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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

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

$ϕ == M I_{1}$

$B_{1} A_{2} = M I_{1}$

$\frac{μ_{o} I_{1}}{2 R_{1}} \times π R_{2}^{2} = M I_{1}$

$M α \frac{R_{2}^{2}}{R_{1}}$

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