The mutual inductance between two planar concentric rings of radii $r_{1} a n d r_{2} (w i t h r_{1} >> r_{2})$ placed in air is given by

$\frac{μ_{0} π r_{2}^{2}}{2 r_{1}}$

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$\frac{μ_{0} π r_{1}^{2}}{2 r_{2}}$

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$\frac{μ_{0} π (r_{1} + r_{2})^{2}}{2 r_{1}}$

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$\frac{μ_{0} π (r_{1} + r_{2})^{2}}{2 r_{2}}$

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Solution

The correct option is A
$\frac{μ_{0} π r_{2}^{2}}{2 r_{1}}$

Magnetic field due to the larger coil at its centre is
$B = \frac{μ_{0} I}{2 r_{1}}$ where I is the current in the larger coil. Flux through the inner coil is
$ϕ = B \times π r_{2}^{2} = \frac{μ_{0} I}{2 r_{1}} \times π r_{2}^{2}$
But $ϕ = M I$ . Therefore
$M = \frac{μ_{0} π r_{2}^{2}}{2 r_{1}}$

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Similar questions

The mutual inductance between two planar concentric rings of radii $r_{1} a n d r_{2} (w i t h r_{1} >> r_{2})$ placed in air is given by

R_{1}

R_{2}

(R_{1} >> R_{2})

R_{1}

R_{2}

R_{1} >> R_{2}

M

R_{1} a n d R_{2}

R_{1} ≫ R_{2}

R_{1}

R_{2}

R_{1}

R_{2}

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