A proton (charge 1.6 × 10^-19 C) moves with a speed of 3.0 × 10^6 m/s in a direction perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the proton?
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Correct answer
A. 9.6 × 10^-14 N
Principle or equation
Lorentz force on a moving charge in a magnetic field: F = qvB sin θ. When velocity is perpendicular to the field, sin θ = 1, so F = qvB.
Why this answer is correct
F = (1.6 × 10^-19) × (3.0 × 10^6) × 0.20 = 1.6 × 3.0 × 0.20 × 10^-13 = 0.96 × 10^-13 = 9.6 × 10^-14 N.
Example
An electron (q = 1.6 × 10^-19 C) moving at 2.0 × 10^6 m/s perpendicular to a 0.5 T field experiences F = 1.6 × 10^-19 × 2.0 × 10^6 × 0.5 = 1.6 × 10^-13 N.
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