Starting from the expression for the energy $\mathit{W}=\frac{1}{2}\mathit{L}{\mathit{I}}^2$, stored in a solenoid of self-inductance L to build up the current I, obtain the expression for the magnetic energy in terms of the magnetic field $\mathrm{B}$, area $\mathrm{A}$ and length $\mathit{l}$ of the solenoid having $\mathit{n}$ number of turns per unit length. Hence show that the energy density is given by $\frac{{\mathit{B}}^2}{2{\mathit{\mu }}_0}$.