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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
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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 electric dipole placed in a non-uniform electric field experiences
- Dependence of intensity of gravitational field (E) of earth with distance(r)
- When 1 kg of ice at 0⁰C melts to water at 0⁰C, the resulting change in its entropy
- A coil of self-inductance L is connected in series with a bulb B and an AC source
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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
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