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
i
r
=
μ
0
2
p
.
With
r
= 20 ft = 6.10 m, we have
B
=
×
⋅
=
×
=
−
4
100
2
33
10
33
6
p
10
p 6.10
-7
T m A
A
m
T
T.
c
hb
g
b
g
.
.
μ
(b) This is about one-sixth the magnitude of the Earth’s field. It will affect the compass
reading.
2. Equation 29-1 is maximized (with respect to angle) by setting
θ
= 90º ( =
π
/
2
rad). Its
value in this case is
0
max
2
4
i ds
dB
R
μ
π
=
.
From Fig. 29-34(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. 29-4 (for the infinite wire) gives
B
∞
= 3.0
μ
T.
3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39
μ
T and must
be directed due south. Since
B
i
r
=
μ
0
2
p
,
i
rB
=
=
×
×
⋅
=
−
2
2
39
10
4
16
0
6
p
p 0.080
p
10
-7
μ
m
T
T m A
A.
b
gc
h
(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.
5. (a) We find the field by superposing the results of two semi-infinite wires (Eq. 29-7)
and a semicircular arc (Eq. 29-9 with
φ
=
π
rad). The direction of
G
B
is out of the page, as
can be checked by referring to Fig. 29-6(c). The magnitude of
G
B
at point
a
is therefore

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