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Math Help - Time Period of a Proton in a Magnetic Field

  1. #1
    Member alexgeek's Avatar
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    Exclamation Time Period of a Proton in a Magnetic Field

    Spent half an hour looking at this problem, but I can't see anyway to get any of the multiple choice answers:
    Quote Originally Posted by AQA Specimen Paper
    Protons, each of mass mand charge e, follow a circular path when travelling perpendicular to a magnetic field of uniform flux density B. What is the time period for on complete orbit?
    • A \frac{2 \pi e B}{m}
    • B \frac{m}{2 \pi e B}
    • C \frac{eB}{2 \pi m}
    • D \frac{2 \pi m}{eB}
    All I know is that it should have something to do with:
     F = m \omega^2 r = m \frac{v^2}{r} = BQv \: , \: \omega = \frac{2 \pi}{T}

    Thanks

    Edit: Oh and the answer is D
    Last edited by alexgeek; January 25th 2011 at 11:45 PM.
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  2. #2
    Super Member Aryth's Avatar
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    You're missing an equation.

    Remember that the tangential velocity of an object in circular motion is: v = \omega r so that:

    F = QvB = m\frac{v^2}{r}

    QB = m\frac{v}{r} = m\omega

    \omega = \frac{QB}{m}

    We know the formula for period, which is:

    T = \frac{2\pi}{\omega}

    T =  \frac{2\pi m}{QB}

    Since we know the charge, we can fill that in to get:

    T = \frac{2\pi m}{eB}

    I'm not sure why D doesn't match up to this, but I'm almost 95% sure this is right.
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  3. #3
    Member alexgeek's Avatar
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    Thanks very much for the reply, will look again when I got some caffeine.
    You were right also, I just had a typo in my LaTeX (\pim instead of \pi m).
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