Geometry of Magnetic field lines as they approach a magnet pole?

[doomer]

Magnetic field lines are loops, so at the surface of a magnet (axially magnetized) cylinder) pole they look like this? there is a dead zone at the center where there is no lines/magnetic field present?

or perhaps they look more like this? lines converge in the middle to a single line or point?

Which picture is correct? also on some googled pictures I can see a single straight line coming from the center of the pole that goes nowhere... I know that lines are really an mathematical artifact, I just want to know how the field is shaped near the surface at middle of the magnetic pole.

 

[vanhees71]

One of the few books which give correct field-line pictures of a bar magnet for both B and H is Sommerfeld, Lectures on theoretical physics, vol. 3.

 

[doomer]

One of the few books which give correct field-line pictures of a bar magnet for both B and H is Sommerfeld, Lectures on theoretical physics, vol. 3.

This? there is a magnetic line in the middle that goes somewhere? or its something else?

 

[marcusl]

Yes, they are loops. The field lines exiting the magnet to the left in these diagrams return into the magnet from the right.

 

[DaveC426913]

And yes, in theory there is a field line that runs through the centre parallel to the axis. Like any other field line it defines points of the same strength outward from the magnetic poles arbitrarily far.

 

[vanhees71]

The field lines give the direction. The magnitude of the depicted field of course changes along the field lines.

 

[DaveC426913]

The field lines give the direction. The magnitude of the depicted field of course changes along the field lines.

?
The field lines dont define points of equal magnitude? Like geographic contour lines? I guess I'd never really thought that through.

 

[vanhees71]

Maybe you confuse it with iso-surfaces for the electrostatic potential? These are of course perpendicular to the electrostatic field: For any line $\vec{x}(\lambda)$ on the surface, defined by
$$\Phi(\vec{x})=C=\text{const}. \; \Rightarrow\; \mathrm{d}_{\lambda} \Phi(\vec{x}(\lambda))=\dot{\vec{x}} \cdot \vec{\nabla} \Phi=-\dot{\vec{x}} \cdot \vec{E}=0.$$

 

[DaveC426913]

Maybe you confuse it with iso-surfaces for the electrostatic potential? These are of course perpendicular to the electrostatic field: For any line $\vec{x}(\lambda)$ on the surface, defined by
$$\Phi(\vec{x})=C=\text{const}. \; \Rightarrow\; \mathrm{d}_{\lambda} \Phi(\vec{x}(\lambda))=\dot{\vec{x}} \cdot \vec{\nabla} \Phi=-\dot{\vec{x}} \cdot \vec{E}=0.$$

Yeah. I think I do.

 

[berkeman]

Magnetic field lines are loops, so at the surface of a magnet (axially magnetized) cylinder) pole they look like this? there is a dead zone at the center where there is no lines/magnetic field present?

It's probably helpful to look at what a COMSOL simulation shows...

https://www.comsol.com/forum/thread/attachment/91571/PM_example-16387.png

 

[marcusl]

?
The field lines dont define points of equal magnitude? Like geographic contour lines? I guess I'd never really thought that through.

If field lines are drawn to each represent the same flux, then the magnetic induction (what we commonly call field strength) is proportional to the density of field lines.

 

[doomer]

And yes, in theory there is a field line that runs through the centre parallel to the axis. Like any other field line it defines points of the same strength outward from the magnetic poles arbitrarily far.

So there is a line and field in the middle? and the other lines are branching out from that center line and make loops too? I wish I could see field in a 3d and without lines, just a 3d gradient model of the field and its shape...

 

[doomer]

It's probably helpful to look at what a COMSOL simulation shows...

View attachment 334348
https://www.comsol.com/forum/thread/attachment/91571/PM_example-16387.png

I don't get this... there is is no field at the center? a deadzone? single center line can't exist there, right? or its a field cone that starts at single point in the middle near the surface?

 

[berkeman]

So there is a line and field in the middle? and the other lines are branching out from that center line and make loops too? I wish I could see field in a 3d and without lines, just a 3d gradient model of the field and its shape...

You can do the experiment yourself with a magnet and some iron filings...

https://www.alamy.com/stock-photo-b...f3339cedbbad69ca4b1230684f045329&searchtype=0

 

[Ibix]

There has to be one field line that doesn't curve round, but just keeps going. The field strength falls to zero at infinity, so $\nabla\cdot\vec B$ remains zero even though the line doesn't close in the usual sense.

 

[doomer]

There has to be one field line that doesn't curve round, but just keeps going. The field strength falls to zero at infinity, so $\nabla\cdot\vec B$ remains zero even though the line doesn't close in the usual sense.

Thats interesting, one center line that does not form a loop but how? this is a magnetic vector potential or something else? I always fought that lines are just a mathematical model, not a physical thing...

Sun particles are trapped inside the earth's magnetic field "cone area" and this causes aurora borealis, the cone area has no magnetic field inside, correct? but the center line is still there?

If I pass a wire over a cylindrical magnet pole so lines are being cut then I would get EMF ? or not?
I'm thinking that perhaps two opposing voltages appear and cancel out so no EMF is induced in wire, afterall lines are loops and wire is cutting them in two areas: in the middle and at sides, the middle lines direction is UP and the side lines direction is pointing DOWN, 2 directions means two opposing voltages generated? HOWEVER if the single center lines exist then there should be one voltage generated that is not canceled out?

 

[DaveC426913]

If I pass a wire over a cylindrical magnet pole so lines are being cut then I would get EMF ?

Look at the image in berkeman's post 14 again.

Lines converging on the North pole all experience the same polarity. A compass placed anywhere there will always point toward the pole. It just happens that, when the compass is exactly opposite the pole, it will point exactly along the length of the magnet, as opposed to deflecting to one side or the other.

 

[marcusl]

The field line on axis is a mathematical abstraction (as are all field lines). Note that the loops extend longer as they near the axis before they curve back, and you also have to travel farther to the left or right in your diagram to see the returns. On the axis, the loop extends to infinity before curving to both left and right and returns infinitely far from the axis on either side. Since you can’t see those return lines (they’re infinitely far away), you are left with the line down the middle.

 

[doomer]

The field line on axis is a mathematical abstraction (as are all field lines). Note that the loops extend longer as they near the axis before they curve back, and you also have to travel farther to the left or right in your diagram to see the returns. On the axis, the loop extends to infinity before curving to both left and right and returns infinitely far from the axis on either side. Since you can’t see those return lines (they’re infinitely far away), you are left with the line down the middle.

Ok so there is a line that goes to infinity, got it. This axis line create unbalance so induction can happened when I cut the line loops? take a look:
The blue lines represent induced voltage vectors in orange (copper) wire that cuts the lines over the pole.
The wire form a loop and the rest of it goes outside of magnetic field, notice the field direction arrows, they determine induced voltages directions.
All generated opposing voltages should cancel out BUT because of the exis line there will be an extra voltage (not drawn on above picture) generated by this line so EMF would appear on scope (and this is what I'm getting when doing this experiment). Is that a correct picture? without this center axis line there will be no EMF?

 

[marcusl]

Well, the axial line is just one of many that contribute to an induced EMF. No matter how large you make the wire loop, there are always field lines that return outside of it and prevent perfect cancellation. (BTW, you don't get an EMF from the static magnet and static fields that you've pictured, but only from changing flux. The setup works if you replace your magnet with a solenoid energized by an AC current.)

 

[doomer]

Well, the axial line is just one of many that contribute to an induced EMF. No matter how large you make the wire loop, there are always field lines that return outside of it and prevent perfect cancellation. (BTW, you don't get an EMF from the static magnet and static fields that you've pictured, but only from changing flux. The setup works if you replace your magnet with a solenoid energized by an AC current.)

No matter how large you make the wire loop, there are always field lines that return outside of it and prevent perfect cancellation

What do you mean? every line that is coming out from pole must be cut in two points; at the middle and at sides, this should lead to cancelation, you are saying that some lines do not form loops when they are far away from magnet?? I can make a loop that is 100 meter long, there is no way that field can reach that far.

nah, flux does not need to change, moving wire "cutting" is enough to cause induction, many generators works this way, rotor wires rotate inside static magnetic field, even in disc homopolar generator.

Also, when I rotate wire over the magnet there is no EMF but when I just pass the wire over the magnet then there is EMF, why?

 

[vanhees71]

Usually it's easier to think about the forces on current conducting (and/or moving) wires in terms of the electromagnetic force, $\vec{F}=q (\vec{E} + \vec{v} \times \vec{B})$.

To understand the homopolar generator, see

https://itp.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf

 

[doomer]

Usually it's easier to think about the forces on current conducting (and/or moving) wires in terms of the electromagnetic force, $\vec{F}=q (\vec{E} + \vec{v} \times \vec{B})$.

To understand the homopolar generator, see

https://itp.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf

Can you simplify things for me?
Why when I rotate wire over the magnet pole there is no EMF induced but when I just pass the wire over the magnet then there is EMF?

 

[marcusl]

What do you mean? every line that is coming out from pole must be cut in two points; at the middle and at sides, this should lead to cancelation, you are saying that some lines do not form loops when they are far away from magnet?? I can make a loop that is 100 meter long, there is no way that field can reach that far.

Yes, field lines "reach" that far. The field is weak our there so the EMF will be tiny, but it exists. Since field lines extend to infinity (conceptually), I can always draw field lines that extend beyond any loop no matter its diameter.

nah, flux does not need to change,

Untrue. Only changes in flux induce an EMF. If you translate a wire loop through a uniform field, there is no induced EMF because the flux in the loop never changes.

moving wire "cutting" is enough to cause induction, many generators works this way, rotor wires rotate inside static magnetic field, even in disc homopolar generator.

True, if the field is inhomogeneous as for your magnet, moving the loop relative to the magnet will suffice. I didn't realize that was your premise--sorry.

Also, when I rotate wire over the magnet there is no EMF but when I just pass the wire over the magnet then there is EMF, why?

There should be an EMF in both cases because the flux in the loop changes.

 

[vanhees71]

Can you simplify things for me?
Why when I rotate wire over the magnet pole there is no EMF induced but when I just pass the wire over the magnet then there is EMF?

I don't understand the question. Can you make a picture of the different situations? In general: if due to the rotation of the wire loop you have an induced EMF if the magnetic flux through the loop changes with time. That's Faraday's Law in integral form in its most general form:
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

 

[doomer]

Yes, field lines "reach" that far. The field is weak our there so the EMF will be tiny, but it exists. Since field lines extend to infinity (conceptually), I can always draw field lines that extend beyond any loop no matter its diameter.

Ok thanks, right, the field become weaker with distance and so the "opposing voltage" thus emf is not canceled, the axis line is not needed for emf.
However:

There should be an EMF in both cases because the flux in the loop changes.

I have some tiny current in spinning wire scenario BUT not as much as in passing wire scenario, I blame it on
wobble and vibrations when wire rotates because my contraption is far from perfect, I will try to make more stable rotation without any runout and other unwanted movement.
Spinning wire loop scenario should not bring any EMF because there is no such thing as Faraday disc generator brushless version (wires soldered to the disc with measuring apparatus - one wire at center and another to the edge) and it does not matter if we use disc with wires or just single wire loop. The argument is made that opposing canceling voltages are being induced in the wires and disc, hence the brushes is the only way to get EMF from Faraday disc, same for rotating single wire loop.

This is how I see it (blue arrows represent induced voltages directions)

 

[doomer]

I don't understand the question. Can you make a picture of the different situations? In general: if due to the rotation of the wire loop you have an induced EMF if the magnetic flux through the loop changes with time. That's Faraday's Law in integral form in its most general form:
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

That's correct, but equal and opposite canceling voltages appear in wire(?), the magnetic field curl, double cut at two different points, take a look:
Two scenarios: spinning wire (brushless faraday disc) EMF=0 and passing wire EMF=1

This is the reason why spinning Faraday disc needs brushes.

 

[marcusl]

Ok thanks, right, the field become weaker with distance and so the "opposing voltage" thus emf is not canceled, the axis line is not needed for emf.
However:

I have some tiny current in spinning wire scenario BUT not as much as in passing wire scenario, I blame it on
wobble and vibrations when wire rotates because my contraption is far from perfect, I will try to make more stable rotation without any runout and other unwanted movement.
Spinning wire loop scenario should not bring any EMF because there is no such thing as Faraday disc generator brushless version (wires soldered to the disc with measuring apparatus - one wire at center and another to the edge) and it does not matter if we use disc with wires or just single wire loop. The argument is made that opposing canceling voltages are being induced in the wires and disc, hence the brushes is the only way to get EMF from Faraday disc, same for rotating single wire loop.

This is how I see it (blue arrows represent induced voltages directions)

Ah, it wasn't clear before that your loop was spinning around the magnet axis. In this case I agree that there will be no EMF, assuming the magnet has perfect azimuthal symmetry, the rotational axis is accurately aligned with the magnet axis, there's no wobble, etc.

 

[doomer]

Ah, it wasn't clear before that your loop was spinning around the magnet axis. In this case I agree that there will be no EMF, assuming the magnet has perfect azimuthal symmetry, the rotational axis is accurately aligned with the magnet axis, there's no wobble, etc.

Yes, just one more thing... I don't understand why there is no EMF in this case? :
The upper section of the wire loop is so wide and high that field lines can't reach it (or at least the induced voltage in this section is very tiny), left side of wire loop (section near the axis) is cut and voltage nr 1 is induced, right side of the wire loop is also cut but experience weaker flux so induced voltage nr 2 is less than voltage nr 1. Voltage nr 1 should overcome opposing voltage nr 2 and current should flow.
I assume that flux near the axis is stronger than flux near the edge of the magnet also the spinning velocity of right section of wire is lower than left section, left section gets all the best juice.

This should work as Faraday brushless generator yet there is no such thing.