Analyzing Magnetic Field of an Infinite Cylinder with Constant Magnetization

In summary, if an infinite slab between z=0 and z=d (thickness d) is uniformly magnetized in the x direction, then the electric field is -2πρd\hat{z}
  • #1
peripatein
880
0
Hi,

Homework Statement


Suppose I have an infinite cylinder with radius R, axis along the z axis and constant magnetization M[itex]\hat{z}[/itex]. I wish to find the magnetic field everywhere. (This is not a HW question per se, yet thought I might get some comments on my attempt at solving it nevertheless.)

Homework Equations





The Attempt at a Solution


As M is constant and [itex]\vec{j}[/itex] = cM[itex]\hat{θ}[/itex], i.e. tangential current, may I consider this to be an infinite solenoid, whose magnetic field is zero everywhere, except for within the solenoid where it is equal to 4πnI/c (using c.g.s), where n is the density of wraps, which is equal to 4πM (in the positive z direction)?
 
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  • #2
That all looks correct to me.
 
  • #3
Okay, and if this were an infinite slab between z=0 and z=d (thickness d), with uniform polarization P[itex]\hat{z}[/itex]? How could I find its electric field?

I do know that for a non-polarized slab it would be:

2πρ(2z-d)[itex]\hat{z}[/itex] for 0≤z≤d; 2πρd[itex]\hat{z}[/itex] for z>d; -2πρd[itex]\hat{z}[/itex] for z<d

And that here:

ρ_bound = -[itex]\nabla[/itex][itex]\cdot[/itex]P = 0?
 
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  • #4
I believe I made a mistake, didn't I? That relation between P and rho was for rho_bound, wasn't it? What about rho_free?
 
  • #5
I do know that:
[itex]\nabla[/itex][itex]\cdot[/itex]E = 4πρfree, right?
 
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  • #6
Nabla and E should be vectors of course. Should/could I use the relation in #5 to find E, or rather discontinuity?
 
  • #7
peripatein said:
Okay, and if this were an infinite slab between z=0 and z=d (thickness d), with uniform polarization P[itex]\hat{z}[/itex]? How could I find its electric field?

I do know that for a non-polarized slab it would be:

2πρ(2z-d)[itex]\hat{z}[/itex] for 0≤z≤d; 2πρd[itex]\hat{z}[/itex] for z>d; -2πρd[itex]\hat{z}[/itex] for z<d

I don't know where you got these expressions. Why would there be any electric field if the slab is not polarized?

And that here:

ρ_bound = -[itex]\nabla[/itex][itex]\cdot[/itex]P = 0?

Yes, for a uniform polarization, ρbound will be zero at all interior points of the slab. But there will be bound surface charges on each surface. The electric field can be found from these surface charges.
 
  • #8
peripatein said:
I do know that:
[itex]\nabla[/itex][itex]\cdot[/itex]E = 4πρfree, right?

[itex]\nabla[/itex][itex]\cdot[/itex]E = 4πρtotal = 4π(ρfree + ρbound)
 
  • #9
But wouldn't rho_bound (volumetric!) be zero here?
 
  • #10
So should -P be equal to the jump in electric field? Or is it P? How may I determine the direction of the normal vector?
 
  • #11
peripatein said:
But wouldn't rho_bound (volumetric!) be zero here?

Yes, that's correct for points interior and exterior to the slab for this problem. ( I thought you were writing a general expression rather than an expression specific to this problem.) Now, what is ρfree for this problem?
 
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  • #12
Other than expressing it via the terms in #8, I am not sure. Here's an attempt - may it also be zero? Are there free charges in this setting?
 
  • #13
peripatein said:
So should -P be equal to the jump in electric field? Or is it P? How may I determine the direction of the normal vector?

You are free to define your own convention for the direction of the normal vector. If you think about the polarization as coming from a bunch of little electric dipoles uniformly spread in the slab, you should be able to see which one of the two surfaces of the slab has a positive surface charge density and which one has a negative charge density.
 
  • #14
peripatein said:
Other than expressing it via the terms in #8, I am not sure. Here's an attempt - may it also be zero? Are there free charges in this setting?

From the way in which you stated the set-up, I assumed that there is no free charge anywhere. Of course, you could add free charge if you want!
 
  • #15
So that implies that E is constant, right? Now, is the following correct: 4*pi*sigma (=discontinuity in electric field) = P = -E? Within the slab, that is. And zero outside it?
 
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  • #16
peripatein said:
So that implies that E is constant, right?
Yes, E will be constant inside the slab.

Now, is the following correct: 4*pi*sigma (=discontinuity in electric field) = P = -E? Within the slab, that is. And zero outside it?
I think you are missing a factor of 4*pi in front of P. I'm not sure what your sign convention means when you write -E. If you mean that the direction of E inside the slab will be opposite to the direction of P, then that's correct.
 
  • #17
Within the slab, E should be equal to 4*pi*Pi in the negative z direction, whereas it would be zero anywhere outside the slab. Right?
I would like to challenge my understanding even further by replacing the polarization with a uniform magnetization, M in the (positive) x direction. Then, for starter, j should be equal to -Mc in the y direction, right?
 
  • #18
peripatein said:
Within the slab, E should be equal to 4*pi*Pi in the negative z direction, whereas it would be zero anywhere outside the slab. Right?
Right.

I would like to challenge my understanding even further by replacing the polarization with a uniform magnetization, M in the (positive) x direction. Then, for starter, j should be equal to -Mc in the y direction, right?

Yes, jy = -Mc on the upper surface of the slab.
 
  • #19
Could I then simply substitute that value of j in the formula for the magnetic field of a slab, viz.:
2*pi*j*d/c in the y direction (for z>d); -2*pi*j*d/c in the y direction (for z<d); 4*pi*j*z/c in the y direction (for z between 0 and d)?
 
  • #20
peripatein said:
Could I then simply substitute that value of j in the formula for the magnetic field of a slab, viz.:
2*pi*j*d/c in the y direction (for z>d); -2*pi*j*d/c in the y direction (for z<d); 4*pi*j*z/c in the y direction (for z between 0 and d)?

No, I don't think that's right. When I wrote jy = -Mc, I meant for jy to represent the surface current density on the upper surface of the slab. There will also be a surface current density on the lower surface. You should be able to see that there will not be any volume current density within the slab.
 
  • #21
There is a jump (discontinuity) in magnetic field here, isn't there? May it assist in finding the magnitude of the field?
 
  • #22
peripatein said:
There is a jump (discontinuity) in magnetic field here, isn't there?
Yes.

May it assist in finding the magnitude of the field?

Well, once you know the surface current densities, the field will be the superposition of the fields from these currents. It is not hard to use Ampere's law to find the field due to a uniform surface current on an infinite plane.
 
  • #23
Well, using Ampere's Law I did find the magnetic field to be that stated above (#19). Could I use that, or would it be to no avail in this set-up?
 
  • #24
Your solution in #19 looks like the solution for a slab with a uniform volume current density. That's not what you have here. You have two surface current densities.
 
  • #25
Alright, so, using Ampere's Law, outside B would be zero, whereas within the slab B would be equal to 4*pi*j*d/c. I am not sure this is correct. Could you please confirm? I used a rectangular Amperian loop with height h > d and side L.
 
  • #26
peripatein said:
Alright, so, using Ampere's Law, outside B would be zero,
Yes.

whereas within the slab B would be equal to 4*pi*j*d/c. I am not sure this is correct.

This is close, but not quite correct. Your answer does not have the right dimensions for B. Remember that here j represents a surface current density with dimensions Amperes/meter.

You will ultimately want to express your answer in terms of the magnetization M.
 
  • #27
There's something here which stymies me. Suppose the loop is placed so that it is perpendicular to the plane and only a portion of it, of area L*d, is inserted within the slab. Wouldn't Ampere's Law be written thus:
B*L = 4*pi*j*L*d/c?
 
  • #28
peripatein said:
There's something here which stymies me. Suppose the loop is placed so that it is perpendicular to the plane and only a portion of it, of area L*d, is inserted within the slab. Wouldn't Ampere's Law be written thus:
B*L = 4*pi*j*L*d/c?

The current passing through the loop would not be j*L*d. Since j is a surface current density, j is the current per unit length passing through a line on the surface that is oriented perpendicular to j.
 
  • #29
You are absolutely right, pardon me :-). OK, hence the expression for the magnetic field would be:
4*pi*M for the lower surface and -4*pi*M for the upper one. Do you agree?
 
  • #30
peripatein said:
OK, hence the expression for the magnetic field would be:
4*pi*M for the lower surface and -4*pi*M for the upper one. Do you agree?

No sure what you're saying here. If you pick an arbitrary point inside the slab, what is the contribution to B at that point due to the current on the upper surface? What is the contribution to B at that same point due to the current on the lower surface?
 
  • #31
I am confused :S. Should the magnetic field within the slab be 2*pi*j/c?
 
  • #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.
 

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