problem25_03

problem25_03 - 25.3: a) vd = I 4.85 A = 28 19 nqA (8.5 10...

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25.3: a) ) ) m 10 05 . 2 )( 4 π )( C 10 6 . 1 )( 10 5 . 8 ( A 85 . 4 2 3 19 28 - - × × × = = nqA I v d s m 10 08 . 1 4 - × = min 110 s 6574 time travel s m 10 08 . 1 m 71 . 0 4 = = = = - × d v d b) If the diameter is now 4.12 mm, the time can be calculated using the formula above or comparing the ratio of the areas, and yields a time of 26542 s =442 min.
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This note was uploaded on 09/29/2009 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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