29. (a) At any instant the displacement current
i
d
in the gap between the plates equals the
conduction current
i
in the wires. Thus
i
max
=
i
d
max
= 7.60
µ
A.
(b) Since
i
d
=
ε
0
(
d
Φ
E
/
dt
),
d
dt
i
E
d
Φ
F
H
G
I
K
J
==
×
×
=×
⋅
−
−
max
.
..
0
6
12
5
760 10
859 10
A
8.85 10
F m
Vms
(c) According to Problem 3213, the displacement current is
iC
dV
dt
A
d
dV
dt
d
0
.
Now the potential difference across the capacitor is the same in magnitude as the emf of
the generator, so
V
=
m
sin
ω
t
and
dV
/
dt
=
ωε
m
cos
t
. Thus,
0m
(/
)
c
o
s
d
iA
dt
=
and
max
0
m
/.
d
d
=
This means
()
( )
( )( )
2
12
3
6
max
8.85 10
F m
0.180 m
130rad s 220 V
3.39 10 m,
7.60 10 A
d
A
d
i
εωε
−
−
−
×π
=
×
×
where
A =
π
R
2
was used.
(d) We use the AmpereMaxwell law in the form
G
G
Bd
s
I
d
⋅=
z
0
, where the path of
integration is a circle of radius
r
between the plates and parallel to them.
I
d
is the
displacement current through the area bounded by the path of integration. Since the
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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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