NTA Abhyas JEE Main2020PhysicsMagnetic Effects of CurrentPractice
A cyclotron's oscillator frequency is 10 M H z and the radius of its dees is 60 c m , then the kinetic energy of the proton beam produced by the accelerator is [ m proton = 1 . 67 × 10 - 27 kg ]
Options
- A9 M e V
- B10 M e V
- C7 M e V
- D11 M e V
Correct answer
C. 7 M e V
Step-by-step solution
KE = q 2 B 2 R 2 2 m Given, f = 10  M H z = 10 7    H z R = 60   c m = 0 .6   m f = q B 2 π m ∴     K E = 2   π 2   m   f 2   R 2 = 2 × 3 .14 2 × 1.67 × 10 - 27 × 10 14 × 0. 36   J ≈ 7  M e V