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.

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 LA 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 byB=μ₀NI/lμ₀=absolute magnetic permeabilityN=total number of turnsMagnetic flux through each turn of solenoidφ=B×area of each turnφ=μ₀NI/l×ATotal flux=flux×total number of turnsNφ=N(μ₀NI/l×A). (1)If L is the coefficient of inductance of solenoidNφ=LI.. (2)from equation 1 and 2LI=N(μ₀NI/l×A)L=μ₀N²A/l(3)The magnitude of emf is given by|e| or e=LdI/dt.. (4)multiplying I to both sides eIdt=LIdIbut I=dq/dtIdt=dqAlso work done (dW)= voltage X Charge(dq)or dW = eXdq = eIdtsubstituting the values in equation 4dW = LIdIBy integrating both sides∫₀ʷdW=∫₀ⁱ₀LIdI W=1/2LI₀²this work done is in increasing the current flow through inductor is stored as potential energy (U) in the magnetic field of inductorU=1/2LI₀²