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HWCh30Solutions

# HWCh30Solutions - PHYSICS 133 P30.3(a B Homework solutions...

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PHYSICS 133 Homework solutions Chapter 30 2007 P30.3 (a)    B = 4 μ 0 I 4 π a cos π 4 - cos 3 π 4   where     a = l 2 is the distance from any side to the center.    B = 4.00 × 10 - 6 0.200 2 2 + 2 2 = 2 2 × 10 - 5  T = 28.3  μ T into the paper (b) For a single circular turn with     4 l = 2 π R ,    B = μ 0 I 2 R = μ 0 π I 4 l = 4 π 2 × 10 - 7 ( 29 10.0 ( 29 4 0.400 (29 = 24.7  μ T into the paper FIG. P30.3 P30.4    B = μ 0 I 2 π r = 4 π × 10 - 7 ( 29 1.00 A ( 29 2 π 1.00 m ( 29 = 2.00 × 10 - 7  T P30.12    dB = μ 0 I 4 π d l × ˆ  r 2    B = μ 0 I 4 π 1 6 2 π a a 2 - 1 6 2 π b b 2 B = μ 0 I 12 1 a - 1 b  directed out of the paper P30.17 By symmetry, we note that the magnetic forces on the top and bottom segments of the rectangle cancel. The net force on the vertical segments of the rectangle is (using Equation 30.11)    F = F 1 + F 2 = μ 0 I 1 I 2 l 2 π 1 c + a - 1 c ˆ  = μ 0 I 1 I 2 l 2 π - a c c + a ( 29 ˆ

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