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45.
(a) We use the result of part (a) in Sample Problem 323:
B
=
µ
0
ε
0
r
2
dE
dt
(for
r
≤
R
)
,
where
r
=0
.
80
R
and
dE
dt
=
d
dt
µ
V
d
¶
=
1
d
d
dt
³
V
0
e
−
t/τ
´
=
−
V
0
τd
e
−
t/τ
.
Here
V
0
= 100 V. Thus
B
(
t
)=
³
µ
0
ε
0
r
2
´
µ
−
V
0
e
−
t/τ
¶
=
−
µ
0
ε
0
V
0
r
2
e
−
t/τ
=
−
(
4
π
×
10
−
7
T
·
m
/
A
)
³
8
.
85
×
10
−
12
C
2
N
·
m
2
´
(100 V)(0
.
80)(16 mm)
2(12
×
10
−
3
s)(5
.
0mm)
e
−
t/
12 ms
=
−
(1
.
2
×
10
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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