A small square loop of wire of each side l is kept inside a large square loop of wire of each side ‘L’ ( $L > > l$ ). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to (current i in the outer loop)

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$\frac{l}{L}$

$\frac{L}{l}$

$\frac{l^{2}}{L}$

$\frac{L^{2}}{l}$

answer is C.

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Detailed Solution

$\begin{array}{l} B = [\frac{μ_{0} i}{4 πd} (\sin α + \sin β)] 4; B = \frac{μ_{0} i}{4 π} \frac{8 \sqrt{2}}{L} \\ ϕ = B_{1} A_{2}; ϕ = \frac{μ_{0} i}{4 π} \frac{8 \sqrt{2}}{L} l^{2} \end{array}$

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