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

Question

Quizplus · Accepted Answer

We can calculate the magnitude of the enclosed current by adding the currents flowing through both the inner and outer conductors.
The cross-sectional area at a radius of 3.25 mm is:
π(0.00325)^2 - π(0.002)^2 = 0.0000296 m^2
The current flowing through the inner conductor is 15 A and the current flowing through the outer conductor is also 15 A. Therefore, the total enclosed current at a radius of 3.25 mm is:
15 A + 15 A = 30 A
So, the magnitude of the enclosed current for use in Ampere's Law at a radius of 3.25 mm is 30 A.

A Coaxial Cable Consists of a Solid Inner Cylindrical Conductor