Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ______.

The magnetic field at a point on the axis of a circular coil of radius $$R$$ at a distance $$x$$ from the centre is given by $$B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$$.

Given that the fields at distances $$x_1 = 0.05$$ m and $$x_2 = 0.2$$ m from the centre are in the ratio $$8 : 1$$, we have:

$$\frac{B_1}{B_2} = \frac{(R^2 + x_2^2)^{3/2}}{(R^2 + x_1^2)^{3/2}} = 8$$

Taking the cube root of both sides (since $$8^{1/3} = 2$$) and then squaring:

$$\frac{R^2 + x_2^2}{R^2 + x_1^2} = 8^{2/3} = 4$$

Substituting $$x_1 = 0.05$$ m and $$x_2 = 0.2$$ m:

$$R^2 + (0.2)^2 = 4(R^2 + (0.05)^2)$$

$$R^2 + 0.04 = 4R^2 + 0.01$$

$$0.03 = 3R^2$$

$$R^2 = 0.01$$

$$R = 0.1$$ m

The correct answer is Option (3): 0.1 m.