Question

Two concentric circular coils, one of small radius ${r_1} $ and the other of large radius ${r_2} $ , such that ${r_1} \ll {r_2} $ , are placed coaxially with centers coinciding. Obtain the mutual inductance of the arrangement.

Answer 1

Hint If we place two circular coils near each other and change the magnetic field of one coil, it will affect electromagnetic changes to the other coil. This way the two coils reciprocate mutually. According to faraday the electromagnetic changes in the second coil will try to oppose the change in the magnetic field of the first coil.

Formula Used
 ${E_M} = - \dfrac{{d\phi }}{{dt}} $
where ${E_M} $is the EMF induced in the second coil, $\phi $ is the flux of the magnetic field through the second coil.

Complete Step-by-step answer
Given, two circular coils primary coil of radius ${r_1} $ and the secondary coil of radius ${r_2} $ placed coaxially with their centers coinciding. Let the lengths of the two coils be $l $ . As both the solenoids are wound each other, all the magnetic field lines of the primary coil pass through the secondary coil.
Let the number of turns in the primary coil be ${N_1} $ and the number of turns in the secondary coil be ${N_2} $ .
If the current primary coil is ${I_1} $ , the strength of at the axial magnetic field will be:
 ${B_1} = \dfrac{{{N_1}}}{l}.{\mu _0}{I_1} $
As there are ${N_1} $ turns in the coil, we multiply ${N_1} $ to the magnetic field of a single turn to get the total magnetic field. ${\mu _0} $is the permeability of free space. For the secondary coil ${S_2} $ , this magnetic field exists only in the area occupied by the primary coil ${S_1} $, which is $\pi {r_1}^2 $. Thus, the flux of the magnetic field ${\phi _2} $ through the secondary coil is:
 ${\phi _2} = {N_2}.{B_1}.\pi {r_1}^2 $
On putting the value of ${B_1} $ in the above equation,
 ${\phi _2} = {N_2}.\left( {\dfrac{{{N_1}}}{l}.{\mu _0}{I_1}} \right)\pi {r_1}^2 $
On further simplifying the equation we get,
 ${\phi _2} = \dfrac{{{\mu _0}{N_1}{N_2}\pi {r_1}^2}}{l}{I_1} $ ............. $\left( 1 \right) $
The induced EMF is given by Faraday’s law:
 ${E_M} = - \dfrac{{d\phi }}{{dt}} $
Using equation $\left( 1 \right) $ we get,
 ${E_M} = - \dfrac{{d\dfrac{{{\mu _0}{N_1}{N_2}\pi {r_1}^2}}{l}{I_1}}}{{dt}} $
 $ \Rightarrow {E_M} = - \dfrac{{{\mu _0}{N_1}{N_2}\pi {r_1}^2}}{l}\dfrac{{d{I_1}}}{{dt}} $
 $ \Rightarrow {E_M} = - M\dfrac{{d{I_1}}}{{dt}} $
 $M $ is called mutual inductance of the coils, given by the expression $M = \dfrac{{{\mu _0}{N_1}{N_2}\pi {r_1}^2}}{l} $

Note The students may mistake the formula for the magnetic field for an inductor as $B = \mu nI $ with the one used above $B = \mu .\dfrac{N}{l}.I $ . You should keep in mind that $n $ is the number of turns per unit length and $N $ is total turns of the coil, i.e., $n = \dfrac{N}{l} $ .