Wire 1:
V = 1000V
P = 500W
I = P/V = 500W / 1000V = 0.5A
R_wire = 2 Ohm
Loss in wire 1 = I^2*R = 0.5A^2 * 2 Ohm = 0.5W
Wire 2:
V = 10000V
P = 500W
I = P/V = 500W / 10000V = 0.05A
R_wire = 2 Ohm
Loss in wire 2 = I^2*R = 0.05A^2 * 2 Ohm = 0.005W
UPDATE:
To work out the exact loss, assuming voltage is measured at start of wire:
simulate this circuit – Schematic created using CircuitLab
Total current, I = V1 / (R1 + R2)
Load power, P2 = 500W = I^2 * R2 = (V1^2 R2) / (R1 + R2)^2
This equation can be rearranged into a quadratic form and R2 solved for. Once R2 is found, the loss in the wire can be found by I^2 * R1
To find it using the P = VI formula, calculate V2 as I * R2, the loss is given by (V1 - V2) * I.
As you've discovered, an electric motor is not well modeled as a resistor, and as such doesn't obey Ohm's law.
A better model for a DC electric motor is there is some resistance in series with a variable voltage source.
Additionally, a battery has some internal resistance, which can be modeled as a series resistor*. A PC power supply also can use this same model, but the series resistance is likely to be smaller. The system then looks like:
simulate this circuit – Schematic created using CircuitLab
We can explain why in the first case your measured voltage is less than the no-load battery voltage because we have a voltage divider. Doing some math,
\begin{align}
V_{emf} = V_+ - I R_m\\
R_s = \frac{V_{bat} - V_+}{I}
\end{align}
You measured \$R_m = 3.5 \Omega\$, \$I = 0.19 A\$, and \$V_+ = 2.9V\$, so \$V_{emf} = 2.24 V\$ and \$R_s = 1.47 \Omega\$.
In the second case, \$V_+ = 4.92V\$ and \$I = 0.28A\$. Thus: \$V_{emf} = 3.94 V\$ and \$R_s = 0.43 \Omega\$.
Notice that \$V_{emf}\$ is different between the two. This is because \$V_{emf}\$ is roughly linearly proportional to how fast the motor is spinning. You should have observed the motor spinning faster when hooked to the 5V supply.
Additionally, how multi-meters measure current is by introducing a series shunt resistance and measuring the voltage across this resistor. This further complicates the analysis, so the measured current and load voltage are not exactly correlated. It's more difficult to do this analysis, but is possible if you know the series shunt resistance. This is sometimes quoted as a "burden voltage" at a rated test current and you can use Ohm's law to recover the shunt resistance.
It is possible to reconstruct what the measured load voltage should be with just a single meter, but it requires more information on how \$V_{emf}\$ behaves which is beyond the scope of this answer.
If you set your meter to the largest current range this will use the smallest shunt resistance, you can minimize the impact of having the meter in series at the cost of losing a bit of accuracy.
*note: Batteries don't have a constant internal resistance, but this is a reasonable approximation. It depends on a ton of factors including but not limited to stored energy, temperature, and load.
Best Answer
Correct.
The smaller the resistance the more current will flow. (That should be fairly obvious - think of the water analogy: a bigger pipe has lower resistance and will have a larger flow for a given head of water.)
Figure 1. Water "current" and flow analogy. Note that the pressure (voltage) is the same in both cases. Source: Sparkfun.
The "load" is the load on the source, the energy supply or, in the case of the water analogy, the water supply or reservoir. The lower the resistance of the load the more current will flow.
Update after question updated:
So far, so good.
Not necessarily. It's the ampere-turns that will determine the strength of the solenoid. If you double the number of turns but keep the wire gauge and voltage the same then the resistance will double and the current will halve. Double turns x half the current gives the same ampere-turns so no improvement.
To increase the ampere-turns you could:
Correct.