HWCh30Solutions

HWCh30Solutions - PHYSICS 133 Homework solutions Chapter 30...

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Unformatted text preview: PHYSICS 133 Homework solutions Chapter 30 2007 P30.3 (a) B = 4 I 4 a cos 4- cos 3 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 T into the paper (b) For a single circular turn with 4 l = 2 R , B = I 2 R = I 4 l = 4 2 10- 7 ( 29 10.0 ( 29 4 0.400 ( 29 = 24.7 T into the paper FIG. P30.3 P30.4 B = I 2 r = 4 10- 7 ( 29 1.00 A ( 29 2 1.00 m ( 29 = 2.00 10- 7 T P30.12 dB = I 4 d l r r 2 B = I 4 1 6 2 a a 2- 1 6 2 b b 2 B = 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 segments of the rectangle cancel....
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This note was uploaded on 06/02/2008 for the course PHY 133 taught by Professor K during the Fall '08 term at Cal Poly Pomona.

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HWCh30Solutions - PHYSICS 133 Homework solutions Chapter 30...

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