A small square loop of wire of side $$l$$ is placed inside a large square loop of wire of side $$L(L = l^2)$$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $$\sqrt{x} \times 10^{-7}$$ H, where $$x$$ = ______.

$$\text{Let current } I \text{ flow through the outer square loop of side } L.$$

$$\text{Distance from center to any side: } r = \frac{L}{2}$$

$$\text{Field due to one side of length } L: B_1 = \frac{\mu_0 I}{4\pi r}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)}\left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I}{\sqrt{2}\pi L}$$

$$\text{Total field at the center due to all 4 sides: } B = 4B_1 = \frac{2\sqrt{2}\mu_0 I}{\pi L}$$

$$\text{Flux linked with the inner small square loop of side } l \text{ and area } l^2: \phi = B \cdot l^2 = \frac{2\sqrt{2}\mu_0 I l^2}{\pi L}$$

$$\text{Mutual Inductance: } M = \frac{\phi}{I} = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}$$

$$\text{Given: } L = l^2 \implies \frac{l^2}{L} = 1$$

$$M = \frac{2\sqrt{2}\mu_0}{\pi} = \frac{2\sqrt{2}(4\pi \times 10^{-7})}{\pi} = 8\sqrt{2} \times 10^{-7}\text{ H} \implies x = 8\sqrt{2} \implies x = 128$$