1
Class Examples for Chapter 26
9.
We note that the radial width
r
= 10
m is small enough (compared to
r
= 1.20 mm)
that we can make the approximation
22
Br
rdr
Br
r r
Thus, the enclosed current is 2
Br
2
r
= 18.1
A.
Performing the integral gives the same
answer.
18.
(a)
i = V
/
R
= 23.0 V/15.0
10
–3
= 1.53
10
3
A.
(b) The crosssectional area is
A
r
D
2
1
4
2
. Thus, the magnitude of the current
density vector is
J
i
A
i
D
4
4 153
10
541
10
2
3
3
2
7
2
.
.
A
6.00
10
m
A / m .
c
h
c
h
(c) The resistivity is
3
3
2
8
(15.0 10
) (6.00 10
m)
10.6 10
m.
4(4.00 m)
RA
L
(d) The material is platinum.
24.
(a)
Since the material is the same, the resistivity
is the same, which implies (by Eq.
2611) that the electric fields (in the various rods) are directly proportional to their
currentdensities.
Thus,
J
1
:
J
2
:
J
3
are in the ratio 2.5/4/1.5
(see Fig. 2625).
Now the
currents in the rods must be the same (they are “in series”) so
J
1
A
1
=
J
3
A
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 Fall '08
 IASHVILI
 Current, Magnetic Field, 2 mM, 3.0 m, current density

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