Recall that electric power is $P_{\text{elec}}=I\text{Δ}V$. As current passes through an inductor, for which $\text{Δ}{V}_{\text{L}}=-L\left(dI\text{/}dt\right)$, the electric power is

$P_{\text{elec}}$ is negative because a circuit with an increasing current is losing electric energy. That energy is being transferred to the inductor, which is storing energy $U_{\text{L}}$ at the rate

where we’ve noted that power is the rate of change of energy.

We can find the total energy stored in an inductor by integrating Equation 30.38 from $I=0$, where $U_{\text{L}}=0$, to a final current I. Doing so gives

The potential energy stored in an inductor depends on the square of the current through it. Notice the analogy with the energy $U_{\text{C}}=\frac{1}{2}C{\left(\text{Δ}V\right)}^2$ stored in a capacitor.

In working with circuits we say that the energy is “stored in the inductor.” Strictly speaking, the energy is stored in the inductor’s magnetic field, analogous to how a capacitor stores energy in the electric field. We can use the inductance of a solenoid, Equation 30.33, to relate the inductor’s energy to the magnetic field strength:

We made the last rearrangement in Equation 30.40 because ${\mu }_{0}NI\text{/}l$ is the magnetic field inside the solenoid. Thus

But Al is the volume inside the solenoid. Dividing by Al, the magnetic field energy density inside the solenoid (energy per ${\text{m}}^3)$ is

We’ve derived this expression for energy density based on the properties of a solenoid, but it turns out to be the correct expression for the energy density anywhere there’s a magnetic field. Compare this to the energy density of an electric field $u_{\text{E}}=\frac{1}{2}{∊}_0{E}^2$ that we found in Chapter 26.