Question

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 :

• A. $\frac{2 \sqrt{2} \mu_0 L ^2}{\pi \ell}$
• B. $\frac{\mu_0 \ell^2}{2 \sqrt{2} \pi L }$
• C. $\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi L }$
• D. $\frac{\mu_0 L^2}{2 \sqrt{2} \pi \ell}$

Answer 1

Answer: (C)

Assuming current I in outer loop magnetic field at centre $=4 \times \frac{\mu_0 i }{4 \pi \times \frac{ L }{2}} \times\left(2 \sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_0 i }{\pi L }$

$ M=\frac{\text { Flux through inner loop }}{ i } $
$ M =\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi L }$