A proton moves at 2.0 × 10⁶ m/s at right angles to a uniform magnetic field of 0.50 T. Take e = 1.6 × 10⁻¹⁹ C. What is the magnitude of the magnetic force on it?
Correct answer C. 1.6 × 10⁻¹³ N
Explanation
F = qvB sin 90° = 1.6 × 10⁻¹⁹ × 2.0 × 10⁶ × 0.50 = 1.6 × 10⁻¹³ N. Omitting the 0.50 T factor gives 3.2 × 10⁻¹³ N.
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