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Chapter 32
3
(a) Use Gauss’ law for magnetism:
H
~
B
¢
d
~
A
=0
. W
r
i
te
H
~
B
¢
d
~
A
=
©
1
+
©
2
+
©
C
,whe
re
©
1
is the magnetic flux through the first end mentioned,
©
2
is the magnetic flux through the second
end mentioned, and
©
C
is the magnetic flux through the curved surface. Over the first end, the
magnetic field is inward, so the flux is
©
1
=
¡
25
:
0
¹
Wb. Over the second end, the magnetic
field is uniform, normal to the surface, and outward, so the flux is
©
2
=
AB
=
¼r
2
B
,whe
re
A
is the area of the end and
r
is the radius of the cylinder. Its value is
©
2
=
¼
(0
:
120 m)
2
(1
:
60
£
10
¡
3
T) = +7
:
24
£
10
¡
5
Wb = +72
:
4
¹
Wb
:
Since the three fluxes must sum to zero,
©
C
=
¡
©
1
¡
©
2
=25
:
0
¹
Wb
¡
72
:
4
¹
Wb =
¡
47
:
4
¹
Wb
:
(b) The minus sign indicates that the flux is inward through the curved surface.
5
Consider a circle of radius
r
(= 6
:
0 mm), between the plates and with its center on the axis of
the capacitor. The current through this circle is zero, so the AmpereMaxwell law becomes
I
~
B
¢
d~
s
=
¹
0
²
0
d
©
E
dt
;
where
~
B
is the magnetic field at points on the circle and
©
E
is the electric flux through the
circle. The magnetic field is tangent to the circle at all points on it, so
H
~
B
¢
d~s
=2
¼rB
.T
h
e
electric flux through the circle is
©
E
=
¼R
2
E
, where
R
(= 3
:
0 mm) is the radius of a capacitor
plate. When these substitutions are made, the AmpereMaxwell law becomes
2
¼rB
=
¹
0
²
0
¼R
2
dE
dt
:
Thus
dE
dt
=
2
rB
¹
0
²
0
R
2
=
2(6
:
0
£
10
¡
3
m)(2
:
0
£
10
¡
7
T)
(4
¼
£
10
¡
7
H
=
m)(8
:
85
£
10
¡
12
Fm)(3
:
0
£
10
¡
3
m)
2
=2
:
4
£
10
13
V
=
m
¢
s
:
13
The displacement current is given by
i
d
=
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View Full Documentwhere
A
is the area of a plate and
E
is the magnitude of the electric field between the plates.
The field between the plates is uniform, so
E
=
V=d
,whe
re
V
is the potential difference across
the plates and
d
is the plate separation. Thus
i
d
=
²
0
A
d
dV
dt
:
Now
²
0
A=d
is the capacitance
C
of a parallelplate capacitor without a dielectric, so
i
d
=
C
dV
dt
:
21
(a) For a parallelplate capacitor, the charge
q
on the positive plate is given by
q
=(
²
0
A=d
)
V
,
where
A
is the plate area,
d
is the plate separation, and
V
is the potential difference between the
plates. In terms of the electric field
E
between the plates,
V
=
Ed
,so
q
=
²
0
AE
=
²
0
©
E
,whe
re
©
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 Fall '98
 Heckman
 Physics, Magnetism

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