An electron travels

Question

An electron travels through a particular point in an experimental apparatus with a velocity of 0.867 × 10^6 m/s at an angle of +54.7 degrees(counter-clockwise) relative to the positive x-axis in the x-y plane. If the magnetic field at this point has a magnitude of 0.01 T and is directed in along the negative y-axis, what is the magnitude of the acceleration of the electron at this point? (It is advisable to sketch a picture of the situation first.)

Answer 1

F = evBsinθ = 1.6·10^-16C·0.867·10⁶m/s·0.01Tsin(90° + 54.7°) = 8.016 ·10^-13N

Vector F directed along positive z-axis and so acceleration.

a = F/m = 8.016·10^-13N/(9.11·10^-31C) = 8.799·10¹⁸m/s²