# Homopolar generator

1. Feb 10, 2014

### Mr.Bomzh

Hi there , I would like to understand more about how exactly current is induced in a homopolar generator aka Faraday disc ? I have read some papers on this subject and so on and as far as I understand in a normal generator current is induced because magnetic field lines travel through a loop of wire or lets just say a coil or vice versa and so current carriers electrons are forced to move by the field , but in a homopolar generator the whole disc is acts like one big very low resistance wire , the very low resistance probably accounts for the quite high amperage one can get out of such a machine , yet how is a PD achieved on such a device ? Does it has to do with the fact that the electrons closer to the center have less angular valocity than those further to the sides were the radius increases , or is it because the one pole of the magnet is located closer to the center of the disc while the other further towards the sides of it , one of these probably has to do with this effect as I can't explain this device with the faraday induction laws and as far as wikipedia goes it says it's not possible at all.

2. Feb 10, 2014

### cabraham

The attached paper gives details on the current and flux paths. I will elaborate if needed.

Claude

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• ###### munley-Faraday's Law.pdf
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3. Feb 11, 2014

### Mr.Bomzh

Well I had a look at the paper and I went through it but it had only formulas and some for me vague explanations because I'm not that good at maths , anyways the paper felt like when youre really hungry but your hands are shaking and everything falls out of the spoon so you can;'t get no satisfaction. :D Ok let's take the simplest case , a rotating disc , a magnet at one point from both sides of the disc and contact points , now the magnetic field line from the magnet are goign through the disc paralell to the center shaft or axis and perpendicular to the surface of the disc , current goes from the center to the outer part of the disc or vice versa depending on the poles or way of rotation , the lorentz force then is 90 degrees shifted from the current or aka perpendicular to both current and magnetic field. If so far is fine but I still can't quite figure out why does there is current induced in the disc , well the only way that I could find out is that maybe one does need to imagine the disc being cut and made from many many smaller seperate wires all starting from the middle going towards the outerside, moving such wires through the field of the magnets would induce current in them , so the only thing that differs here is that instead of many wires we have a one solid disc but we could effectively treat the part which passes the magnets at each given time as as sinbgle wire or piece of conductor through which the current moves right? and more the current can only move between the contacts so if the ciontacts are made at the same position at which the magnet is located then everytime a part of that disc moves through those magnets and the field they make in that part a current is induced which runs towards those contacts is they form a circuit , is that abour right?

I found such an explanation on youtube , sounds about right , also I think I'm getting this stuff , can you verify ? Thank you

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4. Feb 11, 2014

### cabraham

Dr. Munley's paper explains this. The current path for the electrons is from the center towards the rim (or vice-versa, but let's go with center to rim for now). But as the electrons transit from center to rim, the disk is rotating. When an electron which starts at the center reaches the contact on the rim, it sweeps out a pie shaped sector of area. The B value is constant, and B is flux density in tesla, or weber/m2. The flux ø is area times B, or A*B. THe area of the pie shaped sector is θ*R2/2, where θ is the angular displacement in radians.

So the area is changing as the disk turns in a manner directly related to θ to the 1st power. But θ is related to angular valocity ω as follows:

θ = ω*t

Here "t" is time. Thus the area of the filamentary current loop is just ω*R2*t/2. Thus the magnetic flux ø is the area times B, or ø = ω*B*R2*t/2. Notice that t is in the 1st power (time).

But we know that v(t), the emf, is -N*dø/dt. What is the derivative of a constant times time to the 1st power? I.e. d(ω*B*R2*t/2)/dt is just ω*B*R2/2. In the derivative, time does not appear, because the derivative of t to 1st power is a constant. Thus the induced emf is indeed zero frequency, or "dc".

I can elaborate if needed. But please re-read Dr. Munley, and hopefully it will make more sense now.

Claude

5. Feb 11, 2014

### Mr.Bomzh

Hey thanks for the response Claude , your math was already too much but anyways :D Ok let me explain how I see this based on what I think and what you wrote. The homopolar generator generates DC because as the rotor turns it turns through a constant magnetic fields so the electrons jsut feels a force and form a current in one direction , the current can probably be increased by two means , either by stronger magnetic field or by faster rpm of the rotor which in both cases would result in the need to input more mechanical energy to the rotor. The only way one could make an AC with a homopolar generator would be to to turn the rotor to one side and then to the other and do this fast enough to form some frequency , although I think in practice one could only make this up to some few tens of Hz no more.So the homopolar generator in it's normal application is a strictly DC machine. the alternator can induce AC because there the rotor consists of magnets and the coils are stationary so when the N pole goes past a certain coil the current is induced in one direction and when the S pole goes past that same coil the current is reversed and so the frequency of the induced AC is proportional to the rpm of the rotor correct?

As for the homopolar generator the electrons move because of the lorentz force which points in the same direction as current just with a 90 degrees angle to it , in a mechanical analogy it would be like a centrifuge where you put a rolling ball at the center but due to centripal force the ball will tend to move as far outward as possible is that correct?

6. Feb 11, 2014

### cabraham

You seem to understand it better. Increasing speed or magnetic flux density will increase induction. As for a centrifuge, I would be cautious w/ analogies. In a centrifuge, the *centripetal* acceleration is inward, not outward. Centrifugal is the outward "force", which is a math construct used in the rotating reference frame. Otherwise, I would say that the main challenge with a rotating disk is understanding the current path. A simple filamentary current path is not intuitively obvious with the disk. Munley covered it in a sketchy way. I would like to expand upon Dr. Munley's paper, but have no time to do so. BR.

Claude

7. Feb 11, 2014

### Mr.Bomzh

But why does the analogy is wrong , all o0bjects according to newton tend to move in straigth lines , in a circle an object is constantly accelerated outwards but if the circle encloses the object then the object exerts a force on the outside of the rim that encloses it because of it's kinetic energy that it gained. Well ok, another question that I actually have thought about , normally one attaches one wire at the shaft and the other at the outermost part of the rim or disc of the homopolar generator , the shaft brush or bearing expierences much smaller wear than the outermost attachment because of the angular speeds involved, that makes a question , would it be possible to attach a current carrying wire to the inside of the outermost part of the ring (assuming the disc middle is empty) the wire would be isolated both electrically and with a screen for EM, and brought through the inside of the sahft towards a small ball bearing at the end of the shaft ? Could such a scenario be possible or would it seriously affect the efficiency and working of the machine? I think it shouldn't because the current carrying wire even though carrying current in the opposite direction than the disc, sits inside the disc + would be enclosed and so , well I;m not sure what do you think? Also can one make the shaft connection connected to the whole chasis via a bearing that would aslo serve as the main bearing , so the whole chasis would become say - and the wire from the outermost part coming out the shaft end being the + and also isolated from the chasis , I ask this because iof the very low voltage involved.?

8. Feb 15, 2014

### Mr.Bomzh

Anyone interested in my last paragraph :D ?