# $\mu \rightarrow \infty$ in magnetized cylinder

Please don't see this as a homework question, for I have already solved the problem and all I'm trying to do is understand the results and make a correct interpretation. In a cylinder made out of magnetic material without free currents, with magnetization $$\boldsymbol{M} = M_0 \frac{r}{a} \boldsymbol{u}_{\phi}$$ one can easily see that the bound currents are given by $$\boldsymbol{J}_b = \frac{2}{a} M_0 \boldsymbol{u}_z$$ inside the cylinder and $$\boldsymbol{k}_b = - M_0 \boldsymbol{u}_z$$ at $$r = a$$, being $$a$$ the radius of the cylinder. In this way, as $$\boldsymbol{\nabla} \times \boldsymbol{B} = \mu_0 \boldsymbol{J}_b,$$ we get that $$\boldsymbol{B} = \frac{\mu_0 r}{a} \boldsymbol{u}_{\phi} = \mu_0 \boldsymbol{M},$$ and vanishes out of the cylinder. This leaves us with $$\boldsymbol{H} = \vec{0}$$, am I right?

So, what happens with $$\mu$$? In order to get a non-trivial solution for $$\boldsymbol{B}$$ in $$\boldsymbol{B} = \mu_0 \left( \boldsymbol{H} + \boldsymbol{M} \right) = \mu \boldsymbol{H}$$ we get $$\mu \rightarrow \infty$$, right? What is this?

In order to get a non-trivial solution for B in $$\bf{B}=μ_0(\bf{H}+\bf{M})=μ\bf{H}$$ we get μ→∞, right?
This is not true in general. This is true when the magnetization is induced by an external field inside a material with a linear magnetic response (standard paramagnetism or diamagnetism). In this case you can indeed write $$\bf{M} = \chi \bf{H}$$ from which you get the relation you have mentioed: $$\bf{B} = \mu_0(\bf{H} + \bf{M} ) = \mu_0(1+\chi) \bf{H} = \mu \bf{H}.$$
In the exercise though you are in a different situation, where the magnetization inside the cylinder is not induced by an external field $$\bf{H}$$, so the relations $$\bf{M} = \chi \bf{H}$$ and $$\bf{B} = \mu \bf{H}$$ are no longer true. In other words the cylinder is ferromagnetic.