Q 12-04-114JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
A circular loop of radius $R$ is carrying current $i$ A. The ratio of magnetic field at the centre of circular loop and at a distance $R$ from the center of the loop on its axis is
Answer: (C) $2\sqrt2:1$
$B_{centre}=\dfrac{\mu_0i}{2R}$, $B_{axis}=\dfrac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}$. With $x=R$:
$$\frac{B_{centre}}{B_{axis}}=\frac{(2R^2)^{3/2}}{R^3}=2\sqrt2$$
Solution by Sreeraj P, M.Sc Physics