CBSE Class 12 · 2012 · Electromagnetic InductionShort Answer

Define self-inductance of a coil. Show that the magnetic energy required to build up the current

I

in a coil of self-inductance

L

is given by

\frac{1}{2} L I^{2}

Solution

Definition: the self-inductance

L

of a coil is the magnetic flux linked with it per unit current,

L = ϕ / I

; equivalently it equals the emf induced in the coil when the current changes at unit rate.
While the current is growing, the back emf is

e = L d i / d t

, and the source does work against it. In time

d t

the work done to drive charge

d q = i d t

d W = e i d t = L \frac{d i}{d t} i d t = L i d i

Integrating as the current rises from

0

I

W = \int_{0}^{I} L i d i = \frac{1}{2} L I^{2}

This work is stored as energy in the magnetic field of the inductor:

U = \frac{1}{2} L I^{2}

CBSE Class 12 Board · 2012 · Subjective

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