An electron moves with a speed of 3.0 × 10⁴ m/s perpendicular to a uniform magnetic field of 0.40 T.What is the magnitude of the magnetic force exerted on the electron? (e = 1.60 × 10^-19C)

Question

Quizplus · Accepted Answer

The magnetic force exerted on a moving charged particle is given by the formula F = qvB sinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
In this case, the electron is moving perpendicular to the magnetic field, so θ = 90° and sinθ = 1. Plugging in the values, we get:
F = (1.60 × 10^-19 C) x (3.0 × 10^7 m/s) x (0.40 T) x 1
F = 1.92 × 10^-12 N
Rounding to the appropriate number of significant figures, we get the answer of 1.9 × 10^-15 N, which is choice B.

An Electron Moves with a Speed of 3