1.

A coaxial cable, whose outer radius is five times its inner radius, is carrying a current of 1.5 A. What is the magnetic field energy stored in a 2 m length of the cable ?

Answer»

Data : b/a = 5, 

I = 1.5A, l = 2m, \(\cfrac{μ_0}{4π}\) = 10-7 H/m 

The total magnetic energy in a given length of a current-carrying coaxial cable,

Um\(\left(\cfrac{μ_0}{4π}\right)\) I2l loge \(\cfrac ba\)

Therefore, the required magnetic energy is

Um = (10-7)(1.5)2 (2)loge 5

= 4.5 × 107 × 2.303 × log10

= 4.5 × 10-7 × 2.303 × 0.6990 = 7.24 × 10-7 J



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