Electric polarization of a dielectric due to a magentic field

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Imagine a dielectric cylinder placed in a uniform magnetic field parallel to its axis.
Then we make it rotate around its axis.
The nuclei and electrons should feel a radial lorentz force because of their rotation around the axis of cylinder.The forces applied to electrons and nuclei are in opposite direction so the atoms should become polarized and there should arise a [itex] \vec{P} [/itex] field in the dielectric.But this [itex] \vec{P} [/itex] field is not due to an electric field but a magnetic field.
I can't find explanations or formulas regarding polarization due to magnetic fields.
Can somebody help?
Thanks
 

Answers and Replies

  • #2
Simon Bridge
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Polarization due to electric field is from the separation of charge since qE acts in opposite directions for + and - charge and atoms are made of + and - charges.

How does a B field interact with atoms?
 
  • #3
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Ok,let me explain precisly.
I will use cylindrical coordinates [itex] (\rho,\phi,z) [/itex]
There is a magnetic field [itex] \vec{B}=B_0 \hat{z} [/itex]
We place a dilelectric cylinder in it with its axis parallel to z axis
Then we make the cylinder rotate around its axis with angular velocity [itex] \omega [/itex]
every atom is then rotating with angular velocity [itex] \omega [/itex] which translates to a linear velocity [itex] v=\rho \omega [/itex]
So the electron cloud and nucleus of the atom will experience a force [itex] \vec{F}=Zer \omega B_0 \hat{\rho} [/itex] which is toward the center of the cylinder for the electron cloud and outward for the nucleus.This force accelerates the nucleus and the electron cloud in opposite directions until its stopped by the coloumb attraction of the two.This will make a electric dipole and since it happens for all of the atoms in the dielectric,it should form a [itex] P [/itex] field.
 
  • #4
Simon Bridge
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OK - so what is the problem?
 
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The problem is that every text about polarization density,just mentions electric field as a source and I can't find formulas and explanations about polarization density due to magnetic fields.
 
  • #6
Simon Bridge
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That's fair enough - texts leave out all kinds of things - you have all the information needed to figure them out for yourself.
In fact it is a fairly common problem given to students. (Since it is so common, I'm being a tad more circumspect that I could be.)

Was there a particular problem you were interested in?
Have you seen:
https://www.physicsforums.com/showthread.php?t=448392
 
  • #7
DrDu
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You could also try to Lorentz transform the magnetic field into the reference frame where the dielectric is at rest. There will be an electric field component in that frame that accounts for polarisation.
 
  • #8
Simon Bridge
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Rereading my last responce I may have been a bit terse.
My thinking is:

The first step to solving a problem by scientific method is to clearly identify the problem.

Shyan's stated problem is that texts don't cover this situation.
I don't think many of us are in a position to do anything about that.

Since it is a common configuration for a teaching exercise, it may be that the problem comes up in the context of homework.

So this is either a homework problem or a discussion about text-book publishing policies.
(Maybe he needs to look at a different text: a GR one covering the Faraday tensor perhaps?)

There are several paths through this as a homework problem - I think I'd like to know more so I can tell what level the suggestions should be pitched at, at least. But I didn't need to be so closed mouthed about it :)
 
  • #9
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That's fair enough - texts leave out all kinds of things - you have all the information needed to figure them out for yourself.
In fact it is a fairly common problem given to students. (Since it is so common, I'm being a tad more circumspect that I could be.)

Was there a particular problem you were interested in?
Have you seen:
https://www.physicsforums.com/showthread.php?t=448392
Well,it was a problem at the end of the "Electromagnetic Induction" chapter of the "Foundations of electromagnetic theory" by Reitz and Milford.

I guess the reason that no formula can be found anywhere,is that there is no general formula for it.
I managed to calculate it and this is my result:
[itex] \vec{P}=\frac{4}{3}e\sqrt{\frac{ke}{\omega R B}} \hat{\rho} [/itex]
in cylindrical coordinates.
I should mention that the above formula is for a cylinder containing hydrogen(well,I couldn't find the coloumb force between the poles for atoms with more electrons and well,we have to think about solid hydrogen!)
 
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  • #10
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You could also try to Lorentz transform the magnetic field into the reference frame where the dielectric is at rest. There will be an electric field component in that frame that accounts for polarisation.
That's correct but not for this particular problem because here the velocity depends on the distance from the center of the cylinder.

Thanks both
 
  • #11
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I should mention that the result I mentioned,is under the assumption that the distance to the center is much greater than the dipole length otherwise I think we should consider the change in lorentz force due to the movement of the nucleus and the electron cloud.
It has several difficulties if analyzed carefully.
 
  • #12
Simon Bridge
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Well,it was a problem at the end of the "Electromagnetic Induction" chapter of the "Foundations of electromagnetic theory" by Reitz and Milford.
I thought it was something like that ... it is a teaching/learning exercize: you learn more by doing it yourself than from me pointing you at a solution.
I guess the reason that no formula can be found anywhere,is that there is no general formula for it.
It is more that this is a special case of an accelerating reference frame ... there are an infinite number of them and listing all the equations, even just for the interesting ones, would make for an expensive book. The general method is covered in other kinds of courses like general relativity.

I managed to calculate it and this is my result:
[itex] \vec{P}=\frac{4}{3}e\sqrt{\frac{ke}{\omega R B}} \hat{\rho} [/itex]
in cylindrical coordinates.
I should mention that the above formula is for a cylinder containing hydrogen(well,I couldn't find the coloumb force between the poles for atoms with more electrons and well,we have to think about solid hydrogen!)
Hmmm... in a rotating cylinder, wouldn't the gas be more dense towards the outside?

For a rotating solid dielectric, the dielectric constant is supposed to take care of these concerns. Did you look at the link I gave you?
I also have found a more rigorous treatment: Kirk T. McDonald's Dielectric Cylinder That Rotates in a Uniform Magnetic Field (2003).

That's [Lorentz transform approach] correct but not for this particular problem because here the velocity depends on the distance from the center of the cylinder.
It is perfectly correct for this problem - it just means you need the Lorentz transformation for a rotating reference frame ... that would be covered in General Relativity.

You could try applying the LT you know already to annular shells thickness dr at radius r... but I think MacDonald's paper is the one you want.
 
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  • #13
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Hmmm... in a rotating cylinder, wouldn't the gas be more dense towards the outside?

For a rotating solid dielectric, the dielectric constant is supposed to take care of these concerns. Did you look at the link I gave you?
I also have found a more rigorous treatment: Kirk T. McDonald's Dielectric Cylinder That Rotates in a Uniform Magnetic Field (2003).
Well,as I mentioned,I was imagining solid hydrogen :biggrin:
Yes,but that link was about a problem in reverse
Thanks,that seems nice,I'll read it
It is perfectly correct for this problem - it just means you need the Lorentz transformation for a rotating reference frame ... that would be covered in General Relativity.

You could try applying the LT you know already to annular shells thickness dr at radius r... but I think MacDonald's paper is the one you want.
Yeah,you're right and then an integration

Thanks
 
  • #14
DrDu
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That's correct but not for this particular problem because here the velocity depends on the distance from the center of the cylinder.

Thanks both
So what? The Lorentz trafo will depend on position. You are also going to a non-inertial system and this will lead in principle to modifications of the dielectric response. The point is that you have to go to the rest frame of the dielectric because the dielectric function is defined in the rest frame.
 

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