A small square loop of wire of side \( l \) is placed inside a large square loop of wire of side \( L \) (\( L \gg l \)). The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{l^{2}}{L} \)
The mutual inductance is the flux linked with the small loop per unit current in the large loop, \( M = \Phi_{2}/I_{1} \). Send a current \( I \) through the large loop and find the field at the common centre.
Each side of the large loop is at a perpendicular distance \( L/2 \) from the centre and subtends \( 45^\circ \) at each end, so by the Biot-Savart law one side gives
\( \displaystyle B_{\mathrm{side}} = \frac{\mu_0 I}{4\pi\,(L/2)}\,(\sin45^\circ + \sin45^\circ) = \frac{\sqrt{2}\,\mu_0 I}{2\pi L} \)
All four sides add in the same direction at the centre:
\( \displaystyle B = 4\,B_{\mathrm{side}} = \frac{2\sqrt{2}\,\mu_0 I}{\pi L} \)
Since \( L \gg l \), this field is almost uniform over the small loop, so the flux through it is \( B\,l^{2} \).
\( \displaystyle M = \frac{\Phi}{I} = \frac{B\,l^{2}}{I} = \frac{2\sqrt{2}\,\mu_0\,l^{2}}{\pi L} \)
Hence \( M \propto l^{2}/L \), which is Option (2).