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28. (a) Noting that the magnitude of the electric Feld (assumed uniform) is given by
E
=
V/d
(where
d
=5
.
0 mm), we use the result of part (a) in Sample Problem 323
B
=
µ
0
ε
0
r
2
dE
dt
=
µ
0
ε
0
r
2
d
dV
dt
(for
r
≤
R
)
.
We also use the fact that the time derivative of sin(
ωt
)(where
ω
=2
πf
π
(60)
≈
377/s in this
problem) is
ω
cos(
). Thus, we Fnd the magnetic Feld as a function of
r
(for
r
≤
R
; note that this
neglects “fringing” and related e±ects at the edges):
B
=
µ
0
ε
0
r
2
d
V
max
ω
cos(
)=
⇒
B
max
=
µ
0
ε
0
rV
max
ω
2
d
where
V
max
= 150 V. This grows with
r
until reaching its highest value at
r
=
R
= 30 mm:
B
max
¯
¯
¯
¯
r
=
R
=
µ
0
ε
0
RV
max
ω
2
d
=
(
4
π
×
10
−
7
H
/
m
)(
8
.
85
×
10
−
12
²
/
m
30
×
10
−
3
m
)
(150 V)(377
/
s)
2(5
.
0
×
10
−
3
m)
=1
.
9
×
10
−
12
T
.
(b) ²or
r
≤
0
.
03 m, we use the
B
max
=
µ
0
ε
0
rV
max
ω
2
d
expression found in part (a) (note the
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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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