CUET · PHYSICS · PYQ PAPER 2023
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 ?
- A 1 MeV
- B 2 MeV
- C 4 MeV
- D 3 MeV
Answer & Solution
Correct Answer
(B) 2 MeV
Step-by-step Solution
Detailed explanation
\(R = \frac{\sqrt{2mKE}}{qB}\) \(\frac{\sqrt{m_p KE_p}}{q_p} = \frac{\sqrt{m_\alpha KE_\alpha}}{q_\alpha}\) \(\frac{\sqrt{m_p (2 \text{ MeV})}}{e} = \frac{\sqrt{4m_p KE_\alpha}}{2e}\) \(\sqrt{2m_p \text{ MeV}} = \sqrt{m_p KE_\alpha}\) \(KE_\alpha = 2 \text{ MeV}\)
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