Two conducting circular loops of radii and are placed in the same plane with their centres coinciding. If , the mutual inductance between them will be directly proportional to [2021]

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Solution

Q.
Two conducting circular loops of radii $R_{1}$ and $R_{2}$ are placed in the same plane with their centres coinciding. If $R_{1} ≫ R_{2}$ , the mutual inductance $M$ between them will be directly proportional to [2021]

1 $\frac{R_{2}^{2}}{R_{1}}$

2 $\frac{R_{1}}{R_{2}}$

3 $\frac{R_{2}}{R_{1}}$

4 $\frac{R_{1}^{2}}{R_{2}}$

Ans.
(1)

Let a current $I_{1}$ flows through the outer circular coil of radius $R_{1} .$

The magnitude field at the centre of the coil is $B_{1} = \frac{μ_{0} I_{1}}{2 R_{1}}$

As the inner coil of radius $R_{2}$ is placed coaxially, therefore, $B_{1}$ may be taken constant over its cross-sectional area.
Hence, flux associated with the inner coil is

$ϕ_{2} = B_{1} π R_{2}^{2} = \frac{μ_{0} I_{1}}{2 R_{1}} π R_{2}^{2} \Rightarrow$ As $M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{0} π R_{2}^{2}}{2 R_{1}} ∴ M \propto \frac{R_{2}^{2}}{R_{1}}$

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