JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Magnetic Effects of Current

In a magnetic field the radius of the circular path is $R=\frac{mv}{qB}$. The proton and the alpha particle enter with the same speed $v$ and the same field $B$, so the radius depends only on the ratio $\frac{m}{q}$ of each particle. Writing the proton's mass and charge as $m$ and $e$, the alpha particle has mass $4m$ and charge $2e$, so $\frac{R_{\alpha }}{R_p}=\frac{m_{\alpha }/q_{\alpha }}{m_p/q_p}=\frac{4m/2e}{m/e}=\frac{2m/e}{m/e}=2$. The radii of the circular paths of the alpha particle and the proton are therefore in the ratio $2:1$.