A proton and an alpha particle moving with same kinetic energy enter in the region of uniform magnetic field perpendicular to it. The ratio of radii of their trajectories will be :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 : 1
A charge moving perpendicular to a uniform magnetic field moves on a circular path with \( q v B = m v^{2}/r \), so \( r = m v/(qB) = p/(qB) \). With kinetic energy \( K \), the momentum is \( p = \sqrt{2mK} \):
\( \displaystyle r = \frac{\sqrt{2mK}}{qB} \)
For the same \( K \) and \( B \), \( r \propto \sqrt{m}/q \).
\( \text{proton: } m_p = m,\ q_p = e \)
\( \alpha\text{ particle: } m_{\alpha} = 4m,\ q_{\alpha} = 2e \)
\( \displaystyle \frac{r_p}{r_{\alpha}} = \frac{\sqrt{m_p}}{q_p}\times\frac{q_{\alpha}}{\sqrt{m_{\alpha}}} = \frac{\sqrt{m}}{e}\times\frac{2e}{\sqrt{4m}} = 1 \)
So the ratio of radii is 1 : 1.
Note: A similar question was also asked in CUET (UG) 2026 on 26-05-2026 (Slot 2), in CUET (UG) 2025 on 29-05-2025 (Slot 2), and in CUET (UG) 2023 on 02-06-2023 (Slot 2).