Unformatted text preview: N â‰ˆ 62 . 8 cm / 1 . 0 mm = 628. Thus L = Âµ N 2 h 2 Ï€ ln Âµ b a Â¶ â‰ˆ ( 4 Ï€ Ã— 10 âˆ’ 7 H / m ) (628) 2 (0 . 02 m) 2 Ï€ ln Âµ 12 10 Â¶ = 2 . 9 Ã— 10 âˆ’ 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 â„¦ / m) = 1 . 0 â„¦ . Thus Ï„ L = L R = 2 . 9 Ã— 10 âˆ’ 4 H 1 . 0 â„¦ = 2 . 9 Ã— 10 âˆ’ 4 s ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Inductance

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