 points
An aluminium wire with a cross-sectional area
of 4
.
5
×
10
−
6
m
2
carries a current of 4
.
1 A.
Find the drift speed of the electrons in the
wire. The density of aluminium is 2
.
7 g
/
cm
3
.
(Assume that two
electrons is supplied by
each atom).
Correct answer: 4
.
7202
×
10
−
5
m
/
s.
Explanation:
Let:
A
= 4
.
5
×
10
−
6
m
2
,
I
= 4
.
1 A = 4
.
1 C
/
s
,
ρ
= 2
.
7 g
/
cm
3
= 2
.
7
×
10
6
g
/
m
3
,
and
n
e
= electrons
/
atom = 2
.
Assuming
n
e
free conduction electrons per
atom, the density of charge carriers is
n
e
times
the density of aluminium atoms:
n
=
n
e
density
mass
/
atom
=
n
e
ρ
M/N
A
=
n
e
N
A
ρ
M
.
Thus the drift speed is
v
d
=
I
n q
e
A
=
I M
n
e
N
A
ρ q
e
A
=
(4
.
1 C
/
s) (26
.
98 g)
(2) (6
.
02
×
10
23
) (2
.
7
×
10
6
g
/
m
3
)
×
1
(1
.
602
×
10
−
19
C) (4
.
5
×
10
−
6
m
2
)
=
4
.
7202
×
10
−
5
m
/
s
.
004
10.0 points
A current of 4 A flows in a copper wire 4 mm
in diameter. The density of valence electrons
in copper is roughly 9
×
10
28
m
−
3
.
Find the drift speed of these electrons.
Correct answer: 2
.
21049
×
10
−
5
m
/
s.
Explanation:
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malsam (wgm329) – hw 12 – turner – (56705)
2
Let :
I
= 4 A
,
r
=
4 mm
2
= 0
.
002 m
,
and
n
e
= 9
×
10
28
m
−
3
.
I
=
j
(
π r
2
) =
e n
e
v
d
(
π r
2
)
v
d
=
I
e n
e
π r
2
=
4 A
(1
.
6
×
10
−
19
C) (9
×
10
28
m
−
3
)
×
1
π
(0
.
002 m)
2
= 2
.
21049
×
10
−
5
m
/
s
=
7
.
95775 cm
/
hours
.

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- Spring '08
- Turner
- Physics, Current, Ratio, electrical energy, Electrical conductor, Esaved
-
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