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A particle moving in an electromagnetic field

  1. Nov 6, 2009 #1
    Dear Friends

    A particle having charge q and mass m moves in a region where
    an electric field
    E = ( 0 , E_0 , 0 )
    and a magnetic one
    B = ( 0 , 0 , B_0 )
    exist.

    For t = 0 its position is
    R = ( 0 , 0 , 0 )
    and its velocity is
    V = ( V_0 , 0 , 0 ).

    I have worked out the equations of motion

    (
    you can see them and their solution in a fine graphical form at
    http://i786.photobucket.com/albums/yy150/simona1989_photo/img.jpg [Broken]

    In text only mode they are:

    m x''[t]==B0 q y'[t]
    m y''[t]==q (E0-B0 x'[t])
    m z''[t]==0

    x[t] = ( B0 E0 q t-m (E0-B0 V) Sin[(B0 q t)/m] ) / ( B_02 q)
    y[t] = ( 2 m (E0-B0 V) Sin[(B0 q t)/(2 m)]2 ) / ( B_02 q )
    z[t] = 0
    )

    but I get a strange and obviously wrong phenomenon in the solution:
    they correctly give the solution of the case E_0 = 0 but they have no meaning for B_0 = 0 (because B_0 appears into denominator of solutions).

    Please note that the equations of motions of the general case ( E_0 and B_0 both not zero ) reduces normally at the right equations of motions for E_0 = 0 and B_0 = 0.

    Please can you suggest an hint for the reason of that, or the wrong step in my reasoning ?

    If you feel confortable with Mathematica, may be you want see, at bottom of this message, the Mathematica code I have used to confirm my hand written solutions.

    Thanks for your attention!

    Warmest regards
    Barbara.Ba

    ------------------------------------------------------------------------------
    (* position at t *)
    pos = {x[t], y[t], z[t]};

    (* velociy' and acceleration t *)
    vel = (D[#, {t, 1}]) & /@ pos;
    ac = (D[#, {t, 2}]) & /@ pos;

    (* the given fields *)
    cE = {0, E0, 0};
    cB = {0, 0, B0};

    (* the Lorentz force for the given fields *)
    F = q (cE + Cross[vel, cB]);

    (* the equations of motions *)
    eq = Table[m ac[[k]] == F[[k]], {k, 1, Length[ac]}];

    (* position and velocity for t = 0 *)
    cond = {x[0] == 0, y[0] == 0, z[0] == 0, x'[0] == V, y'[0] == 0, z'[0] == 0};

    (* this solves the equations of motions *)
    sol = DSolve[{eq, cond}, pos, t] // Simplify;

    (* output *)
    Print["\n\n--- equations of motions ---\n\n", eq // TableForm];
    Print["\n\n--- solutions ---\n\n", Transpose[sol] // TableForm];
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 6, 2009 #2
    This is a good question. A crossed field (crossed dc E and B fields) is used to filter out charged particles with different velocities in a particle beam with defined momentum. This is called a Wein filter. See
    http://titan.triumf.ca/equipment/beamline/wienfilter.shtml
    (correct spelling is Wein). Also search web.
    Also, crossed fields are used in magnetron microwave tubes, like the ones used in microwave ovens. The electrons are accelerated away from the filament by the E field, and rotate around the cavity, deflected into a circular orbit by the B field.

    Bob S
     
  4. Nov 10, 2009 #3
    Sorry for mistake.

    Despite B_0 into the denominator the whole functions doesn't diverge to infinity.
    I got the right limit by means of Mathematica.

    Warmest Regard
    Barbara
    (Rome)
     
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