An electron moves with a speed of 8.0 × 10⁶ m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5 T, directed at an angle of 60° to the +x-axis and lying in the x-y plane. Calculate the magnitude of the magnetic force on the electron. A) 2.8 × 10^-10 N B) 3.2 × 10^-10 N C) 2.8 × 10^-12 N D) 3.2 × 10^-12 N E) 0 N

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by the equation F = qvBsinθ, 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 direction of the velocity and the magnetic field.

In this case, the electron has a charge of -1.6 x 10^-19 C, a velocity of 8.0 x 10^6 m/s, and is moving at an angle of 60° to a magnetic field of 2.5 T.

Using the equation above, we get:

F = (-1.6 x 10^-19 C)(8.0 x 10^6 m/s)(2.5 T)sin(60°)
F = 2.8 x 10^-12 N

Therefore, the magnitude of the magnetic force on the electron is 2.8 x 10^-12 N, which is closest to option C, 2.8 x 10^-12 N.

An Electron Moves with a Speed of 8