Homework 9 Solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

Ask a homework question - tutors are online