When creating a magnetic field with a coil in the vicinity of a static magnet, where does the potential energy of the magnet come from? Imagine I have a coil/solenoid with an iron core, just a typical eletromagnet. A rod magnet is located at some distance to it. Imagine they are positioned and oriented like this:
      _[V]_
     |     |
     |     |        rod magnet
(S)  =|||||= (N)        [ N | S ]

       ^^^
    coil (+iron core)   
    connected to voltage source

The coil is wired so that the electromagnet's polarity is such that the rod magnet will be repelled. (Imagine that the rod magnet is perfectly aligned with the electromagnet and it can only move in the horizontal direction, not rotate, like fixed on a rail or inside a plastic tube)
Imagine the state in which the electromagnet is powered on, but the magnetic rod is held in place. In this state, within the electric field of the coil, the rod has a potential energy, which is converted to kinetic energy once it is released and free to move and accelerate, similar to a weight being released from height in earth's gravity field accelerating as its potential energy is converted into kinetic energy.
However, where does the potential energy come from here? Is it "put in" when the electromagnet is powered on? Will it consume more energy to power up the electromagnet In the presence of the rod magnet then it would were the magnet not present? Or does the repel movement induce a back-current, to work against which the voltage source must afford additional energy?
 A: If the coil were alone in the system, without the nearby bar magnet, there would be an energy cost in setting up the current in the coil. This is because, as the current in the coil increases, so does the magnetic field generated by the coil. This changing magnetic field induces an electric field that tends to oppose the increase in current, and the voltage source must do work against that back emf in order to continue increasing the current. The amount of energy it takes to set up a current of $I$ in the coil is given by $\frac{1}{2}LI^2$, where $L$ is the inductance of the circuit.
Now, if there's a current of $I$ in the circuit already, and the nearby bar magnet in the configuration shown by the OP, then when the bar magnet is released, there will be a change in the magnetic flux through the coil. In particular, if we use a sign convention where flux is positive in the positive x-direction (parallel to the coil's magnetic field) then

*

*the part of the flux through the coil that's due to the coil's own field is positive;

*the part of the flux through the coil due to the bar magnet is negative;

and as the bar magnet moves away, the latter contribution becomes less negative, which means it's increasing. The induced electric field tends to oppose this change, i.e., it tries to decrease the flux. This means that the emf opposes the current that's already flowing in the circuit (which we can infer from the fact that the current in the circuit produces positive flux). Consequently, the current in the circuit will decrease from $I$ to some smaller value $I'$, and the circuit will hold $\frac{1}{2}L(I^2 - I'^2)$ less energy. This is where energy is taken from to balance the kinetic energy acquired by the bar magnet.
