Question

A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L(L>>l)$. The loops are co-planar and their centers coincide. The mutual inductance of the system is proportional to

• A. $\frac{l}{L}$
• B. $\frac{l^2}{L}$
• C. $\frac{L}{l}$
• D. $\frac{L^2}{l}$

Answer 1

The correct option is B $\frac{l^2}{L}$

Magnetic field produced by a current $i$ in the larger square loop at its center,

$B\propto \frac{i}{L}\Rightarrow B=\frac{Ki}{L}$

Now, magnetic flux linked with the smaller loop,

$\varphi =BA=\frac{Ki}{L}\times l^2=\frac{Ki{l}^2}{L}$

Further, the mutual inductance,

$M=\frac{\varphi }{i}=\frac{\frac{Ki{l}^2}{L}}{i}=\frac{K{l}^2}{L}$

$\Rightarrow M\propto \frac{l^2}{L}$

Hence, option $\left(B\right)$ is the correct answer.