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Magnetic Force Part 2

  1. Jul 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Oops, I forgot to add in the second part of the problem. Here's the original question plus the second part of the question.

    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 driving east, how fast would you have to drive in order for the magnetic force on your head to equal 210 N (enough to knock you over)?

    3. The attempt at a solution
    Here's what I did but it seems too simple. Could anyone tell me if i'm actually doing this the right way or not? Thanks.
    If force (F) = 210N
    Then the formula for the velocity is
    v = F / (Bq)
    v = 210 N/ (5e-5 * 3e-9)
    v = 1.4e15 m/s
  2. jcsd
  3. Jul 27, 2007 #2


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    Homework Helper

    Assuming that the magnetic field is pointing straight up, your solution appears "correct" minus one little detail, that is, the speed of light is 3e8 m/s, and knowing that it is impossible to bypass this limit, the answer would be impossible.
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