An electron moves with a speed of 8.0 * 10^{6} 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 xy-plane. What is the magnitude of the magnetic force of the electron?

The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

Explanation:

Given;

speed of the electron, v = 8.0 * 10⁶ m/s

magnetic field strength, B = 2.5 T

angle of inclination of the field, θ = 60°

The magnetic force experienced by the electron in the magnetic field is given as;

F = qvBsinθ

where;

q is charge of electron = 1.6 x 10⁻¹⁹ C

B is strength of magnetic field

v is speed of the electron

Substitute the given values and solve for F

F = (1.6 x 10⁻¹⁹) (8.0 * 10⁶) (2.5) sin60

F = 2.77 x 10⁻¹² N

Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N