An inductor of inductance L carries an electric current i. How much energy is stored while establishing the cu

III. Long Answer Questions · Q10

Q. An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?

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Concept understanding — Energy Stored in an Inductor

Energy Stored in an Inductor

Think about what happens when you push a heavy door shut against a spring. You do work — you apply a force through a distance — and that work gets stored as elastic potential energy in the spring. Later, the spring can release that energy to push the door back open.

An inductor does something similar, but with a magnetic field instead of a spring.

When you push current through an inductor, you are not just fighting ordinary resistance. You are building up a magnetic field around the coil. That field takes energy to create, and once it exists, the energy is stored in the field itself. If you later try to stop the current, the inductor pushes back, using that stored energy to keep the current flowing for a while.

The Intuition: Work Done Against Induced EMF

Here is the key physical picture. When you start current flowing in an inductor, the changing current produces a changing magnetic flux, which induces a back EMF. This back EMF opposes the change in current — Lenz's law. So to increase the current from zero to some final value $I$ , the battery must do work against this opposing EMF.

That work does not disappear. It is stored as magnetic potential energy in the inductor's field.

If you think of the inductor as a "magnetic spring," the current is like the spring's displacement. The more current you push, the more energy you store. And just like a spring, the inductor can give that energy back when you let the current drop.

The Precise Derivation

Let the current at some instant be $i$ , and let the self-inductance be $L$ . The instantaneous back EMF is

\mathcal{E} = -L \frac{di}{dt}

To keep the current increasing, the source must supply a voltage equal and opposite to this EMF. The instantaneous power delivered to the inductor is

P = v\, i = L \frac{di}{dt} \cdot i

The total work done to raise the current from 0 to $I$ is the integral of power over time:

U = \int_0^I P\, dt = \int_0^I L i \, \frac{di}{dt} dt = \int_0^I L i \, di

That integral is straightforward:

U = L \int_0^I i \, di = L \left[ \frac{i^2}{2} \right]_0^I = \frac{1}{2} L I^2

U = \frac{1}{2} L I^2

This is the magnetic energy stored in an inductor carrying a steady current $I$ .

What the Formula Tells You

Energy depends on $I^2$ , not on $I$ . Doubling the current quadruples the stored energy. That makes sense: a stronger current means a stronger magnetic field, and the field energy density is proportional to $B^2$ .
Energy depends on $L$ . A larger inductance means more magnetic flux per ampere, so more energy is stored for the same current.
The $\frac12$ factor is the same one that appears in the energy of a capacitor ( $\frac12 CV^2$ ) and the energy of a spring ( $\frac12 kx^2$ ). It is a signature of a linear storage system — one where the "effort" (voltage, force) is proportional to the "displacement" (current, extension).

Tip

Compare with a capacitor: $U_C = \frac12 C V^2$ . The inductor stores energy in a magnetic field; the capacitor stores energy in an electric field. Both have the same $\frac12$ factor because both are linear devices.

Where Does the Energy Live?

It is tempting to say the energy is "in the inductor," but more precisely, it is stored in the magnetic field that the current creates. For a long solenoid of cross-sectional area $A$ and length $\ell$ , the inductance is $L = \mu_0 n^2 A \ell$ (where $n$ is turns per unit length), and the magnetic field inside is $B = \mu_0 n I$ . Substituting into $U = \frac12 L I^2$ gives

U = \frac12 (\mu_0 n^2 A \ell) I^2 = \frac12 \frac{B^2}{\mu_0} (A \ell)

The volume of the field is $A\ell$ , so the energy density (energy per unit volume) in a magnetic field is

$u_B = \frac{B^2}{2\mu_0}$ …

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Step 1. Building up current i against the self-induced back-emf $\varepsilon=-L\,di/dt$ requires external work at every instant.

Step 2. Moving a small charge $dq=i\,dt$ against this opposition costs work $dW=-\varepsilon\,dq=-\varepsilon\,i\,dt$ ; substituting $\varepsilon=-L\,di/dt$ gives $dW=Li\,di$ .

Step 3. Integrating from zero current to the final current i: $W=\int_0^i Li\,di = \dfrac{1}{2}Li^2$ . …

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Q.An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?

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