Question

Considering the case of magnetic field produced by air-filled current carrying solenoid, show that the magnetic energy density of a magnetic field \( B \) is \( \frac{B^2}{2\mu_0} \).

Hint:

The energy density of a magnetic field in a solenoid can be derived using the energy stored in the magnetic field and the volume over which the field exists.

Answer 1

The magnetic field inside a solenoid is given by: \[ B = \mu_0 n I \] where \( n \) is the number of turns per unit length, \( I \) is the current, and \( \mu_0 \) is the permeability of free space. The energy density \( u \) of a magnetic field is given by: \[ u = \frac{U}{V} \] where \( U \) is the total energy stored in the magnetic field and \( V \) is the volume. The energy stored in a magnetic field is given by: \[ U = \frac{1}{2\mu_0} \int B^2 \, dV \] For a uniform magnetic field inside the solenoid, we have: \[ u = \frac{B^2}{2\mu_0} \] Thus, the magnetic energy density of the magnetic field is \( \frac{B^2}{2\mu_0} \).