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Direction of electron in magnetic field?

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    An electron moves with speed 4.00 105 m/s in a uniform magnetic field of 3.2 T, pointing south. At one instant, the electron experiences an upward magnetic force of 1.00 10-14 N. In what possible directions might the electron be moving at that instant? Give your answers as angles clockwise from south (from 0° to 360°), in increasing degrees.

    (I have to find two answers in degrees, both clockwise from south).

    2. Relevant equations

    F = qvBsin(theta)

    3. The attempt at a solution

    F = qvBsin(theta)

    sin(theta) = F / qvB

    = (1x10-14) / [(1.6x10-19)(4x105)(3.2)]

    = .0488

    theta = sin-1(.0488)

    theta = 2.8

    So my first angle was approximately 2.8. I subtracted that from 270 degrees ("south") to get my final angle of 267.2 degrees. Then I subtracted 2.8 from 180 to get my second angle, which was approximately 177.2. I subtracted THAT number from 270 degrees to get my final second angle of approximately 92.8 degrees. I thought that 267.2 and 92.8 were my answers, but they are coming up as incorrect. I also tried rounding to a whole number, which didn't work either.
     
  2. jcsd
  3. Mar 1, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I wouldn't "subtract from 270".
    The 2.8 is already the angle from south.
    According to the left hand rule, it has a component to the left, so it must be 2.8 degrees counterclockwise from south. That would 357.2 degrees clockwise from south.
     
  4. Mar 2, 2009 #3
    It was 182.8 and 357.2. Thanks!
     
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