Question

Find the mutual inductance of the system of a small square loop of wire of side $l$ placed inside a large square loop of wire of side $L\left(\right.>>l\left.\right)$ in the same plane as shown in the figure.

• A. $2\sqrt{2}\frac{µ_{0}}{\pi }\frac{l^{2}}{L}$
• B. $8\sqrt{2}\frac{µ_{0}}{\pi }\frac{l^{2}}{L}$
• C. $2\sqrt{2}\frac{µ_{0}}{2 \pi }\frac{l^{2}}{L}$
• D. $2\sqrt{2}\frac{µ_{0} L^{2}}{\pi l}$

Answer 1

Answer: (A)

Let the current $I$ be flowing in the larger loop.
The larger loop is made up of four wires each of length $ \, L$ , the field at the center $i.e.,$ at a distance $\frac{L}{2}$ from each wire, will be
$B=4\times \frac{\left(\mu \right)_{0} I}{4 \pi \left(\frac{L}{2}\right)}\left(\right.sin \left(45\right)^{o} + sin ⁡ \left(45\right)^{o}\left.\right)$
$=4\times \frac{\mu _{0}}{4 \pi }\frac{2 I}{L}\frac{2}{\sqrt{2}}=2\sqrt{2}\frac{\mu _{0}}{\pi }\frac{I}{L}$
Flux linked with smaller loop
$\phi_{2}=BA_{2}=2\sqrt{2}\frac{\mu _{0}}{\pi }\frac{I}{L}\times l^{2}$
Hence, $M=\frac{\phi_{0}}{I}\Rightarrow M=2\sqrt{2}\frac{\mu _{0}}{\pi }\frac{l^{2}}{L}$