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Magnetic moment of a particle suspended in a magnetic field?

  1. Oct 15, 2014 #1
    Derive the following expressions for ux, uy and uz:
    ux(t) = ux(0) cos(wo t) + uy(0) sin(wo t)
    uy(t) = - ux(0) sin(wo t) + uy(0) cos(wo t)
    uz(t) = uz(0)
    wo is angular frequency.
    wo = g Bo

    2. Relevant equations
    wo of this precession is proportional to Bo
    u and J have the same orientation:
    u = g J, where g is gyromagnetic ratio.
    du/dt = u X g B (where X is cross product)

    3. The attempt at a solution
    u
    = ux i + uy j + uz k
    B
    = B k
    u
    X B = (ux i + uy j + uz k) X (B k) = -ux B j + uy B i
    u
    X g B = -ux g B j + uy g B i = uy g B i - ux g B j
    du/dt = u X g B
    du/dt = uy g B i - ux g B j
    Therefore,
    dux/dt = uy g B
    duy/dt = -ux g B
    duz/dt = 0

    I do not know how to proceed. From duz/dt = 0, we get uz = constant = uz(0)
    But how to derive the expressions for ux and uy?
     
  2. jcsd
  3. Oct 20, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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