A small square loop of wire of side $l$ is placed inside a large square loop of side $L(L>>l)$. If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to :

• A

  $\frac{L}{l}$
• B

  $\frac{l}{L}$
• C

  $\frac{L^2}{l}$
• Correct answer
  D

  $\frac{l^2}{L}$

Solution

Considering the larger loop to be made up of four rods each of length $L$, the field at the centre, i.e., at a distance $\left(\frac{L}{2}\right)$ from each rod, will be
$B=4 \times \frac{\mu_0}{4 \pi} \frac{l}{d}[\sin \alpha+\sin \beta]$ i.e., $\quad B=4 \times \frac{\mu_0}{4 \pi} \frac{I}{\left(\frac{L}{2}\right)} \times 2 \sin 45$ i.e., $\quad B_1=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{L} I$ So, the flux with smaller loop $\phi_2=B_1 S_2=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{L} l^2 I$ and hence, $M=\frac{\phi_2}{I}=2 \sqrt{2} \frac{\mu_0}{\pi} \frac{l^2}{L}$ or $\quad M \propto \frac{l^2}{L}$

Hence, the correct answer is option: 'D'.