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The reason I feel this isn't right is V=-dE/dr so, when there is an electric field in the wire, there is a potential drop across it too, however, I've been taught that the potential drop for an ideal current-carrying wire IS zero, Why is my reasoning wrong?

math and physics forever
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    V=I*R. Solve for V when R=0. – user1850479 Jul 04 '22 at 18:41
  • Because the electric field is generated by charges. And if there is a difference in potentials, the charges will flow through the conductor until there is no difference. – Eugene Sh. Jul 04 '22 at 18:41
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    @user1850479 Ohm's law is an abstraction/model of higher level, it is not explaining the phenomenon but using it. – Eugene Sh. Jul 04 '22 at 18:42
  • See https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant – Eugene Sh. Jul 04 '22 at 18:45
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    You're assuming that dE is not zero. In the steady state, in a superconductor, it is zero. – Simon B Jul 04 '22 at 19:35
  • This is a physics question. Superconductivity is a strange phenomenon, but it does indeed exist. The effect is electrons pairing up reduces their collision to zero (for some electrons in the material, anyway). They still have momentum, which you can think of as given at one end of the conductor, and so just that initial push need be given. Which basically means an inductive effect (voltage drop decreases over time, for same current), which consumes no DC power, which is fine. – Tim Williams Jul 04 '22 at 19:54

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Since it does not take energy to move charges across superconductors, their potential is unchanged by moving them. So dE/dr is zero. Apart from their magnetic field and time difference consequences, the ends of a superconductor are exchangeable for charges. With regard to electrical fields, it is a wormhole.