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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
- A copper disc of radius 0.1 m is rotated about its centre, with 10 revolutions
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- An alternating voltage given as, V=100√2 sin100t V is applied to a capacitor
- Two planets at mean distances d₁ and d₂ from the sun have their frequencies n₁ and n₂
- A particle of mass m moves with constant speed along a circular path of radius r under
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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