Also, it is obvious that the work done is due to force of "magnetic field" on "moving electrons".
This part is problematic, as you probably already know (you've put the quotes). There is macroscopic magnetic force on wire 1 due to wire 2 given by $\int \mathbf j_1 \times \mathbf B_2 \,d^3\mathbf x$. Often this force is present but balanced by other forces so the wire does not move and there is no work involved.
When wire moves and kinetic energy increases, work is done on the wire. However this work is not due to the above force, since it is everywhere perpendicular to the motion of the charges it acts on, as you wrote above.
The only force that can do work on the wire and increase its kinetic energy is electric force. Since there are no external electric fields, it can be only the electric field of the wire 1 itself. This is always present, since there is current inside the wire, but usually does not move the wire in a noticeable way since mechanical equilibrium is easily established in common circuits. When the electric and magnetic forces cease to be counteracted by contact and mechanical forces (say, attached wire gets loose), the wire 1 will have non-zero acceleration due to magnetic force of wire 2, the power of this force being always zero. However, as soon as the wire 1 accelerates, this produces change in its own electric field. The changed electric field will now work on the wire itself and give it kinetic energy. This will happen at expense of the energy of magnetic field of the wires (and the source maintaining the current).
EDIT
The above explanation does not seem correct to me now, because one can extract work from the system of current carrying conductors very slowly, in which case the induced electric field will be negligible, while the force doing work is still great and given by the formula like $\int \mathbf j_1 \times \mathbf B_2\,d^3\mathbf x$. Please remove the green sign of acceptance. The real answer to your question requires more insight.
I assume you are referring to the method of electrostatic images : http://en.wikipedia.org/wiki/Method_of_image_charges
If so then a charged particle over a perfect conductor can be simplified for calculations
If the particle has a charge $+q$ and is at a distance $d$ from the surface of the perfect conductive plane then the image charge will be at a distance of $-d$ for the perfect conductor and its charge will be $-q$
So yes it depends on the distance $d$, if the charged particle goes farther away from the conductive plane its image will also go away from it
The image charge of a thin wire will be a charged thin wire as seen through a mirror across the conductive interface.
If the charge of your wire is $q=0$ then it's current is also $I = 0$ as:
$$I = \frac{dQ}{dt}$$
In that case, the image has also $Q' = 0$
Best Answer
The work comes from the battery that is driving the current through the wire.
Even if the wire were stationary, the battery would be supplying work at a rate $I^{2}R$. But with the wire moving, the battery would need to be supplying extra work at a rate $\mathscr{E}I$ in order to overcome the emf generated by the moving wire.
Now, $\mathscr{E}$ is equal to the rate at which the wire cuts magnetic flux so $\mathscr{E}=BLv$ (in which $v=\frac{d}{t}$), so the extra rate of doing work has to be $\mathscr{E} I=BLvI=BLdI / t $. And this is equal to the rate of mechanical work done on the wire!