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Charged Particle Moving in a Magnetic Field

  1. Jun 13, 2014 #1
    Hello. Can you please check out the attachment?

    Solving (a) is easy. The correct answer to (b) is "the period for the return trip is unchanged" (that is, T = 130 ns).

    Okay, fine: the attachment shows why T only depends on m, q, and B.

    How on Earth, though, do you reconcile this with the inverse relationship between velocity and period, given the initial and return arc radius values are the same???

    In other words, if KE is doubled, the speed increases, and (again for a constant radius) the period must DECREASE!

    What am I missing? Thank you!

    Attached Files:

  2. jcsd
  3. Jun 13, 2014 #2


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    If the speed of a charged particle in a magnetic field increases, the radius of the circular trajectory also increases. If the speed doubles the radius and so the circumference of the circle doubles and the time remains constant.
  4. Jun 13, 2014 #3
    Thank you!

    And, right, that's the rub... the problem statement (taken from Halliday & Resnick) specified:

    "If the particle is sent back though the magnetic field (along the same initial path)..."

    My guess this "same initial path" specification is incorrect.

  5. Jun 13, 2014 #4
    Why is that? The same initial path is the downward arrow on the right. It goes that way once, then again with twice the energy.
  6. Jun 13, 2014 #5
    Twice the energy means 1.4 times the speed, which means a 1.4 times greater radius for a given period, T.
  7. Jun 13, 2014 #6
    Ok, is there a problem with that? I have not checked your numbers, but assume that you did it fine. Note that the path through the magnetic field is not the initial path. Is that what is tripping you up?
  8. Jun 13, 2014 #7
    The stated problem text said: "... along the same initial path" which I take to mean the initial path and return path are one and the same?

  9. Jun 13, 2014 #8
    lol, why would you take it like that? Initial path means initial path, the path the on the right. Initial doesn't mean initial and return...
  10. Jun 13, 2014 #9


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    It is better learning to work it out as you have, but you could also get the answer from the equation for the gyrofrequency (or cyclotron frequency).

    The gyrofrequency is the (angular) frequency for full revolution, but your problem is half a circle, so you should be able to see that it is just double the frequency. And you can take the inverse to get period. It doesn't depend on velocity at all since the velocity cancels out (in the nonrelativistic regime).
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