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Electrodynamics equation help

  1. Nov 2, 2006 #1
    Could anybody help me solve this equations ( i'm sorry for my english)
    [tex]m\frac{d^{2}{x}}{dt^{2}}=qB\frac{{dy}}{{dt}}[/tex]

    [tex]m\frac{d^{2}{y}}{dt^{2}}=qEy-qB\frac{{dx}}{{dt}}[/tex]

    [tex]m\frac{d^{2}{z}}{dt^{2}}=qEz[/tex]


    X(0)=Y(0)=Z(0)=0
    [tex]\frac{dx}{dt}(0)=Vx[/tex]
    [tex]\frac{dy}{dt}(0)=0[/tex]
    [tex]\frac{dx}{dt}(0)=Vz[/tex]
     
  2. jcsd
  3. Nov 2, 2006 #2
    Use the first equation to write the second equation in terms of

    [itex]\frac{dx}{dt}=v_x[/itex]

    Then make a substitution:

    [itex]u_x = v_x - \frac{E_y}{B}[/itex]
     
  4. Nov 3, 2006 #3
    Using the equations in my post, the second equation becomes:

    [itex]\frac{d^2u_x}{dt} + \frac{qB}{m}u_x = 0[/itex]

    That differential equation hopefully looks familiar. Once you have solutions for [itex]u_x(t)[/itex] you should then be able to write down [itex]v_x(t)[/itex]. Then you should be able to use that to solve the first equation.
     
  5. Nov 3, 2006 #4
    I'm sorry i have problems with tex :), I think i understand thank you
     
    Last edited: Nov 3, 2006
  6. Nov 3, 2006 #5

    dextercioby

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    Differentiate wrt "t" in the second eq and then use the first one.

    Daniel.
     
  7. Nov 3, 2006 #6
    if your familiar with laplace transforms, you can use one of those on the first two equations, then just do algebraic manipulations until you have X(s) and Y(s) then use the inverse laplace transform on each to get x(t) and y(t)
     
  8. Nov 3, 2006 #7

    HallsofIvy

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    Obviously the z equation can be integrated separately from the x and y equations. One way to do the x and y equations is to differentiate the y equation again, getting
    [tex]m\frac{d^3y}{dt^3}= qE\frac{dy}{dt}- qB\frac{d^2x}{dt^2}[/tex]
    [tex]= qE\frac{dy}{dt}- \frac{qB}{m}(qB\frac{dy}{dt})[/tex]
    or
    [tex]m\frac{d^3y}{dt^3}= \left(qE- \frac{q^2B^2}{m}\right)\frac{dy}{dt}[/tex]
    which is easy to solve.

    I'm a bit concerned about your equations, however. They don't look quite right to me. Are you sure that isn't
    [tex]m\frac{d^{2}{y}}{dt^{2}}=qE-qB\frac{{dx}}{{dt}}[/tex]
    and
    [tex]m\frac{d^{2}{z}}{dt^{2}}=qE[/tex]

    I don't recall the electric force being proportional to a distance!
     
  9. Nov 6, 2006 #8
    Ey,z - I understand like constant electric field in y,z direction, y,z are the index
     
  10. Nov 7, 2006 #9

    dextercioby

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    Then use the "_{}" command

    [tex] E_{y}, \ E_{z} [/tex]

    Daniel.
     
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