A proton and an α particle enter a magnetic field in a direction perpendicular to it. If the force acting on the proton is twice that acting on the α- particle, then what is the ratio of their velocities? 

4:1

The correct answer is Option 1: 4:1

For a charged particle moving perpendicular to a magnetic field, the magnetic force is given by: $F = qvB$

For the proton: $F_p = e v_p B$

For the $\alpha$-particle: $F_\alpha = 2e , v_\alpha B$

Given: $F_p = 2F_\alpha$

$e v_p B = 2(2e v_\alpha B)$

$v_p = 4v_\alpha$

$v_p : v_\alpha = 4 : 1$