A small square loop of wire of side l is placed inside a large square loop of wire L (L >> l). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is :

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$\frac{2 \sqrt{2} μ_{0} ℓ^{2}}{πL}$

$\frac{μ_{0} ℓ^{2}}{2 \sqrt{2} πL}$

$\frac{μ_{0} L^{2}}{2 \sqrt{2} πℓ}$

$\frac{2 \sqrt{2} μ_{0} L^{2}}{πℓ}$

answer is C.

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Detailed Solution

Assuming current 'i' in outer loop magnetic field at centre $= 4 \times \frac{μ_{0} i}{4 π \times \frac{L}{2}} \times (2 \sin 45^{\circ}) = \frac{2 \sqrt{2} μ_{0} i}{πL}$

$\begin{array}{l} M = \frac{Flux through inner loop}{i} \\ M = \frac{2 \sqrt{2} μ_{0} ℓ^{2}}{πL} \end{array}$

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