A small square loop of side $$a$$ and one turn is placed inside a larger square loop of side $$b$$ and one turn $$(b \gg a)$$. The two loops are coplanar with their centres coinciding. If a current $$I$$ is passed in the square loop of side $$b$$, then the coefficient of mutual inductance between the two loops is:

We have a large square loop of side $$b$$ carrying a current $$I$$. The small square loop of side $$a$$ lies at the centre, in the same plane and with only one turn. To obtain the mutual inductance, we must first calculate the magnetic flux linked with the small loop when the current $$I$$ flows in the large one, and then divide that flux by the current.

The magnetic flux through the small loop is $$\Phi = BA$$ because the field is essentially uniform over the tiny area (remember that $$b \gg a$$). Here $$B$$ is the magnetic field at the centre of the big loop and $$A = a^{2}$$ is the area of the small loop. So we need $$B$$ at the centre of a square loop.

For one straight, finite conductor carrying current $$I$$, the magnetic field at a point lying a perpendicular distance $$r$$ from its midpoint is given by the Biot-Savart result

$$B = \frac{\mu_{0} I}{4\pi r}\left(\sin \theta_{1} + \sin \theta_{2}\right),$$

where $$\theta_{1}$$ and $$\theta_{2}$$ are the angles that the lines joining the point to the two ends of the wire make with the wire.

Now consider one side of the square loop. The centre of the square is at a perpendicular distance $$r = \dfrac{b}{2}$$ from this side, and the ends of the side subtend equal angles. Each angle is clearly $$\theta_{1} = \theta_{2} = 45^{\circ}$$ because the half-length of the side is also $$\dfrac{b}{2}$$, forming an isosceles right triangle with the centre.

Hence for one side,

$$\sin \theta_{1} = \sin \theta_{2} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}.$$

Substituting these values into the Biot-Savart expression, we get for a single side

$$\begin{aligned} B_{\text{one side}} &= \frac{\mu_{0} I}{4\pi \left(\dfrac{b}{2}\right)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) \\ &= \frac{\mu_{0} I}{4\pi \left(\dfrac{b}{2}\right)}\left(\frac{2}{\sqrt{2}}\right) \\ &= \frac{\mu_{0} I}{4\pi}\left(\frac{2}{b}\right)\left(\sqrt{2}\right) \\ &= \frac{\mu_{0} I}{4\pi}\frac{2\sqrt{2}}{b}. \end{aligned}$$

The square loop has four identical sides, and the directions of the fields from all sides at the centre are the same (into or out of the page depending on the current direction). Therefore the total magnetic field at the centre is

$$\begin{aligned} B_{\text{total}} &= 4 \times B_{\text{one side}} \\ &= 4 \times \frac{\mu_{0} I}{4\pi}\frac{2\sqrt{2}}{b} \\ &= \frac{\mu_{0} I}{4\pi}\frac{8\sqrt{2}}{b}. \end{aligned}$$

Now we turn to the flux through the small loop. Its area is $$A = a^{2}$$, so

$$\Phi = B_{\text{total}} \, A = \left(\frac{\mu_{0} I}{4\pi}\frac{8\sqrt{2}}{b}\right)(a^{2}) = \frac{\mu_{0} I}{4\pi}\frac{8\sqrt{2} \, a^{2}}{b}.$$

The mutual inductance $$M$$ is defined by $$\Phi = M I$$ (one turn only), so dividing the flux by the current we obtain

$$\begin{aligned} M &= \frac{\Phi}{I} \\ &= \frac{\mu_{0}}{4\pi}\frac{8\sqrt{2} \, a^{2}}{b}. \end{aligned}$$

This expression exactly matches Option A.

Hence, the correct answer is Option A.