Where does the energy go if the current is induced in a perfect conductor?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.