Question

Consider two inductance coils P and S. Let ${L_1}$ and ${L_2}$ be the self-inductance of P and S respectively. If the coupling between them is ideal, show that the mutual inductance between these coils is $M = \sqrt {{L_1}{L_2}} $.

Answer 1

Hint: Inducing an emf in a coil by changing its flux as a varying current passes through the same coil refers to self-inductance. Self-inductance of a coil depends on its dimensions. Mutual inductance, however, refers to inducing an emf in a secondary coil when a varying current passes through the primary coil and changes the magnetic flux of the secondary. We assume the two coils to have the same dimensions.

Formula Used:
1) Self-inductance of a coil is given by, $L = \dfrac{{{\mu _0}{N^2}A}}{l}$ where $N$ is the number of turns on the coil, $A$ is the area of cross-section of the coil, $l$ is the length of the coil and ${\mu _0}$ is the permeability of a vacuum.
2) Mutual inductance of a secondary coil is given by, $M = \dfrac{{{N_s}{\phi _s}}}{I}$ where ${\phi _s} = \dfrac{{{\mu _0}{N_p}IA}}{l}$ is the magnetic flux produced in the secondary coil with the number of turns ${N_s}$ when a varying current $I$ passes through the primary coil with the number of turns ${N_p}$ .

Complete step by step answer:
Step 1: List the parameters of the two coils.
We have, ${L_1}$ and ${L_2}$ as the self-inductance of the coil P and S respectively.
Let the dimensions of the two coils be the same i.e., $l$ is the length of each coil and $A$ is the cross-sectional area of each coil.
Let ${N_1}$ be the number of turns on the coil P and ${N_2}$ be the number of turns on the coil S.
Step 2: Express the self-inductance of the two coils.
The Self-inductance of a coil is given by, $L = \dfrac{{{\mu _0}{N^2}A}}{l}$ -------- (1)
where $N$ is the number of turns on the coil, $A$ is the area of cross-section of the coil, $l$ is the length of the coil and ${\mu _0}$ is the permeability of a vacuum.
Then using equation (1) we can express the self-inductance of the coil P as ${L_1} = \dfrac{{{\mu _0}N_1^2A}}{l}$ ----- (2)
as ${N_1}$ is the number of turns on P, $A$ is its area of cross-section and $l$ is its length.
Similarly, the self-inductance of the coil S, with the same dimensions as that of P but with ${N_2}$ number of turns, can be expressed as ${L_2} = \dfrac{{{\mu _0}N_2^2A}}{l}$ ---------- (3).
Step 3: Find the mutual inductance between the two coils.
Mutual inductance of a secondary coil when a varying current $I$ passes through the primary coil and produces a magnetic flux ${\phi _s}$ in the secondary coil of number of turns ${N_s}$ is given by,
$M = \dfrac{{{N_s}{\phi _s}}}{I}$ ------- (4)
 Here, let a varying current $I$ pass through the coil P which then produces a magnetic flux ${\phi _2} = \dfrac{{{\mu _0}{N_1}IA}}{l}$ in the coil S.
Then the mutual inductance between the two coils can be expressed using the equation (4) as $M = \dfrac{{{N_2}{\phi _2}}}{I}$
Substituting for ${\phi _2} = \dfrac{{{\mu _0}{N_1}IA}}{l}$ in above equation we get, $M = \dfrac{{{\mu _0}{N_1}{N_2}A}}{l}$ -------(5)
Using equations (2) and (3) we can express $\sqrt {{L_1}{L_2}} $ as $\sqrt {{L_1}{L_2}} = \sqrt {\dfrac{{{\mu _0}N_1^2A}}{l} \times \dfrac{{{\mu _0}N_2^2A}}{l}} $
This then becomes, $\sqrt {{L_1}{L_2}} = \dfrac{{{\mu _0}{N_1}{N_2}A}}{l}$ -------- (6)
Finally, comparing equations (5) and (6) we get, $M = \sqrt {{L_1}{L_2}} $

Note: When a varying current passes through a coil P a magnetic field is generated in it and changes the magnetic flux of the coil S. The magnetic flux of the coil S is the dot product of its area and the magnetic field in the coil P.