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Magnetic Field, potential difference and speed

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm. Find the speed of the electron as it enters the magnetic field.

    2. Relevant equations

    r=mv/(qB)
    F=qvB

    3. The attempt at a solution

    I though of relating the potential difference (3000V) with the speed using U=E=mv^2 but I dont think its right. I dont know how else to approach the problem because in r=mv/qB I have two unknowns and in F=qvB I have three.
    The answer key says: 3.25 x 107 m/s
    Thanks in advance!
     
  2. jcsd
  3. Feb 26, 2012 #2

    I like Serena

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    Homework Helper

    Hi physgrl! :smile:

    You're on the right track.
    You only need that the energy of the electron when accelerated by a potential difference V is charge of the electron times the potential difference.
     
  4. Feb 26, 2012 #3
    Thanks!! :)
     
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