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Homework Help: Lorentz equation?

  1. Feb 12, 2004 #1
    The motion of a charged particle in an electromagnetic field can be obtained from the Lorentz equation for the force on a particle in such a field. If the electric field vector is E and the magnetic field vector is B, the force on a particle of mass m that carries a charge q and has a velocity v is given by
    F=qE+qv X B
    (The X is the cross product.)
    If there is no electric field and if the particle enters the magnetic field in a direction perpendicular to the lines of magnetic flux, show that the trajectory is a circle with radius
    where omega=qB/m, which is the cyclotron frequency.

    Okay, if there's no electric field, then I drop the qE term, meaning F=qv X B. I set this equal to F=ma.
    qv X B = ma
    I think I should make a=dv/dt, and solve for v, then take the derivative to get the trajectory. But the cross product is really confusing me, and I don't really know how to simplify from there. Am I going in the right direction?
  2. jcsd
  3. Feb 12, 2004 #2

    Tom Mattson

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    After you take the cross product, note that the force is perpendicular to the velocity. What kind of motion does that give rise to?

    (Hint: consider a ball on a string, being swung around parallel to the ground--you have the same type of situation).

  4. Feb 12, 2004 #3
    If the velocity and force are perpendicular, it's centripetal motion. But I don't really know what to do with that. I know velocity is the cross product of the radius and omega (d.theta/dt). Because I'm trying to find the radius, I can't use this, I think?

    Otherwise, when I try to do the math, I get a natural log, which I know I shouldn't be getting.

    qvxB=qvB*sin 90=qvB
    (qB/m)t=ln v

    I'm not using the fact that it's centripetal motion, but I don't really know how to tie this in.
  5. Feb 12, 2004 #4


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    Homework Helper

    Did you mean "antiderivative?"
  6. Feb 12, 2004 #5
    Yes, sorry, that's what I meant. And that's what I'm trying to do, I wasn't typing carefully. I have to integrate v to get r.
  7. Feb 13, 2004 #6


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    Greetings !
    If the qE component is zero, you still have
    circuilar motion due to the magnetic field.
    I can't tell you how to solve for polar coordinates
    but for cartesian coordinates all you need to do
    is write the equations for two axes, and you get:
    ( qB/m ) Vy = ax
    -( qB/m ) Vx = ay (the second minor is negative)
    Then you use the derivative of either equation and
    put the result in the other.
    The solutions, depending on the enitial conditions
    at t=0 (giving you the enitial angle = q and x0 and y0)are:
    x = x0 + Rx sin(wt+q)
    y = y0 + Ry cos(wt+q)

    Live long and prosper.
  8. Feb 13, 2004 #7
    Doesn't the fact centripetal acceleration = v^2/r make this exercise a whole lot easier?:smile:
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