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Define self-inductance of a coil. Derive the expression for the energy required to build up a current \(I\) in a coil of self-inductance \(L\).

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The energy required to build up a current in a coil is proportional to the square of the current and the self-inductance. This energy is stored as magnetic potential energy in the coil.
Updated On: Jan 13, 2026
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Solution and Explanation

The self-inductance \(L\) of a coil quantifies its opposition to changes in current. A current \(I\) through a coil generates magnetic flux. Self-inductance is defined as the ratio of the induced electromotive force (emf) to the rate of current change. Mathematically, \( L = \frac{N \Phi}{I} \), where \(N\) is the number of turns, \(\Phi\) is the magnetic flux per turn, and \(I\) is the current. To derive the energy required to establish a current \(I\) in a coil of self-inductance \(L\), consider the work \(dW\) to increase the current by \(dI\). This work is \(dW = \mathcal{E} \cdot dI\). From Faraday's law, \( \mathcal{E} = -L \frac{dI}{dt} \). Thus, \( dW = -L \frac{dI}{dt} \cdot dI \). Integrating to find the total work \(W\) from 0 to \(I\): \[ W = \int_0^I L \, I \, dI \]. Solving this yields \( W = \frac{1}{2} L I^2 \). The energy stored in the coil's magnetic field when a current \(I\) is established is therefore: \[ \boxed{W = \frac{1}{2} L I^2} \]

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