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Homework Help: Insight needed into electromagnetism problem

  1. Nov 29, 2005 #1
    The problem (From Electromagnetism by Pollack and Stump, exercise 11.34)
    A stationary charge of charge e and mass m encounters a electromagnetic wave with vector potential

    [tex]\vec{A}=\vec{j}f(x-ct)[/tex]

    where
    [tex]\vec{j}[/tex]

    is the unit vector in the y-direction. The scalar potential is zero.
    What are the components of the velocity, as a function of time,

    [tex]v_{x}, v_{y},v_{z}[/tex] ?

    What I´m asking for is some insight into why the y-velocity is not a more complicated function. The y-velocity is:

    [tex]v_{y}=\frac{e}{m}f(x-ct)[/tex]

    More specifically, when the charge is stationary, it will begin to move in the y-direction because of the changing vector potential but the motion in the y-direction will result in the magnetic field causing a motion in the x-direction which in turn should (?) effect the y-velocity, but apparently it does not. Why?
    I wrote the equations of motion but they don´t seem to give any insight into the y-motion. I determined the equations of motion by starting with the following equations:

    [tex]\vec{E}=-\frac{\partial A}{\partial t}[/tex]

    (In the equation above, A is a vector)

    [tex]\vec{B}=\nabla X \vec{A} [/tex]

    [tex]\vec{F}=e(\vec{E}+\vec{v}X\vec{B})[/tex]

    The function f(x-ct) isn´t a problem if we use the chain rule

    w = x -ct

    [tex]\frac{\partial f}{\partial t}=\frac{df}{dw}\frac{\partial w}{\partial t}=-c\frac{df}{dw} [/tex]

    likewise,

    [tex]\frac{\partial f}{\partial x}=\frac{df}{dw}\frac{\partial w}{\partial x}=\frac{df}{dw} [/tex]

    Note, I´ve been a little sloppy about notation regarding partials and full derivatives.
    The resulting equations of motion are:

    [tex]\frac{dv_{x}}{dt}=\frac{e}{m}v_{y}\frac{df}{dw}[/tex]

    [tex]\frac{dv_{y}}{dt}=\frac{e}{m}(c\frac{df}{dw}-v_{x}\frac{df}{dw})[/tex]

    [tex]\frac{dv_{z}}{dt}=0[/tex]

    Thus the z-component of the velocity is zero since it started at rest.

    The equations do give the right answer for the x-velocity if I assume the value of the y-velocity. But they aren´t helpful in giving me the y-velocity. Note that the two equations involving vy and vx have an additional unknown df/dw. I´ve tried everything, even eliminating df/dw and getting an equation that appears to have just x-dependence on one side and y-dependence on the other which one could solve by setting each to a constant but that doesn´t seem to work either.
     
    Last edited: Nov 30, 2005
  2. jcsd
  3. Nov 29, 2005 #2

    Physics Monkey

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    Science Advisor
    Homework Helper

    You have your vector potential pointing in the z direction so the electric field points in the z direction also, yes? How is the acceleration in the z direction zero then? Also, the magnetic field is in the y direction, so the y velocity can't be affected by either the electric or magnetic field, right?

    I think you have some confusion about which way everything is pointing though I suspect this is just a typographical error.

    Here is a hint for when you sort it all out: what is [tex] \frac{d}{dt} f(x(t)-ct) [/tex]?
     
    Last edited: Nov 29, 2005
  4. Nov 30, 2005 #3
    Thank you so much for your reply, Physics Monkey.

    You are correct about the direction of the vector potential, I made a mistake in its direction. It should be in the y-direction and I've edited my original post to correct it.

    I've also gone back and replaced full derivatives which should have been partial derivatives. I think the derivatives for the velocity components should be full derivatives.

    I have a feeling that you may be right about considering the x-dependence of time because when the charge does move in the x-direction, it no longer "feels" the same field and this might account for the charge not experiencing further effects in the y-direction due to motion in the x-direction.

    Thanks again.
     
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