1
Class examples from Chapter 32
2.
(a)
The flux through the top is +(0.30 T)
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.
6.
The integral of the field along the indicated path is, by Eq. 3218 and Eq. 3219, equal
to
00
2
enclosed area
(4.0 cm)(2.0 cm)
(0.75 A)
52 nT m
total area
12 cm
d
i
.
7.
(a) Inside we have (by Eq. 3216)
2
01
/2
d
B
i r
R
, where
1
0.0200 m,
r
0.0300 m,
R
and the displacement current is given by Eq. 3238 (in SI units):
12
2
2
3
14
0
(8.85 10
C /N m )(3.00 10
V/m s)
2.66 10
A
E
d
d
i
dt
.
Thus we find
7
14
19
22
(4
10 T m/A)(2.66 10
A)(0.0200 m)
1.18 10
T
2
2 (0.0300 m)
d
ir
B
R
.
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 Fall '08
 IASHVILI
 Ratio, 10°C, 4.0 cm, 10 K, 30°C, 25 km

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