Is the energy needed for a current through a straight and a coiled wire different? When you add current to a straight piece of wire does it use less electricity than if it was coiled? The power wire on telephone pole's are curved while buried cables are pretty strait in comparison. Does the curve of the wire and proximity to itself make a difference in energy consumption? 
 A: Okay I'll have a go at answering this, although it may be a make belief scenario.
Looking at where you got your inspiration from, he stated using a magnet as a core for an electromagnet, so your curiosity must have piqued from the idea that the wire carrying electric could itself be magnetic. Because of polarization and magnetization, all the magnetic dipoles in the wire would align with the potential of the wire, which would send you on a kind of wild goose chase, but hey lets do some maths anyway.
The steps to find the efficiency would be this:


*

*Find the general force of the electrons in the wire through the
magnetism of the wire.  

*Find the force of the electron through the magnetic field generated
    by their current density

*Take the ratio of these forces, multiply by the distance the
        electrons have traveled and find its difference from one.


1
The force on the electron from the wire itself
$$ F_{wire} = q(\hat{v_{e^{-}}} \times \hat{B_{wire}}) $$
2
The force of the electron generated by the rotation of it's own current density (lol)
$$ \hat{j} = nq\hat{v_{e}}A $$
$$ \nabla \times \hat{j} = \hat{M} $$
$$ \frac{\hat{M}}{\mu} = \hat{B_{j}} $$
It would be important to maintain the unit vector of the magnetization such that you know the direction of the magnetic field. 
$$ F_{e^{-}} = q(\hat{v_{e^{-}}} \times \hat{B_{j}})  $$
3
The work:
$$ W = F_{wire} \cdot \hat{r} $$
$$ W = F_{e^{-}} \cdot \hat{r} $$
Where I guess r would be the integral of the electron's velocity vector.
The efficiency would then be, with:
$$ W_{1} \leqslant W_{2} $$
$$ \epsilon = 1 - \frac{W_{1}}{W_{2}} $$
A: I may be wrong, but I think Lenz's Law might provide an answer.
The circuit with the straight wire takes in current i(suppose) once the switch is closed.
The one with the looped wire, will having a changing flux through it once the switch is closed. Since any change is to be opposed, the current drawn this time will be less,(assuming the dimensions of the wire loops do not change).
NOTE: In both cases, we observe and measure the current very soon after the switch is closed.
The situation is quite like an inductor. The current grows slowly(compared to a normal scenario). However, as the equations show, after a sufficiently long time, both should draw the same current.
