A proton is moving in a circular path of radius R in a uniform magnetic field B. It carries a kinetic energy of 2 MeV. What should be the energy of an \( \alpha \)-particle to move in a circle of same radius in the same field ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 MeV
A charged particle moving perpendicular to a uniform magnetic field moves in a circle because the magnetic force \( qvB \) provides the centripetal force.
\( \displaystyle qvB = \frac{mv^{2}}{R} \;\Rightarrow\; R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \)
Here the momentum is written in terms of the kinetic energy, \( mv = \sqrt{2mK} \). Solving for \( K \):
\( \displaystyle K = \frac{q^{2}B^{2}R^{2}}{2m} \;\Rightarrow\; K \propto \frac{q^{2}}{m} \quad (\text{same } R \text{ and } B) \)
For the alpha particle \( q_\alpha = 2e \) and \( m_\alpha = 4m_p \):
\( \displaystyle \frac{K_\alpha}{K_p} = \frac{q_\alpha^{2}}{q_p^{2}}\times\frac{m_p}{m_\alpha} = 4\times\frac{1}{4} = 1 \)
\( \displaystyle K_\alpha = K_p = 2\,\mathrm{MeV} \)