Mutual Inductance Between Circular and Square Concentric Loops

To find the mutual inductance $M$ between the square loop and the concentric circular loop, we use the principle of mutual inductance reciprocity: $M = \frac{\Phi}{I}$ where $\Phi$ is the magnetic flux linked with the inner circular loop of radius $R$ due to a current $I$ flowing through the outer square loop of side length $L$.

Since $L \gg R$, the magnetic field produced by the square loop within the region of the circular loop can be assumed to be uniform and equal to the magnetic field $B$ at the center of the square loop.

The magnetic field at the center of the square loop is the sum of the magnetic fields produced by its four equal straight sides.

For a straight current-carrying wire of length $L$, the perpendicular distance from the center of the square to each of its sides is $d = \frac{L}{2}$. Each side subtends an angle of $45^\circ$ on either side of the perpendicular drawn from the center.

The magnetic field at the center due to one side of the square loop is: $B_1 = \frac{\mu_0 I}{4\pi d} \sin(45^\circ)$

Substituting $d = \frac{L}{2}$: $B_1 = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{2\sqrt{2}\pi L}$

Since the square loop consists of 4 such identical segments, the total magnetic field $B$ at the center is: $B = 4 \times B_1 = 4 \times \frac{\mu_0 I}{2\sqrt{2}\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}$

The total magnetic flux $\Phi$ passing through the circular loop of area $A = \pi R^2$ is: $\Phi = B \cdot A = \left(\frac{2\sqrt{2}\mu_0 I}{\pi L}\right) \times (\pi R^2) = \frac{2\sqrt{2}\mu_0 R^2 I}{L}$

Therefore, the mutual inductance $M$ between the circular loop and the square loop is given by: $M = \frac{\Phi}{I} = \frac{2\sqrt{2}\mu_0 R^2}{L}$

This corresponds to Option D.