Q 12-04-140JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
The magnetic field at the centre of a circular coil of radius $r$, due to current $I$ flowing through it, is $B$. The magnetic field at a point along the axis at a distance $\dfrac r2$ from the centre is:
Answer: (C) $\left(\dfrac{2}{\sqrt5}\right)^3 B$
$B = \dfrac{\mu_0 I}{2r}$ and on the axis $B_x = \dfrac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}}$, so
$$\frac{B_x}{B} = \frac{r^3}{(r^2+x^2)^{3/2}} = \frac{r^3}{\left(\frac54 r^2\right)^{3/2}} = \left(\frac{2}{\sqrt5}\right)^3$$
Solution by Sreeraj P, M.Sc Physics