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
p1
0
p6
.10
7
TmA
A
m
TT
.
ch
b
g
b
g
..
(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
θ
= 90º ( =
π
/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
Bi
r
=
0
2
p
,
i
rB
==
×
=
−
2
23
9
1
0
4
16
0
6
p
p 0.080
0
7
mT
A.
bg
(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 semiinfinite wires (Eq. 297)
and a semicircular arc (Eq. 299 with
φ
=
rad). The direction of
G
B
is out of the page, as
can be checked by referring to Fig. 296(c). The magnitude of
G
B
at point
a
is therefore
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View Full DocumentCHAPTER 29
1128
7
3
000
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
iii
B
RR
R
μμ
π
μ
ππ
−
−
×⋅
⎛⎞
=+
=
+
=
+
=
×
⎜⎟
4
⎝⎠
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
00
(4
10 T m/A)(10 A)
28
.
0
1
0
T
.
2
(0.0050 m)
b
ii
B
−
−
==
=
=
×
π
(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
ˆˆ
ij
.
rs
R
=−
+
G
Since
ˆ
i
ds
ds
=
G
, then
.
ds r
Rds
×=
GG
Therefore, with
22
R
,
Eq. 293 gives
0
3
/
2
4(
)
iR ds
dB
sR
=
+
.
(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
=
.
(b) We want to find the
s
value such that
max
/10
dB
dB
=
. This is a nontrivial algebra
exercise, but is nonetheless straightforward. The result is
s
=
10
2/3
−
1
R
. If we set
2.00 cm,
R
=
then we obtain
s
= 3.82 cm.
7. (a) Recalling the
straight sections
discussion in Sample Problem — “Magnetic field at
the center of a circular arc of current,” we see that the current in the straight segments
collinear with
P
do not contribute to the field at that point. Using Eq. 299 (with
φ
=
θ
)
and the righthand rule, we find that the current in the semicircular arc of radius
b
contributes
μθ
0
4
ib
p
(out of the page) to the field at
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 Fall '08
 schuller
 Magnetism, Work, Magnetic Field, Wire

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