Q. A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
A) 2 MeV
B) 1 MeV
C) 0.5 MeV
D) 4 MeV
Explanation
[Solution from source] r = √(2mK)/(qB), so for the same r and B, K ∝ q²/m. For an α-particle q²/m = (2e)²/(4m_p) = e²/m_p, the same as the proton, so it needs the same 1 MeV. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.