The answer is C I just need someone to run me through the math..thanks . A coaxi

Question

The answer is C I just need someone to run me through the math..thanks


. A coaxial cable consists of a solid inner cylindrical conductorof radius 2 mm and an
outer cylindrical shell of inner radius 3 mm and outer radius 3.5mm. A current of 15 A
flows down the inner wire and an equal return current flows in theouter conductor. If
we assume that the currents are uniform over the cross section ofthe conductors, then
calculate the magnitude of the enclosed current for use in Ampere'sLaw at a radius of
3.25 mm.

A) 7.2 A
B) 3.8 A
C) 7.8 A
D) 11.2 A
E) 7.5 A

Explanation / Answer

Total area of shell A = (3.52 -32) = 3.25 Area of shell at 3.25 mm   A1 = (3.252 - 32) = 1.5625 A1 / A = .481     the currentin this fraction of the shell cancels the corresponding currentin the solid core Current cancelled = 15 * .481 = 7.21 amps Current remaining to be included in Ampere's Law   I= 15 - 7.21 = 7.79 amps

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