# Charge moving in uniform circular motion

If a charge $$+Q$$ is moving in a uniform circular motion with angular velocity $$\omega$$ the equivalent current produced is $$Q\omega /2\pi$$. (We consider this as a circular loop and put a counter to check the rate at which the charge is passing) Is this average current or both average and instantaneous?

According to me, it should be average current except when the charge moves with infinite angular velocity, because in every $$dt$$ second the charge being passed is not the same. $$dq/dt$$ is sometimes $$0$$ and sometimes infinite as sometimes the charge passing through that counter is $$0$$ and sometimes a finite value.

If we were to distribute the same charge on a circular plate or ring uniformly and rotate it with the same angular velocity, both the instantaneous and average current will be the same: $$Q\omega /2\pi$$. This is because in $$dt$$ seconds, when the ring /plate rotates by $$d\theta$$ =$$\omega dt$$, the charge $$dq$$ passing through that counter is $$Qd\theta/2\pi$$ and hence the current will be $$Q\omega /2\pi$$. Is my reasoning correct ?

Thinking loudly.

Current is the amount of charge passing through a section per second, i.e. Amperes = Coulombs/seconds.

Then, the average current on a circle would be: $$\text{Average Current}=\frac{\text{Total Amount of Charges Completing the Loop}}{\text{Time Required For This to Happen}}$$ which is:

$$I_{avg} = \frac{Q}{\frac{2\pi}{w}} = \frac{Q\cdot w}{2\pi}$$

So, this should be the formula for the average current of a point charge rotating on a circle; and to me, it is not the same thing with the instantaneous current on a specific location. But, things might be different in the quantum world.

The same formula seems to be applicable for the disk, too. In that case, due to the geometrical symmetry and equal charge distribution, the average current on any section should be equal to the instantaneous current on the same section.