Question

Obtain an expression for the energy density of a magnetic field.

Answer 1

Consider a short length ¡ near the middle of a long, tightly wound solenoid, of cross-sectional area A, number of turns per unit length n and carrying a steady current I. For such a solenoid, the magnetic field is approximately uniform everywhere inside and zero outside. So, the magnetic energy U_m stored by this length l of the solenoid lies entirely within the volume Al.

The magnetic field inside the solenoid is B = µ₀nI …………… (1)

and if L be the inductance of length l of the solenoid,

L = µ₀n² lA …………… (2)

The stored magnetic energy,

U_m = \(\cfrac12\) LI² …………. (3)

and the energy density of the magnetic field (energy per unit volume) is

U_m = \(\cfrac{U_m}{AI}\) = \(\cfrac12\)\(\cfrac{LI^2}{Al}\)......(4)

Substituting for L from Eq. (2)

Equation (6) gives the magnetic energy density in vacuum at any point in a magnetic field of induction B, irrespective of how the field is produced.

[Note : Compare Eq.(6) with the electric energy density in vacuum at any point in an electric field of intensity

e, u_e = \(\cfrac12\) ε₀e² . Both u and u_m are proportional to the square of the appropriate field magnitude.]