An electron traveling with a velocity of 6.0 * 10⁵ m/s along the direction enters a uniform magnetic field of 0.60 T pointing in the direction.The magnitude of the force on the electron is

Question

Quizplus · Accepted Answer

The magnitude of the force on a moving charge (F) in a magnetic field (B) is given by the formula F = qvB sin(theta), where q is the charge of the particle, v is its velocity, and theta is the angle between the velocity vector and the magnetic field. In this case, the electron is moving perpendicular to the magnetic field, so sin(theta) = 1. Plugging in the given values, we get:

F = (1.6 x 10^-19 C)(6.0 x 10^5 m/s)(0.60 T)(1) = 5.76 x 10^-14 N

Rounding to two significant figures, the answer is B) 5.8 x 10^-14 N.

An Electron Traveling with a Velocity of 6