Question

A short solenoid of radius a, number of turns per Unit length $n_1$. and length L is kept coaxially inside a very long solenoid of radius b, the number of turns per Unit length $n_2$. What is the mutual inductance of the system?

• A. ${\mu }_0\pi b^2{n}_1{n}_{2}L$
• B. ${\mu }_0\pi a^2{n}_1{n}_2{L}^2$
• C. ${\mu }_0\pi a^2{n}_1{n}_{2}L$
• D. ${\mu }_0\pi b^2{n}_1{n}_2{L}^2$

Answer 1

The correct option is C ${\mu }_0\pi a^2{n}_1{n}_{2}L$

Let $L$ be the length of each solenoid $S_1$ and $S_2$ having radius a and b respectively.

$n_1$ and $n_2$ be the number of turns per unit length of $S_1$ and $S_2$.

And $I$ be the current through solenoid $S_2$

Magnetic field in $S_2=B_2={\mu }_0{n}_{2}I$

Magnetic flux linked with each turn of $S_1=B_2\times$ area of each turn $=B_1\pi a^2$

Total magnetic flux linked with $S_1=B_2\pi a^2{n}_{1}L$

$\therefore {\varphi }_1=\left({\mu }_0{n}_{2}I\right)\pi a{}^2{n}_{2}L={\mu }_0{n}_1{n}_2\pi a^{2}IL$

But magnetic flux linked with $S_1$ is due to $I$

$\therefore {\varphi }_1\propto I$ or ${\varphi }_1=MI$

Where $M$ is the mutual inductance of $S_2$ and $S_1$

$\therefore MI={\mu }_0{n}_1{n}_2\pi a^{2}IL$

$\therefore M={\mu }_0{n}_1{n}_2\pi a^{2}L$