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Electron velocity through earth's magnetic field

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Electron is travelling in line with the equator, perpendicular to earth's magnetic field of 6E-5 Tesla. What velocity is required for the electron to stay in a straight line path, such that it's weight is exactly matched by the magnetic force against it.

    2. Relevant equations

    F = q.(v×B)

    3. The attempt at a solution

    I've actually done a whole module on electromagnetism before, right through to Maxwell's equations etc. However I've changed to a new university and am required to go through it all again, because that's just the way their course goes. The work is trivial enough but the answer I've got here is strangely small so I need to double check to see if I've missed something.

    Electron weight needs to match the Magnetic force. So...

    F = mg = q.(v×B) = qvB (because of perpendicularity so can 'ignore' cross-product formalities)

    v = mg/qB = (9.11E-31 * 9.81) / 1.60E-19 * 6E-5)

    v = 9.31E-7 m/s

    Now this seems incredibly slow to me. I understand electrons are very, very small, but I've never come across anything where such an incredibly slow velocity would be of any use at all.

    So, am I correct or have I made some glaring error somewhere?

    Thanks.
     
  2. jcsd
  3. Oct 8, 2015 #2
    It means that the gravity acting on the electron is really weak and doesn't affect much to the motion of electron.
     
  4. Oct 8, 2015 #3
    Great stuff, I take it there's no problem with the answer then :smile:
     
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