I A magnetostatics problem of interest 2

Homework Helper
Gold Member
2018 Award
Summary
The problem of a uniformly magnetized cylinder of finite length is of much interest. There are basically two methods of solution for the magnetic field
$\vec{B}$
that arrive at the same result: The magnetic pole method and the magnetic surface current method.
A very important problem in magnetostatics is the uniformly magnetized cylinder of finite length. Permanent cylindrical magnets can be modeled as having approximately uniform magnetization, and it is of much interest, given such a uniformly magnetized cylinder, to be able to calculate the magnetic field both outside and inside the magnetized cylinder.
The solution of the magnetic field $\vec{B}$ can be computed by both the magnetic pole method and by the magnetic surface current method. Both methods yield identical results for the vector $\vec{B}$. We will summarize the two methods below:
1) Magnetic pole method: Magnetic pole density (fictitious) is given by $\rho_m=-\nabla \cdot \vec{M}$. The result for uniform magnetization $\vec{M}=M_o \hat{z}$ is simply a magnetic surface charge density at the end faces of the cylinder given by $\sigma_m=\vec{M} \cdot \hat{n}$ resulting in + and - poles at the end faces of the cylinder. The vector $\vec{H}$ is then computed everywhere as $\vec{H}(\vec{x})=\int \frac{ \sigma_m (\vec{x}-\vec{x}')}{4 \pi \mu_o |\vec{x}-\vec{x}'|^3} \, dA$ and $\vec{B}=\mu_o \vec{H}+\vec{M}$. This last equation takes some work to prove in detail, but we will simply use it in the solution for $\vec{B}$. It has in fact been proven.
2) Magnetic surface current method: Magnetic surface current density $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$ . For a uniform $\vec{M}=M_o \hat{z}$ the result is a surface current density per unit length $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o}$ on the outer surface of the cylinder, in the same geometry as a solenoid. The magnetic field $\vec{B}$ is then found everywhere inside and outside the cylinder using Biot-Savart as $\vec{B}(\vec{x})=\frac{\mu_o}{4 \pi} \int \frac{ \vec{K}_m \times (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3 } \, dA$, where this $dA$ is over the cylindrical surface, unlike the previous $dA$, where the integration is over the end faces. $\\$
This is an important problem in magnetostatics, and it is hoped that the E&M (electricity an magnetism) students at the upper undergraduate level and higher find it of interest.

• Paul Colby
Related Classical Physics News on Phys.org

Homework Helper
Gold Member
2018 Award
Working the case of the magnetic field $\vec{B}$ inside the uniformly magnetized cylinder of infinite length is perhaps the easiest one to compute by both methods. $\\$ For the pole method case, there are no poles, so $\vec{H}=0$, and $\vec{B}=\mu_o \vec{H}+\vec{M}=\vec{M}$. $\\$ For the surface current method, Ampere's law (in place of Biot-Savart) applies to get the same answer. $\oint \vec{B} \cdot dl=\mu_o I$ gives $\vec{B}=\vec{M}$ because $I=K_m L =M L/\mu_o$.

Last edited:

Homework Helper
Gold Member
2018 Award
Note for the above: The case of the magnetic surface currents for the uniformly magnetized cylinder of infinite length has the same geometry as that of a solenoid of infinite length with surface current per unit length $K_m=\frac{M_o}{\mu_o}$ replacing $nI$ in the formula for the magnetic field $\vec{B}=\mu_o n I \hat{z}$, where $n$ is the number of turns per unit length. Thereby $\vec{B}=M_o \hat{z}$.

Last edited:

Homework Helper
Gold Member
2018 Award
A minor correction to post 1, but I don't want to attempt to edit it, the $dA$ in both cases should be $dA'$.

Meir Achuz

Homework Helper
Gold Member
A full derivation of this insight for finite length is given in the answer (given in the book) to problem 8.3 in Franklin, "Classical Electromagnetism", Vol. II.

• Homework Helper
Gold Member
2018 Award
A full derivation of this insight for finite length is given in the answer (given in the book) to problem 8.3 in Franklin, "Classical Electromagnetism", Vol. II.
I'm glad to see there is at least one E&M textbook that has it. IMO this should be an important part of the undergraduate physics curriculum.

vanhees71

Gold Member
Hm, I have it for the simpler case of a homogeneously magnetized (hard ferromagnet) sphere using both the scalar potential for $\vec{H}$ and the vector potential for $\vec{B}$ and accordingly both the method using the magnetization directly and the surface-current formulation.

I wonder, whether there's a closed solution for the homogneously magnetized cylinder of finite length. I haven't tried yet. It seems to be more complicated, because the magnetic potential depends on both $\rho$ (radial cylinder coordinate) and $z$.

• Homework Helper
Gold Member
2018 Award
@vanhees71 The homogeneously magnetized sphere is also another magnetostatics problem that I think the upper level undergraduate physics students would do well to have studied, and if they really want to challenge themselves, to be able to work it routinely. I do think the universities would do well to place increased emphasis on the advanced E&M courses for the undergraduate physics majors. $\\$ And could you perhaps post a "link" to your solution(s) of the magnetized sphere. It would help to make this thread more complete. Thanks. $\\$ And I don't think a closed-form solution exists for the magnetized cylinder of finite length, but the very interesting thing is that both the "magnetic pole" method and "magnetic surface current" method give the exact same result for $\vec{B}$. They are two very different methods, and the agreement, although shown by mathematical proof to be the same, is quite remarkable.

Last edited:
• vanhees71

vanhees71

Gold Member
The homogeneously magnetized sphere is in my German lecture notes on E&M for highschool-teacher students. This manuscript is matching the needs of this special audience. Unfortunately we have had the Bologna reform in Europe and thus "module plans". For the standard lectures thus we are forced to teach them in a given order, i.e., I'm not free to do "relativity first", because that's part of another module. That's why these lecture notes follow the old-fashioned 19th-century approach to E&M. Otherwise I think they are not too bad:

The homogeneously magnetized sphere can be found in Sect. 3.3.3 (worked out in both ways discussed here).

• Homework Helper
Gold Member
2018 Award
@vanhees71 I worked through a good part of your section 3.3.3 including equations 3.3.18 and 3.3.19 leading up to some of the calculations of section 3.3.3. My German is just good enough that I am able to work through your calculations.
I haven't gone through your entire solution yet, but it really appears to be an E&M work of art. Thank you very much. Edit: The problem is precisely analogous to the dielectric sphere with uniform polarization. Originally, our physics class in college on 1975 was shown the Legendre polynomial solution, but since then Griffith's E&M has a solution to the problem, and the surface polarization charge problem was also done in closed form. Let me see if I can find a couple of "links"... See post 13 of https://www.physicsforums.com/threads/find-the-electric-field-of-a-uniformly-polarized-sphere.877891/#post-5514160 And see post 9 of "Discuss in the Community" of https://www.physicsforums.com/threads/the-homopolar-generator-an-analytical-example-comments.932758/ And see post 2 of https://www.physicsforums.com/threads/electric-field-inside-a-uniformly-polarised-cylinder.942173/#post-5959482 Griffiths works the problem for a cylinder oriented sideways, but no doubt this technique would also work for the uniformly polarized sphere.

Last edited:

Homework Helper
Gold Member
2018 Award
And a follow-on: For those physics students with a keen interest in E&M, the solutions in the "links" of post 10 and post 11 might be of much interest. $\\$ IMO, for the magnetostatic problem, the magnetic surface current method really provides the better solution from a physics sense. The magnetic pole solutions are mathematically quite elegant, and can be shown to give the exact same answer for $\vec{B}$, but IMO, they hide and even confuse the underlying physics.

Last edited:
• JD_PM

Meir Achuz

Homework Helper
Gold Member
The underlying physics is that ferromagnetism comes from the polarization of the intrinsic magnetic moments of electrons, and not from electric currents.

Homework Helper
Gold Member
2018 Award
The underlying physics is that ferromagnetism comes from the polarization of the intrinsic magnetic moments of electrons, and not from electric currents.
I disagree with this interpretation. Many who were taught the magnetic "pole model" subscribe to this. The pole model uses the equation $B=\mu_o H+M$, and assigns a "local" contribution to $B$ that results from the magnetization $M$. It also treats $H$ and $B$ as two different types of magnetic fields. The magnetic surface current method is able to avoid these (magnetic pole) explanations, which I believe are incorrect. In the magnetic surface current explanation, the $H$ is shown to be a mathematical construction, and does not represent an actual magnetic field. In addition, the $M$ in the equation $B=\mu_o H+M$ is shown to come from the contribution of the surface currents that result from any finite distribution of magnetization $M$. In the absence of currents in conductors, for the magnetic field $B$ in the material, the $H$ is simply a correction that appears for shapes other than cylinders of infinite length. $\\$ Outside the material, the $H$ from the magnetic poles does give the magnetic field strength $B$, (depending on the system of units there may be a proportionality constant), but the reasons why this happens to work are open to debate. For any uniformly magnetized cylinder of finite length, (outside the material), the $H$ from the poles (using inverse square law) completely agrees with the $B$ from the magnetic surface currents (using Biot-Savart)... Inside the material (inside the cylinder), a local $M$ needs to be added artificially to the $B$. This is where the pole model, IMO, fails to explain the underlying physics. In the magnetic surface current explanation, the contribution to $B$ in the amount of $M$ actually comes from the surface currents. $\\$ The uniformly magnetized cylinder of finite length can be considered to be the building block for any arbitrary distribution of magnetization. If we look at the building block, we can say the $B$ needs to have added to the $H$ an amount of value $M$ to give the $B$ its proper continuity etc., and that there may, in fact, be a "local" contribution of $M$ to $B$. This certainly could be a possibility. The alternative would be that the agreement between magnetic pole model and magnetic surface current model is simply mathematical coincidence.

Last edited:

vanhees71

Gold Member
No, @Meir Achuz is right. There's not only fundamental electric charge but also fundamental magnetic dipole moments. Ferromagnetism on the microscopic level is a quantum phenomenon (exchange interaction of electrons and spontaneous symmetry breaking).

• Homework Helper
Gold Member
2018 Award
No, @Meir Achuz is right. There's not only fundamental electric charge but also fundamental magnetic dipole moments. Ferromagnetism on the microscopic level is a quantum phenomenon (exchange interaction of electrons and spontaneous symmetry breaking).
I agree that the magnetic dipole is fundamental. My consideration here is with solving the macroscopic problem. In Grififith's E&M textbook, he begins with an arbitrary distribution of microscopic magnetic dipoles, and he derives an equation for the vector potential $A$ that shows there is an "equivalent" magnetic surface current that can be considered as the source of this vector potential. $\\$ Even though I know that it works, and I have even proven that it works, the complete prescription for the magnetic pole method, with the equation $B=\mu_oH+M$, is not completely intuitive for me.

vanhees71

Gold Member
Of course, concerning the macroscopic theory you are right. There we average out all quantum effects and have no clue about the inner workings of the microscopic consistuents.

Here, we obviously consider a "hard ferromagnet", i.e., we describe it as something with a constant magnetization $\vec{M}(\vec{x})$ (with the magnet at rest of course). We have a case of magnetostatic then, i.e.,
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{H}=\vec{j}_{text{free}}=0.$$
The constitutive equation in this case simply is
$$\vec{B}=\mu_0 (\vec{H}+\vec{M}),$$
where I use the standard definition of $\vec{M}$ concerning factoring out $\mu_0$, used in the SI (which is always very confusing, but that's due to historical misunderstandings and the inertia of physicists to make consistent definitions after gaining new knowledge, but that's a minor obstacle).

There are now two ways to proceed to solve the above equations. The mathematical more convenient one starts from the curl-freeness of $\vec{H}$, which allows to introduce a magnetic potential,
$$\vec{H}=-\vec{\nabla} \phi_m.$$
From $\vec{\nabla} \cdot \vec{B}=0$ we find
$$\vec{\nabla} \cdot \vec{H}=-\Delta \phi_m=-\vec{\nabla} \cdot \vec{M}=\rho_m.$$
Here $\rho_m$ is an effective "magnetic charge". If $\vec{M} \simeq \text{const}$ what you get in fact is a magnetic "surface charge", but that's a bit cumbersome to treat in general (the div on the rhs. of the previous equation gives a $\delta$-distribution like singularity with support along the surface, but that's indeed not very convenient to evaluate for general shapes). That's why we use a trick since we solve our equations anyway by integrals.

Using the Green's function of the D'Alembert operator we have
$$\phi_m(\vec{x})= \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho_m(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}. \qquad (1)$$
Now we can write
$$\frac{\rho_m(\vec{x}')}{|\vec{x}-\vec{x}'|}=-\frac{\vec{\nabla}' \cdot \vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|} = \vec{M}(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}-\vec{\nabla}' \frac{\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Plugging this into the previous equation, the integral over the total divergence vanishes due to Gauss's theorem (with the boundary surface at inrinity and a magnet of finite extent). This leads to
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}') \cdot (\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}=-\vec{\nabla} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
In the final step we have used
$$\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|}=-\vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}$$
and than carried the nabla wrt. $\vec{x}$ out of the integral.

I think this is what you call the "magnetic-pole method". As the derivation shows there are of course no real magnetic (mono)poles, but $\vec{M}$ occurs as a magnetic-dipole distribution with its potential, as can also immediately seen from the 1st expression for $\phi_m$.

The other method is to use $\vec{\nabla} \cdot \vec{B}$, which is generally valid (in contradistinction to the pole method which is restricted to the here discussed static case with vanishing free currents) and introduce the vector potential
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Since there's gauge freedom we can make life easier by imposing the Coulomb-gauge condition (almost always most convenient in the static case),
$$\vec{\nabla} \cdot \vec{A}=0.$$
Then
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A}.$$
On the other hand you have
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{\nabla} \times(\vec{H}+\vec{M}) = \mu_0 \vec{\nabla} \times \vec{M}=\mu_0 \vec{j}_m.$$
Here for $\vec{M} \simeq \text{const}$ you get an effective surface current.

[Note in proof:]
It's related to the boundary condition
$$\vec{k}_m=\frac{1}{\mu_0} \vec{n} \times (\vec{B}_>-\vec{B}_<),$$
i.e., the jump of the magnetic field across the boundary of the magnet.

Last edited:

Homework Helper
Gold Member
2018 Award
The constitutive equation in this case simply is

→B=μ0(→H+→M),B→=μ0(H→+M→),​

\vec{B}=\mu_0 (\vec{H}+\vec{M}),
I don't know that I am willing to accept this as a constitutive equation, without proof. Biot-Savart's equation follows from Maxwell's $\nabla \times B=\mu_o J_{total}$. I do of course accept Maxwell's equations as constitutive.$\\$ The equation $B=\mu_o (H+M)$ can be shown to be a result of Maxwell's equations and Biot-Savart, (as I myself have done independently). I do think they would do well to derive $B=\mu_o (H+M)$ in the textbooks from Biot-Savart. $\\$ Some textbooks base the equation $B=\mu_o(H+M)$ on the analogous $D=\epsilon_o E+P$, (there are different definitions for $M$: Some books use $B=\mu_o H+M$ ), but I think this is a little bit of a stretch. $\\$ I should add here what $H$ represents is this equation: it is a combination of what is basically the $B$ (divided by $\mu_o$), from currents in conductors, along with a magnetic pole contribution=fictitious magnetic charge density $\rho_m=-\nabla \cdot M$, with $H$ computed from the inverse square law. The "local" contribution to $B$ in the amount of $\mu_o M$, when derived from Biot-Savart and the surface currents, turns out to be a non-local contribution, with $H$ from the poles as a correction factor for geometries other than long cylinder. $\\$ One correction in @vanhees71 's last line above : Surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$.

Last edited:

Meir Achuz

Homework Helper
Gold Member
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ${\bf A}$) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.

Homework Helper
Gold Member
2018 Award
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ${\bf A}$) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.
I think we all (@vanhees71 included here) have a pretty good understanding that the magnetic surface currents are more of a mathematical resultant of a distribution of microscopic magnetic dipoles than an actual current that can be measured with an ammeter. We have previously discussed this in detail in other threads.

Meir Achuz

Homework Helper
Gold Member
I apologize for restating the obvious. I guess I was still replying to your first three posts without looking in detail at later posts.

• Homework Helper
Gold Member
2018 Award
I apologize for restating the obvious. I guess I was still replying to your first three posts without looking in detail at later posts.
@vanhees71 is our in-house E&M expert. I like to bounce ideas off of him=sometimes I get it right, and other times he makes corrections to my inputs. In any case, I do think they would do well to place more emphasis on some of the fundamental magnetostatics problems in the present physics curriculum for the undergraduate students. It is only once in a great while these days that I see a homework post on Physics Forums that addresses some of these very fundamental magnetostatic calculations, such as the one in this thread.

vanhees71

Gold Member
I don't know that I am willing to accept this as a constitutive equation, without proof. Biot-Savart's equation follows from Maxwell's $\nabla \times B=\mu_o J_{total}$. I do of course accept Maxwell's equations as constitutive.$\\$ The equation $B=\mu_o (H+M)$ can be shown to be a result of Maxwell's equations and Biot-Savart, (as I myself have done independently). I do think they would do well to derive $B=\mu_o (H+M)$ in the textbooks from Biot-Savart. $\\$ Some textbooks base the equation $B=\mu_o(H+M)$ on the analogous $D=\epsilon_o E+P$, (there are different definitions for $M$: Some books use $B=\mu_o H+M$ ), but I think this is a little bit of a stretch. $\\$ I should add here what $H$ represents is this equation: it is a combination of what is basically the $B$ (divided by $\mu_o$), from currents in conductors, along with a magnetic pole contribution=fictitious magnetic charge density $\rho_m=-\nabla \cdot M$, with $H$ computed from the inverse square law. The "local" contribution to $B$ in the amount of $\mu_o M$, when derived from Biot-Savart and the surface currents, turns out to be a non-local contribution, with $H$ from the poles as a correction factor for geometries other than long cylinder. $\\$ One correction in @vanhees71 's last line above : Surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$.
These are constitutive equations. They follow from microscopic electrodynamics (many-body QFT) in terms of linear-response (Green-Kubo) theory, and they are very much simplified, but amazingly successful to describe a lot of phenomena (at least) qualitatively right.

To get the systematics straight you usually put all "free charges, currents and magnetizations" in the equations for the field components $\vec{D}$ and $\vec{H}$. The only idiosyncrazy of the common convention is how the susceptibilities (or in the most simple approximation with just constant permittivities and permeabilities) are distributed. It's in fact a convention, but systematically was to flip $\mu$ to the other side, but this would confuse people more than anything, and that's why one leaves it as is.

The most "physical system of units" are Heaviside-Lorentz units. There you have the Maxwell equations

$$\vec{\nabla} \times \vec{B} +\frac{1}{c} \partial_t \vec{B}=0,\\ \vec{\nabla} \cdot \vec{B}=0 \\ \vec{\nabla} \times \vec{H} -\frac{1}{c} \partial_t \vec{D} = \frac{1}{c} \vec{j}_{\text{free}},\\ \vec{\nabla} \cdot \vec{D}=\rho.$$
The constitutive equations in the most simple form then read
$$\vec{D}=\epsilon \vec{E}, \quad \vec{B}=\mu \vec{H}.$$

Then I'm a bit puzzled about your remarks concerning the magnetic current. For the discussed case you have (in SI units) always
$$\vec{j}_m=\vec{\nabla} \times \vec{M}.$$
If you have a body in a region $V$ and you assume a homogeneous isotropic $\vec{M}=\text{const}$ across $V$ (it's $\vec{M}(\vec{x})=0$ for $\vec{x} \notin V$ of course). Then $\vec{j}$ becomes a Dirac $\delta$-distribution with support on the boundary $\partial V$ of the body. That's effectively a surface current.

Take the example of a sphere of radius $a$. Then you have
$$\vec{M}(\vec{x})=M \Theta(a-r) \vec{e}_3, \quad r=|\vec{x}|.$$
The curl is (using spherical coordinates when convenient)
$$\vec{j}_m=\vec{\nabla} \times \vec{M} = -M \vec{e}_3 \times \vec{\nabla} \Theta(a-r)=M \vec{e}_3 \times \vec{e}_r \delta(a-r)=M \sin \vartheta \vec{e}_{\varphi} \delta(r-a).$$
This means there's a magnetic surface current
$$\vec{k}_m=M \sin \vartheta \vec{e}_{\varphi}$$
along the surface of the sphere.

From
$$\Delta \vec{A}=-\mu_0 \vec{j}_m$$
you get
$$\vec{A}(r)=\int_{0}^{\pi} \mathrm{d} \vartheta' \int_{0}^{2 \pi} \mathrm{d} \varphi' a^2 \sin \vartheta \frac{\vec{k}_m(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Here $$\vec{x}'=a (\cos \varphi' \sin \vartheta',\sin \varphi' \sin \vartheta ',\cos \vartheta').$$
Of course it's more convenient to solve the differential equation in spherical coordinates using the obvious ansatz
$$\vec{A}(\vec{x})=\vec{e}_{\varphi} A_{\varphi}(r,\vartheta),$$
which automatically fulfills the Coulomb gauge condition. Also be careful to use
$$-\Delta \vec{A}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A}),$$
because you must not naively calculation the Laplace operator of a vector field in non-Cartesian coordinates.

For the rest of the calculation see my (German) lecture notes, pp. 91ff

• vanhees71

Gold Member
"In addition, the MM in the equation B=μoH+MB=μoH+M is shown to come from the contribution of the [surface currents] that result from any finite distribution of magnetization M."
That is NOT CORRECT in a permanent ferromagnet.
There is a term (in derivations for ${\bf A}$) that reads
$${\bf A(r')=\int} d^3r\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$
That looks and acts like a surface current, but IS NOT an actual surface current for a permanent ferromagnet.
Ferromagnetism comes from the intrinsic dipole moments of electrons in the iron atom.
It is this misunderstanding that has you confused.
Actually neither method is 'wrong'. Each is a useful mathematical method of finding B and H for a permanent ferromagnet. I find the dipole method easier if it is not confounded by SI.
I'm not sure what you mean by this formula. I guess it's the following idea: From
$$\Delta \vec{A}=-\mu_0 \vec{\nabla} \times \vec{M}$$
you get
\begin{equation*}
\begin{split}
\vec{A}(\vec{x}) &=\mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{\nabla}' \times \vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|} \\
&=+ \mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{M}(\vec{x}') \times \vec{\nabla}' \frac{1}{4 \pi |\vec{x}-\vec{x}'|} \\
&=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{M}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.
\end{split}
\end{equation*}

Homework Helper
Gold Member
2018 Award
It may be worthwhile for me to present a derivation of the equation $B=\mu_o (H+M)$ that I did in 2012-2013 using Biot-Savart and $\nabla \times M=J_m$ along with some vector identities. It is done for the case where there are no currents in conductors. The contribution from currents in conductors is simply an add-on to both sides of the equation. For the case at hand, $\vec{H}$ comes from contributions from the magnetic poles with magnetic charge density $\rho_m=-\nabla \cdot \vec{M}$. The proof is somewhat lengthy, so let me put it in post 26. It might take me couple of tries to get the Latex right, because there are quite a few equations.

• vanhees71

"A magnetostatics problem of interest 2"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving