ThatIn the situation of a magnet moving towards a flat sheet of copper; why doesn't the copper gain angular moment as the electrons move in concetric circles? The electrons are experiencing friction as they move through the copper.
It does! Easy to demonstrate, too. Set up a flat copper sheet in a vertical plane. A hinge pin through the plate allows it to swing freely in that same vertical plane. Drag a strong magnet near it and set it swinging.In the situation of a magnet moving towards a flat sheet of copper; why doesn't the copper gain angular moment as the electrons move in [...]?
My reasoning for the copper plate to rotate is because the currents induced in the plate are encountering resistance as they move in concentric circles. If the copper heats up due to joule heating; why wouldn't it also rotate?If I understand your question correctly you are moving a bar magnet towards a copper plate that is stood / hung up at right angles to the magnet?
You ask wether the copper plate should gain angular momentum from the flux lines emanating from the magnet. I would comment thus. Copper is non magnetic and doesn't respond to magnet forces like iron. Secondly copper is an extremely good conductor it has a very low resistivity however an electric field and therefore a magnetic field magnetic field will be induced in the copper as the magnet moves, the field will be in opposition to the magnet and will try to cancel it out, this phenomenon is Lenz's Law. For more take a look at the effects of a magnet falling through a copper pipe. I don't see any reason for the copper plate to rotate.
Yes, exactly. Think of dropping a magnet perfectly perpendicular to a copper plate. Why doesn't the plate experience a torque? The eddy currents heat up the conductor as the charges collide with impurities and ions. If we placed this plate on a very low friction surface; why wouldn't it be like spinning tires on ice?I guess you are asking about eddy currents? Can you sketch the set up.
Not if we drop a magnet with it's poles perpendicular to the surface of the copper plate.I think you will find that in part of the copper plates the eddy currents will circulate in one direction, and in other parts they will circulate in the opposite direction, so no angular momentum is transferred.
In your example; we can go the opposite direction. If the electron drift velocity is on the order of microns per second; why wouldn't the ring start spinning in the direction of current flow? It must require less energy to rotate the ring at such a low speed compared to the amount of energy converted to heat by the resistance of the ring. If we found the electric drift velocity for this ring and this amperage; would spinning the ring at this rate reduce it's resistance? If electrons are encountering resistance on it's path through a conductor; why wouldn't the resistance decrease with the fewer collisions that would theoretically occur in a conductor moving in the same direction of the electrons in relation to the current source?Try a situation. Suppose you have a ring 2 cm diameter. If 1 amp of current is running through it, what does that equate to electrons per second. But you need to find the velocity of the electrons in the conductor. I think you will find any angular momentum due to the electrons will be tiny.
Thatsheet of copperthing that the electrons rub against - have you consider the fact that it's extremely positive?
Maybe that causes some kind of force to be exerted on it?
I think you're wrong. Look at the attached diagram. The magnetic field lines from the magnet form closed loops. Inside some radius, the magnetic field lines are pointing upward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has one sign. Further out, the magnetic field lines are pointing downward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has the opposite sign. I think if you calculate it out you will find that the total angular momentum transfer integrates to zero.Not if we drop a magnet with it's poles perpendicular to the surface of the copper plate.
The total net flux does not sum to zero for any finite size plate. (Consider a long solenoid and a plate the diameter of the solenoid)I think you're wrong. Look at the attached diagram. The magnetic field lines from the magnet form closed loops. Inside some radius, the magnetic field lines are pointing upward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has one sign. Further out, the magnetic field lines are pointing downward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has the opposite sign. I think if you calculate it out you will find that the total angular momentum transfer integrates to zero.
View attachment 239998
If this is correct the outer current loops would accelerate the magnet towards the sheet of copper. Actually, no force would act on the magnet because the magnetic fields of the induced currents would cancel. We can use the right hand rule to find the direction of current flow. The flux lines that are facing down are also growing perpendicularly to the sheet and out from the magnet. If I use my right index finger in place of the black arrow; my thumb is the direction the magnetic field is growing. My middle finger points in the direction of the emf. In this case; the current will flow counter clockwise.I think you're wrong. Look at the attached diagram. The magnetic field lines from the magnet form closed loops. Inside some radius, the magnetic field lines are pointing upward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has one sign. Further out, the magnetic field lines are pointing downward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has the opposite sign. I think if you calculate it out you will find that the total angular momentum transfer integrates to zero.
View attachment 239998
All you need is the charge to mass ratio for the electron (1.75E+11C/kg) so the electrons in 1 meter of wire at 1A carry 5.7E-12kgm/s of linear momentum. Small indeed.Try a situation. Suppose you have a ring 2 cm diameter. If 1 amp of current is running through it, what does that equate to electrons per second. But you need to find the velocity of the electrons in the conductor. I think you will find any angular momentum due to the electrons will be tiny.
I really don't like this argument. I do like Faraday's concept of lines generally, but people take them too literally. They don't really exist. There aren't seven of them attached to this magnet, nor any other number of them. (*)You could draw a unique line through any point in space and for any two points in space, you could find a point, say, midway between them to get another line. Or you could not bother to draw any lines! The field would still be there.I think you're wrong. Look at the attached diagram. The magnetic field lines from the magnet form closed loops. Inside some radius, the magnetic field lines are pointing upward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has one sign. Further out, the magnetic field lines are pointing downward, so as the magnet moves closer to the plate, [itex] \frac{\partial B}{\partial t}[/itex] has the opposite sign. I think if you calculate it out you will find that the total angular momentum transfer integrates to zero.
View attachment 239998
When I asked the original question I assumed I understood the effect of a magnet moving towards a conductor. Now I'm not so sure. I always thought that the field lines can only be used when also identifying the direction the magnetic field is growing. In the diagram Phyzguy made; the down arrows aren't propagating towards the sheet. They are radiating perpendicularly out from the magnets center. That means all the eddy currents will flow in the same direction. Is this incorrect?I really don't like this argument. I do like Faraday's concept of lines generally, but people take them too literally. They don't really exist. There aren't seven of them attached to this magnet, nor any other number of them. (*)You could draw a unique line through any point in space and for any two points in space, you could find a point, say, midway between them to get another line. Or you could not bother to draw any lines! The field would still be there.
Now in your diagram you get your effect, IMO, by being selective in which lines you choose to draw. Put another line outside your largest magnetic loop and that field links with the current ring in the correct manner. All your field lines do not link with the larger circuit and so do not induce any emf.
If you consider just that one magnetic line of force (the largest closed line you show) as being in a thin iron core of a transformer. (Grey)
The primary coil of the transformer wound round it in the position of the permanent magnet. (Red)
Two secondary coils positioned wher you show the two current loops. (Red)
View attachment 240014
In which coil do you think an emf (and, since it's a conducting loop, a current) would be induced?
a) Both secondaries, in opposite directions
b) Both secondaries, in the same direction
c) In the smaller secondary coil, but not the larger
d) In the larger secondary coil, but not the smaller
e) In neither
f) (and just for good measure) None of the above
Then the same question if we used a larger "line" of magnetism as the core instead. (Blue)
View attachment 240015
If we used both cores, would the induced emfs be in the same direction or in opposite directions? (Or any of the other combinations.)
I think it's all irrelevant owing to hutchphd's light mobile carriers and heavy fixed carriers of opposite polarity, but I just don't like Faraday's lovely litlle lines being taken advantage of when he's not here to defend them.
(*) Edit: I've just realised I've said this in a way that isn't true. I intended to imply that there are an unlimited number of lines to draw, but was careless in my description. If anyone wants to argue, I'll come back with a more careful wording, but I hope most will accept it without.
b) Both secondaries, in the same direction
Could this however be due to a net torque applied about the hinge/pivot axis due to magnetic forces & not just angular momentum of the electrons?It does! Easy to demonstrate, too. Set up a flat copper sheet in a vertical plane. A hinge pin through the plate allows it to swing freely in that same vertical plane. Drag a strong magnet near it and set it swinging.
The magnet exerts a torque on the sheet. I presume it's due to an interaction involving the electrons in the copper plate. It also works with an aluminum sheet, but it won't work with a wooden sheet.Could this however be due to a net torque applied about the hinge/pivot axis due to magnetic forces & not just angular momentum of the electrons?
Yes, that would be Eddy currents which are induced by the motion of the magnet and generate a magnetic moment to interact with the magnet. Eddy currents flow best in high-conductivity material and practically not at all in wood.The magnet exerts a torque on the sheet. I presume it's due to an interaction involving the electrons in the copper plate. It also works with an aluminum sheet, but it won't work with a wooden sheet.