An electron moves at 4.0 × 10⁵ m/s at 30° to a uniform magnetic field of 0.20 T. Take e = 1.6 × 10⁻¹⁹ C. What is the magnitude of the magnetic force on it?
Correct answer D. 6.4 × 10⁻¹⁵ N
Explanation
F = qvB sin θ = 1.6 × 10⁻¹⁹ × 4.0 × 10⁵ × 0.20 × sin 30° = 1.28 × 10⁻¹⁴ × 0.5 = 6.4 × 10⁻¹⁵ N. Using cos 30° gives 1.1 × 10⁻¹⁴ N.
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