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Homework Help: Cyclotron Frequency Formula

  1. Dec 29, 2008 #1


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

    I am currently writing about Cyclotrons, and want to show how the period is related to the magnetic field.

    2. Relevant equations

    [tex] T = \frac{2 \pi}{\omega} [/tex]

    [tex] \omega = \frac{v}{r} [/tex]

    3. The attempt at a solution

    I have rearranged the above formula to give:

    [tex] T = \frac{2 \pi}{\frac{v}{r}} = \frac{2 \pi r}{v} [/tex]

    Now a website now goes onto to say that this is equal to:

    [tex] \frac{2 \pi r}{v} = \frac{2\pi mv}{qBv} [/tex]

    Now I don't want to just write this down (for a start is plagarism, but anyway), so could anyone recommend some useful laws that mat come in useful?

    Thanks in advanced,

  2. jcsd
  3. Dec 29, 2008 #2


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    Staff: Mentor

    I start with the Lorentz force, and apply that to circular motion (either UCM or accelerated CM) of a charge in a combination of electric and magnetic fields:


    Hope that helps.
  4. Dec 29, 2008 #3


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    The charge is held in a circular orbit in a cyclotron because the magnetic force (look up "Lorentz force law") produces an acceleration that is equal to the "centripetal acceleration for uniform circular motion". Look up those.
  5. Dec 30, 2008 #4


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    Thanks, So Lorentz Force:

    [tex] F_{mag} = qv \times B [/tex]

    Since these are perpendicular to each other:

    [tex] F_{mag} = qvB [/tex]

    This is then equal to the Centripetal acceleration:

    [tex] a = \frac{v^2}{r} [/tex]

    thus centripetal force is:

    [tex] F_{cent} = \frac{mv^2}{r} [/tex]

    These can thgus be equated when the particle is expirienceing constant circular motion (in the case of the cyclotron in the two 'D' sections):

    [tex] F_{mag} = F_{cent} \therefore qvB = \frac{mv^2}{r} [/tex]

    From before,

    [tex] \omega = \frac{v}{r} [/tex]

    this can rearrnge for v,

    [tex] v = \omega r [/tex]

    Put into equation:

    [tex] q \omega r B = \frac{m(\omega r)^2}{r} [/tex]

    [tex] q \omega r B = \frac{m \omega^2 r^2}{r} [/tex]

    [tex] q \omega r B = m \omega^2 r [/tex]

    And, again:

    [tex] T = \frac{2 \pi}{\omega} [/tex]


    [tex] \omega = \frac{2 \pi}{T} [/tex]


    [tex] q \omega r B = m (\frac{2 \pi}{T})^2 r [/tex]

    [tex] q \omega r B = m \frac{4 \pi^2}{T^2} r [/tex]

    Rearrange to get T:

    [tex] T^2 q \omega r B = m 4 \pi^2 r [/tex]

    [tex] T^2 = \frac{m 4 \pi^2 r}{q \omega r B} [/tex]

    [tex] T = sqrt{\frac{m 4 \pi^2 r}{q \omega r B}} [/tex]

    does this look right? I didn't think there was a square root?

  6. Dec 30, 2008 #5


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    You seem to be taking algebraic steps that aren't getting you closer to what you want. You are also not cancelling variables when they appear on both sides and keeping way to many related variables around. You were almost done when you wrote qvB=mv^2/r. T=2*pi*r/v. Just solve the first equation for v and substitute into the second. Don't forget to cancel the 'v'. Don't write v=sqrt(rqvB/m). If you do that you haven't really 'solved' for v. There is still one on both sides.
  7. Dec 30, 2008 #6


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    Thank's, so:

    [tex] qvB = \frac{mv^2}{r} [/tex]

    this can cancel down to:

    [tex] qB = \frac{mv}{r} [/tex]

    rearrange for v:

    [tex] mv = rqB [/tex]

    [tex] v = \frac{rqB}{m} [/tex]

    So now we substitute this into:

    [tex] T = \frac{2 \pi r}{v} [/tex]

    [tex] T = \frac{2 \pi r}{\frac{rqB}{m}} [/tex]

    This can be rearranged into:

    [tex] T = \frac{2 \pi r m}{rqB} [/tex]

    Cancel down the r

    [tex] T = \frac{2 \pi m}{qB} [/tex]

    This looks a lot better.


  8. Dec 30, 2008 #7


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    Staff: Mentor

    Nice work TFM. Keep in mind, though, that you simplified the Lorentz force right away to assume no E field. That's okay, but may not be true in the case of the cyclotron (I don't know enough about cyclotrons to know whether that's true or not).
  9. Dec 30, 2008 #8


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    Well, the Electric field would only be in the gap between the two D sections, which is what accelerates the particles, not in the D sections themselves, so the circular motion in the D sections should not be affected. It also matches up with the formula from here;


    Thanks again,

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