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27. (a) Let ∆
T
be the change in temperature and
κ
be the coeﬃcient of linear expansion for copper. Then
∆
L
=
κL
∆
T
and
∆
L
L
=
κ
∆
T
=(1
.
7
×
10
−
5
/
K)(1
.
0
◦
C) = 1
.
7
×
10
−
5
.
This is equivalent to 0
.
0017%. Since a change in Celsius is equivalent to a change on the Kelvin
temperature scale, the value of
κ
used in this calculation is not inconsistent with the other units
involved. Incorporating a factor of 2 for the twodimensional nature of
A
, the fractional change in
area is
∆
A
A
=2
κ
∆
T
=2(1
.
7
×
10
−
5
/
K)(1
.
0
◦
C) = 3
.
4
×
10
−
5
which is 0
.
0034%. For small changes in the resistivity
ρ
,length
L
,andarea
A
of a wire, the change
in the resistance is given by
∆
R
=
∂R
∂ρ
∆
ρ
+
∂R
∂L
∆
L
+
∂R
∂A
∆
A.
Since
R
=
ρL/A
,
∂R/∂ρ
=
L/A
=
R/ρ
,
∂R/∂L
=
ρ/A
=
R/L
,and
∂R/∂A
=
−
ρL/A
2
=
−
R/A
.Fu
r
the
rmo
re
,∆
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 Fall '08
 SPRUNGER
 Physics

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