It depends. There are two main kind of losses in the motors, copper losses and iron losses. Copper losses = I^2*R_{phase}*constant

If my understanding is correct, this constant is basically the integral of the three sine waves…

In FOC the graph of the phase currents looks the same as the graph of a three phase AC. The integral of a sine wave under the period of pi is 2. As the period of a full sine wave is 2*pi, the absolute value of the area under a full sine wave is 4. We have 3 of these sine waves under the period of 2*pi, so this constant should equal 3*4/(2*pi) which equals 6/pi

So we can get the copper losses by I^2*R_{phase}*6/pi.

If we move to a bigger motor while keeping the same KV, the resistance goes down, therefore you will have less copper losses. With the same input power, lower losses mean a higher output power due to a higher efficiency. Note: the difference in output power is nowhere near enough that you can feel it.

The other type of losses is iron losses. Iron losses equal the idle current of motor (I_{0}) times the BEMF which it is spinning at, which is nearly the same as the voltage applied to the coils. This BEMF roughly equals V_{battery}*duty_cycle. When current flows through the motor, the actual voltage varies a bit, usually not more than by 2V, so lets just leave this out not to overcomplicate the equations.

So iron losses equal I_{0}*V_{battery}*duty_cycle. The idle current depends on the quality of the motors construction, better magnetic properties leading to a lower idle current. It also depends on the motor’s size, higher size means higher idle currents. And it’s also influenced by the rotational speed of the motor, by how much, I don’t exactly know right now.

However, what we can say for sure, is that if we take the exact same motor, and just scale it up in size, *keeping the quality of the construction as a constant,* then the idle current and thus the iron losses increase. Meaning, that a bigger motor has higher iron losses all else being equal. Which means for the same input power, we get less output power. But again, the difference is so small, that you won’t feel the difference.

This was my take on it, if there are some electrical engineers here, feel free to correct me if i messed something up, I am not an electrical engineer *yet*. (although I am currently enrolled in a university course as a first semester EE bachelor student)