I'm not positive I understand the relation of the inductor and wire in your first question, but here's an attempt to give the physics.
Current flowing through a wire creates a magnetic field around the wire. If the current changes, then the magnetic field also changes in time. However, time-changes in magnetic fields create electro-motive forces (EMF) on charged particles, as a fundamental part of Maxwell's equations. These forces oppose the change in the current. This effect will occur in any circuit, regardless of whether an intentionally inductive coil is included.
What happens to the energy? The time-changing fields also lead, by Maxwell's equations, to electromagnetic radiation. That carries energy away.
You can see that this particular way of energy leaking away only happens for ac, not dc. Yes, you're correct that superconductors also radiate energy this way. For dc currents, this effect is absent and so an ideal superconductor (lacking internal ways of leaking energy) doesn't lose energy.
(published on 09/20/2010)