Magnetic force on a moving charge?

Question

A magnet produces a 0.30 T field between its poles, directed to the east. A dust particle with charge q = -8.0 10-18 C is moving straight down at 0.30 cm/s in this field. What is the magnitude and direction of the magnetic force on the dust particle?

Answer 1

F = q v x B where the "x" means the vector cross product. The scalar version is

F = q v B sin(q) where q = angle between v and B

q = 90 deg in the above problem. SO

F = -8x10^-18 C* 0.3 x 10^-2 m/s x 0.3T= 7.2 x10^-20 N

DIrection follows right hand rule --> direction is south