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Homework Help: An electron in a magnetic field

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The acceleration of an electron in a magnetic field of 87 mT at a certain point is 1.268×1017 m/s2. Calculate the angle between the velocity and magnetic field.

    2. Relevant equations
    F = q dot v cross B
    F = qBsin(theta)
    F/m = a

    3. The attempt at a solution

    I calculated the force on the electron using F/m = a rearranged to F = ma, then plugged F into F = qBsin(theta). I came up with sin-1(8243000) = theta, which is obviously incorrect.

    Am I making a calculation error, or is there an error with my formulas?

    In advance, any help is much appreciated and thank you.
  2. jcsd
  3. Mar 16, 2009 #2
    There is a mild error in your formula, at least a bit. You know F = ma, you know the mass of the electron (it's a constant) and you know a since it's given. You also know F = qB sin([tex]\theta[/tex]) and you know B (given) and q, constant. So you get [tex]sin^{-1}(\frac{ma}{qB}) = \theta[/tex]

    Make sense?
  4. Mar 16, 2009 #3
    Hmm, that is what I have above, at least I think.

    Because when you substitute F = ma into F = qBsin(theta), you end up with ma = qbSin(theta), which can be arranged to sin-1(ma/qb) = Theta.

    The values I used:
    m = 9.1E-31 kg
    q = 1.609E-19 C
    B = 87E-3 T
    a = 1.268E17

    I still end up calculating 8243000, and when I try to take the inverse sine of that I get a domain error.
    Last edited: Mar 16, 2009
  5. Mar 16, 2009 #4
    I get a very similar number for ma/(qb). Is it possible you have a or B written down incorrectly?
  6. Mar 16, 2009 #5
    I just checked again, and those are the correct values for a and B.
  7. Mar 16, 2009 #6
    OH! F = qvBsin([tex]\theta[\tex]). You missed the v which should account for the 107[\SUP] factor.
  8. Mar 16, 2009 #7
    Ah alright, that does make sense then. How would I go about calculating the velocity though?

    qvb = F
    F = ma
    qvb = ma
    v = 8243000

    qvB/ma = sin(theta)

    qvB/ma = 1, giving you theta = 90 degrees.

    This unfortunately is incorrect.. =\
  9. Mar 16, 2009 #8
    qvb = F? That's missing a sin(theta) isn't it, that would then explain how you get sin(theta) = 1.

    I think there's an equation we're not looking at that we should be because otherwise we've two unknowns and only one question. Or perhaps velocity should be a given (which is how I see the question posed elsewhere online).
  10. Mar 16, 2009 #9
    It mentions in an earlier part of the problem that the velocity is 9.40×106, however it doesn't specify if that remains constant throughout the problem.

    Assuming for a moment that it does, lets try to plug that into the equation:

    F = qvBsin(theta)
    F = 1.609E-19 * 9.40×106 * 87E-3 (I'm assuming 87 mT = 87 E-3 T)

    And, assuming that I can use F = ma (which I'm not sure about)

    I can substitute ma = 9.1E-31 * 1.268E17, divide that by qvB

    End up with sin-1(.8769)
    = 61.27 deg

    Which is the correct answer! Thanks a lot for your help, sorry I missed the velocity, probably would have helped from the beginning..
  11. Mar 16, 2009 #10
    No worries; for all the online things I've worked with like that all the givens are constant unless stated otherwise.

    Welcome, and good work!
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