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Unformatted text preview: 3rad ) ˆ
k = −(1.7 × 10−6 T)k
4π ( 0.050 m )
4π ( 0.040 m )
or | B | = 1.7 ×10−6 T .
(b) The direction is − k , or into the page.
(c) If the direction of i1 is reversed, we then have
B=− μ0 ( 0.40 A )(π rad ) ˆ μ0 ( 0.80 A ) ( 2π / 3rad ) ˆ
k = −(6.7 × 10−6 T)k
4π ( 0.050 m )
4π ( 0.040 m ) or | B | = 6.7 × 10−6 T.
(d) The direction is − k , or into the page.
16. Using the law of cosines and the requirement that B = 100 nT, we have
⎛ B12 + B2 − B 2 ⎞
⎟ = 144° ,
⎝ −2 B1 B2 ⎠ θ = cos −1 ⎜ where Eq. 29-10 has been used to determine B1 (168 nT) and B2 (151 nT).
17. Our x axis is along the wire with the origin at the right endpoint, and the current is in
the positive x direction. All segments of the wire produce magnetic fields at P2 that are
out of the page. According to the Biot-Savart law, the magnitude of the field any
(infinitesimal) segment produces at P2 is given by
dB = μ 0i sin θ
4p r 2 dx 1132 CHAPTER 29 where θ (the angle between the segment and a line drawn from the segment to P2) and r
x 2 + R 2 and sin θ with (the length of that line) are functions of x. Replacing r with
R r=R x 2 + R 2 , we integrate from x = –L to x = 0. The total field is B= μ 0iR
4p ∫ dx 0 −L ( 4p ×10
-7 2 +R ) 2 32 = μ 0iR 1
4p R 2 (x x
2 +R T ⋅ m A ) ( 0.693 A ) 4p ( 0.251 m ) 0 ) 2 1 2 −L = μ 0i L 4pR L2 + R 2 0.136m
(0.136m) + (0.251m)
2 2 = 1.32 × 10−7 T. 18. In the one case we have Bsmall + Bbig = 47.25 μT, and the other case gives Bsmall – Bbig
= 15.75 μT (cautionary note about our notation: Bsmall refers to the field at the center of
the small-radius arc, which is actually a bigger field than Bbig!). Dividing one of these
equations by the other and canceling out common factors (see Eq. 29-9) we obtain
(1/ rsmall ) + (1/ rbig )
(1/ rsmall ) − (1/ rbig ) = 1 + (rsmall / rbig )
1 − (rsmall / rbig ) =3 . The solution of this is straightforward: rsmall = rbig /2. Using the given fact that the
rbig = 4.00 cm, then we conclude that th...
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