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, then what is the ratio of their velocities?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4 : 1
A charge \( q \) moving with speed \( v \) perpendicular to a magnetic field \( B \) feels the magnetic force \( F=qvB \). The proton has charge \( e \), and the alpha particle (a helium nucleus) has charge \( 2e \).
\( F_p=e\,v_p\,B \)
\( F_\alpha=2e\,v_\alpha\,B \)
Taking the ratio of the two forces:
\( \begin{aligned} \frac{F_p}{F_\alpha} &= \frac{e\,v_p\,B}{2e\,v_\alpha\,B} \\ &= \frac{v_p}{2v_\alpha} \end{aligned} \)
Putting \( F_p/F_\alpha=2 \):
\( \displaystyle \frac{v_p}{2v_\alpha}=2 \)
\( \Rightarrow\; v_p:v_\alpha=4:1 \)