Magnetic flux is the same if we apply the Biot Savart?

[vanhees71]

The magnetic surface currents are present even in the static case, and occur also in a permanent magnet. The magnetic field for any magnet can be computed from Biot-Savart (or Ampere's law) if the magnetization $\vec{M}$ is known, so that the surface currents can be computed. (In complete detail, the magnetic current density $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$. The result is that there are surface currents at boundaries given by surface current per unit length $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o}$, but there can also be magnetic current densities in the bulk material in regions of non-uniform $\vec{M}$). $\\$ For a somewhat detailed example of the magnetic surface currents, see https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/

[Later note:] There seems to be varying conventions in the physics and electrical-engineering literature: I've just checked Simonyi's comprehensive textbook on "theoretical electric engineering" (the title of my German edition is "Theoretische Elektrotechnik", I don't know, whether this amazing book is available in English; it's simply great!) for another question preparing my lecture notes. There I stumbled over the surprising fact that he defines the magnetization as part of $\vec{B}$. As I write below, for me as a physicist used to interpreting everything from the point of view of the fundamental principles, the magnetization should clearly belong to $\vec{H}$, because it's an external source for the magnetic field $\vec{B}$. There seems to be more historical balast in E&M than I ever dreamt of ;-))). Of course, at the end both conventions lead to the correct result, the only difference is that $\vec{M}_{\text{engineers}}=\mu_0 \vec{M}_{\text{physicists}}$, i.e., the constitutive equation reads
$\vec{B}=\mu_0 \vec{H} + \vec{M}_{\text{engineers}} \equiv \mu_0 (\vec{H}+\vec{M}_{\text{physicists}}$ ;-)).

Just a hint. I'm teaching E&M right now to high school-teacher students, and there I've of course to stick to the SI units. So I'm a bit thinking how to explain to the students the somewhat idosyncratic terminology of permittivities and permeabilities occurring in opposite ways (and even occurring for the vacuum just because of the SI units), because in the early days of electromagnetism the physicists somehow mixed up $\vec{H}$ and $\vec{B}$. It's very important not to confuse where the $\mu_0$ (or $\mu=\mu_0 \mu_{\text{r}}$ in para- and diamagnetic substances) occur.

For hard ferromagnetics you only give the magnetization (magnetic-dipole density per unit volume) and assume full saturation. The magnetization belongs to the "excitation quantities", i.e., to the "magnetic excitation" $\vec{H}$ rather than the "magnetic field" $\vec{B}$, where I use the modern naming scheme following from the physical meaning and relativity, where $\vec{E}/c$ and $\vec{B}$ together form an antisymmetric four-tensor as well as $\vec{D}$ and $\vec{H}$. Physically $\vec{E}$ and $\vec{B}$ are the fields occurring in the Lorentz force law, while $\vec{D}$ and $\vec{H}$ are auxilliary quantities to separate the external (or "free") sources from the sources due to the presence of matter from bound charges (i.e. the electric and magnetic dipole moments induced by the perturbation through the fields produced by the free charges and current densities).

Thus for hard ferromagnetics the basic static equations read
$$\vec{\nabla} \times \vec{H}=\vec{j}_f, \quad \vec{\nabla} \cdot \vec{D}=\rho_f, \quad \vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
and the constitutive equations
$$\vec{D}=\epsilon_0 \epsilon_r \vec{E}, \quad \vec{B}=\mu_0 (\vec{H}+\vec{M}),$$
where $\vec{M}$ is the magnetization unchanged by external influence due to the presence of hard ferromagnets. The electric and magnetic fields decouple completely, and the electric part is just as in electrostatics. So we can concentrate on the magnetic fields.

In this most general case for static fields, i.e., with currents in addition to permanent magnetization present, the most general way to solve the equations is to introduce the vector potential for $\vec{B}$. For the case without currents, $\vec{H}$ is curl-free and you can simplify the task somewhat by introducing a magnetostatic potential for $\vec{H}$.

The first method, which applies everywhere in the most general static case, you get formally vacuum equations for $\vec{B}$ with an effective current composed by the free current and the curl of the magnetization. This becomes clear by taking the curl of the 2nd constitutive equation and using Ampere's Law for $\vec{H}$:
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{\nabla} \times (\vec{H}+\vec{M}) = \mu_0 (\vec{j}_e + \vec{j}_m), \quad \vec{j}_m=\vec{\nabla} \times \vec{M}.$$
Here $\vec{j}_e$ and $\vec{j}_m$ are usually considered to be given. If one assumes $\vec{M}$ to be constant within the magnet and (of course) jumps instantaneously to 0 at the boundary of the magnet, its curl leads to a singular $\delta$-function like $\vec{j}_m$ which is equivalent to a surface current along the boundary surface of the magnet. This is in some sense in accordance with Ampere's old molecular-current hypothesis, which of course today must be substituted by the proper quantum-mechanical treatment. For the phenomenology it's just a calculational trick to substitute the approximate analysis with homogeneous magnetization within the magnet by the equivalent surface current given also in this case by $\vec{\nabla} \times \vec{M}$, leading however to a $\delta$-distribution due to the jump at the boundary.

In any case the solution lies in the introduction of the vector potential. Due to gauge freedom we can also impose a gauge constraint, which in this static case is most conveniently chosen as the Coulomb gauge, i.e., we set
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{\nabla} \cdot \vec{A}=0,$$
and for Cartesian components we can thus write
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A}=\mu_0 \vec{j}, \quad \vec{j}=\vec{j}_f + \vec{\nabla} \times \vec{M}$$
with the Biot-Savart-like solution
$$\vec{A}(\vec{r}) = \mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\vec{j}(\vec{r}')}{|\vec{r}-\vec{r}'|},$$
or for the field
$$\vec{B}(\vec{r}) =\nabla \times \vec{A}(\vec{r})=\mu_0 \int_{\mathbb{R}^3} \mathrm{d^3} r' \vec{j}(\vec{r}') \times \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}.$$
This is Biot-Savart's Law in its usual form but with the total current density including the effective magnetization current.

If somebody is interested in my German manuscript for these lectures (unfinished, but quickly growing ;-)), here they are:

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

 

[Charles Link]

For hard ferromagnetics you only give the magnetization (magnetic-dipole density per unit volume) and assume full saturation.

Normally, the only quantities of interest for permanent magnets are $\vec{M}$, and $\vec{B}$, but the relation $\vec{B}=\mu_o(\vec{H}+\vec{M})$ can still be used. The vector $\vec{H}$ is useful in the "magnetic pole model" of computing $\vec{B}$ both inside and outside of the permanent magnet.$\\$ For permanent magnets, there is no longer a linear relationship between $\vec{H}$ and $\vec{M}$. An equation can still be written as before $\vec{M}=\chi_m \vec{H}$, but for this case, $\chi_m$ is no longer a constant. The vectors $\vec{H}$ and $\vec{M}$ can at times even point in opposite directions, (making $\chi_m$ negative), depending on the location of the operating point of the magnetic material on the hysteresis curve. $\\$ The vector $\vec{H}$ is also useful in calculations with a permanent magnet if you want to determine the magnitude of $\vec{H}$ from the currents of a solenoid that surrounds that magnet that would be required to reverse the direction of magnetization. Even though in most cases the permanent magnet isn't placed inside a solenoid, it can still be a useful scenario to consider. The same equation $\vec{B}=\mu_o(\vec{H}+\vec{M})$, that is often used in calculations with linear magnetic materials, is still applicable for permanent magnets. $\\$ In the transformer core problem given in the OP by @mertcan , the very interesting feature is how the magnetic field $\vec{H}$ that is generated by the current in the windings is totally reshaped and amplified by the magnetic core, but the magnetic field $\vec{B}$ and the magnetization $\vec{M}$ are kept at a very stable and controlled amount by the presence of the air gap. The magnetic circuit theory solution, which is presented in Feynman's lectures, is equivalent to a Biot-Savart/Ampere's law solution, as discussed in posts 28, 29, and 30 above.

 

[jim hardy]

Even though in most cases the permanent magnet isn't placed inside a solenoid, it can still be a useful scenario to consider.

These folks have a small production facility just around the corner from me

https://azind.com/permanent-magnets/
I usually stop by their annual 'yard sale' because they sell really strong magnets .. It's fun to watch the ladies magnetize them . They use a solenoid about a foot tall and half as wide made of huge wire. It's amazing to see an inert blank turned into a giant 'fridge magnet' that tries to pull your carkeys out of your pocket when you walk by.

I found this tutorial on their site . It appears to be quite practical and surprisingly thorough - maybe it'll be of help to @vanhees71 with his presentations for lay people.
https://azind.com/wp-content/uploads/2014/11/Permanent-Magnet-Guidlines-MMPAPMG-88.pdf
I saved a copy to my PF folder for reference so i can read it off line.

I really should learn more about the science behind modern magnetic materials.

In the transformer core problem given in the OP by @mertcan , the very interesting feature is how the magnetic field →HH→ \vec{H} that is generated by the current in the windings is totally reshaped and amplified by the magnetic core, but the magnetic field →BB→ \vec{B} and the magnetization →MM→ \vec{M} are kept at a very stable and controlled amount by the presence of the air gap. The magnetic circuit theory solution, which is presented in Feynman's lectures, is equivalent to a Biot-Savart/Ampere's law solution, as discussed in posts 28, 29, and 30 above.

Well, thinking like a circuits fellow that's what iron does , it short circuits free space for magnetic flux. Air gap in a core becomes the controlling reluctance and iron's nonlinear and other peculiarities are minimal in comparison (unless you have enough mmf available to saturate it despite the gap.) .

I leave that vector calculus to you fellows. Please understand i do envy your fluency with it and that's not a dismissal .

old jim

 

[vanhees71]

Normally, the only quantities of interest for permanent magnets are $\vec{M}$, and $\vec{B}$, but the relation $\vec{B}=\mu_o(\vec{H}+\vec{M})$ can still be used. The vector $\vec{H}$ is useful in the "magnetic pole model" of computing $\vec{B}$ both inside and outside of the permanent magnet.$\\$ For permanent magnets, there is no longer a linear relationship between $\vec{H}$ and $\vec{M}$. An equation can still be written as before $\vec{M}=\chi_m \vec{H}$, but for this case, $\chi_m$ is no longer a constant. The vectors $\vec{H}$ and $\vec{M}$ can at times even point in opposite directions, (making $\chi_m$ negative), depending on the location of the operating point of the magnetic material on the hysteresis curve. $\\$ The vector $\vec{H}$ is also useful in calculations with a permanent magnet if you want to determine the magnitude of $\vec{H}$ from the currents of a solenoid that surrounds that magnet that would be required to reverse the direction of magnetization. Even though in most cases the permanent magnet isn't placed inside a solenoid, it can still be a useful scenario to consider. The same equation $\vec{B}=\mu_o(\vec{H}+\vec{M})$, that is often used in calculations with linear magnetic materials, is still applicable for permanent magnets. $\\$ In the transformer core problem given in the OP by @mertcan , the very interesting feature is how the magnetic field $\vec{H}$ that is generated by the current in the windings is totally reshaped and amplified by the magnetic core, but the magnetic field $\vec{B}$ and the magnetization $\vec{M}$ are kept at a very stable and controlled amount by the presence of the air gap. The magnetic circuit theory solution, which is presented in Feynman's lectures, is equivalent to a Biot-Savart/Ampere's law solution, as discussed in posts 28, 29, and 30 above.

For the case of no free currents, i.e., just the permanent magnet there are two possible equivalent ways to solve the problem. The equations then take the form
$$\vec{\nabla} times \vec{H}=0, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{B}=\mu_0(\vec{H}+\vec{M}).$$

1st strategy: Scalar potential for $\vec{H}$

Since $\vec{H}$ is vortex free, there's a scalar potential
$$\vec{H}=-\vec{\nabla} \phi_m.$$
Then we know that we also need its sources, and these are readily calulated using Gauss's Law for $\vec{B}$ and the constitutive equation
$$\vec{\nabla} \cdot \vec{H}=\vec{\nabla} \cdot (\vec{B}/\mu_0-\vec{M})=-\vec{\nabla} \cdot \vec{M}=\rho_m.$$
We have an effective magnetic charge, given by $\rho_m=-\vec{\nabla} \cdot \vec{M}$, and the solution is analogous to electrostatics
$$\phi_m(\vec{r})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho_m(\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|}.$$
If there are jumps of $\vec{M}$ at the boundary, you get the usual "jump condition"
$$\Sigma_m=\vec{n} \cdot (\vec{H}_{>}-\vec{H}_{<}),$$
where $\Sigma_m$ is an effective magnetic surface charge. For the simplifying model of having $\vec{M}=\text{const}$ jumping at the surface of the magnet to 0, you only have this surface charge. This is also clear from the definition of the effective magnetic-charge density since taking the divergence of $\vec{M}$ leads to a $\delta$-distribution at the surface.

2nd stragey: Vector potential for $\vec{B}$

This way is working for all cases, i.e., even including free current densities. But let's stick to just the magnet without external currents. Then from Gauss's Law for $\vec{B}$ it's clear that there is a vector potential, fullfilling the Coulomb-gauge constraint
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{\nabla} \cdot \vec{A}=0,$$
and I need the vortices of $\vec{B}$ to get the sources for $\vec{A}$. Now we use the constitutive equations and the curl-freeness of $\vec{H}$ to get
$$\vec{\nabla} \times \vec{B}=-\Delta \vec{A}=\mu_0 \vec{\nabla} \times (\vec{H}+\vec{M})=\mu_0 \vec{\nabla} \times \vec{M}=\mu_0 \vec{j}_m.$$
The solution is Biot-Savart's Law for $\vec{B}$ with the effective magnetic current density $\vec{j}_m=\vec{\nabla} \times \vec{M}$ (see my previous posting). The boundary conditions are this time for $\vec{B}$
$$\vec{n} \cdot (\vec{B}_{>}-\vec{B}_{<})=0, \quad \vec{n} \times (\vec{B}_{>}-\vec{B}_<)=\mu_0 \vec{k}_m,$$
where $\vec{k}_m$ is an effective magnetic surface current (if present).

For the model with $\vec{M}=\text{const}$ jumping at the magnet's surface you get the surface current again by taking the curl of $\vec{M}$ which is then 0 everywhere, and you only have a surface current, given by a $\delta$-distribution valued $\vec{j}_m$.

Right now I'm preparing a section in my manuscript, where everything is analytically analyzed for the homogeneously magnetized sphere. Stay tuned :-).

The constitutive equation for a ferromagnet cannot be described by linear response theory anymore. You have a case of spontaneous symmetry breaking, and the constitutive relation depends on the history of the material. Heating it up without magnetic fields present, at a certain point the material will loose all magnetization. Then you can cool it down, and still there's no magnetization, but that's a metastable state since the magnetized state is energetically favored. Bringing the material in a magnetic field, the magnetization will grow from 0 to a certain saturation value and then stay there, even when switching the external magnetic field off. That's your permanent magnet. To demagnetize it you need a minimum external magnetic field, i.e., the constitutive law $\vec{M}=\vec{M}(\vec{H})$ is not going throuh 0 at $\vec{H}=0$. That's the famous hysteresis curve, shown in many textbooks. The most simple phenomenological theory is of course Ginzburg-Landau theory, i.e., the mean-field approximation for the many-body QFT describing the ferromagnetic medium.

 

[Charles Link]

For the first method above, where we have $\vec{B}=\mu_o(\vec{H}_m+\vec{M})$ without any free currents, this method can be expanded to include free currents. The $\vec{H}_{free}$ from the free currents is computed from Biot-Savart/Ampere's law and has for this contribution that $\vec{B}=\mu_o \vec{H}_{free}$. Thereby, if $\vec{H}$ is redefined to include the contribution from free currents, so that $\vec{H}=\vec{H}_m+\vec{H}_{free}$, this contribution to $\vec{H}$ can be added to both sides of the equation $\vec{B}=\mu_o(\vec{H}+\vec{M})$ and the equation $\vec{B}= \mu_o (\vec{H}+\vec{M})$ remains valid. In this case though, we no longer have $\nabla \times \vec{H}=0$, because $\nabla \times \vec{H}_{free}=\vec{J}_{free}$. $\\$ Basically the free current solution for $\vec{H}_{free}$ can be superimposed upon the solution for $\vec{H}_m$ that comes from a scalar potential $\vec{H}_m=-\nabla \phi_m$. (The scalar potential method requires $\nabla \times \vec{H}=0$ in order to work). $\\$ The reason why this solution with the $\vec{H}_{free}$ can be important, is that it is then possible to use the " magnetic pole model" method with the scalar potential $\phi_m$ to solve the problem that the OP has of a transformer with an air gap.

 

[vanhees71]

Yes, indeed. See my posting #31.

 

[jim hardy]

where →MM→\vec{M} is the magnetization unchanged by external influence due to the presence of hard ferromagnets.

So M then is remanent magnetism ?
Trying to nail down the terminology in my alleged mind.

THANKS - old jim

 

[Charles Link]

So M then is remanent magnetism ?
Trying to nail down the terminology in my alleged mind.

THANKS - old jim

Here $\vec{M}$ can be either a remanent=permanent magnetization that is virtually unchanged by any applied $\vec{H}$ field, or it can be the type that is found in a transformer where $\vec{M}=\chi_m \vec{H}$. $\\$ The equation $\vec{B}=\mu_o(\vec{H}+\vec{M})$ works for both types. The equation $\vec{B}=\mu_r \mu_o \vec{H}$ only works for the transformer type of magnetization. In that case $\mu_r=1+\chi_m$.

 

[vanhees71]

In general, ferromagnets can not be described by linear-response theory as can para- and diamagnets. The latter have
$$\vec{M}=\chi_m \vec{H},$$
where the magnetic susceptibility is a dimensionless number, with its modulus usually much smaller than one. A paramagnet had by definition $\chi_m>0$, a diamagnet $\chi_m<0$. A special case is a superconductor, where formally $\chi_m=-1$, i.e., it's a perfect diamagnet. The consitutive equation is in these cases linear, i.e.,
$$\vec{B}=\mu_0 (\vec{H}+\vec{M})=\mu_0 \mu_{r} \vec{H}, \quad \mu_r=1+\chi_m.$$
There's no magnetic field (traditionally called magnetic induction) $\vec{B}$ in the interior of a superconductor (which is an idealization of course).

For ferromagnets the relation between $\vec{H}$ and $\vec{M}$ is non-linear and not even unique. If you start with an unmagnetized piece of material, you get a curve starting from $0$ at $\vec{H}=0$. Beginning with large enough $|\vec{H}|$ the magnetization saturates, and you have $\vec{M}(\vec{H})=\text{const}$. When you now switch off $\vec{H}$, however, the magnetization doesn't go away completely again. For "hard ferromagnets" it stays at (nearly) at the full saturate value as long as $\vec{H}$ doesn't become too large. That's your usual permanent magnet. For soft ferromagnets the magnetization goes away at least partially, and that's why you like to use them in transformers and electro magnets, but usually you still cannot use the linear-response approximation.

Varying $\vec{H}$ and the ferromagnetic material was magnetized before leads to a more complicated curve $\vec{M}=\vec{M}(\vec{H})$ with $\vec{M}(0) \neq 0$. This is called the "hysteresis loop". A very nice phenomomenological explanation for this behavior using semiclassical arguments can be found in the excellent textbook

J. Schwinger, Classical Electromagnetism

Of course if you make the $\vec{H}$ field strong enough you can demagnetized the material of flip the magnetization to another direction. At high enough temperatures the material becomes paramagnetic. That's a typical case of a 2nd-order phase transition, which can be phenomenologically described by Ginzburg-Landau theory.

 

[mertcan]

@Charles Link @vanhees71 @jim hardy there is something scratch my mind...in the link https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section5-Magnetostatics.pdf and equation 5.73 we can see that magnetic moment or magnetic moment density depends on a kind of current density. I believe it is not the free charge density but what kind of current density is this what results in that current density??

 

[Charles Link]

@Charles Link @vanhees71 @jim hardy there is something scratch my mind...in the link https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section5-Magnetostatics.pdf and equation 5.73 we can see that magnetic moment or magnetic moment density depends on a kind of current density. I believe it is not the free charge density but what kind of current density is this what results in that current density??

The current density can be free moving electrical charges, but often times, the magnetic current density given by $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$ is the source of the magnetic moment in these equations. The magnetization $\vec{M}$ is the magnetic moment density. When there are gradients in this quantity, magnetic currents $\vec{J}_m$ are the result=(similar to moving electrical charges, but here any electrical charge motion usually consists of bound quantum states). When the magnetization $\vec{M}$ is uniform but encounters a boundary of the magnetic material, the result by Stokes' theorem is a magnetic surface current per unit length given by $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o }$. (Here I'm defining $\vec{M}$ by $\vec{B}=\mu_o \vec{H}+\vec{M}$, instead of by $\vec{B}=\mu_o (\vec{H}+\vec{M}')$). $\\$ You might also find this Insights article that I authored of interest: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/ You might also find in the section "Discuss in the Community" portion of this "link", posts 32 and 33 of some interest. $\\$ Post 32 is a "link" to a posting that appeared on Physics Forums of an interesting homework laboratory on the Curie temperature. Let me provide that "link" here as well: https://www.physicsforums.com/threa...perature-relationship-in-ferromagnets.923380/ You might find the experimental technique and calculations to determine the magnetization $\vec{M}$ of the permanent magnet of considerable interest. $\\$ Post 33 of "Discuss in the Community" mentions some of the complications that arise from the exchange effect, so that these simple linear theories are, in general, not going to be exact, and may be only somewhat good approximations that don't work in all cases.

 

[jim hardy]

Sometimes a picture helps.
This thought experiment leads to those scary looking formulae that describe it.
https://www.physics.wisc.edu/undergrads/courses/phys202fall96/Phys202_Magnetism.html

old jim

 

[Charles Link]

@jim hardy This is all completely consistent with the calculation that says the magnetic surface current per unit length is given by $\vec{K}_m=\frac{\vec{M} \times \hat{n} }{\mu_o}$. A very good "link" indeed. $\\$ For the illustration, I like to consider a square-shaped small current loop, so that it clearly cancels the current of the adjacent loop.

 

[mertcan]

The current density can be free moving electrical charges, but often times, the magnetic current density given by $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$ is the source of the magnetic moment in these equations. The magnetization $\vec{M}$ is the magnetic moment density. When there are gradients in this quantity, magnetic currents $\vec{J}_m$ are the result=(similar to moving electrical charges, but here any electrical charge motion usually consists of bound quantum states). When the magnetization $\vec{M}$ is uniform but encounters a boundary of the magnetic material, the result by Stokes' theorem is a magnetic surface current per unit length given by $\vec{K}_m=\frac{\vec{M} \times \hat{n}}{\mu_o }$. (Here I'm defining $\vec{M}$ by $\vec{B}=\mu_o \vec{H}+\vec{M}$, instead of by $\vec{B}=\mu_o (\vec{H}+\vec{M}')$). $\\$ You might also find this Insights article that I authored of interest: https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/ You might also find in the section "Discuss in the Community" portion of this "link", posts 32 and 33 of some interest. $\\$ Post 32 is a "link" to a posting that appeared on Physics Forums of an interesting homework laboratory on the Curie temperature. Let me provide that "link" here as well: https://www.physicsforums.com/threa...perature-relationship-in-ferromagnets.923380/ You might find the experimental technique and calculations to determine the magnetization $\vec{M}$ of the permanent magnet of considerable interest. $\\$ Post 33 of "Discuss in the Community" mentions some of the complications that arise from the exchange effect, so that these simple linear theories are, in general, not going to be exact, and may be only somewhat good approximations that don't work in all cases.

@Charles Link as far as I have observed in your links and @jim hardy links and my own search, the current density I mentioned at post 40 in given link and equation which affect the magnetic moment density is the circular current around proton or nucleus. Am I right??

I am asking to understand the model of permanent magnets deeply

 

[Charles Link]

@Charles Link as far as I have observed in your link and my own search, the current density I mentioned at post 40 in given links and equation which affect the magnetic moment density is the circular current around proton or nucleus. Am I right

The magnetization $\vec{M}$ , (vector sum of the magnetic moments per unit volume), can actually be due to spin magnetic moment rather than than orbital angular momentum, and I believe that is what happens in the case of iron. $\\$ In the case of spin, the magnetic moment is $\vec{\mu}_s=\frac{g_s \, \mu_B \, \vec{S}}{\hbar}$ , where $\mu_B$ is the Bohr magneton. $\\$ In c.g.s. units, the Bohr magneton $\mu_B=\frac{e \hbar}{2 m_e c }$. $\\$ The constant $g_s \approx 2.0$, and results from the Dirac equation that are beyond my level of expertise give a result that $g_s =2.0032...$ $\\$ For the orbital angular momentum, for which classical models of current loops apply, (with an electron in orbit about the nucleus, etc.), $\vec{\mu}_L=\frac{g_L \, \mu_B \, \vec{L}}{\hbar}$, with $g_L=1.000$. $\\$ The equation $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$, and a related equation $\vec{A}(\vec{r})=\frac{\mu_o}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|} \, d^3 \vec{r}'$, still apply, regardless of the origin of the magnetization.

 

[jim hardy]

For the illustration, I like to consider a square-shaped small current loop, so that it clearly cancels the current of the adjacent loop.

You are not alone. I found a lot of those before finding that circular one.
http://farside.ph.utexas.edu/teaching/302l/lectures/node77.html

I like the round ones because i think of atoms as round and electron orbits as circles - that's all.

old jim

 

[mertcan]

The magnetization $\vec{M}$ , (vector sum of the magnetic moments per unit volume), can actually be due to spin magnetic moment rather than than orbital angular momentum, and I believe that is what happens in the case of iron. $\\$ In the case of spin, the magnetic moment is $\vec{\mu}_s=\frac{g_s \, \mu_B \, \vec{S}}{\hbar}$ , where $\mu_B$ is the Bohr magneton. $\\$ In c.g.s. units, the Bohr magneton $\mu_B=\frac{e \hbar}{2 m_e c }$. $\\$ The constant $g_s \approx 2.0$, and results from the Dirac equation that are beyond my level of expertise give a result that $g_s =2.0032...$ $\\$ For the orbital angular momentum, for which classical models of current loops apply, (with an electron in orbit about the nucleus, etc.), $\vec{\mu}_L=\frac{g_L \, \mu_B \, \vec{L}}{\hbar}$, with $g_L=1.000$. $\\$ The equation $\vec{J}_m=\frac{\nabla \times \vec{M}}{\mu_o}$, and a related equation $\vec{A}(\vec{r})=\frac{\mu_o}{4 \pi} \int \frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|} \, d^3 \vec{r}'$, still apply, regardless of the origin of the magnetization.

You are not alone. I found a lot of those before finding that circular one.
http://farside.ph.utexas.edu/teaching/302l/lectures/node77.html

View attachment 225059

I like the round ones because i think of atoms as round and electron orbits as circles - that's all.

old jim

Another interesting question gentlemen : Let's think that we put the iron in a steady state magnetic field environment and we know that iron is going to behave like magnet, points in iron are going to have different pole strength at the end, but HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field at the end?? Can we calculate that situation??
Also until reaching the steady magnetic field at the end, while time passes HOW does the magnitude of specific point's magnetic field change?? Can we calculate also that situation??

 

[Charles Link]

Another interesting question gentlemen : Let's think that we put the iron in a steady state magnetic field environment and we know that iron is going to behave like magnet, points in iron are going to have different pole strength at the end, but HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field at the end?? Can we calculate that situation??
Also until reaching the steady magnetic field at the end, while time passes HOW does the magnitude of specific point's magnetic field change?? Can we calculate also that situation??

@mertcan This is basically a frequency response question, and in post 20, @jim hardy actually showed some examples of frequency response with and without laminations, where the laminated iron layers have better frequency response because it doesn't have the Faraday EMF's and eddy currents to contend with, with their effect increasing with increasing frequency. $\\$ The question you are asking I think could basically be answered by knowing the amount of phase shift that occurs between the primary and secondary. The experiment might be a little complicated by the items mentioned in post 16. In any case, I don't really know what type of frequency response might be expected.

 

[jim hardy]

A lot goes on in a simple piece of iron. I don't think you can write a general equation. From Sylvanus Thompson's 1896 edition of "Dynamo Electric Machinery"

oops sorry for overlapping snips.

I have an original 1901 edition and it's a delight .
Your question would i think be best answered by experiment on a specimen of the particular alloy, heat-treatment, shape, size, temperature, and recent magnetization that interests you.

A search on phrase "sylvanus thompson retardation of magentization" also turns up the 1903 edition of his book "Design of Dynamos" with some similar paragraphs. Some of his books are reprinted in India, to the credit of educators there. You might find one an interesting addition to your library.

old jim

 

[jim hardy]

in post 20, @jim hardy actually showed some examples of frequency response with and without laminations,

Actually the three images in post #20 are the same non-laminated core at three different frequencies 3hz 10hz and 60 hz. That core was hopeless at 60 hz.

 

[mertcan]

A lot goes on in a simple piece of iron. I don't think you can write a general equation. From Sylvanus Thompson's 1896 edition of "Dynamo Electric Machinery"

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oops sorry for overlapping snips.

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I have an original 1901 edition and it's a delight .
Your question would i think be best answered by experiment on a specimen of the particular alloy, heat-treatment, shape, size, temperature, and recent magnetization that interests you.

A search on phrase "sylvanus thompson retardation of magentization" also turns up the 1903 edition of his book "Design of Dynamos" with some similar paragraphs. Some of his books are reprinted in India, to the credit of educators there. You might find one an interesting addition to your library.

old jim

you say that in short there is no equation related to required time to reach steady state magnetic field and how magnitude of specific point's magnetic field changes until reaching its final magnetic field magnitude(steady)??

 

[jim hardy]

you say that in short there is no equation related to required time to reach steady state magnetic field and how magnitude of specific point's magnetic field changes until reaching its final magnetic field magnitude(steady)??

your question stipulated presence of iron

Let's think that we put the iron in a steady state magnetic field environment and we know that iron is going to behave like magnet, points in iron are going to have different pole strength at the end, but HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field at the end?? Can we calculate that situation??

and experimenters find iron to be not a simple substance.

This is described as "The Classic Text" on ferromagnetism
https://www.barnesandnoble.com/w/ferromagnetism-richard-m-bozorth/1101204822?ean=9780780310322

i seem to have misplaced my copy. I hope it surfaces...
IEEE press is reprinting the 1978 edition
if your technical library has a copy or can get one i suggest perusing it. You will find modern mathematical treatments for all the effects noted in Thompson's late 19th century textbooks. But no single equation.

 

[mertcan]

@jim hardy @Charles Link , initially thanks again for your nice responses. But although jim shared a couple of sources, I do not have a chance now to attain those sources for instance the book is not open source and I do not have it in library. Thus would you mind sharing much more sources with me?? (links, videos, pdf files...)

I am really eager to understand HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field strength at the end when we put the iron in a steady state magnetic field environment and until reaching the steady magnetic field strength at the end, while time passes HOW does the magnitude of specific point's magnetic field or pole strength change

 

[jim hardy]

Try here for Thompson
https://archive.org/details/dynamoelectricma00thomrich
https://www.forgottenbooks.com/en/books/DynamoElectricMachinery_10842070
https://www.bookdepository.com/Dyna...nics-Thompson-Silvanus-Phillips/9781110351046

this page might help you find Bozorth in a library nearby
http://www.worldcat.org/title/ferromagnetism/oclc/29017070&referer=brief_results

I am really eager to understand HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field strength at the end when we put the iron in a steady state magnetic field environment and until reaching the steady magnetic field strength at the end, while time passes HOW does the magnitude of specific point's magnetic field or pole strength change

There's no time term in Ampere's law.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html

So the delays are a property of the material present.Grow your understanding like a garden.
Put your search engine to work. Find articles then search on terms in them.

 

[Charles Link]

@jim hardy @Charles Link , initially thanks again for your nice responses. But although jim shared a couple of sources, I do not have a chance now to attain those sources for instance the book is not open source and I do not have it in library. Thus would you mind sharing much more sources with me?? (links, videos, pdf files...)

I am really eager to understand HOW MUCH time is required for a given specific point in iron to reach its steady magnetic field strength at the end when we put the iron in a steady state magnetic field environment and until reaching the steady magnetic field strength at the end, while time passes HOW does the magnitude of specific point's magnetic field or pole strength change

For a small laminated transformer, (3" diameter), I believe the time constant is quite fast: Just a guess is that it is in the 1-2 millisecond range, but I would need futher research to confirm. @jim hardy Might this number $\tau \approx 2 \, msec$ seem reasonable?

 

[jim hardy]

To what degree of precision ?
Thickness of laminations is the antidote for delay due to eddy current.

Your estimate sounds reasonable for a 60 hz power transformer
a 400 hz transformer has thinner laminations so is faster

but there's still 'creep' and Barkhausen to account for.

i don't know much about high fidelity audio transformers that go to tens of kilohertz .

Ferrite is of course way faster .

That's why I'm reluctant to say "yes", for fear it'll be taken for an absolute answer.
One needs to appreciate the shortcomings of any simple mental model; and be able to put a number on its residual error.

That last few per-cent is difficult to characterize..

I appreciate your desire to give mertcan an answer so he can proceed with his inquiry. In that spirit,
"Sure, in a couple milliseconds most of the little transformers you are likely to encounter will be ~98% of the way to steady state.
But do not forget about the early pioneers and their giant railway dynamos that took minutes to settle."

One of those things that's easy to envision but difficult to calculate. You have to characterize the medium.

Going back to this image - triangle wave current (constant di/dt) should produce square wave voltage , but the corners get rounded off because of lengthy time response.

observe lower trace - that stainless steel bar was not yet settled at 50 miliseconds.
With air core the voltage was textbook perfect square wave. Difference was the magnetic medium.
That little experiment taught me a lot.
That image is the 3 hz response . If you go back to post #50 and look at the 10hz and 60 hz traces , they never get past the rounded corner.

I encourage @mertcan to get some i/o gear for his computer and experiment . We need more students interested in magnetics.

 

[Charles Link]

I encourage @mertcan to get some i/o gear for his computer and experiment . We need more students interested in magnetics.

I agree. I do think for a number of years the basic E&M (electricity and magnetism) has actually been de-emphasized in the college curriculum.

 

[jim hardy]

I do think for a number of years the basic E&M (electricity and magnetism) has actually been de-emphasized in the college curriculum

Yes, we've shortchanged a lot of students.