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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: - A particle of mass M is suited at the centre of a spherical shell of same mass and radius
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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 particle of mass M is suited at the centre of a spherical shell of same mass and radius
- A long straight wire carries a certain current and produces a magnetic field
- Two charged particles have charges and masses in the ratio 2:3 and 1:4 respectively.
- The ionisation energy of hydrogen is 13.6 eV. The energy of the photon released
- At what distance from the centre of earth, the value of acceleration due to gravity
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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