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Force on a current-carrying wire


Danijel Gorupec

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Help needed:

 

post-31066-0-58390500-1483996239_thumb.png

 

I have a C-shaped permanent magnet as on the picture A). I sketched magnetic field lines between magnet poles... very close to the magnet, there is a current-carrying wire (perpendicular to your monitor plane). Some of the magnetic field can reach the wire and so I know the wire 'feels' a force F.

 

In the picture B) there is the same setup, only now I plugged the gap with some magnetically permeable iron insert. My question is will the wire now feel a force F'? Is this force higher, smaller or equal to F? Can you explain what is happening?

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Will there be magnetic field there?

I don't know. For sure there will be some because there is always some leakage of the flux. But I cannot understand if, after the iron plug is inserted, the leakage becomes weaker (because it is now better channeled within iron core) or stronger (because now there is probably increased flux within core, so maybe leakages also increase) or stays about the same.

 

But even if I suppose there are no leakages at all, the following puzzles me: When I draw the magnetic field created by current in the wire I see it expands around the wire, and because the wire is very close to the magnetic core it enters the core too - there it interacts with the flux of the magnet. Is this enough to create force on the wire? The answer, I think, is YES, but it troubles me.

 

post-31066-0-66698300-1484032866.png

 

[The picture shows how I imagine what is happening - the picture C) shows two magnetic fields; the picture D) shows their addition]

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Ideally there is no field around the wire, as it would be contained in the plug. If there is any field, it would be smaller than before. The number of flux lines leaving the magnet would be fixed, and more of those are in the plug.

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Ideally there is no field around the wire, as it would be contained in the plug. If there is any field, it would be smaller than before. The number of flux lines leaving the magnet would be fixed, and more of those are in the plug.

The conclusion is right that the field and thus the force at the wire will be much less. However, the number of flux lines leaving a magnet is not fixed and depends on the magnetic reluctance and thus on the presence of an air gap. So there will be a larger flux in the magnetic loop, but less around the now none-existent air gap.

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After reading your answers (thanks), and rethinking, I now believe that:

- if the magnet core was already near saturation, then inserting the iron plug would significantly decrease the leakage flux in that area (because the flux within the magnet cannot increase much and almost all of the flux will be channeled through the iron plug)

- if the magnet core is far from saturation and the flux strength was regulated only by the air gap, then inserting the iron plug would decrease the leakage flux only slightly (because the flux within magnet will greatly increase)

- in any case, the leakage flux will not increase.

 

Unfortunately, it does not answer me what happens with the force:

- if the force only depends on the leakage flux, then it will decrease

- but if the force depends also on the flux within the magnet (because two flux fields interacts) then the force can also increase under some circumstances - that is, in the case when the flux internal to magnet increases a lot after inserting the plug.

 

What do you think?

 

(Note: whenever I read about the force excreted on a current-carrying wire, the text always talk about flux density the wire is immersed into. But from the beginning I am not sure what the word 'immersed' actually means - does it include nearby magnetic fields that are not really 'touching' the wire but are within the reach of the wire-generated magnetic field?)

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- even if the core is far from saturation, the leakage will decrease drastically, since the increase in flux is relatively moderate and the decrease in leakage could be orders of magnitude

 

- the force only depends on the flux density at the wire, the flux density anywhere else has no influence.

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The conclusion is right that the field and thus the force at the wire will be much less. However, the number of flux lines leaving a magnet is not fixed and depends on the magnetic reluctance and thus on the presence of an air gap. So there will be a larger flux in the magnetic loop, but less around the now none-existent air gap.

 

 

How does reluctance play into this? We have a permanent magnet.

 

The distinction here is between H and B. The net value of H is unchanged. B increases in the iron plug.

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- the force only depends on the flux density at the wire, the flux density anywhere else has no influence.

Here is my motivation for the question:

 

post-31066-0-67681200-1484057935.png

 

The picture above shows a part of a slotted motor rotor. Current-carrying wires are inserted into slots. The problem is that almost all the flux goes through teeth, very small amount of flux actually goes through slots. These teeth in a way carry the flux away from the wires. Wires are actually immersed into a low-density flux.

 

If you are correct, then shouldn't it be inefficient, generating small torque? Would then it be better to have iron wires, instead of copper wires, in order to increase flux density through them?

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Here is my motivation for the question:

 

attachicon.gifmagnets3.png

 

The picture above shows a part of a slotted motor rotor. Current-carrying wires are inserted into slots. The problem is that almost all the flux goes through teeth, very small amount of flux actually goes through slots. These teeth in a way carry the flux away from the wires. Wires are actually immersed into a low-density flux.

 

If you are correct, then shouldn't it be inefficient, generating small torque? Would then it be better to have iron wires, instead of copper wires, in order to increase flux density through them?

 

 

It's not the B inside the wire that matters, it's the B outside the wire. That's what's in the equation — the ambient field.

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It's not the B inside the wire that matters, it's the B outside the wire. That's what's in the equation — the ambient field.

 

Yes, this makes sense to me... This would mean that also in my original 'experiment' (the picture B) in OP) the force can still increase despite the fact that the leakage flux decreased. It can increase because the flux outside the wire (within magnet body) can increase a lot after inserting the iron plug.

 

There are various texts (all that I read, Wikipedia included) where wording is not clear and the claim can be understood as the wire is only effected by the flux that actually flows through the wire.

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Short answer: in an iron core electromotor, the torque is produced by a minimisation of the magnetic energy. This happens more efficiently when the flux is going through the iron core and the wires are essentially air gaps. They are only present to generate the interacting magnetic fields.

 

 

 

How does reluctance play into this? We have a permanent magnet.

 

The distinction here is between H and B. The net value of H is unchanged. B increases in the iron plug.

Inside the magnet, H increases as the air gap increases. This results in a decrease in B. I can provide more details upon request and when not posting from my phone. Edited by Bender
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Yes, this makes sense to me... This would mean that also in my original 'experiment' (the picture B) in OP) the force can still increase despite the fact that the leakage flux decreased. It can increase because the flux outside the wire (within magnet body) can increase a lot after inserting the iron plug.

 

 

But that's the leakage flux. If it increases it's because you're concentrating the leakage flux as it goes into the plug. Where the wire is located is likely to have a reduced flux.

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Short answer: in an iron core electromotor, the torque is produced by a minimisation of the magnetic energy.

 

I am not sure what do you exactly mean by 'minimization of the magnetic energy'. Because how I understand it, the force can exist without any change in magnetic energy:

 

post-31066-0-47293600-1484084527.png

 

In the above setup a constant-current-carrying-wire is placed between two 'indefinitely' wide magnet poles. The force on the wire exist despite the fact that magnetic energy remains the same after the wire is moved laterally (although, moving the wire will draw some energy from the current source).

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That is not the mechanism that drives an iron core motor, because as you deducted correctly, the magnetic field doesn't go through the wires. (or not a significant amount of it anyway)

Edited by Bender
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Those only interested in the main topic can safely ignore this, but I felt I had to address this more properly.

How does reluctance play into this? We have a permanent magnet.

 

The distinction here is between H and B. The net value of H is unchanged. B increases in the iron plug.

Inside the magnet:

[math]B=\mu_0 (M+H_{PM})[/math]

with M the magnetisation of the magnet, (and assuming it is a rare earth magnet or another magnet with constant M in the absence of other sources of H)

In the air gap:

[math]B=\mu_0 H_{gap}[/math]

In the iron; H is negligible. Most likely, the C-shaped magnet is a bar magnet on the right and then some iron to guide the magnetic field around.

 

Now Ampere's law in the absence of currents or changing electrical fields:

[math]\oint_{l}Hdl=0[/math]

so:

[math]H_{PM}l_{PM}+H_{gap}l_{gap}=0[/math]

Assuming all flux goes through the air gap and, for simplicity, that the magnet and the air gap have the same surface area:

[math]\mu_0 (M+H_{PM})=\mu_0 H_{gap}[/math]

Combining the last two equations:

[math]H_{PM}=-M \cdot \frac{l_{gap}}{l_{gap}+l_{PM}}[/math]

So as the air gap increases (and thus the magnetic reluctance), the field strength H in the magnet increases and from the first equation you can see that B will decrease.

[math]B=\mu_0 M \cdot(1- \frac{l_{gap}}{l_{gap}+l_{PM}})[/math]

e.g. for an air gap the same size as the magnet, B will halve.

Edited by Bender
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In the air gap:

[math]B=\mu_0 H_{gap}[/math]

In the iron; H is negligible. Most likely, the C-shaped magnet is a bar magnet on the right and then some iron to guide the magnetic field around.

 

 

 

H is the same in both the air gap and the iron.

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As my studying of the problem continues, I learned to accept the fact that the wire on the picture B) (of the original post) will 'feel' smaller force, likely significantly smaller, than the wire on the picture A).

 

1. I believe now that a current-carrying wire always 'feels' a force that only depends on the B field that actually crosses the wire; the surrounding B field has no effect on the produced force. This is so from the definition of the B field - the B field is defined by the force/torque it produces on current-carrying wires and moving charges. For example, saying that there is no B field in certain region is equivalent to say that there will be no force/torque excreted on moving charges there.

 

2. Because in an electric motor wires are often placed inside slots, I believe (as Bender confirmed) that these wires are actually immersed into considerably lower B field than the surrounding iron. Therefore force/torque excreted on motor wires is relatively small... But as we know that motors are not at all weak, it seems to me that the motor torque is mostly generated inside iron.

 

3. The fact that motor torque is generated in iron seems quite obvious to me when I look at the motor macroscopic-ally, from perspective of magnetic energy minimization. However, I keep failing to understand how the force/torque is generated inside iron at microscopic level - so I must continue my studies until everything fits.

 

If you have any tip or find any error in my thinking, let me know.

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You could look at it as two opposite poles of magnets attracting and two similar poles repelling. In the case of an electromotor, one or both of the magnets are replaced by elektromagnets which can change polarity in different ways.

Edited by Bender
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