Question

# Define self-inductance of a coil. Derive the expression for magnetic energy stored in the inductor of inductance L carrying current I.

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Hint: When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given by
$\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}$

Complete step by step solution:
It is the property of a coil by the virtue of which it possesses the growth or decay of induced current in a single coil. Mathematically, Magnetic flux changes in the coil is directly proportional to the induced current produced in the coil.
Now derivation of the expression for magnetic energy stored in the inductor of inductance L and current carrying I

When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given by
$\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}$
$= - L\dfrac{{dI}}{{dt}}$
Now the rate of doing work is given by
$\dfrac{{dW}}{{dt}} = \left| \varepsilon \right|I$
Solving the given equation, we get,
$= LI\dfrac{{dI}}{{dt}}$

Now integrating we get,
$W = \int {dW} \\ = \int\limits_0^I {LIdI} \\$

Thus, the total work done in establishing current from 0 to I is
$\dfrac{1}{2}L{I^2}$ .