The energy stored in an inductor is in the form of _____.

The correct answer is Option (2) → magnetic energy

When a current builds up in an inductor, work is done against the back emf. This work is stored as the energy \( U = \frac{1}{2}LI^{2} \), which resides in the magnetic field of the inductor.

    \( \displaystyle U = \frac{1}{2}LI^{2} = \frac{B^{2}}{2\mu_{0}} \times \text{volume (for a solenoid)} \)

NCERT (2026-27):
Self-Inductance – “… The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf \( (\varepsilon) \) in establishing the current. This work done is stored as magnetic potential energy. For the current \( I \) at an instant in a circuit, the rate of work done is

    \( \displaystyle \frac{\mathrm{d}W}{\mathrm{d}t} = |\varepsilon|\,I \)

If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.14),

    \( \displaystyle \frac{\mathrm{d}W}{\mathrm{d}t} = L\,I\,\frac{\mathrm{d}I}{\mathrm{d}t} \)

Total amount of work done in establishing the current \( I \) is

    \( \displaystyle W = \int \mathrm{d}W = \int_{0}^{I} L\,I\,\mathrm{d}I \)

Thus, the energy required to build up the current \( I \) is,

    \( \displaystyle W = \frac{1}{2}LI^{2} \qquad (6.17) \)

This expression reminds us of \( mv^{2}/2 \) for the (mechanical) kinetic energy of a particle of mass \( m \), and shows that \( L \) is analogous to \( m \) (i.e., \( L \) is electrical inertia and opposes growth and decay of current in the circuit).”