A proton carrying 1 MeV kinetic energy is moving in a circular path of radius

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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:

1. A solid cylinder of mass 50kg and radius 0.5 m is, free to rotate about the horizontal axis
2. Which is correct relation?
3. Young’s modulus of perfectly rigid body material is
4. A Gaussian surface in the cylinder of cross-section πa² and length L is immersed
5. Ohm’s law is not applicable except to

Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)

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