Question 128 of 153

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius $R$ is placed inside a large square loop of wire of side $L(L≫R)$. The loops are coplanar and their centres coincide :

A$M=\dfrac{\sqrt{2}{\mu }_{0}{R}^{2}}{L}$B$M=\dfrac{2\sqrt{2}{\mu }_{0}R}{{L}^{2}}$C$M=\dfrac{2\sqrt{2}{\mu }_{0}{R}^{2}}{L}$D$M=\dfrac{\sqrt{2}{\mu }_{0}R}{{L}^{2}}$

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Answer: (C) $M=\dfrac{2\sqrt{2}{\mu }_{0}{R}^{2}}{L}$

The magnetic field due to the rectangular loop at its centre is:
$B=4\times \dfrac{{\mu }_{0}i}{4\pi \left(\dfrac{L}{2}\right)}\left[\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right]=\dfrac{2\sqrt{2}{\mu }_{0}i}{\pi L}$
Therefore, the flux through the circular loop $\phi =\pi {R}^{2}\times B=\dfrac{2\sqrt{2}{\mu }_{0}{R}^{2}i}{L}$
If M is the mutual inductance, then
$M\dfrac{di}{dt}=\dfrac{d\phi }{dt}=\dfrac{2\sqrt{2}{\mu }_{0}{R}^{2}}{L}\dfrac{di}{dt}$
$\Rightarrow M=\dfrac{2\sqrt{2}{\mu }_{0}{R}^{2}}{L}$