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Particle physics -- Cyclotron

  1. Dec 13, 2014 #1
    • Member warned about not posting with homework template
    According to the equation F=BQ/2pieM the frequency depends on the magnetic field and not the speed or radius of the particle. Can someone please explain why?

    B= Magnetic field strength Q=Charge M=Mass

    I think it's because of the force felt due to magnetic field so the time it reaches the dees depends on how large the force is eventhough, there's an electric field that accelerates the particle.
     
    Last edited by a moderator: Dec 13, 2014
  2. jcsd
  3. Dec 13, 2014 #2

    haruspex

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    It's an example of SHM. The frequency of SHM does not depend on the amplitude.
     
  4. Dec 14, 2014 #3
    So why does it depend on the magnetic field strength? Is the reason similar to I said?
     
  5. Dec 14, 2014 #4

    haruspex

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    I didn't understand that part. What are 'dees' here?
     
  6. Dec 14, 2014 #5
     
  7. Dec 14, 2014 #6

    haruspex

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    Interesting video, but there is one serious error in it. Where it says va is the voltage, it means va is the frequency of the voltage.
    So your question here is why does the natural frequency depend on the magnetic field strength but not on the electric field strength.
    Again, by SHM analogy, the magnetic field supplies the 'restoring force', like the modulus of a spring. But you need to be careful with the analogy because the perturbation being restored is the particle's velocity, not its position.
    The reversing electric field is not required to produce the SHM. Once circulating, a charged particle would tend to cycle around perpendicularly to the magnetic field at constant speed and frequency. The electric field only serves to raise the amplitude each cycle, and, as discussed, the amplitude does not change the period.
    Alternatively, we can throw the analogies away as potentially misleading and just look at the dynamics. If the particle is circling with speed v at radius r then the centripetal force required is mv2/r. The radial force supplied is qBv. Equating these gives v/r = qB/m. Since the path length is 2 pi r, the period is 2 pi r/v = 2 pi m/(qB).
     
  8. Dec 16, 2014 #7
    Sorry for the long response. Now that I think of that way it does make sense. Thanks.
     
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