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Unformatted text preview: 64. (a) The diﬀerence between this and Sample Problem 6 is that the current in wire 2 is reversed form
what is shown in Fig. 3059(a). Thus, we replace i → −i in the expression for B2 (x) and add the
ﬁelds:
µ0 (−i)
µ0 ix
µ0 i
+
=−
B1 (x) + B2 (x) =
2π (d + x) 2π (d − x)
π (d2 − x2 )
which is equivalent to the desired result.
(b) As remarked in that Sample Problem, this expression does not apply within the wires themselves.
If we assume the wires have nearly zero thickness, then the expression applies over nearly all of the
range −0.02 < x < 0.02 (with SI units understood). To be deﬁnite about this issue, we have picked
a small wire radius (.005 m) and graphed the ﬁeld over the range −.0195 ≤ x ≤ 0.0195.
0.004 B_0.002 –0.015 –0.01–0.005 0
–0.002 –0.004 0.005 0.01 0.015
x ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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