Question

Two coils of self-inductances 2mH and 8mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these two coils is:
$
  {\text{A}}{\text{. 10mH}} \\
  {\text{B}}{\text{. 6mH}} \\
  {\text{C}}{\text{. 4mH}} \\
  {\text{D}}{\text{. 16mH}} \\
$

Answer 1

Hint: To find the mutual inductance between these two coils, we find the mutual inductance caused by one coil on another and the self-inductances of the coils using the respective formulae. When effective flux of one coil is completely linked with the other, maximum inductance occurs.

Complete Step-by-Step solution:
Given data, Self-Inductances are 2mH and 8mH respectively.
Mutual Inductance due a coil A on coil B is given as ${{\text{M}}_{{\text{AB}}}} = \dfrac{{{{\text{N}}_{\text{B}}}{\phi _{\text{B}}}}}{{{{\text{i}}_{\text{A}}}}}$, where ${{\text{N}}_{\text{B}}}$ is the number of turns in coil B, ${\phi _{\text{B}}}$ is the flux in coil B and ${{\text{i}}_{\text{A}}}$ is the current in coil A.
Inductance of a coil is given as ${\text{L = }}\dfrac{{{\text{N}}\phi }}{{\text{i}}}$, where N is the number of turns in the coil, ϕ is the flux and i is the current passing through the coil.
Now let us consider two coils, Coil – 1 and Coil – 2 with number of turns ${{\text{N}}_1}$ and ${{\text{N}}_2}$ of each coil respectively, ${\phi _{\text{1}}}{\text{ and }}{\phi _2}$ be the flux of each coil and ${{\text{i}}_1}{\text{ and }}{{\text{i}}_2}$ be the currents flowing in them respectively. When placed next to each other each coil causes a mutual inductance in the other and each coil will have its own self-inductance as well.
The figure depicting both the coils is as shown below, each coil has its own self-inductance as well.
Self-Inductance is caused by the coil on itself, in an electric conductor it is the tendency to oppose the change in current flowing in it. The flow of current causes a magnetic field around the coil.
Mutual Inductance is defined as the inductance caused in one coil due to another coil which is placed near it.
Now the self-inductance of coil – 1 is given as ${{\text{L}}_1}{\text{ = }}\dfrac{{{{\text{N}}_1}{\phi _1}}}{{{{\text{i}}_1}}}$and the self-inductance of Coil – 2 is given as ${{\text{L}}_2}{\text{ = }}\dfrac{{{{\text{N}}_2}{\phi _2}}}{{{{\text{i}}_2}}}$.
And the mutual inductance caused on coil – 2 due to the presence of coil – 1 is given as ${{\text{M}}_{12}} = \dfrac{{{{\text{N}}_2}{\phi _2}}}{{{{\text{i}}_1}}}$ and the mutual inductance caused on coil – 1 due to the presence of coil – 2 is given as ${{\text{M}}_{21}} = \dfrac{{{{\text{N}}_1}{\phi _1}}}{{{{\text{i}}_2}}}$.

Now when the two coils are placed so close together that the effective flux in one coil is completely linked with the other, the electric flux of both coils, as well as the mutual inductances becomes equal and maximum inductance occurs.
$ \Rightarrow {\phi _1} = {\phi _2}$
$ \Rightarrow {{\text{M}}_{12}} = {{\text{M}}_{21}} = {\text{M}}$
Therefore the maximum inductance is given as $
   \Rightarrow {{\text{M}}_{{\text{max}}}}{\text{ = }}{{\text{M}}_{12}} \times {{\text{M}}_{21}} = \dfrac{{{{\text{N}}_1}{\phi _{\text{1}}}{{\text{N}}_2}{\phi _2}}}{{{{\text{i}}_1}{{\text{i}}_2}}} \\
   \Rightarrow {{\text{M}}_{{\text{max}}}}{\text{ = }}{{\text{M}}^2} = \dfrac{{{{\text{N}}_1}{\phi _{\text{1}}}}}{{{{\text{i}}_1}}} \times \dfrac{{{{\text{N}}_2}{\phi _2}}}{{{{\text{i}}_2}}}{\text{ - - - - - }}\left( {\because {{\text{L}}_1}{\text{ = }}\dfrac{{{{\text{N}}_1}{\phi _1}}}{{{{\text{i}}_1}}}{\text{ and }}{{\text{L}}_2}{\text{ = }}\dfrac{{{{\text{N}}_2}{\phi _2}}}{{{{\text{i}}_2}}}} \right) \\
   \Rightarrow {{\text{M}}^2} = {{\text{L}}_1} \times {{\text{L}}_2} \\
   \Rightarrow {\text{M = }}\sqrt {{{\text{L}}_1}{{\text{L}}_2}} \\
$
Given ${{\text{L}}_1} = 2{\text{mH and }}{{\text{L}}_2} = 8{\text{mH}}$
Therefore ${\text{M = }}\sqrt {2 \times 8} = \sqrt {16} = 4{\text{mH}}$.
The mutual Inductance between these two coils is 4mH. Option C is the correct answer.

Note – In order to solve this type of questions the key is to know that the effective flux in one coil is completely linked with the other means both the fluxes are equal and the effective mutual inductance becomes the product of both the individual mutual inductances (which is also the product of their self-inductances).
The units of self-inductance and mutual inductance are the same as they are constituted of similar terms, the S.I unit is: ${\text{Kg}}{\text{.}}{{\text{m}}^2}.{{\text{s}}^{ - 2}}.{{\text{A}}^{ - 2}}$ or Henry (H).