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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?
Options
(a) 2 MeV
(b) 1 MeV
(c) 0.5 MeV
(d) 4 MeV
Correct Answer:
1 MeV
Explanation:
According to the principal of circular motion in a magnetic field
Fc = Fm ⇒ mv²/R = qvB
⇒ R = mv/qB = P/qB = √2m.k/qB
Rα = √2(4m)K’/2qB
R/Rα =√ K/K’
but R =Rα (given) Thus K=K’=1 MeV
Related Questions: - An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- Huygen’s wave theory could not explain
- MA⁻¹T⁻² is the dimensional formula of
- A gas is suddenly expanded such that its final volume becomes 3 times
- A block is moving up an inclined plane of inclination 60° with velocity of 20 ms⁻¹
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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