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only region in which their fields might cancel is between them. Thus, if the point at
which we are evaluating their field is r away from the wire carrying current i and is d – r
away from the wire carrying current 3.00i, then the canceling of their fields leads to μ 0i
μ (3i )
2π r 2π (d − r ) r= d 16.0 cm
= 4.0 cm.
4 (b) Doubling the currents does not change the location where the magnetic field is zero.
13. Our x axis is along the wire with the origin at the midpoint. The current flows in the
positive x direction. All segments of the wire produce magnetic fields at P1 that are out of
the page. According to the Biot-Savart law, the magnitude of the field any (infinitesimal)
segment produces at P1 is given by
dB = μ 0i sin θ
4p r 2 dx where θ (the angle between the segment and a line drawn from the segment to P1) and r
(the length of that line) are functions of x. Replacing r with
R r=R x 2 + R 2 and sin θ with x 2 + R 2 , we integrate from x = –L/2 to x = L/2. The total field is B= μ 0iR
4p ∫ dx L2 −L 2 ( 4p × 10
= -7 (x = μ 0iR 1 4p R ( x
T ⋅ m A ) ( 0.0582 A )
2 +R 2 32 2 2p ( 0.131 m ) x
2 L2 +R ) 2 1 2 −L 2 = μ 0i L 2pR L2 + 4 R 2 0.180m (0.180m) + 4(0.131m)
2 2 = 5.03 × 10−8 T. 14. We consider Eq. 29-6 but with a finite upper limit (L/2 instead of ∞). This leads to
B= μ0i L/2 2πR ( L / 2) 2 + R 2 . In terms of this expression, the problem asks us to see how large L must be (compared
with R) such that the infinite wire expression B∞ (Eq. 29-4) can be used with no more
than a 1% error. Thus we must solve
B∞ – B
B = 0.01. 1131 This is a nontrivial algebra exercise, but is nonetheless straightforward. The result is
L= 200 R
≈ 14.1R ⇒
≈ 14.1 .
201 15. (a) As discussed in Sample Problem — “Magnetic field at the center of a circular arc
of current,” the radial segments do not contribute to BP and the arc segments contribute
according to Eq. 29-9 (with angle in radians). If k designates the direction “out of the
μ ( 0.40 A )(π rad ) ˆ μ0 ( 0.80 A ) ( 2π /...
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