Question

The self inductance of two coils $P$ and $S$ are $L_1$ and $L_2$. The coupling between them is ideal. Show that the mutual inductance between these coils is $M=\sqrt{L_1{L}_2}$.

Answer 1

Relationship between mutual inductance between two coils and their self inductance:
Let $N_1$ be the number of turns in one solenoid $P$. Its length is $l$ and its area of cross section is $A$. On passing current $I$ in it, the magnetic flux linked with it is
$I{\varphi }_1=$ magnetic field $\times$ effective area
$=\frac{{\mu }_0{N}_{1}I}{l}\times \left(N_{1}A\right)$
$\therefore$ Self inductance of solenoid $P$ is $L_1=\frac{{\varphi }_1}{I}=\frac{{\mu }_0{N}_1^{2}A}{l}$ .......(1)
Similarly, if $N_2$ is the number of turns of the other solenoid $S$ and its length is $l$, its area of cross section is $A$, then
Self inductance of the solenoid $S$ is $L_2=\frac{{\mu }_0{N}_2^{2}A}{l}$ .....(2)
Now on passing current $I$ in the primary solenoid $P$, the magnetic flux linked with the secondary solenoid $S$ is ${\varphi }_s=\frac{{\mu }_0{N}_{1}I}{l}\times \left(N_{2}A\right)$
Hence mutual inductance between the two solenoids is
$M=\frac{{\varphi }_s}{I}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$ .....(3)
From equations (1) and (2)
$\sqrt{L_1{L}_2}=\sqrt{\frac{{\mu }_0{N}_1^{2}A}{l}\times \frac{{\mu }_0{N}_2^{2}A}{l}}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$ .......(4)
Hence from equations (3) and (4)
$M=\sqrt{L_1{L}_2}$