Find the mutual inductance in the arrangement, when a small circular loop of wire of radius $$R$$ is placed inside a large square loop of wire of side $$L(L \gg R)$$. The loops are coplanar and their centres coincide: 

Solution :

Magnetic field at the centre due to one side of the square loop is :

$$B_1 = \frac{\mu_0 I}{4\pi (L/2)}(\sin45^\circ + \sin45^\circ)$$

$$= \frac{\mu_0 I}{2\pi L}\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)$$

$$= \frac{\mu_0 I}{\sqrt2\pi L}$$

Magnetic field due to all four sides :

$$B = 4B_1$$

$$= \frac{4\mu_0 I}{\sqrt2\pi L}$$

$$= \frac{2\sqrt2\mu_0 I}{\pi L}$$

Since \(L \gg R\), magnetic field over the small circular loop is nearly uniform.

Magnetic flux linked with circular loop :

$$\phi = BA$$

$$= \frac{2\sqrt2\mu_0 I}{\pi L}\times \pi R^2$$

$$= \frac{2\sqrt2\mu_0 IR^2}{L}$$

Mutual inductance :

$$M = \frac{\phi}{I}$$

$$= \frac{2\sqrt2\mu_0 R^2}{L}$$

Final Answer :

$$M = \frac{2\sqrt2\mu_0 R^2}{L}$$