An electron is traveling east with an instantaneous velocity of 3.3 * 10⁵ m/s when it enters a uniform magnetic field of 0.25 T that points X degrees north of east. (Take east as to the right of the paper and north as towards the top of the paper, i.e. both in the plane of the paper.) If the magnitude of the force on the electron is 5.5*10^-15N, then calculate the angle X and whether the electron moves up out of or down into the plane of the page, or otherwise. A) 26° and up out of the page B) 65° and down into the page C) 26° and down into the page D) 65° and up out of the page E) 65° and south in the plane of the paper

Question

Quizplus · Accepted Answer

First, we can use the formula F = qvBsinθ to find the angle θ between the magnetic field and the velocity of the electron:

θ = sin^(-1)(F / qvB) = sin^(-1)((5.5 x 10^-15 N) / ((-1.6 x 10^-19 C) x (3.3 x 10^7 m/s) x (0.25 T))) ≈ 26°

The negative sign in the formula indicates that the force is perpendicular to both the velocity and the magnetic field, so the electron will move in a circular path perpendicular to the plane of the page. The direction of the force can be determined using the right-hand rule, with the fingers pointing in the direction of the velocity, the thumb pointing in the direction of the magnetic field, and the resulting force pointing towards the palm. Since the force is downward in this case, the electron will move down into the page. Therefore, the answer is choice C.

An Electron Is Traveling East with an Instantaneous Velocity of 3.3