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I am working in a project where I am using different shaped permanent magnets for levitation of diamagnets. I am facing problem while calculating the magnetic field around these permanent magnets. To be specific I am using cylindric magnet and cubic ($\rm NdFeB$) magnets. And I want to find the magnetic field around these shapes of magnet. Only the remanence and the size of the magnet of the are provided. I have come across a formula for cylindrical magnet where it calculates the magnetic field along the symmetry axis of the magnet. But I want to get the magnetic field at any point in space around the cylindrical magnet. Here is the formula:

$$B=\frac{B_r}{2}\left ( \frac{D+z}{\sqrt{{R}^2+{(D+z)}^2}} -\frac{z}{\sqrt{R^2+z^2}}\right )$$

Where $B_r$ is the Remanence field, D is the height of cylinder, R is the radius of cylinder, z is the distance from a pole face on the symmetry axis.

I tried first finding out the scalar potential Φaxis(z) and expanding it to Φ(r, θ) using Legendre polynomial but it doesn’t work. I have already come across $$B_Z,B_X,B_Y (x,y,z)$$ for the cube from one of the paper means now I can easily find $$B(x,y,z)=\sqrt{B_x^2+B_y^2+B_z^2}$$ Can anyone help me to find B(x,y,z) for the cylinder ?

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  • $\begingroup$ One can calculate the magnetic field of an arbitrary current distribution by integration alone, but as soon as magnetic materials are in play one has to solve a self-consistent set of equations that include the magnetization of the material. This is best done numerically. There is plenty of software around for that purpose. $\endgroup$ Commented Oct 8, 2022 at 0:43

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The field of a cylindrical magnet can be viewed as originating from a sheet of current flowing around the perimeter of the bar (the dipole moment/unit volume should be the same as that of the magnet). The field at any point can be approximated by doing a numeric integration (for each component) of the contributions from each little segment of that current. Think of it like a current carrying solenoid. (I'm thinking that this approach could also be used with the cube.)

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