Reilly (mar3978) – ohw13 – turner – (56725)
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
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before answering.
001 (part 1 of 6) 10.0 points
Consider two conductors made out of the
same material;
i.e.
, they have the same re
sistivity. The material is ohmic.
V
1
vector
E
1
I
1
ℓ
1
r
1
V
2
vector
E
2
I
2
ℓ
2
r
2
If
A
=
π r
2
,
ρ
2
=
ρ
1
,
r
2
= 2
r
1
,
ℓ
2
= 2
ℓ
1
,
and
V
2
=
V
1
, what is the ratio
E
2
E
1
, where
E
1
and
E
2
are the electric fields inside of
conductor 1 and of conductor 2, respectively?
1.
E
2
E
1
=
1
4
2.
E
2
E
1
=
1
8
3.
E
2
E
1
= 1
4.
E
2
E
1
= 2
5.
E
2
E
1
=
1
2
correct
6.
E
2
E
1
= 4
7.
E
2
E
1
= 8
Explanation:
We know that the potential across a dis
tance
l
with a constant electric field
E
is
V
=
E l
. The ratio is
E
2
E
1
=
V
2
ℓ
2
V
1
ℓ
1
=
ℓ
1
ℓ
2
=
1
2
.
002 (part 2 of 6) 10.0 points
What is the ratio
J
2
J
1
of the current densities?
1.
J
2
J
1
= 8
2.
J
2
J
1
=
1
4
3.
J
2
J
1
=
1
2
correct
4.
J
2
J
1
= 1
5.
J
2
J
1
= 4
6.
J
2
J
1
=
1
8
7.
J
2
J
1
= 2
Explanation:
vector
J
=
σ
vector
E
The material is ohmic, and
σ
1
=
σ
2
, so
J
2
J
1
=
σ
2
E
2
σ
1
E
1
=
1
2
003 (part 3 of 6) 10.0 points
What is the ratio
v
d,
2
v
d,
1
of the magnitudes of
the drift velocities?
1.
v
d,
2
v
d,
1
= 1
2.
v
d,
2
v
d,
1
=
1
4
3.
v
d,
2
v
d,
1
=
1
2
correct
4.
v
d,
2
v
d,
1
=
1
8
5.
v
d,
2
v
d,
1
= 2
6.
v
d,
2
v
d,
1
= 4
7.
v
d,
2
v
d,
1
= 8
Explanation:
vector
J
=
n qvectorv
d
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Reilly (mar3978) – ohw13 – turner – (56725)
2
Since the conductors are made out of the same
material, the density of charge carriers must
be the same (
n
1
=
n
2
), so
v
d,
2
v
d,
1
=
J
2
n
2
q
J
1
n
1
q
=
J
2
J
1
=
1
2
.
004 (part 4 of 6) 10.0 points
What is the ratio
σ
2
σ
1
of the conductivities?
1.
σ
2
σ
1
= 8
2.
σ
2
σ
1
= 1
correct
3.
σ
2
σ
1
=
1
4
4.
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 Spring '10
 TURNER,J
 SEPTA Regional Rail, Jaguar Racing, Series and parallel circuits, Highways in Slovakia, equivalent resistance, Reilly

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