Question
A circular current loop of radius R is placed inside square loop of side length $L(L \gg R)$ such that they are co-planar and their centers coincide. The permeability of free space is $\mu_{0}$. The mutual inductance between circular loop and square loop is $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Flux through the small loop from the square's central field $B$; $M = \Phi/I$.]
Magnetic field due to square
$${B = 4 \times \frac{\mu_{0}I}{4\pi\frac{\text{ }L}{2}}\left\{ 2sin45^{\circ} \right\} }{= \frac{4\mu_{0}I}{\pi L} \times \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}\mu_{0}I}{\pi L} }{Q = B \times A_{\text{circle}\text{~}} }{Q = \frac{2\sqrt{2}\mu_{0}I}{\pi L} \times \pi R^{2} }{Q = \frac{2\sqrt{2}\mu_{0}R^{2}}{L}I}$$
Inductance $= \frac{2\sqrt{2}\mu_{0}R^{2}}{L}$
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