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malsam (wgm329) – hw 12 – turner – (56705) 1
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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:

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 .