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Reviewed CSCA Physics question · Standard

A proton (charge 1.6 x 10^-19 C) moves at 2.0 x 10^6 m/s perpendicular to a magnetic field of 0.5 T. What is the magnitude of the magnetic force on the proton?

  1. 1.6 x 10^-13 N
  2. 3.2 x 10^-13 N
  3. 1.0 x 10^-13 N
  4. 0 N
Show the answer and explanation

Correct answer

A. 1.6 x 10^-13 N

Principle or equation

Lorentz force: F = q v B sin θ, with θ = 90° so sin θ = 1.

Why this answer is correct

F = (1.6 x 10^-19)(2.0 x 10^6)(0.5) = 1.6 x 10^-13 N.

Example

For electron (same charge magnitude) at 1e6 m/s in 0.2 T, F = 1.6e-19 * 1e6 * 0.2 = 3.2e-14 N.

This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.

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