P27_016 - 16. (a) i = V /R = 23.0 V/15.0 103 = 1.53 103 A....

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16. (a) i = V/R =23 . 0V / 15 . 0 × 10 3 Ω=1 . 53 × 10 3 A . (b) The cross-sectional area is A = πr 2 = 1 4 πD 2 . Thus, the magnitude of the current density vector is J = i A = 4 i πD 2 = 4(1 . 53 × 10 3 A) π (6 . 00 × 10 3 m) 2 =5 . 41 × 10 7 A / m 2 . (c) The resistivity is
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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.

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