Homework 9 Solutions

100 m wire a and the wire along y rb 0050 m wire b

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Unformatted text preview: × 10−6 T given in this case) we can solve for i1. We obtain i1 = (45/16π) Α ≈ 0.90 A. (b) With loop 2 at y = 0.06 m (see Fig. 29-73(b)) we are able to determine i2 from μ0i1 R 2 2( R 2 + L2 )3/ 2 = μ0i2 R 2 2( R 2 + y 2 )3/ 2 . We obtain i2 = (117 13 /50π) Α ≈ 2.7 A. 61. (a) We denote the large loop and small coil with subscripts 1 and 2, respectively. B1 = μ 0i1 2 R1 c4π × 10 = −7 b hb g = 7.9 × 10 T ⋅ m A 15 A g 2 012 m . −5 T. (b) The torque has magnitude equal to τ = | μ2 × B1 | = μ2 B1 sin 90° = N 2i2 A2 B1 = π N 2i2 r22 B1 = π ( 50 )(1.3A ) ( 0.82 ×10−2 m ) ( 7.9 ×10−5 T ) 2 = 1.1×10−6 N ⋅ m. 62. (a) To find the magnitude of the field, we use Eq. 29-9 for each semicircle (φ = π rad), and use superposition to obtain the result: B= ⎞ μ0iπ μ0iπ μ0i ⎛ 1 1 ⎞ (4π ×10−7 T ⋅ m/A) ( 0.0562A ) ⎛ 1 1 + = + ⎜ ⎟ ⎜ + ⎟= 4π a 4π b 4 ⎝a b⎠ 4 ⎝ 0.0572m 0.0936m ⎠ = 4.97 ×10−7 T. (b) By the right-hand rule, B points into the paper at P (see Fig. 29-6(c)). (c) The enclosed area is A = (π a 2 + π b 2 ) / 2, which means the magnetic dipole moment has magnitude 1148 CHAPTER 29 |μ| = πi 2 (a 2 + b 2 ) = π (0.0562A) 2 [(0.0572m) 2 + (0.0936m) 2 ] = 1.06 ×10−3 A ⋅ m 2 . (d) The direction of μ is the same as the B found in part (a): into the paper. 63. By imagining that each of the segments bg and cf (which are shown in the figure as having no current) actually has a pair of currents, where both currents are of the same magnitude (i) but opposite direction (so that the pair effectively cancels in the final sum), one can justify the superposition. (a) The dipole moment of path abcdefgha is ( ) μ = μbc f gb + μ abgha + μ cde f c = ( ia 2 ) ˆ − ˆ + ˆ = ia 2ˆ jii j 2 j j = ( 6.0 A )( 0.10 m ) ˆ = (6.0 × 10−2 A ⋅ m 2 ) ˆ . (b) Since both points are far from the cube we can use the dipole approximation. For (x, y, z) = (0, 5.0 m, 0), μ0 μ (1.26 ×10−6 T ⋅ m/A)(6.0 ×10−2 m 2 ⋅ A) ˆ j = = (9.6 × 10−11 T ) ˆ . B (0, 5.0 m, 0) ≈ j 3 3 2π y 2π (5.0 m) 64. (a) 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 page" then μ i (7π / 4 rad) ˆ μ0i (7π / 4 rad) ˆ BP = 0 k− k 4π (4.00 m) 4π (2.00 m) → ^ → where i = 0.200 A. This yields B = −2.75 × 10−8 k T, or | B | = 2.75 × 10−8 T. ˆ (b) The direction is −k , or into the page. 65. Using Eq. 29-20, ⎛ μi ⎞ | B | = ⎜ 0 2 ⎟r , ⎝ 2π R ⎠ we find that r = 0.00128 m gives the desired field value. 66. (a) We designate the wire along y = rA = 0.100 m wire A and the wire along y = rB = 0.050 m wire B. Using Eq. 29-4, we have 1150 CHAPTER 29 68. We take the current (i = 50 A) to flow in the +x direction, and the electron to be at a point P, which is r = 0.050 m above the wire (where “up” is the +y direction). Thus, the field produced by the current points in the +z direc...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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