A horizontal magnetic field B is produced across a narrow gap between the two square iron pole pieces, A closed square loop of side a. mass m and resistance R is allowed to fall with the top of the loop in the field. The loop attains a terminal velocity equal to:
Force on wire $i\overrightarrow{l}\times \overrightarrow{B}$
area of$\varphi =xa$
$\frac{dx}{dt}=-V$
$\varphi =Bax$
$e=\frac{-d\varphi }{dt}=BaV\Rightarrow i=\frac{BaV}{R}$
the magnetic forces on AD and BC are cancelled by each other. Magnetic field applies upward force on CD wires
$F=iBa=\frac{B^2{a}^{2}V}{R}$
which is equal to mg
$\frac{B^2{a}^{2}V}{R}=mg$
$V=\frac{Rmg}{B^2{a}^2}$

Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is self-induced.The self-inductance of the coil depends on its geometry and on the permeability of the medium.

Consider 2 plane circular coils close to each other having same axis. The radius of first coil is $r_1$ and turns $n_1$ and secondary coil be $r_2$ and turns be $n_2$.
When current I flows through primary coil then magnetic field at centre be 
$B=\frac{{\mu }_o{n}_{1}I}{2r_1}$
Due to magnetic field the flux linked with secondary coil be 
$\varphi =n_{2}BA$
$\varphi =n_2\frac{{\mu }_o{n}_{1}I}{2r_1}\pi r{}_2^2$
but $\varphi =MI$
$M=\frac{{\mu }_o{n}_1{n}_2\pi r{}_2^2}{2r_1}$

Example:
A long wire carries a current $5A$. Find the energy stored in the magnetic field inside a volume $1mm$${}^3$ at a distance $10cm$ from the wire.
Solution:
$U$ (energy per unit volume)$=\frac{B^2}{2{\mu }_0}$ and 

Energy $U=\frac{B^2}{2{\mu }_0}\times vol.$
$B=\frac{{\mu }_{0}I}{2\pi d}$

$U={\left(\frac{{\mu }_{0}I}{2\pi d}\right)}^2\times \frac{1}{2{\mu }_0}\times vol.$

$=\frac{{\mu }_0{I}^2}{8{\pi }^2{d}^2}\times vol.$

$=\frac{4\pi \times 10^{-7}\times 25\times 10^{-9}}{8\times 10(10^{-2})}=\frac{\pi }{8}\times 10^{-13}J$