Question

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

• A. $4:1$
• B. $1:4$
• C. $1:2$
• D. $2:1$

Answer 1

The correct option is A $4:1$

We know that Force $\left(F_m\right)$ on a particle with charge "q" moving with a velocity of v at an angle $\theta$ to the magnetic field $\left(B\right)$ can be expressed as

$F_m=qvB\sin \theta$

Given, $\frac{F_m\left(p\right)}{F_m\left(\alpha \right)}$=2

$q_{\alpha }=2\times q_p$

$\Rightarrow \frac{q_p{v}_{p}B}{q_{\alpha }{v}_{\alpha }B}=2$

We get

$\frac{v_p}{v_{\alpha }}=4$