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 ${\Phi }_1$
The magnetic field due to current $I_1$ in $S_1$ is given by
$B_1={\mu }_0{n}_1{I}_1$
Flux linked with each coil of second solenoid,
${\Phi }_2=B_1{A}_1$
$\therefore$ Total flux linked with $N_2$ turns,
${\Phi }_2=B_1{A}_1{N}_2$
${\Phi }_2=\mu n_1{I}_1{A}_1{N}_2...\left(i\right)$
Also ${\Phi }_2=M_{21}{I}_1...\left(ii\right)$
Comparing,
$M_{21}={\mu }_0{n}_1{N}_2{A}_1$
$M_{21}={\mu }_0{n}_1{n}_{2}l{A}_1[\because N_2=n_{2}l]$
Similarly $M_{21}={\mu }_0{n}_1{n}_{2}l{A}_1$