# Magnetic scalar potential far above a magnetic film

The situation I am looking at is a magneto-static problem of a finite magnetic film with magnetization $\bf{M}$. I would like to find the the magnetic field far above the plate. My expectation is that far above the plate, the field should approach the dipolar field which scales as $1/r^3$. However my numerical calculations are yielding $1/r^4$.

I am using Jackson's book (3rd ed.) as a guide. I assume discontinuous magnetization at the surface so I can use Jackson's Eq. 5.100 which is the magnetic scalar potential:

$$\Phi_M (\bf{x}) = \frac{1}{4 \pi} \oint_S \frac{\bf{n}' \cdot \bf{M}(\bf{x}')}{|\bf{x}-\bf{x}'|}da'$$ assuming that the magnetization $\bf{M}(\bf{x}')$ is uniform within the volume. The top and bottom surfaces of the film will contribute to the potential. If I focus just on the top surface and assume a normalized magnetization $\bf{M} = M \hat{z}$:

$$\Phi_M (\bf{x}) = \frac{1}{4 \pi} \oint_{top} \frac{ M}{|\bf{x}-\bf{x}'|}da'.$$

I do not see where to go from here or how the expected $1/r^2$ dependence will arise.

The field you are interested in is always calculated as some derivative of the potential. I don't have Jackson handy, but I'm guessing in the magnetostatic case, you take ${\bf H} = -\nabla \Phi_M$ to calculate the fields. The gradient of a $1/r$ quantity is something like a $1/r^2$ quantity. You can figure this out exactly by taking the gradient in the appropriate coordinate system. In any case, you'll get a term like $\frac{\partial}{\partial r}\frac{1}{r}=-\frac{1}{r^2}$

To do a "far away" expansion, start by figuring that if your surface is small compared to how far away the observation point $\bf x$ will be, then $|{\bf x-x'}|\approx|{\bf x}|$. This probably isn't good enough to get an answer, so let's look at one order of approximation of the effect of the surface, so say your surface is in the $z=0$ plane, kind of centered on the $z$-axis, then

$$|{\bf x-x'}| = \sqrt{(x-x')^2+(y-y')^2+z^2}= z\sqrt{1+\frac{(x-x')^2+(y-y')^2}{z^2}}$$ Then approximate your integrand using the first order series $\left(1+\epsilon\right)^{-1/2}\approx 1-\epsilon/2$ $$\frac{M}{|{\bf x-x'}|}\approx \frac{M}{z}\left(1-\frac{(x-x')^2+(y-y')^2}{2z^2}\right)$$

The integral over the primed coordinates will depend on the specifics of your surface, but you can already see the leading order $1/z$ behavior in the potential. There is a $1/z^3$ correction term as well. It's clear then that near the $z$-axis, far away, $z\approx r$ and you get potentials that fall of as the inverse distance. Taking the gradient of this should get you the appropriate $1/r^2$ field fall-off, and $1/r^4$ correction term that matters near the plate. You'd of course have to crunch a simple integral over the primed coordinates to get these near field specifics, and even then, within the range of validity of the approximation we made, which is that $(x-x')^2+(y-y')^2\ll z^2$, i.e. near the $z$-axis.

The geometry of the plate doesn't really matter in this, as long as it's finite, meaning we can get far away above it so it looks "small" when we look back at it.

As for higher order field fall-off, I know these kinds of effects can occur when two types of sources are near each other in such a way that they cancel to a first order approximation. This could be the case with your second "bottom" surface; if the magnetization on the bottom surface is the opposite of the top, I think the first order $1/z$ dependece of the potentials will cancel, leaving you $1/z^3$ potentials. Taking the gradients of these will give $1/z^4$ fall off directly over the finite plate.

Finally, given the potential formulation you are using, I would not expect dipole like $1/r^3$ field falloff. The potential has to be like $1/r^2$ to get $1/r^3$ field fall off, whereas the potential you are using is $1/r$, like a magnetic monopole. Is this the right formulation for your magnetization? I'm not sure, but Wiki has the magnetostatic scalar potential as falling off as $1/r^2$ to begin with, which would give you your expected dipole-like $1/r^3$ field.

Your numerical simulation must be wrong. The long distance field of any uniformly magnetised material approaches that of a dipole. For your case at hand it is easy to see how this happens: Your thin film can be seen as a collection if small dipoles, the "positive magnetic charges" on the upper face of the film and the negative ones in the lower face. We can treat these charges, methematically as electric charges to compute the scalar magnetic potential. Now we have a problem in electrostatics. It is also well known that in electrostatics there is a multipole expansion of the field

point total charge + dipole + cuadrupole + ....

The total charge is 0 on account of $\nabla \cdot B =0$ everywhere. Now, for the dipole contribution to vanish one would need a total dipole moment zero, in this case the cuadrupole dominates and you get a $1/r^4$ for the field.

But this is not your case. You do have a total magnetic moment. It is ${\bf p} =Ad\bf M$

$A$ is the area of the film. Your potential at large distance is

$\Phi_M (r, \theta)= p\cos \theta /r^2$ This result can also be derived from the formula in Jackson:

Consider integration on both faces, parametrise the vector position of the lower face as ${\bf x}'' = {\bf x}' - \hat{z} d$ where $d$ is the thickness of the film and the $z$ axis is normal to the film's large surface, in this way you can carry the integration on both faces simultaneously:

$$\Phi_M = M \int_{upper ~face} \left( \frac{1}{|{\bf x} -{\bf x}'|} - \frac{1}{|{\bf x} -{\bf x}' + d \hat{z}|}\right) d{\bf a'}$$

Now use Taylor expansion considering $d <<|{\bf x}-{\bf x}'|$ the first term cancels, and the leading term is

$$\Phi_M = M d \int_{upper ~face} \frac{({\bf x}-{\bf x}')\cdot \hat{z}}{|{\bf x}-{\bf x}'|^3} d{\bf a'}$$

For a very large $|{\bf x}-{\bf x}'| =r$ the integral is approximated, the vector ${\bf x}-{\bf x}'$ forms an almost constant angle $\theta$ with the $z$ axis, so that $({\bf x}-{\bf x}')\cdot \hat{z} \simeq r \cos \theta$ and the area interal gives $A \cos \theta / r^2$ which results into

$$\Phi_M (r, \theta) \simeq \frac{MAd\cos \theta}{r^2}$$

This will give fields that fall off as $1/r^3$, not $1/r^4$.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

Maybe I've a confusion about the question. Let's the film rest in $\verts{z} < d/2$ and the magnetization $\vec{\rm M}\pars{\vec{r}}$ is uniform inside the film: $\vec{\rm M}\pars{\vec{r}} = M_{0}\Theta\pars{d/2 - \verts{z}}\hat{z}$. Then, the 'magnetization current' $\vec{\rm J}_{\rm M}\pars{\vec{r}}$ is given by: $$\vec{\rm J}_{\rm M}\pars{\vec{r}} = c\nabla\times\vec{\rm M}\pars{\vec{r}} = cM_{0}\,\delta\pars{{d \over 2} - \verts{z}}\bracks{-\sgn\pars{z}} \quad\overbrace{\color{#ff0000}{\Large\hat{z}\times\hat{z}}}^{{\Large =\ \vec{0}}}$$ Then, the vector potential vanishes out and there is no magnetic field $\pars{~\mbox{when}\ \vec{\rm M}\pars{\vec{r}} \parallel \hat{z}~}$.