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Energy of a proton around a magnetic field

  1. Jul 16, 2008 #1
    A proton moves in a circular path perpendicular to a 4.30 T. magnetic field. The radius of its path is 7.5 cm. Calculate the energy of the proton in electron-volts.


    I'm not sure which equation to use. I tried using m=rqB/v and found the velocity. However, I'm not sure how to relate that to energy. Mass of proton equals 1.67E-27. Q of proton = 1.6E-19. 1eV=1.6E-19 J.
     
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  3. Jul 16, 2008 #2

    Hootenanny

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    Welcome to PF,
    What type of energy would you say the proton possesses?
     
  4. Jul 16, 2008 #3
    Thank you, I love House by the way. I would say it has kinetic energy. SO would I use KE=1/2mv^2? But then I always get confused on what work equals, because I think I need to find work to solve the equation. I think W=KE+PE, and I don't think there is any Pe. Would the work just equal the kinetic energy? Nothing I submit seems to be right.
     
  5. Jul 16, 2008 #4

    Hootenanny

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    I can't wait for the next season :biggrin:
    Correct! Assuming of course that the proton is non-relativistic.
    Correct again, there is no potential energy and hence the total energy of the proton is simply it's kinetic energy.
     
  6. Jul 16, 2008 #5
    So velocity is v=.075m X 1.6X10^-19C X 4.30T / 1.67X10^-27. Which equals 3.09X10^7. I then used KE=.5 X 1.6X10^-19 X (3.09X10^7)^2 and came up with 7.637X10^-5J. I multiplied that by 1eV/1.6X10^-19J and got 4.77X10^14. But it was still wrong. Any other suggestions?
     
  7. Jul 16, 2008 #6

    Hootenanny

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    Why have you multiplied by the charge here, instead of the mass? The blue section is also incorrect and is unnecessary if you simply use the mass.
     
  8. Jul 16, 2008 #7
    Oh thank you. I knew it had to be something stupid. It was right!
     
  9. Jul 16, 2008 #8

    Hootenanny

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    A pleasure.
     
    Last edited: Jul 16, 2008
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