12.3: Magnetization and Susceptibility

The $H$-field inside a long solenoid is $nI$. If there is a vacuum inside the solenoid, the B-field is $\mu_o H = \mu_o nI$. If we now place an iron rod of permeability $\mu$ inside the solenoid, this doesn't change $H$, which remains $nI$. The B-field, however, is now $B=\mu H$. This is greater than $\mu_oH$, and we can write

$$B = \mu_o(H+M) \label{12.3.1}$$

The quantity $M$ is called the magnetization of the material. In SI units it is expressed in A m^-1. We see that there are two components to $B$. There is the $\mu_o H = \mu_o nI$, which is the externally imposed field, and the component $\mu_oM$, originating as a result of something that has happened within the material.

It might have occurred to you that you would have preferred to define the magnetization from

$$B = \mu_0H + M \nonumber$$

so that the magnetization would be the excess of $B$ over $\mu_0H$. The equation $B = \mu_0H + M$, would be analogous to the familiar

$$D= \epsilon_oE + P \nonumber$$

and the magnetization would then be expressed in tesla rather than in A m^-1. This viewpoint does indeed have much to commend it, but so does

$$B = \mu_o (H+M). \nonumber$$

The latter is the recommended definition in the SI approach, and that is what we shall use here.

The ratio of the magnetization $M$ ("the result") to $H$ ("the cause"), which is obviously a measure of how susceptible the material is to becoming magnetized, is called the magnetic susceptibility $\chi_m$ of the material:

$$M = \chi_m H. \label{12.3.2}$$

On combining this with Equation $\ref{12.3.1}$ and $B = mH$, we readily see that the magnetic susceptibility (which is dimensionless) is related to the relative permeability $\mu_r = \mu/\mu_o$ by

$$\mu_r = 1+ \chi_m \label{12.3.3}$$