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Magnetic Field Force

  1. Oct 2, 2005 #1
    Here is the problem word for word:

    An electron in a vacuum is first accelerated by a voltage of 61500 V and then enters a region in which there is a uniform magnetic field of 0.477 T at right angles to the direction of the electron's motion. The mass of the electron is 9.11e-31 Kg and its charge is 1.60218e-19 C. What is the magnitude fo the force on the electron due to the magnetic field? Answer in units of N.

    The problem is easy in the fact that the formula to use is F=qBvsin0 (0 = theta). However, I cannot figure out how to get the velocity, as it is the unknown here. I know that v=E/B; however, I cannot figure out how to get E from the information provided in the problem.

    Can anyone clue me in on how to get E from the information above?
  2. jcsd
  3. Oct 2, 2005 #2


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    Use an energy approach, since clearly, you are given the energy imparted to the electron, because it's being accelerated through a voltage difference. Then equate that energy to the kinetic energy to find the speed v.
  4. Oct 2, 2005 #3
    you dont need E to get the velocity

    [tex]E_{el}=\frac{1}{2}U\cdot q\hspace{2cm} E_{kin}=\frac{1}{2}m\cdot v^2[/tex]

    [tex]E_{el}=E_{kin}\hspace{2cm}\rightarrow v=\sqrt{\frac{U\cdot q}{m}} \hspace{2cm}\rightarrow F=q\cdot B\cdot \sqrt{\frac{U\cdot q}{m}}[/tex]

  5. Oct 2, 2005 #4
    Thanks for your help. I came up with something close to Gellman's solution using Galileo's hint. What I came up with is:

    U = Vq (potential energy)
    1/2 mv^2 (kinetic energy)

    Vq = 1/2 mv^2 (then solve for v)
    v = SQRT(2qv/m)

    F = qB(SQRT(2qv/m))sin0 (0 = theta)
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