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Magnetic Field Direction

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A 0.49 m copper rod with a mass of 0.13 kg carries a current of 10 A in the positive x direction. Let upward be the positive y direction.
    1) What is the magnitude of the minimum magnetic field needed to levitate the rod?
    2) What is the direction of the minimum magnetic field needed to levitate the rod?
    pos x direction, neg x, pos y, neg y, pos z, neg z

    2. Relevant equations

    Right Hand Rule?

    3. The attempt at a solution
    I figured out part 1 but not sure on part two. any guidance would be appreciated. I'm thinking it's positive y direction. Part 1 answer was B = .26 T fwiw
  2. jcsd
  3. Feb 25, 2010 #2
    Yup, simply use the right hand rule. You know the current is going to the right (positive x) and to levitate the rod you would need a force going up (positive y). So by using the right hand rule you should be able to find the direction of the magnetic field.
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