Find the magnetic field inside the cylinder

[CptXray]

Homework Statement

There's a very long cylinder with radius $R$ and magnetic permeability $\mu$. The cylinder is placed in uniform magnetic field $B_{0}$ pointed perpendicularly to the axis of cylinder. Find magnetic field for $r < R$. Assume there's a vacuum outside the cylinder.

Homework Equations

Boundary conditions:
$$B_{2} \cdot \hat{n} = B_{1} \cdot \hat{n}$$
$$B_{2} \times \hat{n} =\frac{\mu_{2}}{\mu_{1}} B_{1} \times \hat{n},$$
where $\hat{n}$ is a unit radial vector.Laplace equation:
$$\Delta \phi_{M} = 0,$$
where $\phi_{M}$ is a scalar magnetic potential.

The Attempt at a Solution

I was trying to solve it but I think something's missing here. I mean, in electrostatics I know that some potential $\phi = 0$ on a conducting electrically neutral surface. Should I assume here that there's some current density on the surface of the cylinder?

 

[Charles Link]

For this geometry, there is a well-known demagnetizing factor $D=\frac{1}{2}$. It can be solved by Legendre methods that will get the same result, but the easiest way is to use the demagnetizing factor. (The demagnetizing factor allows for the computation of a known uniform "magnetic field" $H_D$ inside the cylinder that results from the magnetic surface charge $\sigma_m=M \cdot \hat{n}$, analogous to the electrostatic problem with the same geometry). $\\$ To show you how the demagnetizing factor works, let's assume units where $B=\mu_o H+M$. Then $H_{int}=H_o +H_D$. $\\$ Next $H_D=-\frac{D M}{\mu_o}$, and $M=\mu_o \chi_m H_{int}$ where $\chi_m=\mu_r-1$, with $B=\mu H=\mu_o \mu_r H$. $\\$ You then solve for $H_{int}$ and $M$. (Basically two equations and two unknowns). You then compute $B_{int}$. $\\$ Edit: Here is a thread where the demagnetizing factor 1/2 is essentially computed for this problem with a unique method by Griffiths. https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930 $\\$ The alternative is to go the complete Legendre route, but it can be somewhat painstaking. $\\$ It might also be of interest that for a spherical geometry $D=\frac{1}{3}$, and for a flat disc $D=1$.