Question

Define the term ‘mutual inductance’ between two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length $l$ and radii ${r_1}$ & ${r_2}$ $({r_2} > > {r_1})$. Total number of turns in the two solenoids are ${N_1}$ & ${N_2}$ respectively.

Answer 1

Hint: Mutual inductance can be between two parallel straight wires, between two wire loops or between a parallel wire and closed loop. It can be calculated using the formula for

Complete step by step Solution: 
Mutual inductance between two coils is defined as the magnetic flux induced in one coil due to change in magnetic flux in the other coil.

When two current carrying loops are close to another, there is change in magnetic flux produced in the individual loops due to the change in current in the neighbouring loop. The induced emf due to this phenomenon is known as mutually induced emf and the phenomenon is termed as mutual inductance.

Considering two pair of long coaxial solenoids kept near at each other of same length l with different radii ${r_1}$ & ${r_2}$ $({r_2} > > {r_1})$. Given the total number of turns for each solenoid are ${N_1}$ & ${N_2}$.

Let ${I_1}$ and ${I_2}$ be the current passing in solenoid 1(outer) and 2(inner). Let the emf induced in coil 1(outer) be ${\varepsilon _1}$ and the emf induced in coil 2 (inner) be ${\varepsilon _2}$

By Faraday's law the emf induced coil is given by:

$\varepsilon = - \dfrac{{d{\phi _1}}}{{dt}}$

Where $d{\phi _1}$ is the change in magnetic flux.

Therefore, the induced emf in each coil is given by ${\varepsilon _1} = -
{N_1}\dfrac{{d{\phi _1}}}{{dt}}$ in coil 1 and $\varepsilon _{_2}^{} = - {N_2}\dfrac{{d{\phi _2}}}{{dt}}$ in coil 2.

The magnetic flux induced in coil 1(inner) due to change in magnetic flux in coil 2(outer) is given by:

${\phi _{12}} = {N_1}(\dfrac{{{\mu _0}{N_2}}}{l} \times {I_2}) \times \pi {r_1}^2$

Which gives, \[{\phi _{12}} = {N_1}{N_2}\dfrac{{{\mu _0}{I_2}}}{l}\pi {r_1}^2\]

So, the mutual inductance obtained in inner coil is given by:

${M_{12}} = \dfrac{{{\mu _0}{N_1}{N_2}}}{l}\pi {r_1}^2$

Similarly, the magnetic flux induced in coil 2(outer) due to change in magnetic flux in coil 1(inner) is given by:

${\phi _{21}} = {N_2}(\dfrac{{{\mu _0}{N_1}}}{l} \times {I_1}) \times \pi {r_2}^2$

Which gives, \[{\phi _{21}} = {N_2}{N_1}\dfrac{{{\mu _0}{I_1}}}{l}\pi {r_2}^2\]

So, the mutual inductance obtained in outer coil is given by:

${M_{21}} = \dfrac{{{\mu _0}{N_2}{N_1}}}{l}\pi {r_2}^2$

Note:The phenomenon of mutual inductance is used in transformers to change the values of alternating voltages and currents. This gives advantage for transmission and distribution in the ease and increasing and decreasing the value of alternating voltages.