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Homework Help: Electric Field from Permanent Magnet

  1. Dec 20, 2009 #1
    Hello,

    I've been extremely stuck on the following problems and was hoping someone could give me a push in the right direction:

    1. The problem statement, all variables and given/known data

    1) Given an infinite slab of permanently magnetized matter of thickness d centered on the xy-plane with uniform magnetization [tex]\textbf{M} = (0, M, 0)[/tex] and velocity [tex]\textbf{v} = (v, 0, 0)[/tex], find the electric field at [tex]\textbf{E}(0, 0, 0)[/tex] and [tex]\textbf{E}(0, y, 0)[/tex] where y > d.

    2) A magnetized sphere with uniform magnetization [tex]\textbf{M} = (0, 0, M)[/tex] and radius r is spinning at a rate of [tex]\textbf{\omega} = (0, 0, \omega)[/tex]. Find the electric field for r' > r. (Hint: Find the equivalent magnetic charge density, [tex]\rho_m[/tex], and equivalent surface current, [tex]\sigma_m[/tex].)

    2. Relevant equations

    I'm not entirely sure (hence the thread)!

    [tex]\sigma_{m, n} = \textbf{M} \cdot \textbf{n}[/tex]

    [tex]\rho_{m} = - \nabla \cdot \textbf{M}[/tex]

    ...these are factors of the integrand that give rise to the magnetic scalar potential, [tex]\Omega[/tex], which in turn yields [tex]\textbf{B}[/tex] via [tex]\textbf{H} = - \nabla \Omega[/tex].

    3. The attempt at a solution

    I'm desperately stuck on these; for both problems I can find [tex]\rho_m[/tex] and [tex]\sigma_m[/tex], but I don't see the connection to the [tex]\textbf{E}[/tex]-field. Any suggestions to get me started would be greatly appreciated.
     
    Last edited: Dec 21, 2009
  2. jcsd
  3. Dec 21, 2009 #2
    Well, after doing some digging I found the following problem (7.60) in Griffiths' Electrodynamics:

    Maxwell's equations are invariant under the following duality transformations

    [tex]\textbf{E'} = \textbf{E} cos(\alpha) + c \textbf{B} sin(\alpha)[/tex]

    [tex]c \textbf{B'} = c \textbf{B} cos(\alpha) - \textbf{E} sin(\alpha)[/tex]

    [tex] c q_{e}' = c q_{e} cos(\alpha) + q_{m} sin(\alpha)[/tex]

    [tex] q_{m}' = q_{m} cos(\alpha) - c q_{e} sin(\alpha)[/tex]

    ...where [tex]c = 1/\sqrt{\epsilon_0 \mu_0}[/tex], [tex]q_m[/tex] is the magnetic charge and [tex]\alpha[/tex] is an arbitrary rotation angle in "[tex]\textbf{E}-\textbf{B}[/tex] space."

    Griffiths' says that, "this means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using [tex]\alpha = \pi / 2[/tex]) write down the fields produced by the corresponding arrangement of magnetic charge."

    Thus, if I were to solve for [tex]\textbf{E}[/tex] in (1) and (2) with a polarization [tex]\textbf{P}[/tex] instead of a magnetization [tex]\textbf{M}[/tex], I could then use a duality transformation to find the solutions to (1) and (2), correct?
     
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