Question

Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current through it.

Answer 1

If the current through a coil is altered then the flux through that coil also changes, and this will induce an e.m.f. in the coil itself. This effect is known self-induction and the property of the coil is the self-inductance (L) of the coil, usually abbreviated as the inductance. The self-inductance can be defined in two ways:(a) NF=LI or

(b) Using the equation for the e.m.f. generated: E = - L(dI/dt)

The induced emf is also called back emf . Self-induction is also call inertia of electricity.

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

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