A magnetostatics problem of interest 2

[Charles Link]

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.

 

[Charles Link]

The above post is really what this Insights article that I authored a couple years ago is all about, but the above post is probably much easier to follow than the Insights article. See: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

 

[Charles Link]

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 $I=K_m L =M L/\mu_o$.

 

[Charles Link]

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 $\vec{B}=M_o \hat{z}$.

 

[Charles Link]

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]

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.

 

[Charles Link]

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]

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 $z$.

 

[Charles Link]

@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.

 

[vanhees71]

The homogeneously magnetized sphere is in my German lecture notes on E&M for high school-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:

https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf
The homogeneously magnetized sphere can be found in Sect. 3.3.3 (worked out in both ways discussed here).

 

[Charles Link]

@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/threa...iformly-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/threa...ormly-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.

 

[Charles Link]

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.

 

[Meir Achuz]

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

 

[Charles Link]

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.

 

[vanhees71]

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).

 

[Charles Link]

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]

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.

 

[Charles Link]

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 $\vec{K}_m=\vec{M} \times \hat{n}$.

 

[Meir Achuz]

"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.

 

[Charles Link]

"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]

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

 

[Charles Link]

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]

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

https://itp.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

 

[vanhees71]

"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*}

 

[Charles Link]

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.

 

[Charles Link]

I'm going to use c.g.s. units for the derivation, and I will show that $B=H+4 \pi M$.
Let $a=-\nabla \frac{1}{|x-x'|}=\frac{x-x'}{|x-x'|^3}$.
Let $b=M(x')$. Then $\nabla' \times b=\nabla' \times M(x')=\frac{J_m(x')}{c}$.
Biot-Savart gives $B(x)=\int (\nabla' \times b) \times a \, d^3x'=-\int a \times (\nabla' \times b)$.
By a vector identity $B=-\int [\nabla'(a \cdot b)-(b \cdot \nabla') a-(a \cdot \nabla') b-b \times \nabla' \times a] \, d^3 x'$.
The first term integrates to zero because of a vector identity similar to Gauss' law that $\int \nabla \psi \, d \tau=\int \psi \, dA$, and for a finite sample size, $b$ vanishes on the large outer surface that surrounds the magnetized sample.
The 4th term is zero because it is the curl of a gradient. This leaves the two middle terms.
The derivation is a little more lengthy than I anticipated, so let me just add a couple of key steps here, and I will try to fill in the pieces at a later time:
$\nabla' \cdot a=-4 \pi \delta(x-x')$, and $\nabla \cdot a=-\nabla' \cdot a=4 \pi \delta(x-x')$.
Now $b \cdot \nabla' a=\nabla' \times (a \times b)+b(\nabla' \cdot a)-a(\nabla' \cdot b)+(a \cdot \nabla')b$.
Again, I hope to fill in all the steps=there are quite a number of them=but in any case, the result of all of this vector calculus is that $B=H+4 \pi M$.
It is worth mentioning one more step in this derivation: $(a \cdot \nabla') b=(b \cdot \nabla')a-\nabla' \times (a \times b) -b (\nabla' \cdot a)+a (\nabla' \cdot b)$.
This last substitution seems to be necessary, because in a later step $(b \cdot \nabla')a=-(b \cdot \nabla)a$ (without a prime on the last gradient operator ) is used to evaluate this term, while evaluation of $(a \cdot \nabla')b$ as is would be problematic.
Much of the rest is just evaluating the terms, and also determining that certain integrals equal to zero, because they equate to surface integrals, etc.
The editor is giving me problems again, so I think I will go to post 27 to add additional steps.

 

[Charles Link]

To add to post 26 (I already put this in post 26, but it lost it):
$\int (b \cdot \nabla)a \, d^3x'=\int [\nabla(a \cdot b)-(a \cdot \nabla)b-a \times \nabla \times b -b \times \nabla \times a] \, d^3x'$
$=\nabla \int(a \cdot b) \, d^3x'$.
The last result follows because the gradient operation here is unprimed and operating on a term ($b$) with a primed coordinate in the second and third terms. Meanwhile, the 4th term is the curl of a gradient, so that it is zero. Since the gradient operator on the first term is unprimed, and the integration is over the primed coordinate, the unprimed gradient can be pulled out in front of the integral.
The remaining steps that I haven't shown in this derivation are, for the most part simpler. I will show one more, and that is in evaluating $\int (b \cdot a) \, d^3x'$, the terms become
$M \cdot (\nabla' \frac{1}{|x-x'|})=\nabla' \cdot (\frac{M}{|x-x'|})-\frac{\nabla' \cdot M}{|x-x'|}$, and the first term integrates to zero using Gauss' law.
Much of the rest is relatively straightforward. I would have liked to show it all, but it is quite a lot harder writing it out in Latex, than it is to write it out by hand.

 

[Charles Link]

To summarize posts 26 and 27, (I'm working in cgs there, but now going back to SI) starting with Biot-Savart, and $\nabla \times M=J_m$, and using a bunch of vector identities, the result follows that $\vec{B}=\mu_o (\vec{H}+\vec{M})$.
Here $\vec{H}$ is defined (cgs) as $\vec{H}(x)=\int \frac{\rho_m(x')(x-x')}{|x-x'|^3} \, d^3x'$, where $\rho_m=-\nabla \cdot \vec{M}$.
This derivation doesn't consider the $H$ that is from currents in conductors, which simply gets tacked onto the above result ($B=\mu_o(H+M)$) by redefining $H$ to include the contribution from currents in conductors using Biot-Savart. $\\$
From these results, it is apparent that, what is basically the result of the magnetic pole model, that $B=\mu_o(H+M)$, goes hand-in-hand with magnetic field calculations that are based on magnetic currents and Biot-Savart. It is open to debate which one is the more fundamental. $\\$ Additional item: The pole model equation $B=\mu_o(H+M)$ follows from Biot-Savart and $\nabla \times M=J_m$.
I don't believe the converse would be the case. $\\$ And an additional comment: In grouping large numbers of the fundamental building block, which is the microscopic magnetic dipole, and making a macroscopic magnetized object, it would appear that in the mathematics, it can be treated as having magnetic charges, (regions of $\rho_m=- \nabla \cdot M$), or regions of magnetic currents, (where $J_m=\nabla \times M$). Perhaps in a sense, both are equally fictitious, but it seems necessary to employ one or the other, and to essentially treat it as real, in order to compute the magnetic field $B$.
@vanhees71 I think you might find the above algebraic derivation of interest.

 

[vanhees71]

Of course, the full Maxwell equations clearly say, it's "currents" rather than "monopoles". Of course, it's no problem to add also magnetic monopoles to the electromagnetic theory (see Dirac's famous paper), but it seems as if Nature doesn't realize this idea ;-)).

 

[Meir Achuz]

The posts are coming so fast, I can't read or respond to them, but I will reply to this
from Vanhees71:

"I'm not sure what you mean by this formula. I guess it's the following idea:

$\Delta \vec{A}=-\mu_0 \vec{\nabla} \times \vec{M}...$"

We are doing things in opposite directions. I develop the potential integral for magnetization, $\bf M$, in the same way as the Coulomb integral is developed from Coulomb's law for a point charge q.

I start with the vector potential for a chunk, $\Delta V,$ of magnetized matter,
$$\Delta{\bf A}=\frac{{\bf M}\Delta V\times {\bf r}}{r^3},$$
where ${\bf M}$, the magnetization is defined as the magnetic moment per unit volume.
I don't see any other way to give a physical definition of $\bf M$.
Going to the integral limit gives
$${\bf A(r)}=\int\frac{{\bf M(r')\times r'}d^3r'}{\bf|r-r'|^3}.$$
Then, some vector manipulation leads to

$${\bf A(r)}=\int\frac{{\bf\nabla'\times M(r')}d^3r'}{\bf|r-r'|}+ \int\frac{\bf M(r')\times dS'}{\bf|r-r'|}.$$

This suggests that $\bf\nabla\times M$ behaves like, but IS NOT, a current; and that
$\bf M\times dS$ behaves like, but IS NOT, a differential surface current.

 

[vanhees71]

I see, now your formulae make sense to me. The second integral, however is over the boundary of the integration volume. If it contains the whole body in its interior this vanishes. The only exception is if $\vec{M}$ itself contains a singular surface-magnetization piece. As your derivation shows, $\vec{M}$ is a volume density, namely magnetization per volume.

 

[Charles Link]

What @Meir Achuz has is the same derivation that Griffith's shows in his E&M book=I believe it is in section 6.2. The numerator of the second term should be recognized as the surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$. $\\$
And I think it is rather fascinating that both a "magnetic charge" description and a "magnetic current" description are able to generate the exact same result for the magnetic field $\vec{B}$. The "magnetic charge" at the end faces of the macroscopic cylinder, and the "magnetic surface current" on the outer surface of the macroscopic cylinder are both rather fictitious, and are both the result of some vector calculus, but the magnetic field $\vec{B}$ can be calculated in either manner.

 

[Meir Achuz]

I will try to correct some misinterpretations.
${\bf H=B-}4\pi{\bf M}$ is the definition of $\bf H$.
A definition cannot be derived. $\bf B$ ALWAYS equals ${\bf H+}4\pi{\bf M}$, for any possible model.

Biot and Savart proposed the "law of Biot-Savart", which had nothing to do with $\bf M$.

 

[Meir Achuz]

"What @Meir Achuz has is the same derivation that Griffith's shows in his E&M book."
That is why I was surprised when you seemed to contradict it, and Vanhees71 questioned my integral from Griffiths.

 

[Charles Link]

@Meir Achuz I think if you read through the derivation in posts 26-28, you would find it of interest. What I am basically showing with this is that the macroscopic "pole" model of magnetism, and a magnetic current description, go hand-in-hand, even though they are so different mathematically. $\\$ In the pole (magnetic charge) model, often the equation $B=H+4 \pi M$ is assumed set in stone because it is analogous to $D=E+4 \pi P$. $\\$ Alternatively, if you begin with a magnetic current description, $\nabla \times M=\frac{J_m}{c}$ (cgs) and Biot-Savart, $B(x)=\int \frac{J_m(x') \times (x-x')}{c|x-x'|^3} \, d^3x'$, then, $\\$ defining $H=B-4 \pi M$, $\\$ the result is $H =H_m+H_c$ where $\\$ $H_m(x)=\int \frac{\rho_m(x')(x-x')}{|x-x'|^3} \, d^3x'$, with $\rho_m=-\nabla \cdot M$,
along with the contribution $H_c$ from currents in conductors using Biot-Savart. $\\$
That's why I actually prefer to derive the result $B=H+4 \pi M$, using the above as definitions for $H_m$ and $H_c$, with $H=H_m +H_c$.

 

[vanhees71]

Argh! Now I see, where my misunderstanding comes from! In Griffiths for the homogeneously magnetized spherical shell, he writes
$$\vec{j}_m=0, \vec{K}_m=\vec{M} \times \hat{n},$$
where $\hat{n}$ is the surface-unit normal vector along the boundary of the magnetized body.

In my interpretation it's included in $\vec{\nabla} \times \vec{M}$, because I take the surface current into account in terms of a Dirac-$\delta$-like contribution from generalized derivatives.

Both methods are equivalent, because the surface contribution can be also defined in terms of the socalled (in German) "Sprungrotation" (I don't know an expression in English for it, literally translated it means "jump curl"). The definition is hard to describe without a figure. So here it is

The blue area is the surface across which a vector field $\vec{V}$ may have singularities (in our case of a homogeneously magnetized body, the magnetization $\vec{M}$ has a jump from some finite value inside and 0 outside the body along its boundary).

Now you define the "Sprungrotation" $\text{Curl} \vec{V}$ along the surface by drawing the red curve with an arbitrary tangent-unit vector $\vec{t}$ along the surface and taking the limit of the red curve of contracting it to the point on the surface you want the operator $\text{Curl}$ define at, which goes as follows
$$\Delta h (\vec{t} \times \vec{n}) \cdot \text{Curl} \vec{V} = \Delta h \vec{t} \cdot (\vec{n} \times \text{Curl} \vec{V})= \Delta h \vec{t} \cdot (\vec{V}_2-\vec{V}_1).$$
Since this construction holds for all tangent unit vectors $\vec{t}$ along the boundary surface at the given point, you get
$$\vec{n} \times \text{Curl} \vec{V}=(\vec{V}_2-\vec{V}_1).$$
The two parts of the path along $\vec{n}$ don't contribute because they cancel in the limit, because there's only a jump across but not along the surface. Thus you have $\mathrm{Curl} \vec{V} \cdot{n}=0$ by definition and thus one gets
$$\vec{n} \times (\vec{n} \times \text{Curl} \vec{V})=-\text{curl} \vec{V} = \vec{n} \times (\vec{V}_2 -\vec{V}_1).$$
The surface current for our case of the homogeneously magnetized sphere (taking region 1 as the exterior, region 2 as the interior of this sphere) thus is
$$\vec{k}_m=\mathrm{Curl} \vec{M} = \vec{n} \times (\vec{M}_1-\vec{M}_2) = \frac{\vec{r}}{a} \times (-M \vec{e}_3)=M \cos \vartheta \vec{e}_{\varphi}.$$
So we can write the solution for the vector potential indeed as
$$\vec{A}(\vec{r})=\mu_0 \int_{\partial V} \mathrm{d}^2 f' \frac{\vec{M}(\vec{r}') \times \vec{n}'}{|4 \pi |\vec{r}-\vec{r}'|}.$$

 

[Charles Link]

@vanhees71 I welcome your input on posts 33-35. $\\$ Also @Meir Achuz I welcome your input to the following: $\\$ Although $H=B-4 \pi M$ is used in some textbooks as the definition, I don't think they should cast it in stone like they do. If they define $H$ by this equation, then they need to show what other equations the entity $H$ obeys. $\\$ Other textbooks will define the $H=H_m +H_c$, but so many fail to tie the pieces together, and then simply assume $B=H+4 \pi M$ without proof. $\\$ Alternatively, the textbooks that define $H=B-4 \pi M$, then use the results (basically the alternative definition) of the other textbooks for $H=H_m+H_c$ without proof. $\\$ In my derivation of posts 26-28, I tie the pieces together. Defining $H_m$ and $H_c$ accordingly, I show/prove that the result $B=H+4 \pi M$ then follows from Biot-Savart along with $J_m=c \nabla \times M$.

 

[Meir Achuz]

${\bf B=H+}4\pi{\bf M}$ IS cast in stone. There is no place where it is not true.
Name one "Other textbooks".

 

[Charles Link]

${\bf B=H+}4\pi{\bf M}$ IS cast in stone. There is no place where it is not true.
Name one "Other textbooks".

J.D. Jackson for one. In general, his Classical Electrodynamics is amazing, but IMO, he never proves, at least to my satisfaction, the magnetostatic equations that he employs. Other E&M textbooks I have worked extensively with are Pugh and Pugh, and Schwarz. $\\$ And I agree, $B=H+4 \pi M$ can be cast in stone, because I have independently proven it . Besides the proof in posts 26-28 that I did in 2012-2013, I also did another proof of it in 2009-2010, that I have previously linked on Physics Forums in other threads. See https://www.overleaf.com/read/kdhnbkpypxfk Until I succeeded in proving it with this proof in 2009-2010, I was very much unwilling to "cast this equation in stone", even though I had to accept it as gospel to pass the exams of the E&M courses that I had in 1975-1980. $\\$
The textbooks I have seen have not proven this equation in a manner that I was satisfied with. $\\$
I have to believe that most likely the physicists around 1880-1900 did a couple of very thorough proofs of the equation $B=H+4 \pi M$, and thereby it became cast in stone ever since, but somehow the modern day textbooks have omitted and/or lost some of the details.

 

[Meir Achuz]

Look at Eq. (5.81) in Jackson III.

 

[Charles Link]

Look at Eq. (5.81) in Jackson III.

Unfortunately, I loaned out my copy of J.D. Jackson about 5 years ago, and never got it back. From what I remember though, J.D. Jackson uses a magnetic potential $\Omega$, and doesn't even introduce magnetic currents and magnetic surface currents. What I found lacking in his magnetic potential case is that although $\nabla^2 \Omega=0$, and $H=-\nabla \Omega$, (because $\nabla \cdot H=0$), there is no indication that any boundary conditions will be satisfied. That is why I took it upon myself to construct the proof in 2009-2010 that starts with a very simple case, and proves the boundary condition is satisfied. I then took it a couple of steps further, and proved $B=H+4 \pi M$ in general. $\\$ For example, if you take the simplest macroscopic case, which is a uniformly magnetized cylinder of semi-infinite length, (with the single end face at $z=0$ in the x-y plane), and you compute the potential $\Omega$ from the single pole end face, what guarantee is there that the $H=-\nabla \Omega$ that is computed will indeed be the correct one? i.e. that it gets the same answer as the surface current computation of the same scenario. Anyway, perhaps the textbooks successfully did this, but they did not do it to my satisfaction... I was skeptical until I actually rigorously proved it... $\\$ And to add to this, for this semi-infinite cylinder, $B_z =0$ by inspection by the pole model everywhere in the x-y plane except for $r<a$. When I did the calculation of this in 2010 with the surface currents, I was thinking I might even find it to be inconsistent with the pole model result. Instead, surprisingly, there was complete agreement !

 

[Meir Achuz]

1. Why do you keep 'proving' a definition?
2. There is 'complete agreement' because they are both correct calculations.
3. Your memory about Jackson is skewed. On page 196, just after using the scalar potential, he treats the vector potential with magnetic volume and surface currents. This comes after his definition of H=B/muzero-M on
page 192.

 

[Charles Link]

1. Why do you keep 'proving' a definition?
2. There is 'complete agreement' because they are both correct calculations.
3. Your memory about Jackson is skewed. On page 196, just after using the scalar potential, he treats the vector potential with magnetic volume and surface currents. This comes after his definition of H=B/muzero-M on
page 192.

I don't have a good answer in responding to the above. Perhaps I had an older version of J.D. Jackson's book=the edition I had used cgs. And yes, they are correct calculations, but IMO the textbooks never seem to prove the pole formalism gives the same result for $B$ as the magnetic current formalism. Again, perhaps I missed something when I studied a couple of different E&M textbooks. I'd be interested in hearing @vanhees71 opinion, on whether the textbooks properly treat the equivalence of the $\vec{B}$ calculated by both methods. For the most part, the textbooks I have seen usually presented the pole method or the current method, but didn't tie the two together. Perhaps a later version of J.D. Jackson did precisely that. And again, I no longer have access to the edition that I had studied from=it is possible I overlooked portions of the discussion.

 

[vanhees71]

I think any good textbook on CED has it. I learned CED from Sommerfeld's textbook. Jackson has it for sure too.

I also don't think that there's a contradiction between my formulae and @Meir Achuz 's. It's only that I took the surface distribution into account by using derivatives in the sense of distributions, i.e., a Heaviside unitstep function gives a Dirac $\delta$ when its derivative is taken. You can of course also use the "Sprungrotation" (I'm still unsure how you call this operation in English, maybe "Surface Curl"?). Both methods are equivalent. The "Sprungrotation method", however, is more physical.

Of course, both magnetostatic methods are treated in the standard textbooks, i.e., a scalar potential for $\vec{H}$ in the absence of free currents and the vector potential for $\vec{B}$. Of course the latter approach is more natural since it's valid in the general case. That's why usually it is preferred compared to the magnetic-scalar-potential approach. Of course, both approaches are completely equivalent.

Right now I'm trying to continue to write on my SRT FAQ article for PF. It's amazing, how much knowledge about the relativistic formulation of classical electrodynamics has been lost with the decades since von Laue and Minkowski.

I also come from relativistic many-body QFT, where QED is the perfect playground to try out ideas. It's also a quite hot topic in relativistic heavy-ion collisions and related astro-nuclear physics (neutron stars/kilonovae and gravitational wave signals!). Now that for about 10-20 years relativistic viscous hydrodynamics is under control, also a renewed interest in relativistic magnetohydrodynamics has become a focus of current research. So I think, it's high time to review the macroscopic relativistic electrodynamics with modern tools, including relativistic kinetic theory derived from the Schwinger-Keldysh real-time-contour formulation. Particularly also vorticity in hydro- as well as magnetohydrodynamics still is a quite hot topic with a lot of new attention in hour heavy-ion community.

 

[Charles Link]

Very good post above @vanhees71. Perhaps the E&M textbooks do, in fact, have a complete treatment of both the magnetic pole method and magnetic surface current methods. In that case, the extra calculations that I did that I mentioned in posts 26-28 and post 39 may have been unnecessary, and the results that I obtained may have been very predictable.$\\$ For me, though, I have to work with what I got out of reading and studying the textbooks=and I did put an enormous effort into the E&M studies= in two advanced undergraduate E&M courses as well as two graduate E&M courses=for me the results of my calculations, particularly the bunch in 2009-2010, came as a surprise=I did not expect that the magnetic surface current method of calculation would yield an identical result to that of the magnetic pole method. $\\$ It has been my experience that very few physics people really have a good handle on the magnetostatics subject. A couple years ago, I showed a couple of these calculations to a physics professor at a major university, and he told me that he had seen these types of calculations in graduate school, but he didn't completely understand the subject at that time, and by now he has forgotten most of it. $\\$ I may have taken some extra, perhaps unnecessary, steps to get a good handle on the subject, but for me, it took these extra steps to sort it all out.

 

[Charles Link]

I came up with an additional problem, a takeoff on what was posed in the OP, that is the kind of thing that anyone with a good understanding of the magnetic pole method and the magnetic surface current method should be able to readily work:

Consider a cylinder of length $L$ and radius $a$ that has uniform magnetization along its axis of magnitude $M_o$. Find the magnetic field strength $B$ at the center of this cylinder.

Show by both the pole and surface current methods that it has value $B=M_o \cos{\theta}$, (in units where $B=\mu_o H+M$), where $\theta=\arctan(a/(L/2))$.It's a good simple exercise with the result that the pole method and surface current method do give identical results for the magnetic field $\vec{B}$, as they always do.

 

[aclaret]

hello! I find this interesting problem, and I try to solve it :)

Lay down cylindrical coordinate chart with origin $\mathscr{O}$ centre of cylinder. Magnetic quantities $M, H, B \in R^3$ do satisfy $B = \mu_0 H + M$ from elementary electromagnetism theory. Denote with $x' \in R^3$ the position vector over the domain of integration. By theorem asserted by user charleslink in posting #1, write$$H = \frac{1}{4\pi \mu_0} \int_{\partial V} \frac{-\langle M, n \rangle x'}{|x'|^3} dA$$where $\langle x, y \rangle$ is standard inner product on $R^3$. Because the cylindrical symmetry, assume on physical basis that $H = h(\rho) e_z$ depend only on radial coordinate $\rho$, for some function $h: R \rightarrow R$ and where $e_z$ is upward pointing basis vector.

The integration trivial ;) ;), clearly you write $\partial V = S_1 \cup S_2 \cup S_3$ where

$S_1 = \{(x,y,z) \in R^3: z = \mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ is top face,
$S_2 = \{(x,y,z) \in R^3: z = -\mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ lower face,
$S_3 = \{(x,y,z) \in R^3: -\mathscr{l} < z < \mathscr{l}, \sqrt{x^2 + y^2} = a \}$ side face,

then define $\mathscr{l} := L/2$,
$$H_z = - 2\times \frac{M_0 \mathscr{l}}{4\pi \mu_0} \int_{S_1} \frac{\rho}{(\rho^2 + \mathscr{l}^2)^\frac{3}{2}} d\rho d\phi = - \frac{M_0 \mathscr{l}}{\mu_0} \left[ \frac{1}{\sqrt{\mathscr{l}^2 + \rho^2}} \right]_0^a = \frac{M_0 \mathscr{l}}{\mu_0} \left(\frac{1}{\sqrt{\mathscr{l}^2 + a^2}} - \frac{1}{\mathscr{l}}\right)
$$Obviously follows that$$\langle e_z, \mu_0 H + M \rangle = M_0 \left(\frac{\mathscr{l}}{\mathscr{l}^2 + a^2} - 1 \right) + M_0 = M_0 \cos{\theta}, \quad \cos{\theta} = \mathrm{arctan}(a/\mathscr{l})$$

 

[Charles Link]

hello! I find this interesting problem, and I try to solve it :)

Lay down cylindrical coordinate chart with origin $\mathscr{O}$ centre of cylinder. Magnetic quantities $M, H, B \in R^3$ do satisfy $B = \mu_0 H + M$ from elementary electromagnetism theory. Denote with $x' \in R^3$ the position vector over the domain of integration. By theorem asserted by user charleslink in posting #1, write$$H = \frac{1}{4\pi \mu_0} \int_{\partial V} \frac{-\langle M, n \rangle x'}{|x'|^3} dA$$where $\langle x, y \rangle$ is standard inner product on $R^3$. Because the cylindrical symmetry, assume on physical basis that $H = h(\rho) e_z$ depend only on radial coordinate $\rho$, for some function $h: R \rightarrow R$ and where $e_z$ is upward pointing basis vector.

The integration trivial ;) ;), clearly you write $\partial V = S_1 \cup S_2 \cup S_3$ where

$S_1 = \{(x,y,z) \in R^3: z = \mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ is top face,
$S_2 = \{(x,y,z) \in R^3: z = -\mathscr{l}, \sqrt{x^2 + y^2} \leq a \}$ lower face,
$S_3 = \{(x,y,z) \in R^3: -\mathscr{l} < z < \mathscr{l}, \sqrt{x^2 + y^2} = a \}$ side face,

then define $\mathscr{l} := L/2$,
$$H_z = - 2\times \frac{M_0 \mathscr{l}}{4\pi \mu_0} \int_{S_1} \frac{\rho}{(\rho^2 + \mathscr{l}^2)^\frac{3}{2}} d\rho d\phi = - \frac{M_0 \mathscr{l}}{\mu_0} \left[ \frac{1}{\sqrt{\mathscr{l}^2 + \rho^2}} \right]_0^a = \frac{M_0 \mathscr{l}}{\mu_0} \left(\frac{1}{\sqrt{\mathscr{l}^2 + a^2}} - \frac{1}{\mathscr{l}}\right)
$$Obviously follows that$$\langle e_z, \mu_0 H + M \rangle = M_0 \left(\frac{\mathscr{l}}{\mathscr{l}^2 + a^2} - 1 \right) + M_0 = M_0 \cos{\theta}, \quad \cos{\theta} = \mathrm{arctan}(a/\mathscr{l})$$

Very good @aclaret :) Next, do it with magnetic surface currents and Biot-Savart, where magnetic surface current density per unit length $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o}$.

 

[aclaret]

aww thank :), I'm happy that you like it! Ok, I try the second method... luckily, I don't think it too hard either ;)

Let me employ the same chart, origin $\mathscr{O}$. I effect cylindrical basis $\mathscr{B} = (e_{\rho}, e_{\phi}, e_z)$ of $R^3$ for analysis. Obvious that $K_m = (M_0/ \mu_0) e_{\phi} \in R^3$, so by theorem of charleslink in posting #1, obtain$$B = \frac{M_0}{4\pi} \int_{S_3} \frac{-e_{\phi} \times x'}{|x'|^3} dA$$Because $e_{\phi} \times x' = -a e_z + z e_{\rho}$, and clearly by the symmetry $e_{\rho}$ component integrate to zero, so now effecting the elementary surface integration$$B = \frac{M_0 a^2}{4\pi} e_z \int_{S_3} (a^2 + z^2)^{-\frac{3}{2}} d\phi dz = M_0 a^2 \left [ \frac{\mathscr{l}}{a^2(a^2 + \mathscr{l}^2)} \right] = M_0 \cos{\theta}$$I think that even quicker than the first method! Thank for teaching me, I always like find as many ways to solve the problem :)

 

[Charles Link]

@aclaret The two methods always give the same answer for $B$. I don't think that result is obvious, (there are a couple others who commented in the thread above who think a proof is unnecessary), but in any case I was able to prove that from first principles. You might also find this Insights article of interest: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/