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Particle in a field

  1. Feb 28, 2006 #1
    Solve the equation of motion for a particle of mass m and charge q moving in a constant electric and magnetic field E0 and B0 tat is solve the equation
    [tex] m \dot{v} - q E_{0} + \frac{q}{C} v \times B_{0} [/tex]
    to determine the velocity v(t) and thence the trajctory r(t) of the particle. Discuss the hsape of the trajectory.

    Hint: Choose the z axis along B0 and the y axis perpendicular to the plane of E0 and B0

    Wel accoridng to the hint Bx and By would be zero
    since Y is perpendicular ot the plane of E an B then
    [tex] E_{x} = E_{0} \cos(t)[/tex]
    [tex] E_{z} = E_{0} \sin(t)[/tex]
    Ey = 0
    does v hasve to be dependant on some angle??

    so does this mean i have to have three separate lagrangians?
    would phi depend on t since the velocity will change ??
    force in the Z direction is
    [tex] m \dot{v_{z}} = q E_{0} \sin(t) + \frac{q}{c} v_{?} \times B_{0} [/tex]
    not sure about the direction of v. I dont think its possible... is it? After i find the force do i have to find the potential V(z)? But nothing in that equation is dependant on z, is it? T is indpednant. However since the velocity is changing isnt v depdnant on z??

    for hte Y direction velocity in the X direction yes?
    [tex] m \dot{v_{x}} = q E_{0} \cos(t) + \frac{q}{C} v_{x} \times B_{0} [/tex]
    again... what about the potnetial and its dependance on t or y??

    for the Y direction
    [tex] m \dot_{y} = \frac{q}{c} (-v_{x}) \times B_{0} [/tex]
    are there right so far?
    Last edited: Feb 28, 2006
  2. jcsd
  3. Mar 1, 2006 #2


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    If the fields are constant, why did you put a time dependence for the electric field strength...?

  4. Mar 1, 2006 #3
    i was meaning to correct that ...
    it should jsut be some angle, phi, shouldnt it???
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