Q 12-04-167JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
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 ______.
Answer: (C) 0.1 m
$B = \dfrac{\mu_0 NIR^2}{2(R^2 + x^2)^{3/2}}$, so
$$\frac{B_1}{B_2} = \left(\frac{R^2 + 0.04}{R^2 + 0.0025}\right)^{3/2} = 8 \Rightarrow \frac{R^2 + 0.04}{R^2 + 0.0025} = 4$$
$$R^2 + 0.04 = 4R^2 + 0.01 \Rightarrow R^2 = 0.01 \Rightarrow R = 0.1\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics