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which yields
Φ
B
/
L
= 1.3
×
10
–5
T·m or 1.3
×
10
–5
Wb/m.
(b) The flux (per meter) existing within the regions of space occupied by one or the other
wires was computed above to be 0.23
×
10
–5
T·m. Thus,
5
5
0.23 10 T m
0.17
17% .
1.3 10 T m
−
−
×⋅
==
(c) What was described in part (a) as a symmetry plane at
x
=
A
/2
is now (in the case of
parallel currents) a plane of vanishing field (the fields subtract from each other in the
region between them, as the righthand rule shows). The flux in the
02
<
<
x
A
/
region is
now of opposite sign of the flux in the
AA
<
<
x
region which causes the total flux (or,
in this case, flux per meter) to be zero.
27. (a) We refer to the (very large) wire length as
L
and seek to compute the flux per
meter:
Φ
B
/
L
. Using the righthand rule discussed in Chapter 29, we see that the net field
in the region between the axes of antiparallel currents is the addition of the magnitudes
of their individual fields, as given by Eq. 2917 and Eq. 2920. There is an evident
reflection symmetry in the problem, where the plane of symmetry is midway between the
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 Spring '08
 Any
 Physics, Current

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