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Unformatted text preview: N 62 . 8 cm / 1 . 0 mm = 628. Thus L = N 2 h 2 ln b a ( 4 10 7 H / m ) (628) 2 (0 . 02 m) 2 ln 12 10 = 2 . 9 10 4 H . (b) Noting that the perimeter of a square is four times its sides, the total length ` of the wire is ` = (628)4(2 . 0 cm) = 50 m, the resistance of the wire is R = (50 m)(0 . 02 / m) = 1 . 0 . Thus L = L R = 2 . 9 10 4 H 1 . 0 = 2 . 9 10 4 s ....
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 Fall '08
 SPRUNGER
 Physics, Inductance

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