Let us rotate a closed circuit made of a copper coil in front of a magnet. This will generate an induced current and all the mechanical energy will be converted into heat through Joule effect. If the coil is made of a perfect conductor, there will be no Joule effect and the rotation of the coil will cost no energy. Right? But

1 - According to Lenz's law, the field of the induced current will still oppose itself to the motion of the magnet. So it WILL cost energy to overcome the force exerted by the field on the magnet.

2 - The acceleration of the electrons back and forth in the coil WILL cost energy too.

So: a - will the rotation of the perfect conductor cost energy or not ? and why. b - if yes, does this mean that not all the energy is converted into heat in an ordinary conductor?

  • 1
    $\begingroup$ "If the coil is made of a perfect conductor, there will be no Joule effect and the rotation of the coil will cost no energy. Right ?" - Why do you think that (your reasoning or conclusion) is correct? I recommend expanding your question so that your reasoning here is made more explicit. $\endgroup$ – Alfred Centauri Aug 13 '18 at 1:56
  • $\begingroup$ Well in fact I am not sure, I am just trying to find out. My reasoning goes this way : since I spend energy to overcome the resistance, then I'll spend no energy if there's no resistance to overcome. Now the problem is that no matter resistance or not, there is still Lenz ! But I think I am satisfied with the answer given by safesphere. $\endgroup$ – Anarchasis Aug 17 '18 at 21:17
  • $\begingroup$ Anarchasis, thanks for the reply. I think a mechanical analogy might help but I'm not sure it's worthy of a full answer. In the absence of friction, it does not take any work to keep an object moving at a fixed speed but it does take work to change the speed of an object due to the object's mass. Likewise, in the absence of resistance, it does not require work to sustain a current but it does take work change the current due to inductance. $\endgroup$ – Alfred Centauri Aug 17 '18 at 22:40

My idea: It will just cost energy initially, to overcome the initial resistance to change of flux by the induced current in the loop, and also to accelerate the weight of the loop. After that, it will become like a pendulum or see-saw: like potential to kinetic energy and vice versa wrt the loop as a classical mechanical object. And also potential to kinetic wrt the electrons/current.

So this:

Assuming there is already current in the coil. There is the coil current to magnet B-field alignment, which produces a force on the coil when they are not perfectly aligned. However, this force on the coil does not cause acceleration to the electrons when the coil is static mechanically, even if there is current, since it is always acted perpendicular to the motion of current, so the force only affects the coil mechanically.

Change in current or acceleration of electrons is caused only by the motion of the coil i.e. change in flux.

Now, assuming there is already existing current, a mechanical change in position of coil that is against the current to B-field alignment will cause an induced current that will add to the existing current even more. However, a change in position of coil that improves the alignment will lessen the existing current (meaning decelerate the electrons), or even eventually change the direction of current.

So we can see, everytime the coil is positioned against the direction of the alignment, there is mechanical force resisting against that direction. Now, if the coil is rotating against that resistive force and moving away from alignment, there is deceleration of the coil. That deceleration or loss of mechanical energy becomes acceleration of the electrons, since in fact it is strengthening the existing current everytime it opposes the resistance.

And when the coil is positioned against the alignment, but rotating toward the alignment, there is acceleration, increase in mechanical kinetic energy of the coil since it is going in the direction of the force. It corresponds to deceleration of electrons and decrease in current, because of the direction the coil is rotating toward alignment.

Now the case when electron deceleration stops, and current eventually changes direction, there will be a new current to B-Field alignment configuration because of the change in current direction. And the things said above still applies. If the coil initially accelerates mechanically while the current decreased, then when current direction changes, the coil will begin to decelerate, and the current will continue to strengthen in the new direction.

So apparently, energy is conserved, assuming the coil is friction-less in rotation. Energy is just converted back and forth, from mechanical energy of coil, to energy due to motion of electrons, and vice versa.


Moving a superconductor in a magnetic field will generate a current whose energy will be stored in the magnetic field produced by the current. The interaction of two magnetic fields will result in a force resisting the movement until it stops. If you apply a stronger force, more current would be generated, more energy stored in the magnetic field, and a stronger force would oppose your movement to stop it.

When you move a regular copper conductor in the magnetic field, your work also is stored as the energy of the magnetic field of the generated current, but then this energy very quickly is converted to heat, so the force opposing your movement is reduced. Ultimately, all your work is wasted as heat, but not without a help of the magnetic field in the process.

  • $\begingroup$ Does this mean no matter the resistance of the circuit, the energy I will spend to put it into motion at a given frequency will always be the same ? $\endgroup$ – Anarchasis Aug 17 '18 at 21:08
  • $\begingroup$ @Anarchasis No. Energy is force by distance. The force depends on the strength of the magnetic field in the conductor resisting the motion. In a normal conductor the energy of the induced current will quickly turn into heat, so the magnetic field will stay the same. Think of a faucet filling a sink with a drain open. The water level (magnetic field) in the sink will be constant. In a superconductor the magnetic field will be increasing with every turn, so the force will be increasing until it stops you. Think of the sink drain closed. The level of water will be increasing until it spills over. $\endgroup$ – safesphere Aug 18 '18 at 4:10

Since the resistance of the loop is assumed to be zero, the current will be limited by the inductance of the loop and a small effective radiation resistance. The phase of the current will be lagging the phase of the induced EMF by close to $90$ degrees.

Half of each cycle, some mechanical energy would have to be put into the loop, while the other half of the cycle, the loop will be returning about the same amount of energy back - the balance lost to radiation. A flywheel would be needed to absorb that energy and very little additional energy would be required to keep it going.

  • $\begingroup$ "while the other half of the cycle, the loop will be returning about the same amount of energy back". Cannot understand why ? $\endgroup$ – Anarchasis Aug 17 '18 at 21:04
  • $\begingroup$ @Anarchasis This what happens in purely reactive (inductive or capacitive) circuits, when the phase shift between voltage and current is $90$ degrees. The power is a product of voltage and current. If you multiply instantaneous voltage by instantaneous current, you'll see that half of each cycle the result is positive (energy goes into the circuit) and have of each cycle it is negative (energy is returned to the source). As a result, the average power consumption is zero. $\endgroup$ – V.F. Aug 17 '18 at 21:22

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