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Suppose we have a magnet and coil at sufficiently large distances apart such that any movement of magnet can not be known instantaneously at the coil, as change in magnetic field travels at the speed of light. Now, if i give magnet some kinetic energy. After moving to some distance at constant velocity , i will stop magnet by storing its kinetic energy in a flywheel (in form of rotational energy). so magnet will come at rest.

During the motion of magnet, it won't feel any resistance (due to lez's law) because signal, that magnet has been moved is not reached at coil yet. When this change in magnetic field reaches to coil and it will generate current opposing motion of magnet. But, magnet is already at rest so when this opposing magnetic field will reach at the magnet, the magnet will gain kinetic energy in opposite direction and this be will in turn produce current in coil again.

In this whole process, our input energy is not lost because magnet have not felt any resistance during its motion at constant velocity and we stopped magnet by storing its kinetic energy into flywheel . But still we are able to produce current in coil. Isn't this violating energy conservation law ?

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  • $\begingroup$ When you store the kinetic energy in the flywheel what have you done with the linear momentum that the magnet had and where did the angular momentum of the flywheel come from? $\endgroup$
    – Farcher
    Commented Oct 17, 2017 at 9:41
  • $\begingroup$ We are just converting motional energy of magnet into rotational energy. so magnet will come at rest and flywheel will have its energy in form of rotational energy. $\endgroup$ Commented Oct 17, 2017 at 9:45
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    $\begingroup$ I honestly don't understand what you have in mind when you write "drawing more output from coil". Suppose I have a loop antenna here and you, in the far-field of the antenna, move the magnet as you describe. How do I draw more output from the antenna? $\endgroup$ Commented Oct 17, 2017 at 16:25
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    $\begingroup$ It seems to me that you're assuming a contradiction that is this: the coil is both closely coupled (essentially all of the changing magnetic field threads the coil) and not closely coupled (the coil is light-seconds from the magnet). If the magnet's start, motion, stop all takes place before the magnetic field at the coil begins to change, the coil is not closely coupled and one must take into account both that (a) the electromagnetic field contains and transports energy and (b) most of the changing magnetic field does not thread the coil (unless you also involve a magnetic material). $\endgroup$ Commented Oct 17, 2017 at 23:53
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    $\begingroup$ "How much power coil will produce that depends on what value of resistance we put in the circuit" - this simply isn't true (it would be if the coil were an ideal voltage source). This comment thread is too long and likely to be moved to chat but I'm about a done here. $\endgroup$ Commented Oct 18, 2017 at 11:54

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No.

Imagine pushing a magnet through the wire when they are close together. You push the magnet, a current is induced in the coil, Lenz' Law is applied, and a resistive force is felt. The overall kinetic energy, however, is the work you put in to move it minus the work imparted by the electromagnetic force.

In your system, you put kinetic energy into the magnet to move it, then transfer it to the flywheel. This requires that the flywheel be considered a part of the overall system in which energy is conserved. The magnet then comes to rest and feels the opposing magnetic force some time later, but the total energy in the overall magnet-coil-flywheel system is the same: the energy of the magnet (now zero) plus the energy in the flywheel (equal to the work initially put in) minus the work done by the electromagnetic force.

Therefore, the total energy is the same as in the first case: the work you put in minus the work done by the electromagnetic force.

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  • $\begingroup$ The only input energy we have applied is kinetic energy of magnet. This kinetic energy is stored in flywheel, so it is not lost in producing electrical energy. In the end you have energy stored in flywheel ( which is equivalent to input K.E that we have given to magnet) + electrical energy . Where as our input was only K.E . $\endgroup$ Commented Oct 17, 2017 at 9:38
  • $\begingroup$ I believe it is a sign issue - the electrical force will be of opposite sign to that of the applied force due to Lenz' Law, pushing it in the opposite direction you are. In the close case, this just makes you push it at a lower acceleration than you would if there was no opposing force (sum the forces: applied force - electromagnetic force). In this case, since you stop applying a force and actively halt the magnet, the effect of the force is to start it moving in the other direction. $\endgroup$ Commented Oct 17, 2017 at 9:50
  • $\begingroup$ We are giving K.E to magnet then it will run at constant velocity since it is not feeling any resistive forces. After transfering all its energy to flywheel , our input energy is stored in flywheel. But still we have got electrical energy. Moreover , when opposing force will reach to Magnet, it will gain again kinetic energy in opposite direction and that inturn again produce current in coil. So where did this electrical energy comes from ? Our input energy is already stored in flywheel. $\endgroup$ Commented Oct 17, 2017 at 10:06
  • $\begingroup$ The purpose of Lenz' Law is to keep consistence with conservation with energy. The idea is that you put work into the system by moving the magnet - now Lenz' Law will act to counter that. It doesn't matter that you moved the energy into the flywheel, the coil will still apply an opposing magnetic force. $\endgroup$ Commented Oct 17, 2017 at 10:17
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    $\begingroup$ "Power is V*I , so if you increase current, power will also be increased" - @HemalPansuriya, it's frustrating in the extreme to the smart people here that have attempted to point out genuine problems with your thought experiment only to see you reply with an elementary fact that no one here disputes, that no one here doesn't already know, and that isn't relevant to the objections raised. $\endgroup$ Commented Oct 19, 2017 at 20:07
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In this whole process, our input energy is not lost because magnet have not felt any resistance during its motion at constant velocity and we stopped magnet by storing its kinetic energy into flywheel. But still we are able to produce current in coil. Isn't this violating energy conservation law ?

You forgot to account for electromagnetic energy of the system, which is commonly understood as energy of EM field distributed in the whole space.

The energy conservation only works in EM theory if energy of EM field is taken into account together with kinetic and rest energy of the matter.

In your thought experiment, there is magnetic field everywhere including the space of the coil. When the magnetic field of the magnet begins to change near the coil and electric current in the coil is induced, the current in the coil will create its own magnetic field near the coil. The total magnetic field near the coil will change from its previously established static value, and an electric field will appear as well. The field energy near the coil will change from its previously established static value.

This change of field energy near the coil has to happen in such a way that it will account for any energy extracted (or dissipated) from the electric current in the coil. This follows from the fact that the law of conservation of energy in EM theory is local. It does not matter that the magnet far away will be influenced later. The change of EM field and its energy near the coil happens at the same time the electric current begins to flow.

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  • $\begingroup$ Field energy has to come from work that we do on magnet. Since in our case, magnet is not feeling any resistance from coil, so we are not doing any work on magnet when it is traveling at constant velocity. static field does not create electrical energy, only change in magnetic field does. when distances are not large, the magnet feels resistance from coil and equal amount of work has to be done on magnet. But in our case, we are not doing work on magnet, but still we will get electrical energy in coil. $\endgroup$ Commented Oct 18, 2017 at 9:14
  • $\begingroup$ Field energy in space near the coil is already there, even before the magnetic field begins to change. It is this local field energy near the coil that will get partly extracted when electric current is flowing in the coil. Work being done on the magnet at the same time has nothing to do with the field energy available near the coil. $\endgroup$ Commented Oct 18, 2017 at 10:17
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    $\begingroup$ If only field is doing work and we don't have to apply any work, than repeating this experiment over and over again we will get arbitrarily large amount of energy. $\endgroup$ Commented Oct 18, 2017 at 16:23
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    $\begingroup$ @HemalPansuriya, (1) the flywheel will have less (rotational) energy than the energy required for the initial acceleration of the magnet. (2) Almost all of that difference is in the electromagnetic field. (3) The rest of the difference has been delivered to the coil. How is that you propose to extract arbitrarily large amounts of energy? $\endgroup$ Commented Oct 18, 2017 at 19:06
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    $\begingroup$ @HemalPansuriya, you're like a broken record; you keeping repeating the same nonsense which is this: "output from coil depends on us". This is plainly and simply false. Question: what is you intention? Do you aim to convince anyone here that you can extract arbitrary amounts of energy? Very well then - do it. That will show us a thing or three. $\endgroup$ Commented Oct 19, 2017 at 2:34

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