ch32-p009 - 9. (a) Inside we have (by Eq. 32-16) B = 0id r1...

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71 4 19 01 22 (4 10 T m/A)(2.66 10 A)(0.0200 m) 1.18 10 T 2 2 (0.0300 m) d ir B R µ π ππ −− ×⋅ × == = × . (b) Outside we have (by Eq. 32-17) 02 /2 d B = where r 2 = 0.0500 cm. Here we obtain 4 19 0 2 (4 10 T m/A)(2.66 10 A) 1.06 10 T 2 2 (0.0500 m) d i B r × = × 9. (a) Inside we have (by Eq. 32-16) 2 d B R = , where 1 0.0200 m, r = 0.0300 m, R =
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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