Homopolar generator question: Magnetic “wake”

[JoviG]

So faradays paradox is often depicted as a circular copper disk on a shaft with a donut or toroidal magnet on one or either side of the disk. Spin the disk and magnets together voltage is produced. Since the magnets are rotating along their poles the orientation of the magnetic field compared the disk never changes.

From what I gather, the theory is the magnetic field doesn’t actually rotate with the magnets, the field is stationary in space because the orientation of the poles is stationary compared to the axis of rotation. Alright… seems to make sense, rotating conductor, “stationary” magnetic field, voltage is produced. Now here’s where I get confused. Take a copper disk on a shaft and on one side of disk place 4 bar magnets, 90° apart, 2 of them will have their N poles towards the shaft, the other 2 will have their south poles facing the shaft. Like this, but their could be numerous configurations:



Rotate the shaft, still makes voltage. Now there’s no way you can say the magnetic field isn’t rotating with the magnets… The magnetic field is absolutely rotating. How do you have voltage when you have no relative motion?

The only way that makes sense to me is if the magnetic field produced by a magnet is really a field induced in the space around the magnet because if it’s atomic alignment and not an intrinsic force produced by the magnet itself. And furthermore that magnetic field induced in the space around the magnet does not instantaneously align itself to a rotating magnet, but i imagine it’s pretty quick.

I guess i kinda think of it like water, when the magnet is moved through space it produces a magnetic wake in its path. If you think about like this, the relative motion is the copper disk rotating thru the magnets “wake”. Doesn’t matter that the magnets are rotating with the disk because the magnetic “wake” is separate entity.

If that’s true and I’m not saying it is, if you took a homopolar generator that isn’t rotating on an airplane would it make voltage? You still have a conductor passing through the “wake” of a magnetic field, just in a linear fashion.

 

[vanhees71]

For a full relativistic treatment for the most simple configuration (homogeneously magnetized rotating sphere), see

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

 

[artis]

Since the magnets are rotating along their poles the orientation of the magnetic field compared the disk never changes. From what I gather, the theory is the magnetic field doesn’t actually rotate with the magnets, the field is stationary in space because the orientation of the poles is stationary compared to the axis of rotation. Alright… seems to make sense, rotating conductor, “stationary” magnetic field, voltage is produced.

So far so good! Indeed the magnetic field can only be categorized by either it's strength or pole orientation if both stay the same it really doesn't matter whether the source is physically rotating or not.

Now here’s where I get confused. Take a copper disk on a shaft and on one side of disk place 4 bar magnets, 90° apart, 2 of them will have their N poles towards the shaft, the other 2 will have their south poles facing the shaft. Like this, but their could be numerous configurations.
Rotate the shaft, still makes voltage. Now there’s no way you can say the magnetic field isn’t rotating with the magnets… The magnetic field is absolutely rotating. How do you have voltage when you have no relative motion?

The Faraday disc is just a expanding wire loop. It is essentially another form of induction. Normally in induction you have a fixed wire loop that experiences a changing magnetic field strength through it. In a Faraday disc you have a static magnetic field but the wire loop geometrically changes it's size to constantly encompass a larger area.
This is the reason why brush contacts are needed.
The B field that the continually expanding wire loop encircles can be either static or changing it doesn't change the induction phenomenon taking place, it only changes the output current waveform.

Think about it like this, the varying polarity magnets are still producing a B field that cuts the rotating conductor just not as efficiently as the larger circular magnet.
In other words you can create all kinds of configurations where you will generate voltage because all you need is a rotating conductor and some contacts and some B field. As long as you have that expanding loop of the disc and some B field you will get output.

The simpler analogy to understand all this is a rectangular wire loop where one side is allowed to expand, this is also often referred to as the linear Faraday motor/generator.

In fact its the way most high power railguns work, check google on railgun homopolar etc searchwords to see what i mean.

 

[JoviG]

Yeah… I’m still confused about where your relative motion between the conductor and the magnetic field come from. Their both rotating together. The magnets and their fields are fixed (glued) to the disk, your just rotating the whole assembly. You have multiple magnetic fields rotating around an axis, the field is moving, but not relative to the disk. That can only mean that the field produced by a magnet is an induced field in the space around it and that moving through space cause ripples or something. Right?

 

[osilmag]

Right. This might be simplifying it too much, however it seems that the copper wire is moving through a static magnetic field. The area in the copper that the field goes through changes, which causes the voltage.

Now when the bar magnets are attached to the disks and they rotate, the field might be distorted (ripples) by the movement of the copper wire. This distortion might be what produces a voltage when they are attached.

 

[artis]

Yeah… I’m still confused about where your relative motion between the conductor and the magnetic field come from. Their both rotating together. The magnets and their fields are fixed (glued) to the disk, your just rotating the whole assembly. You have multiple magnetic fields rotating around an axis, the field is moving, but not relative to the disk. That can only mean that the field produced by a magnet is an induced field in the space around it and that moving through space cause ripples or something. Right?

The multiple bar magnet case is harder to talk about because whenever you rotate magnets around there is induction also in the lead wires because the field is physically changing all the time.

As for the Faraday disc itself just think of it as a moving conductor. Lorentz force law states that every charged particle that moves through a magnetic field at an angle to the field which is not zero experiences a force perpendicularly to it's movement.
The force depends on the angle to the field lines and the speed and mass of the particle.
The conducting disc is nothing more than a sea of electrons bound to the metal but also free to move within it.
The field deflects them as they move through it. The brush simply collects them, but there is a catch , the brush cannot move at the same speed as the disc. In an expanding loop which the Faraday disc is you need some relative movement between the different parts of the loop.
Otherwise it's just a stationary loop rotating in a static B field and that doesn't generate voltage as we know.

 

[vanhees71]

In principle, this should help understand it. The only difference to the here discussed setup is that I simplified the case as much as I could by using a spherical homogeneously magnetized permanent magnet:

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

 

[neilparker62]

Don't really understand why any confusion about this the simplest of electric motors/generators. Rotating metallic disk consists of a vast 'sea' of electrons moving perpendicular to a magnetic field. Therefore the Lorentz force acts perpendicular to the magnetic field and to the direction of motion driving electrons from centre to edge of disk creating an emf (generator). Alternatively if current flows radially from edge of disk to centre, the 'sea' of electrons experiences a Lorentz force perpendicular to the radius generating spin. Perhaps that is easier to visualise if we think of a spoked wheel/disc rather than a solid copper disc.

$$\vec{F}=q\vec{v} \times \vec{B}$$
https://en.wikipedia.org/wiki/Barlow's_wheel
https://en.wikipedia.org/wiki/Homopolar_generator