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A current of 2.1 A is flowing in a coaxial cable whose outer radius is three...

A current of 2.1 A is flowing in a coaxial cable whose outer radius is three times its inner radius. What is the magnetic field energy (in J) stored in a 4.4 m length of the cable?

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Answer #1

Let us consider a length d = 4.4 m of the cable.

Outher radius of the cable be rout and the inner one be rin.

Given, rout / rin = 3.

Hence, mutual inductance of the cable = L = ( od / 2 ) ln ( rout / rin ).

Putting values : L = ( 4 x 10-7 x 4.4 / 2 ) ln 3 H

or, L = 9.67 x 10-7 H.

Current in the cable is I = 2.1 A.

Hence, the magnetic field energy stored in a 4.4 m length of the cable is : E = LI2 / 2.

Substituting values : E = 9.67 x 10-7 H x ( 2.1 A )2 / 2 = 2.13 x 10-6 J.

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