How Does Earth's Magnetic Field Affect a Charged Object in Motion?

[kiwikahuna]

Homework Statement

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?

Homework Equations

F = qvB

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.

 

[marlon]

The Earth's magnetic field is directed perpendicular to the equator and goes from the north to the south pole.

marlon

 

[m2003]

I would like to confirm that your calculation is correct. The direction of the magnetic force can be determined by using the right hand rule, which states that if you point your thumb in the direction of the velocity (to the west in this case), and your fingers in the direction of the magnetic field (north to south in this case), then the direction of the force will be perpendicular to both (upwards in this case). This is because the force on a positive charge will be in the direction of the thumb, while the force on a negative charge will be in the opposite direction. Therefore, the force on your head would be upwards due to the positive charge acquired from rubbing the balloon in your hair.