Magnetization - Aetherwisp

Magnetization is the phenomenon where a paramagnet or diamagnet that is subject to a magnetic field will develop an magnetic dipole moment for as long as the field is active.

A measure of magnetization can be given as

\mathbf{M}\equiv\text{magnetic dipole moment per unit volume}\qquad\left[ \frac{\text{A}}{\text{m}^{2}} \right]

Magnetic field of a magnetized object#

The magnetic vector potential of a Magnetic dipole of moment $\mathbf{m}$ is

\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi} \frac{\mathbf{m}\times \hat{\boldsymbol{\mathfrak{r}}}}{\mathfrak{r}^{2}}

In the magnetized object, each volume element $d\tau'$ carries a moment $\mathbf{m}=\mathbf{M}d\tau'$ so

\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int \frac{\mathbf{M}(\mathbf{r}')\times \hat{\boldsymbol{\mathfrak{r}}}}{\mathfrak{r}^{2}}\ d\tau'

We could find the Curl of this and get the field itself, but we can use

\nabla'\left( \frac{1}{\mathfrak{r}} \right)=\frac{\hat{\boldsymbol{\mathfrak{r}}}}{\mathfrak{r}^{2}}

to cast this integral in a more illuminating form. We get

\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int\left[ \mathbf{M}(\mathbf{r}')\times\nabla'\left( \frac{1}{\mathfrak{r}} \right) \right]\ d\tau'

Integrating by parts we get

\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\left\{ \int \frac{1}{\mathfrak{r}}[\nabla'\times \mathbf{M}(\mathbf{r}')]\ d\tau' -\int \nabla'\times\frac{[\mathbf{M}(\mathbf{r}')]}{\mathfrak{r}}\ d\tau' \right\}

The second integral can be turned into a surface integral, so

\mathbf{A}(\mathbf{r})= \frac{\mu_{0}}{4\pi}\int \frac{1}{\mathfrak{r}}[\nabla'\times \mathbf{M}(\mathbf{r}')]\ d\tau' -\frac{\mu_{0}}{4\pi}\oint \frac{1}{\mathfrak{r}}[\mathbf{M}(\mathbf{r}')\times d\mathbf{a}']

The first term is essentially the potential of a volume current density $\mathbf{J}$ :

\boxed{\mathbf{J}_{b}=\nabla\times\mathbf{M}}

and the second one looks like the potential of a surface current density $\mathbf{K}$ :

\boxed{\mathbf{K}_{b}=\mathbf{M}\times \hat{\mathbf{n}}}

so if we plug these in

\boxed{\mathbf{A}(\mathbf{r})=\frac{\mu_{0}}{4\pi}\int_{\mathcal{V}} \frac{\mathbf{J}_{b}}{\mathfrak{r}}\ d\tau'+ \frac{\mu_{0}}{4\pi}\int_{\mathcal{S}} \frac{\mathbf{K}_{b}}{\mathfrak{r}}\ da'}

The currents $\mathbf{J}_{b}$ and $\mathbf{K}_{b}$ are called the bound currents and are completely analogous to the bound charges caused by dielectric polarization.

>Now, the magnetic field produces by a uniform surface charge $\sigma$ corresponds with the current surface density >$$\mathbf{K}=\sigma \mathbf{v}=\sigma \omega R\sin \theta \hat{\boldsymbol{\phi}}

but that means that the field of a uniformly magnetized sphere is equal to that of spinning charged sphere with $\mathbf{M}=\sigma R\boldsymbol{\omega}$ . We already know what the field of a spinning sphere is, so we get

>inside the sphere, whereas outside it is equal to that of a perfect magnetic dipole with >$$\mathbf{m}=\frac{4}{3}\pi R^{3}\mathbf{M}

Bound currents#

Analogously to dielectric polarization, bound currents arise as the collective excitation of many atoms and molecules, each of which acts as a magnetic dipole. Each, then, is a tiny loop of current perpendicular to the axis of magnetization, bordering all the other loops next to it. An edge of a loop will have current flowing in the opposite direction of the adjacent loop. In the case of uniform magnetization, the intensity of this dipole current is identical for all dipoles, so they cancel out. This occurs across the entire chunk of material, except at the edges. There, there is no adjacent dipole, so some of the current is not cancelled out. When you take every single edge dipole into consideration, this creates an effective ring or ribbon of current traversing the sides of the material, with the same intensity as that of a singular dipole.

Say each of the loops has area $a$ and thickness $t$ , under a magnetization $\mathbf{M}$ . It's magnetic dipole moment is $m=Mat$ . The circulating current is $m=Ia$ , so together we get $I=Mt$ , so the surface current is $K_{b}=\frac{Mt}{t}=M$ . In vector form, this current revolves around the dipole moment vector, which is easily expressed by taking the normal vector $\mathbf{\hat{n}}$ from the area $a$ and using the cross product:

\mathbf{K}_{b}=\mathbf{M}\times \mathbf{\hat{n}}

Say now the magnetization is not uniform (and on the $z$ axis). The internal cancellation is thus not perfect. On the surface the touch (say, on the $x$ axis), the current is

I_{x}=[M_{z}(y+dy)-M_{z}(y)]dz=\frac{ \partial M_{z} }{ \partial y } dydz

and the corresponding volume current density is

J_{b,x}=\frac{ \partial M_{z} }{ \partial y }

Similarly, a non-uniform magnetization on the $y$ axis would create add a current $-\frac{ \partial M_{y} }{ \partial z }$ , for a total of

J_{b,x}=\frac{ \partial M_{z} }{ \partial y } -\frac{ \partial M_{y} }{ \partial z }

The same can be applied on each edge, which returns the familiar formula for the cross product components. Thus

\mathbf{J}_{b}=\nabla\times\mathbf{M}

Notably, this abides by charge conservation $\nabla \cdot\mathbf{J}_{b}=0$ as the Divergence of curl is always zero.

In linear media#

Most substances exhibit a linear proportionality between the magnetic field applied onto them and the magnetization they show. This relation is usually expressed in terms of the auxiliary field $\mathbf{H}$ instead of $\mathbf{B}$ :

\mathbf{M}=\chi_{m}\mathbf{H}

where $\chi_{m}$ is the magnetic susceptibility. Reversing the relation to get $\mathbf{B}$ from $\mathbf{H}$ we get:

\mathbf{B}=\mu_{0}(\mathbf{H}+\mathbf{M})=\mu_{0}(1+\chi_{m})\mathbf{H}=\mu \mathbf{H}

with $\mu$ the permeability of the substance.

The volume bound current density in a homogeneous linear material is proportional to the free current:

\mathbf{J}_{b}=\nabla\times\mathbf{M}=\nabla \times(\chi_{m}\mathbf{H})=\chi_{m}\mathbf{J}_{f}

In particular, unless the free current flows inside the material, then there will be no bound current.