Current in wire -movement- more current?

AI Thread Summary
The discussion revolves around the concept of self-induction in a rotating wire loop connected to a current source. When the battery is disconnected, the current will not sustain itself due to the wire's resistance and inductance, leading to energy loss as heat. The participants explore the differences between self-induction and traditional induction from external magnetic fields, questioning why the wire's own magnetic field does not induce a continuous current like a permanent magnet would. They conclude that while the rotating wire generates a magnetic field, it cannot maintain a constant current without external energy input, particularly due to resistive losses. The conversation highlights the complexities of electromagnetic theory and the need for mathematical modeling to fully understand the phenomena involved.
Salvador
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This situation got me confused, this is not a physical setup just a thought.There is a loop of wire , I have that wire on a rotor that I can rotate, I now start to rotate the wire and at the same time attach a current source to the loop like a small battery, current now flows through the wire creating a magnetic field around it ,at the same time I also continue to rotate the wire , assuming the wire is at the " correct" angles to the magnetic field , could I continue to generate current further If I disconnect the battery but keep the wire rotating ?This puzzles me , because if I use a permanent magnet close to the wire and then rotate the wire with hand I get induced current , but can the current in the wire continue to make current from it's own magnetic field if there is energy supplied for the wire to rotate in that field?
 
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Well as soon as you remove the battery, there will be a short time of self induction process. But this process will continue for very short time and even if you keep rotating the wire you cannot keep the self induction process alive unless you have a superconducting wire. If the wire has a finite ohmic resistance R and finite self inductance L, as soon as you disconnect the battery all the energy of the current will be transformed to heat energy and the current will eventually drop to zero. This conclusion follows by doing the math on a RL circuit. The fact that you ll keep rotating the wire doesn't affect the math alot, it only adds some additional small self inductance process on the RL circuit.

However if you rotate the wire at relativistic speeds and/or the wire is superconducting to tell you the truth i am not sure what can happen.
 
A generator does continue to induce current as long as energy is supplied to rotate the shaft.
The thing that puzzles me is the magnetic field , when you have a magnet nearby and then rotate the coil or loop current is continuously induced , in my case i have the same conditions input rotation loop only difference the magnetic field coming from the same wire that is also the load.
Why would you calculate in terms of RL circuit , as much as I know an RL circuit is just an inductor and a resistor or resistance of the wire forming the inductor , but it doesn't assume rotation.
Any ordinary generator has coils that have some inductance because their coils.

the question is why would there be a difference between the field of a permanent magnet and the field from a coil that ahs dc current through it ,I think there is no difference , so why should it not work , I'm not saying it will work I just want some explanation because until this point I have talked about this with some people and no one seems to know for sure.
 
A generator is not a self inductance case. In a generator we have a coil that rotates within the field of another coil or of a permament magnet.

The wire here is a self inductance case because it interacts with its own magnetic field. When you rotate the wire you just rotate the magnetic field generated by the current of the wire and thus -in the stationary frame of reference- you produce an aditional term dB1/dt due to the rotation which adds to the other term dB2/dt due to the magnetic field of the time varying current of the wire. Because it seems to be self inductance in both the terms, that's why i treat it like a RL-circuit and that's why i don't think that the induced current can persist forever.

However to tell you the absolute truth we must do the math for a rotating magnetic field to get a definite answer.
 
But what if the field doesn't rotate ? Much like in a round permanent magnet , actually think of the faraday paradox , you can rotate the magnet at the same speed as the disc or just the disc the outcame is the same due to lenz law current is produced.In both cases the field gradient doesn't change so my question becomes how does the wire tell the difference between a magnetic field from a separate magnet and that of it's own current , assuming the current is static and producing a similar field to that of the round permanent magnet?
 
Salvador said:
...how does the wire tell the difference between a magnetic field from a separate magnet and that of it's own current , assuming the current is static and producing a similar field to that of the round permanent magnet?...

Well there would be no difference if you make that assumption. But what happens in reality is that the current can't be static due heat on the ohmic resistances and that the energy supplied by rotation can't keep the current constant (not sure, but don't ask me to get into trouble writing equations about a rotating magnetic field).

One major thing that you might misunderstood about self induction is that the current generates a magnetic field which magnetic field affects the current that generates it (and even if we have rotation it still is the same -rotating now- magnetic field that affects the current that generates it). While in normal induction the magnetic field is generated by an outside source and it affects the current on another object foreign to the source of the magnetic field.
 
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