Engineering Board Exam Reviewer · Physics · hard
An electron moves at 1 × 107 m/s perpendicular to a uniform magnetic field of 0.01 T. What is the radius of its circular path? (me = 9.109 × 10-31 kg, e = 1.602 × 10-19 C)
- 56.9 mm
- 5.69 mm
- 0.18 mm
- 2.84 mm
Answer
B. 5.69 mm
Solution
r = mv/(qB) = (9.109 × 10-31)(107)/[(1.602 × 10-19)(0.01)] = 5.686 × 10-3 m = 5.69 mm.
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