Energy Stored In An Inductor

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An ideal inductor of inductance 0.4 H carries a steady current of 5 A supplied by a battery. How much magnetic energy is stored in the inductor under these steady-state conditions?

RankGuru · Accepted Answer

An inductor carrying current stores energy in the magnetic field built up within and around its windings, and this stored energy equals the work the source did against the back-EMF while the current grew. The expression is $U = \tfrac{1}{2}LI^2$, which holds in steady state once the current is constant and no further EMF opposes it. Substituting $L = 0.4$ H and $I = 5$ A gives $U = 0.5 \times 0.4 \times 25 = 5$ J. The option 2 J results from forgetting to square the current. The option 10 J omits the factor of one-half. The option 1 J both omits the square and mishandles the constant. This is the energy relation derived in NCERT by integrating the instantaneous power $Li\,di/dt$ delivered to the coil. A plausibility check confirms the units H·A^2 reduce to joules, and a few joules is a sensible store for a modest inductor at several amperes; importantly, this energy returns to the circuit when the current is switched off.

Energy Stored In An Inductor

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