NTA Abhyas JEE Main2020PhysicsMagnetic Effects of CurrentPractice
An electron of mass 9 .0 × 10 - 31 kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 × 10 6 m / s . If a proton of mass 1 .8 × 10 - 27 kg has to move in a circle of same radius and in the same magnetic field, then its speed must be
Options
- A1 .5 × 10 3 m / s
- B3 × 10 6   m / s
- C6 × 10 4   m / s
- D2 × 10 8   m / s
Correct answer
A. 1 .5 × 10 3 m / s
Step-by-step solution
Magnetic force will provide the necessary centripetal force.       B q v = m v 2 r B q r = m v For electron and proton, the magnetic field B , charge q and radius r are same. So, m v = constant i.e. m e v e = m p v p v p = m e m p v e = 9 × 10 - 31 1.8 × 10 - 27 × 3 × 10 6 v p = 1.5 × 10 3   m / s