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Show the drift velocity is ExB/B^2

  1. Mar 26, 2016 #1
    • HW Template missing as it was moved from another forum
    A charged particle drifts in uniform, constant magnetic and electric fields. The electric field, E, is perpendicular to the magnetic field, B.

    Show that the drift velocity is given by vd = (EƗB)/B2

    Heres where I get to:
    F=e(E+vxB)=0 as v is uniform.


    Therefore E+vxB=0.

    Take vector product of B with both sides.

    BxE +Bx(vxB)=0.

    Using identity Ax(BxC) = B(A.C)-C(A.B)

    I get BxE+v(B.B)-B(B.v)=0
    Then I don't know where to go from here.
     
  2. jcsd
  3. Mar 26, 2016 #2

    mfb

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    You can assume that the drift happens in two dimensions, so you can say something about the relative direction of v and B which simplifies (B.v).
    In three dimensions the total velocity does not have to follow the initial equation, so you need that assumption.
     
  4. Mar 26, 2016 #3
    Why are you allowed to assume the drift is in two dimensions though?
     
  5. Mar 26, 2016 #4

    mfb

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    The component in the third dimension is not a drift.
     
  6. Mar 26, 2016 #5
    Would that be an acceleration then?
     
  7. Mar 26, 2016 #6

    mfb

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    A motion in the third direction would stop quickly in matter. In vacuum, the charged particle could freely keep moving in that direction, without influencing the two-dimensional motion in the other directions.
     
  8. Mar 26, 2016 #7
    Can I ask what it means by a drift velocity then?
     
  9. Mar 26, 2016 #8

    mfb

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    The velocity in the two-dimensional plane I mentioned.
     
  10. Mar 26, 2016 #9
    yes but what does it specifically mean by 'drift'. You said the velocity in the 3rd dimension is not a drift.
     
  11. Mar 26, 2016 #10

    mfb

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    I guess it is just convention to call that drift.
     
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