1127
Chapter 29
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a
distance
r
from the wire, is given by
B
0
i
2
r
.
With
r
= 20 ft = 6.10 m, we have
B
4
10
7
T
m A
100A
2
6.10m
3.3
10
6
T
3.3
T.
(b) This is about onesixth the magnitude of the Earth’s field. It will affect the compass
reading.
2. Equation 291 is maximized (with respect to angle) by setting
= 90º ( =
/
2
rad). Its
value in this case is
0
max
2
4
i ds
dB
R
.
From Fig. 2934(b), we have
12
max
60
10
T.
B
We can relate this
B
max
to our
dB
max
by
setting “
ds
” equal to 1
10
6
m and
R
= 0.025 m.
This allows us to solve for the current:
i
= 0.375 A.
Plugging this into Eq. 294 (for the infinite wire) gives
B
= 3.0
T.
3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39
T and must
be directed due south. Since
B
i
r
0
2
,
i
2
rB
0
2
0.080m
39?10
6
T
4p?10
7
T
m A
16A.
(b) The current must be from west to east to produce a field that is directed southward at
points below it.
4. The straight segment of the wire produces no magnetic field at
C
(see the
straight
sections
discussion in Sample Problem — “Magnetic field at the center of a circular arc
of current”). Also, the fields from the two semicircular loops cancel at
C
(by symmetry).
Therefore,
B
C
= 0.
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CHAPTER 29
1128
5. (a) We find the field by superposing the results of two semiinfinite wires (Eq. 297)
and a semicircular arc (Eq. 299 with
=
rad). The direction of
B
is out of the page, as
can be checked by referring to Fig. 296(c). The magnitude of
B
at point
a
is therefore
7
3
0
0
0
1
1
(4
10
T m/A)(10 A)
1
1
2
1.0
10
T
4
2
2
2(0.0050 m)
2
a
i
i
i
B
R
R
R
upon substituting
i
= 10 A and
R
= 0.0050 m.
(b) The direction of this field is out of the page, as Fig. 296(c) makes clear.
(c) The last remark in the problem statement implies that treating
b
as a point midway
between two infinite wires is a good approximation. Thus, using Eq. 294,
7
4
0
0
(4
10
T m/A)(10 A)
2
8.0 10
T.
2
(0.0050 m)
b
i
i
B
R
R
(d) This field, too, points out of the page.
6. With the “usual”
x
and
y
coordinates used in Fig. 2937, then the vector
r
pointing
from a current element to
P
is
?
i
j.
r
s
R
Since
ˆ
i
ds
ds
, then


.
ds
r
Rds
Therefore, with
2
2
r
s
R
,
Eq. 293 gives
0
2
2
3/2
4
(
)
iR ds
dB
s
R
.
(a) Clearly, considered as a function of
s
(but thinking of “
ds
” as some finitesized
constant value), the above expression is maximum for
s
= 0.
Its value in this case is
2
max
0
/ 4
dB
i ds
R
.
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 Spring '11
 Current, Magnetic Field, Wire, Bnet

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