Just as energy is stored in the electrostatic field of a charged capacitor, energy is stored in the magnetic field of a current carrying inductor.

When the circuit of a coil is switched on, due to the self induction of the coil, an emf is induced. The induced emf opposes the rise of current in the coil and hence the current flowing against the induced emf does work. When the current becomes steady, no emf is induced and hence no more work is done. The total work done in bringing the current to its final steady value is stored in the magnetic field of the inductor as **magnetic potential energy**.

## Expression for Energy Stored in an Inductor

Let $\frac{dI}{dt}$ be the rate at which the current through the inductor rises. If $L$ be the self inductance of the inductor, then back emf across the inductor is given by

\[\varepsilon=L\frac{dI}{dt}\text{ __(1)}\]

Let $I$ be the value of current at that time, then the rate at which the work is done agains the back emf called electric power is given by

\[P=\varepsilon I\text{ __(2)}\]

From $\text{(1)}$ and $\text{(2)}$,

\[P=L\frac{dI}{dt}I\]

\[P=LI\frac{dI}{dt}\]

Let the steady value of the current be $I_0$. Then the total work done in bringing the current from zero to $I_0$ is given by

\[W=\int_0^{I_0}Pdt=\int_0^{I_0}LI\frac{dI}{dt}dt\]

\[=L\int_0^{I_0}IdI\]

\[=L\left[\frac{I^2}{2}\right]_0^{I_0}\]

\[=\frac{1}{2}LI_0^2\]

This gives the energy stored in the magnetic field of an inductor when the current is $I_0$.

\[\therefore\text{Magnetic potential energy }(U)=\frac{1}{2}LI_0^2\]

This equation is similar to the equation from the energy stored in the electric field in a capacitor where stored energy is $\frac{1}{2}CK^2$. In both cases, work has to be done to establish the field.

In an ideal inductor where there is no resistance, the energy is conserved without any dissipation into other forms of energy. The energy flows into the inductor when the current increases and the energy is released only when the current decreases.

When the current decreases from $I_0$ to zero, then the same amount energy $\frac{1}{2}LI_0^2$ is released and flows to the external circuit. Now, the inductor acts as a source of energy.

There is no flow of energy when the current has reached its final steady value $I_0$ i.e. rate of change of current is zero i.e. $\frac{dI}{dt}=0$.

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