JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Magnetic Effects of Current

For a particle of kinetic energy $K$ moving perpendicular to a field $B$, the path radius is $r=\frac{\sqrt{2mK}}{qB}\propto \frac{\sqrt{m}}{q}$ when $K$ and $B$ are equal for both. For the $4$ amu, $+2e$ ion, $\frac{\sqrt{m}}{q}=\frac{\sqrt{4}}{2}=1$; for the $16$ amu, $+3e$ ion, $\frac{\sqrt{16}}{3}=\frac{4}{3}$. The lighter ion has the smaller radius, and a smaller radius means a tighter curve, so the lighter ion is deflected more than the heavier ion.