At one instant an electron (charge = -1.6 *10^-¹⁹C) is moving in the xy plane, the components of its velocity being v_x = 5 *10⁵ m/s and v_y = 3 *10⁵ m/s. A magnetic field of 0.8 T is in the positive x direction. At that instant the magnitude of the magnetic force on the electron is: A) 0 B) 3.8 *10^-¹⁴ N C) 5.1 * 10^-¹⁴ N D) 6.4 * 10^-¹⁴ N E) 7.5*10^-¹⁴ N

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by the equation F = qvB sin(theta), where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and theta is the angle between the velocity and the magnetic field. In this case, the electron has a charge of -1.6 x 10^-19 C, a velocity of 5 x 10^5 m/s in the x direction and 3 x 10^5 m/s in the y direction, and the magnetic field has a strength of 0.8 T in the positive x direction. The angle between the velocity and the magnetic field is 90 degrees, so sin(theta) = 1. Plugging these values into the equation, we get F = (-1.6 x 10^-19 C) * (5 x 10^5 m/s) * (0.8 T) * 1 = 3.8 x 10^-14 N.

At One Instant an Electron (Charge = -1