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Chapter 29
7
(a) If the currents are parallel, the two magnetic fields are in opposite directions in the region
between the wires. Since the currents are the same, the net field is zero along the line that runs
halfway between the wires. There is no possible current for which the field does not vanish. If
there is to be a field on the bisecting line the currents must be in opposite directions. Then the
fields are in the same direction in the region between the wires.
(b) At a point halfway between the wires, the fields have the same magnitude,
¹
0
i=
2
¼r
.T
h
u
s
the net field at the midpoint has magnitude
B
=
¹
0
i=¼r
and
i
=
¼rB
¹
0
=
¼
(0
:
040 m)(300
£
10
¡
6
T)
4
¼
£
10
¡
7
T
¢
m
=
A
=30A
:
15
Sum the fields of the two straight wires and the circular arc. Look at the derivation of the
expression for the field of a long straight wire, leading to Eq. 29–6. Since the wires we are
considering are infinite in only one direction, the field of either of them is half the field of an
infinite wire. That is, the magnitude is
¹
0
i=
4
¼R
,whe
re
R
is the distance from the end of the
wire to the center of the arc. It is the radius of the arc. The fields of both wires are out of the
page at the center of the arc.
Now find an expression for the field of the arc at its center. Divide the arc into infinitesimal
segments. Each segment produces a field in the same direction. If
ds
is the length of a segment,
the magnitude of the field it produces at the arc center is (
¹
0
i=
4
¼R
2
)
ds
.I
f
μ
is the angle
subtended by the arc in radians, then
Rμ
is the length of the arc and the net field of the arc is
¹
0
iμ=
4
¼R
. For the arc of the diagram, the field is into the page. The net field at the center, due
to the wires and arc together, is
B
=
¹
0
i
4
¼R
+
¹
0
i
4
¼R
¡
¹
0
iμ
4
¼R
=
¹
0
i
4
¼R
(2
¡
μ
)
:
For this to vanish,
μ
must be exactly 2 radians.
19
Each wire produces a field with magnitude given by
B
=
¹
0
i=
2
¼r
,whe
re
r
is the distance from
the corner of the square to the center. According to the Pythagorean theorem, the diagonal of
the square has length
p
2
a
,so
r
=
a=
p
2and
B
=
¹
0
i=
p
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View Full Document fields due to the wires at the upper right and lower left corners both point toward the upper left
corner. The horizontal components cancel and the vertical components sum to
B
net
=4
¹
0
i
p
2
¼a
cos 45
±
=
2
¹
0
i
¼a
=
2(4
¼
£
10
¡
7
T
¢
m
=
A)(20 A)
¼
(0
:
20 m)
=8
:
0
£
10
¡
5
T
:
In the calculation cos 45
±
was replaced with 1
=
p
2. In unit vector notation
~
B
=(8
:
0
£
10
¡
5
T)
ˆ
j.
21
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.
 Fall '98
 Heckman
 Physics, Current

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