Question

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.

Answer 1

Mutual inductance is numerically equal to the induced e.m.f in the secondary coil when the current in the primary coil changes by unity.

Consider two long co-axial solenoids each of length $l$. We denote the area of the inner solenoid $S_1$ by $A_1$ and the number of turns per unit length by $n_1$.

The corresponding quantities for the outer solenoid $S_2$ are $A_2$ and $n_2$ respectively.

Let $N_1$ and $N_2$ be the total number of turns of coils $S_1$ and $S_2$ respectively.

When a current $I_2$ is set up through $S_2$, it in turn sets up a magnetic flux through $S_1$. Let us denote it by ${\Phi }_2$
The magnetic field due to current $I_2$ in $S_2$ is given by
$B_2={\mu }_0{n}_2{I}_2$
Flux linked with each coil of second solenoid,
${\Phi }_2=B_2{A}_1$
$\therefore$ Total flux linked with $N_1$ turns,
${\Phi }_2=B_2{A}_1{N}_1$
${\Phi }_2=\mu n_2{I}_2{A}_1{N}_1...\left(i\right)$
Also ${\Phi }_2=M_{12}{I}_2...\left(ii\right)$

Where $m_{12}$ is the mutual inductance of solenoid 1 with respect to solenoid 2.
Comparing,
$M_{12}={\mu }_0{n}_2{N}_1{A}_1$
$M_{12}={\mu }_0{n}_1{n}_{2}l{A}_1[\because N_1=n_{1}l]$

Similarly the mutual inductance of solenoid 2 with respect to solenoid 1 is given by,

$M_{21}={\mu }_0{n}_1{n}_{2}l{A}_1=M_{12}$.