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).
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.