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Cyclotron Question

  1. Mar 2, 2013 #1

    CAF123

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

    If a cyclotron with a maximum chamber radius of 110 cm can accelerate ##_{4}^{2+}He##, (alpha particles) up to a energy of 140 MeV, calculate

    A) The magnetic field and comment on your answer.

    If the accelerating voltage is 3 kV and the source of alpha particle is from a radioactive source that emits alpha particles with an energy of 4MeV, calculate

    B) time from emission from the source to exiting the cyclotron.

    3. The attempt at a solution
    A)Max energy of the alpha particles: 140 MeV, which means ##v = \sqrt{2E/m} \approx 8.2 \times 10^{7}m/s > 0.1c## Should I be doing a relativistic calculation?

    The cyclotron moves a particle in a circular fashion, so F points towards the centre and v tangential => B is into the page with magnitude mv/qR = 1.55T.

    It says I should comment on this - perhaps something like the B field is strong enough that the particle is accelerated to a speed closer than 10% of the speed of light?

    B) Total energy: 4Mev + qV = E => v ≈ 1.4 x 107m/s > 0.1c. Proceeding gives ##T = 2 \pi R/v = 4.9 \times 10^{-7} s##. I feel this may be incorrect because I don't think I have taken into account how a cyclotron actually works.

    Many thanks.
     
    Last edited: Mar 2, 2013
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  3. Mar 2, 2013 #2

    TSny

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    With a speed close to 30% c, I would do the relativistic calculation. I think it will make about a 4% difference in your answer.
    If you want to include relativistic effects, then mv should be replaced by the relativistic momentum. [EDIT: Looks like you're off by a factor of 2 somehow in the calculation, but the formula is basically correct except for using the relativistic momentum.] [Edit #2: My mistake, you're not off by a factor of 2.]
    You'll need to read up a bit on how the cyclotron works. T is the time for one orbit, but the ion will make many orbits before reaching 140 MeV. See http://en.wikipedia.org/wiki/Cyclotron
     
    Last edited: Mar 2, 2013
  4. Mar 2, 2013 #3

    CAF123

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    Hi TSny,

    So the angular frequency of the particle is ##\omega = 2 \pi f \Rightarrow T = \frac{2 \pi}{\omega} = 2 \pi \frac{m}{qB}##.

    The ion starts with an initial energy of 4 MeV and every half turn it is accelerated by 3kV. We want to know how many times the particle has to go around the circle until its energy is 140MeV. So I then set up a linear eqn:$$4MeV + x(3kV q) = 140MeV \Rightarrow x = 45333 $$ This is the approximate number of half turns. To get the number of full turns, divide by 2 to get approx 22666 turns.

    So I should multiply the result above for T by 22666 and I attain an answer of the order of a millisecond. This seems very fast!

    Many thanks.
     
  5. Mar 2, 2013 #4

    TSny

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    That all looks good. (It was my mistake earlier when I thought you were off by a factor of 2 in B.)
     
  6. Mar 2, 2013 #5

    CAF123

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    Thanks TSny,
    For part 1), I was asked to comment about my answer of 1.55T. What can I say? The only thing I can think of is that we are nearing the case where we would have to apply relativistic calculations. (since a particle in a B field producing that strength of teslas has an energy such that it's ejection speed is close to c).
     
  7. Mar 2, 2013 #6

    TSny

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    I'm not sure what the question is getting at here. The energy of the ion depends on both B and R, so a large B doesn't necessarily imply large energy if R is small. Perhaps they just want you to note that 1.55 T is a large magnetic field that is not easy to produce uniformly over the cyclotron. I don't know. [See http://en.wikipedia.org/wiki/Magnetic_field#Measurement for largest B field produced in lab.]
     
  8. Mar 2, 2013 #7

    CAF123

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    Maybe they just want us to note that 1.55T is quite a high strength field.
    Thank you TSny.
     
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