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Magnetic Force of the earth

  1. Jul 25, 2007 #1
    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.
     
  2. jcsd
  3. Jul 26, 2007 #2
    The earth's magnetic field is directed perpendicular to the equator and goes from the north to the south pole.

    marlon
     
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