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

Question

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 :

Aicharya · Accepted Answer

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

A small square loop of wire of side $l$ is placed inside a large square loop of side $L(L>>l)$. If the loops…

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