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Electromagnetism - Lorentz Force

  1. Feb 13, 2007 #1

    A particle of charge q starts from rest at the origin of coordinates in a region where there is a uniform electric field of strenth E parallel to the x-axis, and a uniform magnetic field B parallel to the z-axis.

    Find the equations of motion, and solve them to show that the coordinates of the particle at a time t later will be:

    x = (E/B*omega)*(1 - cos(omega*t))

    y = - (E/B*omega)*(omega*t - sin(omega*t))

    z = 0

    where omega = q*B/m. (The path of the circle is a cycloid.)

    2. Relevant equations

    The parametric equation of a cycloid:

    x = constant*(1 - cos(omega*t))

    y = constant*(omega*t - sin(omega*t))

    The force acting on the particle:

    F = q*E + q*vxB

    3. The attempt at a solution

    I've done some work on this problem and so far the equations of motion that I've got for the particle are as follows:

    1) F(x) = q*E + q*v(y)*B -> x[double-dot] = q*E/m + omega*y[dot]

    2) F(y) = -q*v(x)*B -> y[double-dot] = -q*B*x[dot]

    I've tried integrating these equations once (eg. integrate 2)) and then substituting this into the other equation. This then gave me:

    x[double-dot] + omega^2*x = E*B

    And this is where I'm stuck. This has the form of a simple harmonic oscillator, except that the r.h.s. isn't zero, so I can't solve it. Also, I'm not even sure if everything that I've done so far is correct.

    Any help on this would be very much appreciated!
  2. jcsd
  3. Feb 13, 2007 #2


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    I found the same equations of motion but i kinda stuck there as well. I`ll check it again tomorrow :)
  4. Feb 14, 2007 #3
    Thanks, that would be really helpful! I'm glad to know that I don't seem to be a complete moron though, or that I've overlooked some vital detail that will make the whole thing ridiculously easy.
  5. Feb 14, 2007 #4


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    Solving the differential equation x[double-dot]+x*ω^2=qE/m (1)

    X=X1+X2=(c1*cosωt+c2*sinωt)+qE/(m*ω^2) , c1,c2=constants

    X1 is the solution of the : x[double-dot]+x*ω^2=0
    X2 is a "special" solution that satisfies (1).

    I think you can solve that now.
    Last edited: Feb 14, 2007

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