Question 129 of 153

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

A$\dfrac{2\sqrt{2}{\mu }_{0}{L}^{2}}{\pi l}$B$\dfrac{{\mu }_{0}{l}^{2}}{2\sqrt{2}\pi L}$C$\dfrac{{\mu }_{0}{L}^{2}}{2\sqrt{2}\pi l}$D$\dfrac{2\sqrt{2}{\mu }_{0}{l}^{2}}{\pi L}$

Show answer and solution▾

Answer: (D) $\dfrac{2\sqrt{2}{\mu }_{0}{l}^{2}}{\pi L}$

Let a current, $I$ pass through the square loop of side, $L$. The magnetic field at the centre $O$,
$B=4\times Magneticfieldduetoeachside$
$\Rightarrow B=4\times \dfrac{{\mu }_{0}}{4\pi }\cdot \dfrac{I}{L/2}\left(\sin {45}^{^{\circ}}+\sin {45}^{^{\circ}}\right)$
$\Rightarrow B=\dfrac{2{\mu }_{0}I}{\pi L}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=\dfrac{2\sqrt{2}{\mu }_{0}I}{\pi L}$Magnetic flux linked with the small square loop,
$\phi =BA=B{l}^{2}=\dfrac{2\sqrt{2}{\mu }_{0}I{l}^{2}}{\pi L}$
Mutual inductance of the coil,
$M=\dfrac{\phi }{I}=\dfrac{2\sqrt{2}{\mu }_{0}{l}^{2}}{\pi L}$.