# Magnetic Force of the earth

1. Jul 25, 2007

### kiwikahuna

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?