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Question

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


Solution

 Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it.
$$L = \frac{{N{\phi _B}}}{I}$$
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 as 
$$\varepsilon  =  - \frac{{d(N{\phi _B})}}{{dt}} =  - L\frac{{dI}}{{dt}}$$
 The self-induced e.m.f is also called back e.m.f. as it opposes any change in current in the circuit So, work needs to be done against back e.m.f. in establishing current This work done is stored as magnetic potential energy. The rate of doing work is given as
$$\frac{{dW}}{{dt}} = \left| \varepsilon  \right|I = LI\frac{{dI}}{{dt}}$$      (neglecting negative sign)
Thus, the total work done in establishing current from 0 to I is 
$$W = \int {dW} \int\limits_0^I {LIdI}  = \frac{1}{2}L{I^2}$$


Physics
NCERT
Standard XII

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