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Magnetic Field around two magnetic boxes

  • Thread starter DaniV
  • Start date
29
3
Problem Statement
Two magnets with magnetization Mz $\hat$ are placed on a surface parallel to
the xy plane, and are almost adjacent to each other, as illustrated. The
length and width of each magnet is w, the height is h, and the distance
between the magnets is d. w >> h >> d, and you can assume that w → ∞.
Find the magnetic field everywhere in space (above and below the magnets,
inside the magnets, and between them).
Relevant Equations
this question using fictive charge method:
the basic equation is $B=\mu_{0}(M\vec+H\vec)$
$\vec H =-\nabla\psi$
$\psi =\frac{1}{4\pi}\intagral \integral\integral\frac{\ro(x)_{fictive}}{|\vec x-\vec x`| }d^3x'$+\frac{1}{4\pi}\integral \frac{\sigma(x)_{fictive}}{|\vec x -vec x'|}dl'$
I tried to look once at the zy axis and saw a two infinite capacitors with fictive charge density of M on the upper plane, and -M in the lower with a distance of h from each other, the two capacitors saparated with d in the y axis,

but when I look in xy axis there was 2 another capcitors the first with the same fictive charge density M on the left and right plane with distance of d between them, the second one sapareted by h in the z axis show another capacitor with -M on the right and left planes

how I preform such a superposition of fields because there are 4 planes and for which we can preform 2 different ways to look on the same problem (different capacitors) to find $\vec H$?
 

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Charles Link

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This is kind of an odd problem. If ## w >>h ## and ## w>>d ##, basically the magnetic field ## \vec{B} ## will be zero everywhere. ## \\ ## The ## H ## from the plus magnetic charges (fictitious) will (approximately) cancel the ## H ## from the minus magnetic charges, except in the material where the ## H=-M ##, so that ## B=\mu_o(H+M)=0 ## also. I know I'm not supposed to furnish the answer, but that is the only way I can supply the necessary feedback here. ## \\ ## Note: on the faces parallel to the x-y plane, ## \sigma_m=\vec{M} \cdot \hat{n} ##, (comes from ## -\nabla \cdot M=\rho_m ## and Gauss' law), and also ## B=\mu_o(H+M ) ## , with ## \nabla \cdot H=- \nabla \cdot M ## gives with ## H_{outside}=0 ## that ## H_{inside}=-M ##, by Gauss' law. ## \\ ## Perhaps someone else may see a different solution, but this is what I get from working it.
 

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