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

[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"

View attachment 225149

oops sorry for overlapping snips.

View attachment 225150

View attachment 225151

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.