A small square loop of wire of side l is placed inside a large square loop of side L. The loops are coplanar and their centres coincide. The mutul inductance of the system is proportional to (a) l /L. (b) l^2/L

Dear Student,

Magnetic Field at the centre of outer square current(I) carrying loop is

$B_1=\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$

Since L>>l we approximate the magnetic field through entire inner coil by B₁
Thus the flux through inner coil

$$\begin{gathered} {\phi }_{21}=B_1{A}_2 \\ =\left(\frac{2\sqrt{2}{\mu }_{0}I}{\pi }\right)\times \frac{l^2}{L} \end{gathered}$$

Mutual Inductance = $\frac{{\phi }_{21}}{I_1}$

$$\begin{gathered} or, M= \left(\frac{2\sqrt{2}{\mu }_0}{\pi }\right)\times \frac{l^2}{L} \\ or, M= const\times \frac{l^2}{L} \\ or, M\propto \frac{l^2}{L} \end{gathered}$$

option(2) CORRECT ANSWER

Regards.