1. The problem statement, all variables and given/known data The strength of Earth's magnetic field at the equator is approximately equal to 5e-5 Tesla. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = qvB, where v is the speed of the particle. The direction of the force is given by the right hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 3 e-9 C. If you are the equator and driving west at a speed of 70 m/s, what is the strength of the magnetic force on your head due to Earth's magnetic field? What is the direction of that magnetic force? 2. Relevant equations F = qvB 3. The attempt at a solution Here's what I did: Given that Charge(q) = 3*10-9C Speed(v) of the charged particle = 70m/s Earth's magnetic field(B) = 5*10-5T Now formula for the magnetic force acting on the charged particle is F = Bqv = ( 5*10-5T)(3*10-9C)(70m/s) **Is this the right way to work this problem?** Also I'm not sure what the direction of the force would be. I was thinking it would have to be upwards due to the right hand rule? Thank you for your help.