Question

A small square loop of wire of side $l$ is placed inside a large square loop of side $L(l<<L).$ The loops are co-planar and their centres coincide. The mutual induction 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 $\frac{l^2}{L}$

Let, current $I_1$ flows through the bigger loop.


The net magnetic field due to $I_1$ at the common centre of the loop's is,

$(B_1{)}_{net}=4\times B_{oneside}$

$=4\times \frac{{\mu }_0{I}_1}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

Here, ${\theta }_1={\theta }_2={45}^{\circ };d=\frac{L}{2}$

$=4\times \frac{{\mu }_0{I}_1}{4\pi \left(\frac{L}{2}\right)}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$

$B_1=\frac{4{\mu }_0{I}_1}{\sqrt{2}\pi L}⨂$

The net magnetic flux linked with smaller loop is,

${\varphi }_2=B_1{A}_2=B_1{l}^2[\because l<<L]$

$=\frac{4{\mu }_0{I}_1{l}^2}{\sqrt{2}\pi L}$

From the principle of mutual inductance,

$M=\frac{{\varphi }_2}{I_1}=\frac{4{\mu }_0{l}^2}{\sqrt{2}\pi L}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.