What is the magnitude of the force on an electron moving with a velocity v = 5.43 * 10⁶ m/s at an angle $\theta$ = 53.4 $\circ$ with respect to a magnetic field B = 7.72 * 10^-2 T?

Question

Quizplus · Accepted Answer

The magnitude of the force on the electron moving with velocity vector v = (5.43 x 10^6 m/s) at an angle of 53.4 degrees with respect to a magnetic field B = (7.72 x 10^-2 T) can be calculated using the formula F = qvBsin(theta), where q is the charge of the electron (-1.6 x 10^-19 C) and theta is the angle between v and B in radians. Converting degrees to radians gives us theta = 0.932 radians. Substituting the values, we get F = |-1.6 x 10^-19 C| x (5.43 x 10^6 m/s) x (7.72 x 10^-2 T) x sin(0.932) = 5.38 x 10^-14 N. Therefore, the best choice is C.

What Is the Magnitude of the Force on an Electron θ\thetaθ

What Is the Magnitude of the Force on an Electron $\theta$