# I Wire loop in a static homogeneous field question

1. May 9, 2018

### girts

I already did a similar question here but got very little response so I will try to reformulate my question into a better one.

So, the basic idea of the original question was whether a Faraday disc aka homopolar generator be made such as to have no sliding contacts and the load being attached to the rotating disc if the correct rules are satisfied, chief among which is to have the B field at the correct angles at every part of the current generating part of the circuit in order to avoid inducing currents that run counter to the main current.

Theoretically a conductor rotating in a homogeneous B field should get a charge imbalance formed on the conductor due to the Lorentz force acting on the "free" charge inside the conductor, much like the plates of a capacitor get charge imbalance due to the work of the E field.
the question is can this charge imbalance be made to run as a current in a loop if every part of the loop is rotating together with the parts that are subjected to the homogeneous B field.

For simplicity lets imagine a simple circular or rectangular wire loop which is surrounded by a B field such that the field lines point from center of the loop to outside at every part of the loop (by the way this is mechanically possible to make in reality), would this loop then have current if it was moving through space?
because from the reference point of the loop the field lines are static and not changing and also from the lab reference point the field lines are static,
in other words I want to confirm what i already suspect that the reason a homopolar generator/motor can work is that it has a fixed current point at some place on the disc and so the complete loop (both the rotating part and stationary part) is cutting field lines but with a loop in which all parts of the loop are in the same reference frame no extra field lines can be cut in a static homogeneous B field so no current can be generated?

so in a way we could say the Faraday disc still obeys induction laws as ordinary generators simply in a bit different manner. In other words the only reason why a Faraday disc can output steady DC and other generators cannot is because if we imagine the current loop, the part which covers the disc surface is sort of dragged over static field lines forming a loop which in theory is constantly expanding?

anyway I hope I made myself as clear as possible because this is really confusing for me because from one point according to Lorentz law there should be current even without slip contacts from the other point I know of no devices which would have worked or have been made without slip contacts in this fashion so where is the catch?

2. May 9, 2018

### girts

if I may add another paradox that comes to mind is that if indeed the faraday disc can be made such that it can have current within a load which rotates together with the disc itself then if we would turn it into a motor say we put a battery and attach that battery to the disc, where would or maybe I should say against what it would accelerate because there is no stator against which it could develop torque so it should sort of "bootstrap" itself but we know that that is impossible.

and while we are at it may I ask where does a standard homopolar motor develop a torque to rotate? because when the magnet rotates with the disc or conductor it sort of becomes a statorless motor in a way that the torque the conductor develops comes from the current which runs through it creating a B field which works against the B field of the magnet so the torque from the disc comes from the disc B field pushing against the B field of the magnet but since they are both the same physical part I fail to see where "action-reaction" comes into play here? or is it that the whole rotating part of the loop pushes against the B field from the static part of the loop aka the wires and brushes etc?

3. May 10, 2018

### olgerm

Yes, if I understood correctly, that B-field is always crosswise to wire and velocity.
$\epsilon=v*B*l$
$I=\frac{\epsilon}{R}$
$l_{circular}=2\pi*r$

$I_{circular}=\frac{v*B*2\pi*r}{R}$

Last edited: May 10, 2018
4. May 13, 2018

### girts

Ok , I understand what you are saying and in theory this should work like that, let me show a simple picture that I made to maybe better illustrate what I am saying.

See there is a rectangular iron core sliced in half much like a transformer, now say that one part of the iron core has a coil on it so that there is a magnetic field produced in the core, the B field lines (shown in red) loop around the core as shown.
Now the core has an airgap in the middle where a circular wire loop is inserted (shown as green) Now imagine we take the whole iron core together with the loop and either rotate around a fixed axis or it moves with some speed in a straight line like in a linear motor.
The B field is such that the wire loop has B field lines coming from opposite directions at opposite sides in order to satisfy the difference in current directions.

Now my question is this would there be current generated in the wire loop if it moved together with the iron core through space (the wire loop itself stays fixed in between the iron core at all times)

5. May 13, 2018

### olgerm

No.

Your drawing is different than that description.

If you only move the wire, but not the magnet that produces B-field, then the wire would have current ,but usefull(in same direction as wire) electromotive force would be induced only in the "sides" on retangel, where velocity is crosswise to direction of wire.
Moving the magnet affects electromagneticfied!
Moving the magnet induces E-field, that creates electromotive force in opposite direction.

Last edited: May 13, 2018
6. May 13, 2018

### girts

you say there should be no current, but we are not talking about induction but about Lorentz force, in a Faraday disc since the B field is static and homogeneous it does not matter whether the magnet rotates or not , all that matters is for the conductor to rotate,
now imagine this drawing from such viewpoint, see that each of the two parts of the wire loop passes along the side of the sliced iron core, (the iron core can be extended to engulf each halfwire more completely than shown in the picture but that is a detail) so each half of the complete wire loop is situated within a B field that is perpendicular to the wire cutting it, one half is cut from bottom up the other half where the flux returns is cut from top to bottom, so that the current adds in each half,
given this explanation why do you think there would be no current if I simply rotated or moved the whole assembly?
there is current generated when a faraday disc moves together with a magnet.

I kind of have a gut feeling about this but I'm really not sure, I think i could explain it with SR but i'll first wait for some responses before trying to make my own rebuttal

7. May 16, 2018

### olgerm

I said it about situation that were on your image.Sum of forces to charge carriers in wire on that picture is 0 if the wire moves along with magnet with constant speed.
I do not know about Faraday disc. Have you taken E-field induced by changing B-field into account?

8. May 16, 2018

### girts

The E field in this case would have negligible effect due to only a few turns of wire
Well I think I need to reformulate my question about this topic better, also I found some interesting sources , I think I will make then a new thread, I would be happy to see you participating in that thread once it will be up.

9. May 16, 2018

### olgerm

Do you mean the situation on your image or the Faraday disc situation?