Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other.

OR

Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of emf.

Mutual inductance, between a pair of coils, equals the magnetic flux, linked with one of them due to a change of unit current flowing in the other.

Expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other:

Let r1 and r2 be radii of inner (let it be primary P) and outer (let it be secondary S) co-axial solenoids respectively, and n1 and n2 be number of turns per unit length of the two solenoids.

Let N1 and N2 be total number of turns in two solenoids and each of length l. Let the secondary solenoid carry current I2.This current sets up magnetic fluxϕ1 through the primary (inner) solenoid.
The total flux linkages with the primary solenoid are given by N₁ϕ₁ = M₁₂I₂

where, M₁₂ is the mutual inductance of the primary (inner) solenoid with respect to the secondary (outer) solenoid. Magnetic field at the centre of the secondary solenoid due to a current I₂ is given by
${\mathrm{B}}_2={\mathrm{\mu }}_0{\mathrm{n}}_2{\mathrm{I}}_2$
The total flux linkages with the primary solenoid are given by
$\begin{array}{l}{\mathrm{N}}_1{\mathrm{\varphi }}_1=\left({\mathrm{n}}_{1}1\right){\mathrm{B}}_2{\mathrm{I}}_1\\ {\mathrm{N}}_1{\mathrm{\varphi }}_1=\left(\mathrm{\mu }0{\mathrm{\pi n}}_1{\mathrm{n}}_2{\mathrm{lr}}_1^2\right){\mathrm{I}}_2\\ \left.{\mathrm{M}}_{12}={\mathrm{I}}_2\mathrm{N}={\mathrm{\varphi }}_1=\mathrm{\mu }0{\mathrm{\pi n}}_1{\mathrm{n}}_2{\mathrm{lr}}_1^2\right){\mathrm{I}}_2\end{array}$
Similarly
${\mathrm{M}}_{21}={\mathrm{I}}_1{\mathrm{N}}_2{\mathrm{\varphi }}_2={\mathrm{\mu }}_0{\mathrm{\pi n}}_1{\mathrm{n}}_2\left({\mathrm{r}}_1^2\right){\mathrm{I}}_2$

OR

Self-inductance of a coil is equal to the magnitude of induced emf produced in the coil when rate of change of current through the coil is unity.

Self-induction of long solenoid of inductance L, A long solenoid is one which length is very large as compared to its cross-section area. the magnetic field inside such a solenoid is
$\mathrm{B}=\frac{\mathrm{\mu }0\mathrm{NI}}{1}$
constant at any point and given by
μ0 = absolute magnetic permeability
N = total number of turns
Magnetic flux through each turn of solenoid
$$\begin{gathered} \mathrm{\varphi }=\mathrm{B}\times \mathrm{area} \mathrm{of} \mathrm{each} \mathrm{turn} \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0\mathrm{NI}}{1}\times \mathrm{A} \end{gathered}$$
Total flux = flux × total number of turns 
If L is the coefficient of inductance of solenoid
$\mathrm{N\varphi }=\mathrm{LI}$ ........(2)
from equation 1 and 2
$\begin{array}{l}\mathrm{LI}=\mathrm{N}\left(\frac{{\mathrm{\mu }}_0\mathrm{NI}}{1}\times \mathrm{A}\right)\\ \mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{A}}{1} .....\left(3\right)\end{array}$
$\text{ le| or }\mathrm{e}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}} .........\left(4\right)$
multiplying I to both sides
$\mathrm{eIdt}=\mathrm{LIdI}$
but $\mathrm{I}=\frac{\mathrm{dq}}{\mathrm{dt}}$
The magnitude of emf is given by
Also work done (dW) = voltage X Charge(dq)
or dW = eXdq = eIdt
Substituting the values in equation 4
dW = LIdt
By integrating both sides
$\begin{array}{l}{\int }_0^{\mathrm{W}} \mathrm{dW}={\int }_0^{{\mathrm{I}}_0} \mathrm{LIdt}\\ \mathrm{W}=\frac{1}{2}{\mathrm{LI}}_0^2\end{array}$
this work done is in increasing the current flow through inductor is stored as potential energy
$\mathrm{U}=\frac{1}{2}{\mathrm{LI}}_0^2$
(U) in the magnetic field of inductor