PSE533090 : Discussion on Why is the magnetic intensity in a material determined only by the external sources even if the material is magnetized?

Guru Vishnu

Feb 28, 2020 08:18

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie Are you free now sir? I have a doubt about Magnetic Intensity $\vec H$.

John Rennie

Yes I'm free.

Guru Vishnu

Ok sir. The following question on the main site. I received an answer, but still I didn't get it:

1

Q: Why is the magnetic intensity in a material determined only by the external sources even if the material is magnetized?

The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Magnetic Properties of Matter", page 281, topic "Magnetic Intensity $H$": Whenever the end effects of a magnetized material can be neglected, the magnetic intensity due to magnetization is zero. This may be the...

John Rennie

Let me see if I can find my copy of Verma ...

Guru Vishnu

Feb 28, 2020 08:25

Ok sir. I think it would be difficult. Do you want me to take a picture of that particular page?

John Rennie

It's OK I have a copy. Let me read that section ...

Guru Vishnu

Ok sir. For reference: It's from Volume 2, page 281, Section 37.3.

John Rennie

OK, so what he's saying is that when you apply an external field $H$ this may induce a magnetisation in the material and that magnetisation produces a field $I$. The total field is then the sum of the applied field and the induced field:

$$ \frac{B}{\mu} = H + I $$

Guru Vishnu

@JohnRennie $\mu_0$ in place of $\mu$?

John Rennie

And in a vacuum no field is induced so this simplifies to:

$$ \frac{B}{\mu} = H $$

@GuruVishnu I'm deliberately being vague because I can never remember why there's a factor of $\mu$ or which $\mu$ it is.

Guru Vishnu

Feb 28, 2020 08:36

@JohnRennie Ok sir. However I think the choice of permeability - relative or of vacuum will have different meanings if applied interchangeably.

The author hasn't supplied any proofs for this equation. It's stated as a definition, so I don't expect any reason for having a $\mu$ term in the above equation. But I might be wrong.

John Rennie

And then what he says is that inside a magnet and far from the ends the effect of the induced magnetisation $I$ is zero. so we still have $B/\mu = H$ even inside something that does exhibit an induced field.

Guru Vishnu

@JohnRennie Yes sir. This is my doubt. Why do we neglect the intensity of magnetisation term $\vec I$ in this case?

And I don't see what does end effects need to do with this.

John Rennie

I have no idea what he means. Sorry.

Guru Vishnu

Ok sir. No problem. However, may I ask doubts from your previous messages? I hope that might help me.

John Rennie

Yes

Guru Vishnu

Feb 28, 2020 08:45

12 mins ago, by John Rennie

OK, so what he's saying is that when you apply an external field $H$ this may induce a magnetisation in the material and that magnetisation produces a field $I$. The total field is then the sum of the applied field and the induced field:

Is $H$ an external field, or a resultant of all magnetic fields?

John Rennie

$H$ is the externally applied field. The resultant of all fields is $H+I$.

Guru Vishnu

Ok sir. Till now, I've been thinking that it was the resultant due to this specific format of that equation:
$$\vec H=\frac{B}{\mu_0}-\vec I$$

John Rennie

I think that's a misleading way of writing the equation, which is why I wrote $B/\mu = H + I$ instead.

Guru Vishnu

Yes sir. I realised it now.

John Rennie

If you're interested I have found some info on calculating the field inside a magnetised material.

Guru Vishnu

Feb 28, 2020 08:51

That's great sir. I'm interested and wish to be clear on this stuff.

Could you please share it here?

John Rennie

Example 9.3.1

I think what that is saying is that the cylinder behaves as if it had a magnetic charge at the ends.

So if it's a long cylinder the middle will be far from the ends and the field will be small simply because it decreases with distance.

Guru Vishnu

I didn't read it completely, but getting some idea of what that article means. However, we know that field lines get accumulated when there is a paramagnetic or ferromagnetic material whereas get dispersed when we have a diamagnetic material in the field. It's surprising how it would not have any impact on the field $H$ as well as $B$.

On the basis of the magnetic pole picture your reasons seems possible. But on the lines of field lines density, it gives the contradictory result, sir.

John Rennie

Shrug

Guru Vishnu

Ok sir. Did you read GiorgioP's answer? My problem wasn't solved, however, maybe I couldn't have understood it fully.

John Rennie

I don't think he has understood your question, and his answer isn't applicable.

Guru Vishnu

Feb 28, 2020 09:04

Ok sir. Thank you.

John Rennie

Feb 28, 2020 09:15

@GuruVishnu I've read the article I linked more thoroughly and it does explain what is going on, but it's complicated. Unless you're really interested I would just accept the result and move on.

Guru Vishnu

Feb 28, 2020 09:56

@JohnRennie Thank you sir. Due to time constraints, I'd prefer to just proceed now. However, I would like to discuss it after my exams. In order to keep this in my memory I've pinned your message.

John Rennie

Actually I think I have an intuitive explanation ...

Guru Vishnu

@JohnRennie Then, could you please proceed sir? I'm happy to understand this one.

John Rennie

Let's use the analogy of electric dipoles. This is simpler because an electric dipole is just two charges separated by a distance $d$.

Guru Vishnu

Ok sir.

John Rennie

Suppose you have an arrangement of positive and negative charges like this:

There is no net field here because the charges are symmetrically distributed. Yes?

Guru Vishnu

Feb 28, 2020 10:04

@JohnRennie Which point are you referring to, sir?

by the term "here"

John Rennie

Some point in the middle far from the edges of the array.

Guru Vishnu

@JohnRennie Then, yes sir.

John Rennie

OK. But actually I lied because these aren't separate charges. They are dipoles arranged like this:

Guru Vishnu

Ok sir.

John Rennie

What I actually have is an array of dipoles like the dipoles induced by the exteral field in the system we were talking about.

So you'd think the dipoles would all add up to give a net field inside the material pointing upwards. Yes?

Guru Vishnu

Feb 28, 2020 10:10

It seems they are perfectly aligned in the direction of the external field. Is my interpretation correct?

John Rennie

Yes. This is intended as an illustration not an actual real system.

Guru Vishnu

@JohnRennie Yes sir.

@JohnRennie Ok sir.

John Rennie

But ... in the two diagrams the arrangement of the charges is the same. So the fields must be the same. And we have already agreed that the net field in the first diagram is zero.

Guru Vishnu

Aha! Yes sir.

John Rennie

So that must mean that in our array of dipoles the field far from the edges must be zero.

Guru Vishnu

Feb 28, 2020 10:13

@JohnRennie Yes but in order to complete the analogy, the field at an internal point must be equal to the external field. Am I right, sir?

John Rennie

If you look at the top edge then we have an excess of red charges, and at the bottom edge we have an excess of blue charges, but everywhere else the red and blue charge densities are the same so they cancel out.

@GuruVishnu ignore the external field for now. This could be a permanent magnet where there is no external field.

Guru Vishnu

@JohnRennie Ok sir.

John Rennie

So what we're saying is that in our dipole array there is only a non-zero field at the top and bottom because everywhere else the charge density still averages to zero.

Guru Vishnu

@JohnRennie Then we're missing some charges in the first and second diagrams. Or are we just neglecting their fields due to increased distance?

John Rennie

In the second diagram the charges have to come in pairs because now they are the two ends of a dipole.

So that must mean that the top and bottom surfaces have a charge imbalance.

So if you compare the two diagrams I've removed the blue charges from the top layer and the red charges from the bottom layer.

Guru Vishnu

Feb 28, 2020 10:18

@JohnRennie Yes sir. So the field at a central point is something like in a charged parallel plate capacitor and is not zero as per our first agreement based on first diagram.

@JohnRennie Yes sir.

John Rennie

Remember that the field in a capacitor is only uniform in the limit of infinite plates i.e. when the plate size is much greater than the separation.

If the separation is greater then or equal to the plate size then the field at the centre will be decreased.

As the ratio of the separation to the plate size goes to infinity the field at the centre goes to zero.

Guru Vishnu

Ok sir.

Guru Vishnu

Feb 28, 2020 10:33

@JohnRennie: Hi sir.

John Rennie

@GuruVishnu hi

Guru Vishnu

I think I made you drift from your explanation. Could you please tell how and which magnetic terms are related to the electric terms in your analogy?

John Rennie

The point is that if we have a uniform density of dipoles then there is no net internal field. This is true regardless of the type of the dipole.

Guru Vishnu

Feb 28, 2020 10:50

Ok sir. And?

John Rennie

So there is no field in the centre of our material with the induced dipoles.

Guru Vishnu

Ok sir. I understood it based on the distance factor of the surface charge density. How do I relate $\vec H$ to electric field $\vec E$ here? I think $\vec H$ doesn't refer to magnetic field but something else.

John Rennie

H is the magnetic field give or take a factor of $\mu_0$.

Guru Vishnu

Ok sir. Thank you.

PSE533090 : Discussion on Why is the magnetic intensity in a material determined only by the external sources even if the material is magnetized?

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