>> It's not that it's bothering me, it's the fact that none of the proponents
>> of 0 resistance so far bothered to answer my question about how to in fact
>> measure 0 Ohm.

> If there's a finite, non-zero resistance R going around the loop, and a
> current I, then by Ohm's law there must obviously be a voltage drop IR
> around the loop.

This is a misunderstanding of Ohm's law. Ohm's law is limited and needs a
number of conditions to work; one of them is the absence of dynamic
magnetic fields. (In a way, Ohm's law talks about moving charges through
the force of a constant electric field. It doesn't talk about the forces
you can exert on charges through magnetic fields.) With a magnetic field,
you can induce circular currents in just about any conductor, no need for a
superconductor here.

The difference is "only" that superconductors don't (measurably) heat up
and that the current, once induced, doesn't (measurably) degrade -- thus
implying that no (measurable) energy gets transformed from electric energy
into thermal energy, thus implying that the superconductor has no
(measurable) resistance. But if you actually measure something, there's
always the limit of precision of measurement, and considering all the
little dashes around the measuring points (you have done that in physics
lab, don't you?), you get to a result zone that may well include 0, but is
a bit bigger than the point 0. So the real affirmation is "smaller than x",
with x depending on your measurement configuration.

Another thing about Ohm's law seems to be mixing up the definition of
resistance and Ohm's law. Ohm's law says that resistance is a constant,
independent of voltage and current, when keeping everything else constant.
We know that this law is helpful in many cases and use it often enough, but
we also (should) know that it is not always applicable. There are many
materials where it is just plain wrong. (Superconductors, for example.)

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