1243
Chapter 32
1. We use
6
1
0
Bn
n
=
Φ=
∑
to obtain
()
5
6
1
1Wb 2Wb 3Wb
4Wb 5Wb
3Wb .
BB
n
n
=
−Φ=
−
−
+
−
+
−
=
+
∑
2. (a)
The flux through the top is +(0.30 T)
π
r
2
where
r
= 0.020 m.
The flux through the
bottom is +0.70 mWb as given in the problem statement.
Since the
net
flux must be zero
then the flux through the sides must be negative and exactly cancel the total of the
previously mentioned fluxes.
Thus (in magnitude) the flux though the sides is 1.1 mWb.
(b) The fact that it is negative means it is inward.
3. (a) We use Gauss’ law for magnetism:
z
⋅=
G
G
BdA
0 . Now,
z
⋅=+
+
G
G
C
ΦΦΦ
12
,
where
Φ
1
is the magnetic flux through the first end mentioned,
Φ
2
is the magnetic flux
through the second end mentioned, and
Φ
C
is the magnetic flux through the curved
surface. Over the first end the magnetic field is inward, so the flux is
Φ
1
= –25.0
μ
Wb.
Over the second end the magnetic field is uniform, normal to the surface, and outward, so
the flux is
Φ
2
=
AB
=
r
2
B
, where
A
is the area of the end and
r
is the radius of the
cylinder. Its value is
Φ
2
2
35
0120
160 10
7 24 10
72 4
=×
=
+
×
=
+
−−
π
..
.
.
.
m
T
Wb
b
g
ch
Since the three fluxes must sum to zero,
ΦΦ
Φ
C
=−
−
=
−
= −
250
72 4
47 4
.
.
Wb
Thus, the magnitude is 
 47.4 Wb.
C
(b) The minus sign in
C
Φ
indicates that the flux is inward through the curved surface.
4. From Gauss’ law for magnetism, the flux through
S
1
is equal to that through
S
2
, the
portion of the
xz
plane that lies within the cylinder. Here the normal direction of
S
2
is +
y
.
Therefore,
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View Full DocumentCHAPTER 32
1244
00
12
l
e
f
t
1
()
2
2
l
n
3
.
22
rr
r
BB
r
ii
L
SS
B
x
L
d
x
B
x
L
d
x
L
d
x
rx
μ
ππ
−−
−
Φ=
=
=
=
−
∫∫
∫
5. We use the result of part (b) in Sample Problem — “Magnetic field induced by
changing electric field,”
2
,
2
R
dE
B
rR
rd
t
με
=≥
to solve for
dE
/
dt
:
( )
(
)
(
)
73
13
2
2
12
2
2
3
2 2.0 10 T 6.0 10 m
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 Fall '08
 schuller
 Magnetism, Work, Flux, Magnetic Field, C2 /N

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