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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: - The device can act as a complete electronic circuit is
- The electric field at a distance 3R / 2 from the centre of a charged conducting
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- The range of voltmeter is 10V and its internal resistance is 50Ω.
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The device can act as a complete electronic circuit is
- The electric field at a distance 3R / 2 from the centre of a charged conducting
- A cord is wound round the circumference of wheel of radius
- Dimensional formula for electrical resistance R is given by
- The range of voltmeter is 10V and its internal resistance is 50Ω.
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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