A proton with charge 1.6 10^-19 C moves at 2.0 10^6 m/s perpendicular to a magnetic field of 0.50 T. What is the magnitude of the magnetic Lorentz force?

Answer and explanation

Correct answer: D. 1.6 * 10^-13 N

  1. Step 1: Use the magnetic Lorentz force formula: F = q v B * sin(theta).
  2. Step 2: With theta = 90 degrees, sin(90) = 1.
  3. Step 3: Substitute values: F = (1.6 10^-19 C) (2.0 10^6 m/s) 0.50 T.
  4. Step 4: Multiply terms: F = 1.6 10^-19 1.0 10^6 = 1.6 10^-13 N.
Answer
1.6 * 10^-13 N.
Why not the others: '3.2 10^-13 N' forgets the 0.50 T factor; '8.0 10^-14 N' divides by 2 twice; '1.6 * 10^-19 N' ignores the particle velocity.

Practise more Mechanics, Waves and Electromagnetism.

Free questions with an explanation for every one, and full timed General Education Diploma (GED Oman) mock tests.

All Mechanics, Waves and Electromagnetism questionsTake a mock test

More Mechanics, Waves and Electromagnetism questions