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
=
This preview has intentionally blurred sections. Sign up to view the full version.
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
©
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '98
 Heckman
 Physics, Magnetism, Magnetic Field, dipole moment, Lorb

Click to edit the document details