# Ohm's law and current due to magnetisation

I'm reading Landau's Electrodynamic of continuous media, specifically the following paragraph of section §29 (The magnetic field of a constant current):

If a conductor carries a non-zero total current, the mean current density in it can be written as $\rho{\bf v}= c\ {\bf curl}\ {\bf M}+{\bf j}$. The first term, resulting from the magnetisation of the medium, makes no contribution to the total current, so that the net charge through a cross-section of the body is given by the integral $\int {\bf j}\cdot d{\bf f}$ of the second term. The quantity ${\bf j}$ is called the conduction current density.$\dagger$ The statements made in §20 apply to this current; in particular, the energy dissipated per unit time and volume is $\bf{E}\cdot\bf{j}$.

In section §20 appears the Ohm's law. My question is: why magnetisation current ${\bf j_M}= c\ {\bf curl}\ {\bf M}$ doesn't contribute to Ohm's law?

In other words: let ${\bf j_t}={\bf j_f}+{\bf j_b}$ be the total current density, with ${\bf j_b}=c\ {\bf curl}\ {\bf M}$ the bounded (magnetisation) current density. Is ${\bf j_f}=\sigma {\bf E}$ the Ohm's law?

• @adiselann because the charges producing the magnetization current $curlM$ do not collide with or scatter off anything, especially not the ions of the conducting material. And of course the external electric field $E$ has no effect on the bound current whatsoever. – hyportnex May 30 '17 at 22:21