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Electromagnetics help

  1. Feb 22, 2005 #1
    Finding E and B fields from forces experinced by a test charge with three diferent velocities


    F1=qEo(ax-ay+az) for v1=Uo*ax
    F2=qEo(ax-ay-az) for v2=Uo*ay
    F3=0 for v3=Uo*az

    I used Lorenz force equation and by reduce E by subtracting equation 1 from 2 and eq. 3 from 2 I got

    Uo(ay-ax) X B = Eo(-2az)
    Uo(ay-az) X B = qEo(ax-ay-az)

    So

    B = C(-2az) X (ax-ay-az) = -C

    Then I solved For C and for B. I not sure that I got right to this point. Can someone help me with that problem. Thanks
     
  2. jcsd
  3. Feb 22, 2005 #2
    I don't understand what you have done. What is the reason for doing F1-F2 and so on.

    I would have used the equation for lorenz force to 3 individual cases.

    F = q( E + v x B)
    qE0 ( a, -a, a) = q(Ex, Ey, Ez) + (a U0, 0, 0) X (Bx, By, Bz) ----------(1)

    Write similar equations for case 2 and 3.

    You have 9 equations to solve for 6 unknowns.
     
  4. Feb 24, 2005 #3
    Sorry, This is a problem.:
    The forces experienced by a test charge q at a point in a region of electric and magnetic fields, respectivlely, are given with three diferent velocities, where Uo and Eo are constants. Find E and B at that point. Use Lorenz force equation.
    given:
    F1=qEo(ax-ay+az) for v1=Uo*ax
    F2=qEo(ax-ay-az) for v2=Uo*ay
    F3=0 for v3=Uo*az

    From Lorenz force equation I have
    qE+q*Uo(ax) X B = qEo(ax-ay+az)
    qE+q*Uo(ay) X B = qEo(ax-ay-az)
    qE+q*Uo(az) X B = 0

    I reduce q and E by subtracting equation 1 from 2 and eq. 3 from 1. I got

    Uo(ay-ax) X B = Eo(-2az)
    Uo(ax-az) X B = Eo(ax-ay+az)

    so if (-2az) and (ax-ay+az) is perpendicular to B then (ay-ax) and (ax-az) is parallel to B so wwe will have

    B = C(-2az) X (ax-ay+az) = -C (is this right?)

    where C isa proportionality constant. I solved For C and for B. I not sure that I got right to this point. I never did your way so I'm trying the way I know it can be solve. I hope You will understand now better. Thanks
     
  5. Feb 25, 2005 #4
    This is wrong. If this is the case then (ay-ax) X B = 0 and (ax-az) X B = 0 which are not true.

    Any way your idea is right.


    This is wrong. Check again. Right hand side should have x and y. This is a cross product.

    When you do this right, subtitute it in your lorentz force equatins. Equating the coefficient of the x, y, z components should give you B, and E.
     
  6. Feb 25, 2005 #5
    So is this should look like this?

    B = C(-2az) X (ax-ay+az) = -C(ax-ay)

    Substituding

    Uo(ay-ax) X -C(ax-ay) = Eo(-2az)

    -UoC(2az) = Eo(-2az)

    C = Eo/Uo

    Then B = (Eo/Uo)(ax-ay) Is this ok? I really need help with that? Thanks
     
  7. Feb 25, 2005 #6
    B = C(-2az) X (ax-ay+az) = -2 a^2 C(x+y). Since 'a' is a constnat, say

    B = D (x+y)

    Substituting

    Uo(ay-ax) X D(x+y) = Eo(-2az)
    Uo a D (-2z) = Eo(-2az)
    D = Eo/Uo

    Therefore B = Eo/Uo (x+y)

    Do similar thing to find E. Don't make mistake when you take the cross product.

    I am getting,

    E = (aEo, -aEo, 0) or in your notation E = (aEox - aEoy)
     
    Last edited: Feb 25, 2005
  8. Feb 26, 2005 #7
    Thanks for help. I was reading some book and they show another way to do it. From these two equations

    Uo(ay-ax) X B = Eo(-2az)
    Uo(ax-az) X B = Eo(ax-ay+az)

    we can use componets and solve

    ax ay az
    Uo -1 1 0 = Eo(-2az)
    Bx By Bz

    Then

    Uo[Bz*ax+By*az - Bx*az+Bz*ay] = Eo(-2az)
    So:
    By-Bx = -Eo/Uo(az) and Bz=0

    Then I will do same thing for second equation. Is this right way to do it too? Is there simplers way to do that kind problems? Thanks
     
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