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There is a conducting ring of radius R. Another ring having current i and radius r (r << R) is kept on the axis of bigger ring such that its center lies on the axis of bigger ring at a distance x from the center of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring is

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$\frac{μ_{0} π R^{2} r^{2}}{16 {(R^{2} + x^{2})}^{3 / 2}}$

$\frac{μ_{0} π R^{2} r^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$

$\frac{μ_{0} π R^{2} r^{2}}{4 {(R^{2} + x^{2})}^{3 / 2}}$

$\frac{μ_{0} π R^{2} r^{2}}{{(R^{2} + x^{2})}^{3 / 2}}$

answer is D.

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Detailed Solution

Let current i flows in the bigger ring, then the magnetic field on its axis

$B = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$

Flux linked with the smaller ring: $ϕ = B π r^{2}$

$ϕ = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}} π r^{2} = M i$

$∴ M = \frac{μ_{0} π R^{2} r^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}}$

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