Question

Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of emf.

Answer 1

Self-inductance of a coil is the property of the coil in which it opposes the change of current flowing through it.

Self induction of long solenoid of inductance L, A long solenoid is one which length is very large as compared to its cross section area. the magnetic field inside such a solenoid is constant at any point and given by

$B=\frac{{\mu }_{0}NI}{l}$

${\mu }_0=absolutemagneticpermeabilityN=totalnumberofturns$

Magnetic flux through each turn of solenoid

$\varphi =B\times areaofeachturn\varphi =\frac{{\mu }_{0}NI}{l}\times Atotalflux=flux\times totalnumberofturnsN\varphi =N(\frac{{\mu }_{0}NI}{l}\times A)..........\left(1\right)$

If L is the coefficient of inductance of solenoid

$N\varphi =LI...............\left(2\right)$

from equation 1 and 2

$LI=N(\frac{{\mu }_{0}NI}{l}\times A)L=\frac{{\mu }_0{N}^{2}A}{l}\left(3\right)$

The magnitude of emf is given by

$|e|ore=L\frac{dI}{dt}...............\left(4\right)multiplyingItobothsideseIdt=LIdIbutI=\frac{dq}{dt}Idt=dq$

Also work done (dW)= voltage X Charge(dq)

or dW = eXdq = eIdt

substituting the values in equation 4

dW = LIdt

By integrating both sides

${\int }_0^{w}dW={\int }_0^{I_0}LIdtW=\frac{1}{2}L{I_0}^2$

this work done is in increasing the current flow through inductor is stored as potential energy (U) in the magnetic field of inductor

$U=\frac{1}{2}L{I_0}^2$