Analyzing Magnetic Field of an Infinite Cylinder with Constant Magnetization

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Homework Help Overview

The discussion revolves around analyzing the magnetic field of an infinite cylinder with a constant magnetization along its axis, as well as exploring related concepts in electrostatics and magnetostatics involving slabs with uniform polarization and magnetization. Participants are attempting to understand the implications of these setups on the resulting fields and currents.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the analogy of the infinite cylinder to an infinite solenoid and question the magnetic field characteristics. They also discuss the electric field in a polarized slab, questioning the relationships between polarization, bound charges, and electric fields. Additionally, they examine the implications of surface current densities and their effects on magnetic fields within and outside the slab.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the relationships between surface current densities and magnetic fields, as well as the implications of polarization on electric fields. However, there is no explicit consensus on the final outcomes or methods to be used.

Contextual Notes

Participants are navigating through assumptions about the presence of free charges and the definitions of surface and volume current densities. There is an ongoing examination of the implications of uniform magnetization and polarization on the fields in question.

  • #31
I am confused :S. Should the magnetic field within the slab be 2*pi*j/c?
 
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  • #32
peripatein said:
I am confused :S. Should the magnetic field within the slab be 2*pi*j/c?

That would be correct if you are considering the contribution to B due to just one of the surfaces.
 
  • #33
Isn't that the expression one obtains by letting the rectangular Amperian loop transect BOTH planes? Because then the equality should be:
2L*B = 4*pi*j*L/c
shouldn't it?
 
  • #34
Please ignore my last post. So the magnetic field within the slab should be 4*pi*j/c?
 
  • #35
peripatein said:
Isn't that the expression one obtains by letting the rectangular Amperian loop transect BOTH planes? Because then the equality should be:
2L*B = 4*pi*j*L/c
shouldn't it?

It can be confusing. We have two approaches:

1. Consider just one of the current sheets and use Ampere's law to find B on each side of the sheet. Then use superposition to find the total B due to the two current sheets.

2. Consider the total system of two sheets and apply Ampere's law to this system. Choose a rectangular Amperian path that has one leg at z > d and the other leg between the two sheets. This rectangle will capture some of the current of the upper sheet. Use the fact that you know B = 0 for z > d for the total system.

Either approach will lead to the correct answer.
 
  • #36
According to what you just posted, should I then simply substitute the appropriate j (in terms of M and c) for each of the two planes with fields 2*pi*j/c?
 
  • #37
peripatein said:
According to what you just posted, should I then simply substitute the appropriate j (in terms of M and c) for each of the two planes with fields 2*pi*j/c?

Yes :smile:
 
  • #38
But won't they cancel out each other (I mean, upon summation of both contributions)? That is, won't the net magnetic field within the slab be zero?
 
  • #39
Are the currents in the two surfaces in the same direction or opposite direction?
 
  • #40
Based on j, opposite.
 
  • #41
Yes, so will the two fields inside the slab from the two currents be in the same direction or opposite direction?
 
  • #42
Using the RHR the fields would form a unified field of 4*p*M is the positive z direction. Do you agree?
 
  • #43
peripatein said:
Using the RHR the fields would form a unified field of 4*p*M is the positive z direction. Do you agree?

Yes, if you meant to say in the positive x direction.
 
  • #44
Why not z?
 
  • #45
Think about it. (Got to go eat dinner now!)
 
  • #46
Hmm... I find it a bit strange, as curling my fingers around a current conducting wire in the y direction, I seem to be getting a magnetic field in the z direction (coming out of the sheet to the left of the wire and entering the sheet to its right)!
 
  • #47
Consider the upper current sheet with current in the negative y direction, as shown. Suppose you want B at point P. Take two symmetrically placed current elements (shown in red). What would be the direction of the net B at P from these two current elements?
 

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  • #48
I do see it, using B-S Law and the RHR, but why didn't my initial application of the RHR get me there? Why did the RHR "seemingly" yielded the wrong result? Also, how could I have convinced myself that there was no volumetric charge density rho and only surface density? Is it because based on Gauss's Law a volumetric charge density would entail an electric field and there is none in this set-up?
 
  • #49
peripatein said:
I do see it, using B-S Law and the RHR, but why didn't my initial application of the RHR get me there? Why did the RHR "seemingly" yielded the wrong result?
I believe it was because you were considering the field of only one straight wire of current rather than the superposition of the fields of all the current elements in the surface.

Also, how could I have convinced myself that there was no volumetric charge density rho and only surface density? Is it because based on Gauss's Law a volumetric charge density would entail an electric field and there is none in this set-up?

ρ would result from a nonuniform electric polarization P. But here we have only magnetic polarization M.
 
  • #50
Could we have combined both M in the x direction (this set up) and P in the z direction (previous set up) into a new set up? Will it make any sense?
 
  • #51
peripatein said:
Could we have combined both M in the x direction (this set up) and P in the z direction (previous set up) into a new set up? Will it make any sense?

In principle I think you could have both P and M. I don't see why not.
 
  • #52
Thank you so, so much, TSny! You were incredibly kind and helpful :-).
 

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