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Unformatted text preview: 2.0 × 10 5 T.
(b) The direction of the field is into the page.
32. Initially we have
Bi = μ0iφ μ0iφ
+
4π R 4π r using Eq. 299. In the final situation we use Pythagorean theorem and write 1139
cosines of some angle. A little trig (and the use of the righthand rule) leads us to
conclude that when wire 2 is at angle θ2 (shown in Fig. 2961) then its components are
B2 x = B2 sin θ 2 , B2 y = − B2 cos θ 2 .
The magnitudesquared of their net field is then (by Pythagoras’ theorem) the sum of the
square of their net xcomponent and the square of their net ycomponent:
2
B 2 = ( B2 sin θ 2 ) 2 + ( B1 − B2 cos θ 2 ) 2 = B12 + B2 − 2 B1 B2 cos θ 2 . (since sin2θ + cos2θ =1), which we could also have gotten directly by using the law of
cosines. We have
μi
μi
B1 = 0 1 = 60 nT, B2 = 0 2 = 40 nT.
2πR
2πR
With the requirement that the net field have magnitude B = 80 nT, we find
2
⎛ B12 + B2 − B 2 ⎞
−1
θ 2 = cos ⎜
⎟ = cos (−1/ 4) = 104°,
2 B1 B2
⎝
⎠
−1 where the positive value has been chosen.
35. Equation 2913 gives the magnitude of the force between the wires, and finding the xcomponent of it amounts to multiplying that magnitude by cosφ = d2
. Therefore,
d12 + d22 the xcomponent of the force per unit length is
Fx
μ0i1i2 d 2
(4π×10−7 T ⋅ m/A)(4.00 × 10−3 A)(6.80 × 10−3 A)(0.050 m)
=
=
L 2π (d12 + d 22 )
2π [(0.0240 m) 2 + (0.050 m) 2 ]
. = 8.84 ×10−11 N/m. 36. We label these wires 1 through 5, left to right, and use Eq. 2913. Then,
(a) The magnetic force on wire 1 is
−7
μ0i 2l ⎛ 1 1 1 1 ⎞ ˆ 25μ0i 2l ˆ 25 ( 4π ×10 T ⋅ m A ) ( 3.00A ) (10.0m) ˆ
++
j=
j
F1 =
⎜+
⎟ j=
2π ⎝ d 2d 3d 4d ⎠
24π d
24π ( 8.00 ×10−2 m )
2 = (4.69 ×10−4 N) ˆ
j.
(b) Similarly, for wire 2, we have 1140 CHAPTER 29 μ0i 2l ⎛ 1 1 ⎞ ˆ 5μ0i 2l ˆ
F2 =
j = (1.88 ×10−4 N) ˆ
j.
+ ⎟j=
⎜
2π ⎝ 2d 3d ⎠ 12π d
(c) F3 = 0 (because of symmetry).
ˆ
(d) F4 = − F2 = ( −1.88 ×10−4 N)j , and
ˆ
(e) F5 = − F1 = −(4....
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.
 Fall '08
 schuller
 Magnetism, Work

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