Q 12-04-026NEETJEE MainMedium
A circular coil of radius $R$ carries a steady current. The ratio of the magnetic field on its axis at a distance $\sqrt{3}R$ from the centre to the field at the centre is
Answer: (C) $\dfrac{1}{8}$
$$\frac{B_{\text{axis}}}{B_{\text{centre}}} = \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{R^3}{(4R^2)^{3/2}} = \frac{R^3}{8R^3} = \frac{1}{8}$$
Solution by Sreeraj P, M.Sc Physics