Two circular conducting loops of radii $R_{1}$ and $R_{2}$ are laying concentrically in the same plane. If $R_{1}$ > $R_{2}$ then the mutual inductance (M) between them will be proportional to :

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

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

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

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

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Solution

The correct option is D $\frac{R_{2}^{2}}{R_{1}}$
Let a current $I_{1}$ flows through.the outer circular coil of radius $R_{1}$ .
The magnetic field at the centre of the coil is
$B_{1} = \frac{μ_{0} I_{1}}{2 R_{1}}$
As the inner coil of radius $R_{2}$ placed co-axially has small radius $(R_{2} < R_{1})$ , therefore $B_{1}$ may be taken constant over its cross-sectional area.
Hence, flux associated with the inner coil is
$ϕ_{2} = B_{1} π R_{2}^{2} = \frac{μ_{0} I_{1}}{2 R_{1}} π R_{2}^{2}$

As, Mutual Inducatance, $M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{0} π R_{2}^{2}}{2 R_{1}}$

∴ $M \propto \frac{R_{2}^{2}}{R_{1}}$

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