What should be the ratio of the velocity of a proton to that of an alpha particle so that they trace paths of the same radius in a perpendicular magnetic field?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 : 1
A charged particle moving perpendicular to a uniform magnetic field moves in a circle, and the magnetic force \( qvB \) provides the centripetal force:
\( \displaystyle qvB=\frac{mv^{2}}{r} \)
\( \displaystyle \Rightarrow\; r=\frac{mv}{qB} \)
For the same radius in the same field, \( v=qBr/m \), so the speed is proportional to \( q/m \). The proton has charge \( e \) and mass \( m_p \); the alpha particle (a helium nucleus) has charge \( 2e \) and mass about \( 4m_p \).
\( \begin{aligned} \frac{v_p}{v_\alpha} &= \frac{q_p/m_p}{q_\alpha/m_\alpha} \\ &= \frac{e/m_p}{2e/4m_p} \\ &= \frac{2}{1} \end{aligned} \)
Therefore, \( v_p:v_\alpha=2:1 \).