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 radial distance of 3.25 mm from the center.

Question

Quizplus · Accepted Answer

The enclosed current at a radial distance of 3.25 mm is the sum of the current flowing through the inner conductor and the current flowing through the outer conductor. The cross-sectional area of the inner conductor is $\pi (0.002)^2 = 1.26 	imes 10^{-5} 	ext{m}^2$ and the cross-sectional area of the outer conductor is $\pi [(0.0035)^2 - (0.003)^2] = 5.81 	imes 10^{-6} 	ext{m}^2$. Since the current is uniform over the cross section of the conductors, the current density can be found using $J = I/A$, giving $J_{inner} = 15/1.26	imes10^{-5} 	ext{A/m}^2$ and $J_{outer} = 15/5.81	imes10^{-6} 	ext{A/m}^2$. The enclosed current is therefore $I_{enc} = J_{inner}(\pi)(0.00325)^2 + J_{outer}(\pi)(0.00325)^2 = 7.77 	ext{A}$. Rounded to one decimal place, the answer is (C) 7.8 A.

A Coaxial Cable Consists of a Solid Inner Cylindrical Conductor