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Current and Resistance
77
(a) From
L
R
A
ρ
=
, we find that
()
8
8
1.70
10
m
3.40
10
m
0.500
A
LL
R
−
−
×Ω
⋅
==
=×
Ω
(
L
Inserting this expression for
A
into equation (1) gives
)
82
7
3
3.40
10
m
1.12
10
m
L
−−
×=
×
, which yields
1.82 m
L
=
(b) From equation (1),
27
1.12
10
m
4
d
A
L
π
−
×
3
, or
73
4
4 1.12
10
m
4 1.12
10
m
1.82 m
2.80
10
m
0.280 mm
d
L
ππ
−
××
=
17.13
From
L
R
A
=
, we obtain
2
4
dL
A
R
, or
( )
4
4 5.6
10
m
2.0
10
m
4
1.7
10
m
0.17 mm
0.050
L
d
R
−
⋅
×
=
Ω
17.14
( )( )
8
2
2
3
41
.7 10 m 15
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.
 Fall '10
 STAFF
 Physics, Current, Resistance

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