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Chapter 30 - 30.72

72. (a) We designate the wire along y = r A =0 . 100 m wire A and the wire along y = r B =0 . 050 m wire

B . Using Eq. 30-6, we have

B net =

B A + B B

=

−

µ 0 i A

2 πr A

k

−

µ 0 i B

2 πr B

k

10 − 6 ˆ kT.

(b) This will occur for some value r B <y<r A such that

×

µ 0 i A

2 π ( r A −

y ) =

µ 0 i B

2 π ( y

−

r B )

.

0 . 081 m.

(c) We eliminate the y<r B possibility due to wire B carrying the larger current. We expect a solution

in the region y>r A where

≈

µ 0 i A

2 π ( y

µ 0 i B

2 π ( y

−

r A ) =

−

r B )

.

Solving, we ﬁnd y =7 / 40

≈

0 . 018 m.

which yields B net =52 . 0

Solving, we ﬁnd y =13 / 160