| ⇦ | 
| ⇨ |
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: - According to Hook’s law, force is proportional to
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- If the half-life of a material is 10 yr, then in what time, it becomes 1/4th part
- When you make ice cubes, the entropy of water
- The pressure at the bottom of a tank containing a liquid does not depend on
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- According to Hook’s law, force is proportional to
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- If the half-life of a material is 10 yr, then in what time, it becomes 1/4th part
- When you make ice cubes, the entropy of water
- The pressure at the bottom of a tank containing a liquid does not depend on
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply