Question

Determine the magnetic energy stored per unit length of a coaxial cable, represented by two coaxial cylindrical shells of radii a (inner) and b (outer), and carrying a current I. Hence derive an expression for the self inductance of the coaxial cable of length l.

Answer 1

Figure (a) shows a coaxial cable represented by two hollow, concentric cylindrical conductors along which there is electric current in opposite directions. The magnetic field between the conductors can be found by applying Ampere’s law to the dashed path of radius r{a < r < b) in figure (a). Because of the cylindrical symmetry, B is constant along the path, and

∮ \(\vec B\). \(\vec{dt}\) = B (2πr) = u₀I 

∴ B = \(\cfrac{μ_0 I}{2πr}\) ……………… (1) 

Coaxial cable represented by two cylindrical conductors

A similar application of Ampere’s law for r > b and r < a, shows that B = 0 in both the regions. Therefore, all the magnetic energy is stored between the two conductors of the cable. The energy density of the magnetic field is

u_m = \(\cfrac{B^2}{2μ_0}\)…………….. (2)

Therefore, substituting for B from Eq. (1) into Eq. (2), the magnetic energy stored in a cylindrical shell of radius r, thickness dr and length l is

dU_m = u_mdV = u_m (2πr ∙ dr ∙ l)

Thus, the total energy of the magnetic field in a length l of the cable is

Hence, the magnetic energy stored per unit length in the coaxial cable is

If L is the self inductance of the coaxial cable of length l,

U_m = \(\cfrac12\)LI²

Equating the right hand sides of Eqs. (4) and (6)