Question

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 1

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,

U_m = \(\left(\cfrac{μ_0}{4π}\right)\) I²l log_e \(\cfrac ba\)

Therefore, the required magnetic energy is

U_m = (10^-7)(1.5)² (2)log_e 5

= 4.5 × 10⁷ × 2.303 × log₁₀ 5 

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