
For mutual inductance in coplanar loops:
• Use the magnetic field from the larger loop and calculate flux through the smaller loop.
• Divide flux by the current to obtain mutual inductance.
\(M = \frac{\sqrt{2} \mu_0 R^2}{L}\)
\(M = \frac{2 \sqrt{2} \mu_0 R}{L^2}\)
\(M = \frac{2 \sqrt{2} \mu_0 R^2}{L}\)
\(M = \frac{\sqrt{2} \mu_0 R}{L^2}\)
\[ \phi = M i \]
\[ \phi = (BA) \]
\[ \phi = \pi R^2 \left( \frac{4\mu_0}{4\pi} \cdot i \cdot \frac{L}{2} \right) \sqrt{2} \]
\[ \implies M = \frac{2\sqrt{2} \mu_0 R^2}{L} \]

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: