Energy Density of a Magnetic Field

This section derives the general energy-density result $u_B=\frac{B^2}{2\mu_0}$ explicitly for the case of a long solenoid, where the field is confined (to a good approximation) entirely to the solenoid's interior and is uniform there. Consider a length l near the middle of a solenoid of cross-sectional area A carrying current i: the associated interior volume is $A\cdot l$, and since essentially the whole magnetic field -- and hence essentially the whole stored energy $U_B$ of that length -- lies within this volume (the field outside a long solenoid being nearly zero), and the field is uniform inside, the energy is uniformly spread through that volume. The energy density is therefore simply $u_B = \frac{U_B}{Al}$.

Substituting $U_B=\frac{1}{2}Li^2$ (from the previous section) gives $u_B=\frac{1}{2}\frac{L}{l}\cdot\frac{i^2}{A}$. Using the solenoid's inductance-per-unit-length result from Self-Inductance, $\frac{L}{l}=\mu_0 n^2 A$ (n = turns per unit length), this becomes $u_B=\frac{1}{2}\mu_0 n^2 i^2$. Finally, since the interior field of a solenoid is $B=\mu_0 n i$, this is exactly $u_B=\frac{B^2}{2\mu_0}$ -- recovering the general result, but now derived concretely rather than merely asserted, and (crucially) shown to hold regardless of HOW the field was actually produced. …