Question

An electron (me = 9.1 x 10-31 kg) traveling south at a constant speed of 5.0 x 106 m/s enters a

region where the downward component of earth's magnetic field is 3.5 x 10-5 T. What is the
magnitude and direction of the acceleration of the electron at this instant?

Answer 1

F = Q V X B

V X B = south X downard = East - deflection is West

Since Q is negative the deflection = West

a = F / m = Q V B / m

a = 1.6E-19 * 5*E6 * 3.5E-5 / 9.1E-31

a = 1.6 * 5 * 3.5 / 9.1 * 10^13 m/s^2 = 3.1 * 10^13 m/s^2