Prove the energy stored in a current inductor, per unit volume is `(B^(2))/(2mu_(0))`, where B is the magnetic field inside the inductor.

Explanation

When a current l flows through a solenoid of self inductance L, the energy stored in the space inside solenoid, is given by `U=(1)/(2)Ll^(2)" …(i)"`
Now, self inductance `L=mu_(0)n^(2)xxAxxl" …(ii)"`
where n is number of turns per unti length, A area of cross section and l is the length of solenoid.
`U=(1)/(2)(mu_(0)n^(2)Axxl)l^(2)" ...(iii)"`
Supposing the solenouid to be long enough, the magnetic field inside the solenoid is given by `B=mu_(0)nl" ...(iv)"`
Thus `U=(1)/(2)(mu_(0)^(2)n^(2)l^(2))xx(1)/(mu_(0))xxAxxl=(B^(2))/(2mu_(0))xxAxxl" ...(v)"`
where `Axxl=v` (Volume of space inside the solenoid)
Thus `(U)/("Volume V")=(B^(2))/(2mu_(0))`