An electron moves at 3 10^6 m/s perpendicular to a 0.002 T field. The radius of its circular path is (m = 9.11 10^-31 kg, q = 1.6 10^-19 C):

  1. 0.0171 m
  2. 8.54 10^-4 m
  3. 0.00854 m
  4. 0.0854 m
Answer

C) 0.00854 m

Bqv = mv^2r Rightarrow r = mvBq = 0.00854 m. SLO P-09-E-33 · Understanding

Physics MCQ · More Magnetism and Electromagnetic Induction MCQs · Practise