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The solution of the magnetic field ## \vec{B} ## can be computed by both the magnetic pole method and by the magnetic surface current method. Both methods yield identical results for the vector ## \vec{B} ##. We will summarize the two methods below:

1) Magnetic pole method: Magnetic pole density (fictitious) is given by ## \rho_m=-\nabla \cdot \vec{M} ##. The result for uniform magnetization ## \vec{M}=M_o \hat{z} ## is simply a magnetic surface charge density at the end faces of the cylinder given by ## \sigma_m=\vec{M} \cdot \hat{n} ## resulting in + and - poles at the end faces of the cylinder. The vector ## \vec{H}## is then computed everywhere as ## \vec{H}(\vec{x})=\int \frac{ \sigma_m (\vec{x}-\vec{x}')}{4 \pi \mu_o |\vec{x}-\vec{x}'|^3} \, dA ## and ## \vec{B}=\mu_o \vec{H}+\vec{M} ##. This last equation takes some work to prove in detail, but we will simply use it in the solution for ## \vec{B} ##. It has in fact been proven.

2) Magnetic surface current method: Magnetic surface current density ## \vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o} ## . For a uniform ## \vec{M}=M_o \hat{z} ## the result is a surface current density per unit length ## \vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o} ## on the outer surface of the cylinder, in the same geometry as a solenoid. The magnetic field ## \vec{B} ## is then found everywhere inside and outside the cylinder using Biot-Savart as ## \vec{B}(\vec{x})=\frac{\mu_o}{4 \pi} \int \frac{ \vec{K}_m \times (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3 } \, dA ##, where this ## dA ## is over the cylindrical surface, unlike the previous ## dA ##, where the integration is over the end faces. ## \\ ##

This is an important problem in magnetostatics, and it is hoped that the E&M (electricity an magnetism) students at the upper undergraduate level and higher find it of interest.