Question

Two different coils of self-inductance ${L_1}$ and ${L_1}$ are placed close to each other so that the effective flux in one coil is completely linked with the other. If M is the mutual inductance between them, then
A. $M = \dfrac{{{L_1}}}{{{L_2}}}$
B. $M = {L_1}{L_2}$
C. $M = \sqrt {{L_1}{L_2}} $
D. $M = {\left( {{L_1}{L_2}} \right)^2}$

Answer 1

Hint: In order to solve this question, we will use the concept of faraday's laws of induction and we will calculate the self-inductance of the coils and since the mutual inductance of second coil is due to effect of rate of change of current in the first coil and vice versa. Using this concept we will calculate the relation between mutual inductance and self-inductance.

Complete step-by-step answer:
As we know that the mutual inductance is given by
$M = \dfrac{{ - {e_2}}}{{d{i_1}/dt}} = \dfrac{{ - {e_1}}}{{d{i_2}/dt}}.................\left( 1 \right)$
Where M is the mutual inductance of the coil and ${e_1}\& {e_2}$ is the emf induced in the coil first and coil second.
${i_1}\& {i_2}$ is the current flowing in the coil first and second respectively.
According to faraday law of induction
${e_1} = - \dfrac{{d{\varphi _1}}}{{dt}} = - {L_1}\dfrac{{d{i_1}}}{{dt}}............\left( 2 \right)$
${e_2} = - \dfrac{{d{\varphi _2}}}{{dt}} = - {L_2}\dfrac{{d{i_2}}}{{dt}}...........\left( 3 \right)$
Where $\varphi $ flux in the coil L is self-inductance and i is the current in the coil.
Multiplying equation (2) and (3), we get
$
   \Rightarrow {e_1}{e_2} = {L_1}{L_2}\left( {\dfrac{{d{i_1}}}{{dt}}} \right)\left( {\dfrac{{d{i_2}}}{{dt}}} \right) \\
   \Rightarrow \dfrac{{{e_1}{e_2}}}{{\left( {\dfrac{{d{i_1}}}{{dt}}} \right)\left( {\dfrac{{d{i_2}}}{{dt}}} \right)}} = {L_1}{L_2} \\
$
 From equation (1) $M = \dfrac{{ - {e_2}}}{{d{i_1}/dt}} = \dfrac{{ - {e_1}}}{{d{i_2}/dt}}$ , substituting this value in the above equation, we get

$
   \Rightarrow \dfrac{{{e_1}{e_2}}}{{\left( {\dfrac{{d{i_1}}}{{dt}}} \right)\left( {\dfrac{{d{i_2}}}{{dt}}} \right)}} = {L_1}{L_2} \\
   \Rightarrow M \times M = {L_1}{L_2} \\
   \Rightarrow {M^2} = {L_1}{L_2} \\
   \Rightarrow M = \sqrt {{L_1}{L_2}} \\
$

Therefore the relation between self-inductance and mutual inductance of a two coils is $M = \sqrt {{L_1}{L_2}} $
Hence, the correct option is C.

Note: As we know that when a wire moves in a magnetic field an emf is induced in the wire. When this emf is induced in the same circuit where the current changes that effect is called Self-induction. However, if the emf is induced into an adjacent coil within the same magnetic field, the effect is said to be mutual induction.