A circular current loop of radius \(R\) is placed inside square loop of side length \(L\) \((L >> R)\) such that they are co-planar and their centers coincide. The permeability of free space is \(\mu_0\). The mutual inductance between circular loop and square loop is _______.

Using the reciprocity theorem, assume a current \(I\) flows through the larger square loop and compute the flux through the smaller circular loop. The mutual inductance is given by \(M = \dfrac{\Phi}{I}\).

Magnetic field at the centre of the square loop:
For a square of side \(L\), the perpendicular distance from the centre to any side is \(d = \dfrac{L}{2}\), and each side subtends angles \(\theta_1 = \theta_2 = 45^\circ\) at the centre.

The field due to one finite straight segment is:
\(B_{\text{segment}} = \dfrac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)\)

\(B_{\text{segment}} = \dfrac{\mu_0 I}{4\pi \cdot \frac{L}{2}}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) = \dfrac{\mu_0 I}{2\pi L} \cdot \sqrt{2} = \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L}\)

All four sides contribute equally in the same direction at the centre, so:
\(B_{\text{total}} = 4 \cdot \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L} = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L}\)

Flux through the circular loop:
Since \(L \gg R\), the field \(B_{\text{total}}\) can be treated as uniform over the area \(A = \pi R^2\) of the circular loop.

\(\Phi = B_{\text{total}} \cdot A = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L} \cdot \pi R^2 = \dfrac{2\sqrt{2}\,\mu_0 R^2 I}{L}\)

Mutual inductance:
\(M = \dfrac{\Phi}{I} = \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}\)

Hence, the correct option is \((4)\ \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}\).