Mutual inductance of two coils is equal to the e.m.f. induced in one coil when rate of change of current through the other coil is unity.


Mutual inductance of two co-axial solenoids : Consider two long co-axial solenoid each of length 1 with number of turns ${\mathrm{N}}_1$ and ${\mathrm{N}}_2$ wound one over the other. Number of turns per unit length in solenoid, $\mathit{n}=\frac{{\mathit{N}}_1}{\mathit{l}}$. If ${\mathrm{I}}_1$ is the current flowing in primary solenoid, the magnetic field produced within this solenoi
${\mathit{B}}_1=\frac{{\mathit{\mu }}_0{\mathit{N}}_1{\mathit{I}}_1}{\mathit{l}}$
The flux linked with each turn of inner solenoid coil is
${\mathit{\varphi }}_2={\mathit{N}}_2\mathit{\varphi }$
$={\mathit{N}}_2{\mathit{B}}_1{\mathit{A}}_2$
$={\mathit{N}}_2\left(\frac{{\mathit{\mu }}_0{\mathit{N}}_1{\mathit{I}}_1}{\mathit{l}}\right){\mathit{A}}_2$
Mutual Inductance,
${\mathit{M}}_{21}=\left(\frac{{\mathit{\mu }}_0{\mathit{N}}_1{\mathit{N}}_2}{\mathit{l}}\right){\mathit{A}}_2$
If ${\mathit{n}}_1$ is number of turns per unit length of outer solenoid and ${\mathit{r}}_2$ is radius of inner solenoid, then
$\mathit{M}={\mathit{\mu }}_0{\mathit{n}}_1{\mathrm{N}}_2\mathit{\pi }{\mathit{r}}_2^2$