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A proton has a speed of 3 x 10^6 m/s

  1. May 28, 2003 #1


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    I need help deriving a formula to solve for this please.

    A proton has a speed of 3 x 10^6 m/s in a direction perpendicular to a uniform magnetic field and the proton moves in a circle radius .2 m. what is the magnitude of the magnetic field?

    If someone can show me the all formulas to use i can manipulate from here.
    Dx :wink:
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. May 28, 2003 #2


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    A particle with mass m, charge q, moving at speed v in a constant magnetic field of magnitude B at right angles to the velocity vector of the particle, will move in a circle with radius mv/(qB)
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