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Cyclotron physics problem

  1. Apr 12, 2006 #1
    The answer is 1.56 x 10^13
    I can't get the answer.

    Fm = F centripetal Force
    qvB = mv^2/r
    v= rqB/m

    Calculate energy

    KE = 1/2 mv^2
    I put in v = rqB/m

    KE= 1/2 (rqB)^2/m

    I plug in all the #, but still can't the answer

    please help, thx so much!
  2. jcsd
  3. Apr 12, 2006 #2


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    Staff: Mentor

    "The answer is 1.56 x 10^13
    I can't get the answer."

    The answer is not 10^13V (not an energy) or 10^13eV (zowies!). What is the answer again?
  4. Apr 12, 2006 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    What are the units of the given solution?

    What are the units of your answer?

    Are they the same?

    What happens if you make them the same?
  5. Apr 12, 2006 #4
    The anwer doesn't have unit. (This is the sample test from my teacher, he doesn't type units for answer.) But in the question, it said volts. The unit I got is Newton x Seconds/ kg. Actually, I don't know how to make it into volts. I don't even know if it's the right method. :(
  6. Apr 12, 2006 #5
    Some misunstanding is here. Sorry about that. The answer my teacher gave is 1.56 x 10^13 V. I can't get the same answer as my teacher from the calculation above.
  7. Apr 14, 2006 #6
    You probably also have to take into account the role of gravitational force in providing centripetal acceleration apart from magnetic force field.
  8. Apr 15, 2006 #7

    Cheez's question doesn't include the mass of the earth, so it seems it should be disregarded. From my perspective it seem the question is asking:

    Through what potential difference should the protons be accelerated through for the protons to just orbit the earth (at the same height)?

    Which would lead to an answer in Volts.
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