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Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self-inductance L is given by \(\frac{1}{2}\) LI2

OR 

Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.

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Self-inductance – Using formula φ = LI, if I = 1 Ampere then L = φ 

Self-inductance of the coil is equal to the magnitude of the magnetic flux linked with the coil, when a unit current flows through it.

Alternatively

Using formula |-ε| = L\(\frac{dI}{dt}\)

If \(\frac{dI}{dt}\) = 1 A/s then L = |-ε|

Self-inductance of the coil is equal to the magnitude of induced emf produced in the coil itself, when the current varies at rate 1 A/s.

Expression for magnetic energy

When a time varying current flows through the coil, back emf (–ε) produces, which opposes the growth of the current flow. It means some work needs to be done against induced emf in establishing a current I. This work done will be stored as magnetic potential energy. 

For the current I at any instant, the rate of work done is

\(\frac{dw}{dt}\) = |-ε|I

Only for inductive effect of the coil |-ε| = L \(\frac{dI}{dt}\) 

∴ \(\frac{dw}{dt}\) =   L \(\Big(\frac{dI}{dt}\Big)I\) ⇒ dW = LI dI

From work-energy thorem

DU = LIdI

U = \(\int_0^I\) LIdI = \(\frac{1}{2}\) LI2

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