A proton moves at 2 10^6 m/s perpendicular to a 0.5 T field. The radius of its circular path is (m = 1.67 10^-27 kg, q = 1.6 10^-19 C):

  1. 0.418 m
  2. 0.00418 m
  3. 0.0418 m
  4. 0.0835 m
Answer

C) 0.0418 m

Bqv = mv^2r Rightarrow r = mvBq = 0.0418 m. SLO P-09-E-33 · Understanding

Physics MCQ · More Magnetism and Electromagnetic Induction MCQs · Practise