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?

Thank you in advance


1 Answer 1


It's all right except the very last part:

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.

You can check this out https://en.wikipedia.org/wiki/Magnetism in the section called "Magnetic fields in a material"

  • $\begingroup$ Thank you very much, I will read on that topic. Thanks a lot!!! $\endgroup$
    – Pablo
    May 30, 2020 at 10:14
  • $\begingroup$ You are wellcome! :) $\endgroup$
    – Matteo
    May 30, 2020 at 10:25

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