Physics · Electromagnetic Induction · NEET
The current does not jump to its final value I instantly; it rises gradually from 0 to I. The work you do at each instant is (back emf) × (charge) = L(di/dt) × i dt = Li di. Adding this up from i = 0 to i = I gives the integral of Li di = ½LI². The ½ comes from averaging over the rise: energy depends on the current squared, and integrating i from 0 to I gives I²/2. It is the same reason kinetic energy is ½mv², not mv².
It is stored in the magnetic field created inside and around the coil, not in the wire itself. When current flows, a magnetic field builds up; that field holds the energy. This is why an inductor with a bigger field (more turns, iron core) stores more energy for the same current.
A capacitor stores energy in an electric field between its plates: U = ½CV² (depends on voltage). An inductor stores energy in a magnetic field: U = ½LI² (depends on current). Capacitor opposes change in voltage; inductor opposes change in current. For NEET, remember: capacitor → E-field → V², inductor → B-field → I².
The magnetic field collapses and its energy is released back into the circuit. If you break the circuit suddenly, this energy can appear as a spark or a high induced voltage (that is why switching off an inductor can cause a spark). The energy is not destroyed; it converts to heat, light (spark), or is fed back.
To current squared. U = ½LI² means if you double the current, the stored energy becomes 4 times larger, not 2 times. This is a very common NEET trap — do not treat energy as directly proportional to I.
The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is:
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = ½LI², where U is energy in joules (J), L is inductance in henry (H), and I is the current in ampere (A).
The joule (J). Since U = ½LI² uses henry and ampere, the result is henry × ampere² = joule.
In a magnetic field. The current sets up a magnetic field around the coil, and that field holds the energy. A capacitor, in contrast, stores energy in an electric field.
Because energy is built up as current rises from 0 to I. Integrating Li di from 0 to I gives ½LI². The current appears squared, so energy grows very fast as current increases.
The energy per unit volume of the magnetic field is u = B²/(2μ₀). This is the field-based form of the same energy; integrating it over the volume of a solenoid also gives U = ½LI².
