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terry (ect328) – oldhomework 14 – Turner – (59130)
1
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001
(part 1 oF 2) 10.0 points
A3
.
2 g wire has a density oF 13
.
5g
/
cm
3
and
a resistivity oF 6
.
2
×
10

8
Ω
·
m. The wire has
a resistance oF 33 Ω.
How long is the wire?
Correct answer: 11
.
2323 m.
Explanation:
Let :
m
=3
.
2 g = 0
.
0032 kg
,
ρ
= 13
.
/
cm
3
= 13500 kg
/
m
3
,
r
=6
.
2
×
10

8
Ω
·
m
,
and
R
= 33 Ω
.
ρ
is density, and
r
is the resistivity oF the
wire, so
m
=
ρ V
V
=
m
ρ
=
A ±
A
=
m
ρ±
Thus the resistance is
R
=
r
±
A
=
r ±
m
=
r ρ ±
2
m
±
2
=
Rm
ρ r
±
=
±
=
±
(33 Ω) (0
.
0032 kg)
(13500 kg
/
m
3
) (6
.
2
×
10

8
Ω
·
m)
=
11
.
2323 m
.
002
(part 2 oF 2) 10.0 points
The wire is made up oF atoms with va
lence 4 and molecular mass 63 kg, where
u
=1
.
6605
×
10

27
kg is the mass oF a hy
drogen atom.
What is the driFt speed oF the electrons
when there is a voltage drop oF 72
.
6 V across
the wire?
Correct answer: 6
.
40579
×
10

5
m
/
s.
Explanation:
Let :
n
con
=4
,
m
μ
= 63 kg
,
V
= 72
.
6V
,
N
A
.
02
×
10
23
,
ρ
= 13
.
/
cm
3
= 13500 kg
/
m
3
,
±
= 11
.
2323 m
,
and
q
e
.
602
×
10

19
C
,
where
m
μ
is atomic mass,
n
con
is the num
ber oF conduction electrons/atom, and
N
A
is
Avogadro’s number.
Since the current
I
is given by
I
=
n q A v
d
,
where
n
is the density oF charge carriers,
q
the
charge on an electron and
v
d
the driFt speed,
we have
v
d
=
I
nq
e
A
=
V
e
AR
.
(1)
The density oF charge carriers is Found From
knowing how many electrons there are per
atom and then the total number oF atoms in
the wire. The mass oF one atom is
m
μ
n
A
and
ρ
=
m
V
, so
n
=
n
con
ρ
²
m
μ
N
A
³
=
n
con
ρ N
A
m
μ
= (4)
(13500 kg
/
m
3
) (6
.
02
×
10
23
)
63 kg
=5
.
16
×
10
26
m

3
.
Since the volume is the crosssectional area
multiplied by the length this tells us that
=
m
ρ
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View Full Documentterry (ect328) – oldhomework 14 – Turner – (59130)
2
A
=
m
ρ±
=
63 kg
(13500 kg
/
m
3
) (11
.
2323 m)
=0
.
000415468 m
2
,
v
d
=
V
nq
e
AR
(1)
=
72
.
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 Spring '10
 Turner

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