do u include only perpendicular to B vector components when solving for magnetic force using F=qvbsin@?

Question

this is the problem...q=2.5x10^-6 moving 500m/s in direction 30degNoE. At location of particle, B=0.4T. Particle experiences instantaneous magnetic Force due to field, F= 1.71x10^-4 N at 60degWoN. What is the size of the angle between the velocity vector and the magnetic field at the location of the particle? What is the direction of magnetic field? I don't know what to do. I am starting with finding the vector components. Is the 500m/s the "r" value for the velocity vector? I'm stuck there. If I start with components of force vector, how can I tell which components of velocity vector are perpendicular to the Force and B vectors. Do I include all vector components in the cross product, or just the perpendicular components? Please help me.

Answer 1

Start by defining a coordinate system. For instance, make East the positive x direction and North the positive y direction. Then figure out what angle your vectors v and B are pointing.

In the formula F = qvb sin(@), @ is the angle between the two vectors v and B. v is the magnitude of the velocity. b is the magnitude of the magnegtic field. q is the magnitude of the charge. F is the magnitude of the force.

You don't need to worrry about perpendicular components. The formula takes care of that