Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.

Explanation

When the current flows through the inductor of inductance L, an e.m.f is induced in it is given by
`epsi=-l, (dI)/(dt)....(1)`
(-ve sign shows that e.m.f opposes the passage of current) Let an infinitesmal charge dq be driven through the inductor. So the work done by the external voltage is given by `dW=epsidq (therefore v=(dW)/(dq))`
`dW=L (dI)/(dt)xxdq(therefore I=(dq)/(dt))`
dW=LI dI .....(2)
Total work done to maintain the maximum current `(I_(0))` is given by
`W=underset(0)overset(i_(0))intdW=underset(0)overset(i_(0))intLI dI=L[(I^(2))/(2)]_(0)^(i_(0))`
`W=1/2LI_(0)^(2)....(3)`
This work done is stored in the form of potential energy in the magnetic field (U)
`therefore U=1/2LI_(0)^(2)....(4)`
In case of solenoid B=`mu_(0)nI_(0) (or) I_(0)=(B)/(mu_(0)n)and L=mu_(0)n^(2)Al`
`U=1/2xxmu_(0)^(2)Al xx ((B)/(mu_(0)n))^(2)`
`U=(B^(2)Al)/(2mu_(0))....(5)`