Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self-inductance I. is given by $\frac { 1 }{ 2 } L{ I }^{ 2 }$
Self-lnduclance is the properly bv which an opposing induced emf is product*! in a coil due to a change in current, or magnetic flux, linked with the coil.
OR
Self-inductance of a coil is numerically equal to the flux linked with the coil when the current through the coil is 1 A.
OK
Self-inductance of a coil is equal to the induced emf developed in the coil when the rate of change of current in the coil is one ampere per second.
Energy stored in an inductor:
Consider a source of emf connected to an inductor L. As the current starts growing, the opposing induced emf is given by
$[\varepsilon =-L\frac{di}{dt}]$
If the source of emf flows a current i through the inductor for a small time dt, then the amount of work done by the source, is given by
$[dW=\left| \varepsilon \right|idt]$
$=Li\frac { di }{ dt } dt$
=Lidi
Hence the total amount of work done (by the source of emf) when the current increases from its initial value (i= 0) to its final value (I) is given bv
$[W=\int\limits_{0}^{1}{Lidi=L\int\limits_{0}^{1}{idi}}=L\left[ \frac{{{i}^{2}}}{2} \right]_{0}^{1}=\frac{1}{2}L{{I}^{2}}]$
This work done gets stored in the inductor in the form of energy.
$U=\frac { 1 }{ 2 } L{ I }^{ 2 }$