Engineering Board Exam Reviewer · Physics · medium
A proton moves at 2 × 106 m/s perpendicular to a magnetic field of 0.1 T. What is the magnetic force on it? (e = 1.602 × 10-19 C)
- 3.204 × 10-12 N
- 2.000 × 105 N
- 3.204 × 10-14 N
- 0.00 N
Answer
C. 3.204 × 10-14 N
Solution
F = qvB = (1.602 × 10-19)(2 × 106)(0.1) = 3.204 × 10-14 N.
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