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The resistance is therefore
2
5
33
(731
m)(1.94 10 m)
9.81 10
(2.00 10 m)(2.30 10 m)
VL
R
ia
b
ρ
π
−
−−
Ω⋅
×
== =
= × Ω
π×
×
Note that if
b = a
, then
R =
L
/
π
a
2
=
L
/
A
, where
A =
π
a
2
is the crosssectional area of
the cylinder.
35. (a) The current
i
is shown in Fig. 2630 entering the truncated cone at the left end and
leaving at the right. This is our choice of positive
x
direction. We make the assumption
that the current density
J
at each value of
x
may be found by taking the ratio
i
/
A
where
A
=
π
r
2
is the cone’s crosssection area at that particular value of
x
. The direction of
G
J
is
identical to that shown in the figure for
i
(our +
x
direction). Using Eq. 2611, we then
find an expression for the electric field at each value of
x
, and next find the potential
difference
V
by integrating the field along the
x
axis, in accordance with the ideas of
Chapter 25. Finally, the resistance of the cone is given by
R = V
/
i
. Thus,
J
i
r
E
==
π
2
where we must deduce how
r
depends on
x
in order to proceed. We note that the radius
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

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