Homework 9 Solutions

# 00 103 a680 103 a0050 m l 2 d12 d 22 2 00240

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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. 29-9. In the final situation we use Pythagorean theorem and write 1139 cosines of some angle. A little trig (and the use of the right-hand rule) leads us to conclude that when wire 2 is at angle θ2 (shown in Fig. 29-61) then its components are B2 x = B2 sin θ 2 , B2 y = − B2 cos θ 2 . The magnitude-squared of their net field is then (by Pythagoras’ theorem) the sum of the square of their net x-component and the square of their net y-component: 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 29-13 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 x-component 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. 29-13. 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.

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