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Uniform circular motion for proton and 2 particles

  1. Nov 28, 2014 #1
    [Note by mentor: This thread does not use the homework template because it was moved here from a non-homework forum.]

    I have a problem with this exercise. I don't know if my solution is right! Could you check it, please?

    A proton with kinetic energy Kp is moving in a uniform magnetic field on a circular orbit.
    I have to calculate the energy of two particles moving on the same orbit knowing that:
    particle 1 : q=+e, m=2u
    particle 2: q=+2e, m=4u

    [itex]Kp = 1/2 mv^2[/itex]
    [itex]v = qBr/m[/itex] (uniform circular motion).


    [itex]Kp = 1/2 q^2B^2r^2/m[/itex]

    proton: [itex]Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}[/itex]

    particle 1: [itex]K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2[/itex]

    particle 2: [itex]K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp[/itex]

    Is that right?
    Thank you
    Last edited by a moderator: Nov 28, 2014
  2. jcsd
  3. Nov 28, 2014 #2


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    Yes, I think it is right.
  4. Nov 28, 2014 #3
    Thank you.
    I'm sorry but I didn't see the Homework section, won't happen again!
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