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Question
Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to ‘I’.
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Solution
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
`in=-(d(Nphi_E))/dt=-L(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
`(dW)/(dt)=|in|I=LI(dI)/dt " (neglecting negative sign)"`
Thus, the total work done in establishing current from 0 to I is
`w=intdw=int_0^1LIdI=1/2LI^2`
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