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I know that for a $0 \;\Omega$ wire of uniform cross sectional area, the potential difference across its ends is zero as no electric field is required to move charge with constant velocity (necessary to keep current constant throughout the circuit) as no resistive force is offered by the wire. But what if the cross sectional area of that $0 \;\Omega$ wire increases on moving? Do we still say potential difference across its ends need to be zero? Yes from Ohm's law it needs to be zero, but zero potential difference means that there is no electric field inside that wire, therefore no acceleration and constant velocity of charges. But if charge continues to move with constant velocity and area increases on moving ahead then current through wire will increase ($I = neAV_d$).

So why there is a contradiction between Ohm's law and my concept?

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The theory perspective

I know that for a $0Ω$ wire of uniform cross sectional area, the potential difference across its ends is zero

Yes but only trivially so. Why should there be any potential difference across you resistor at all? Is it in a circuit? Is there a potential across it? Is there a current flowing through it? You haven't mentioned any of this. A $0$ ohm resistor lying disconnected from any sources can have whatever potential we want at its ends depending on ambient electrical environment.

as no electric field is required to move charge with constant velocity (necessary to keep current constant throughout the circuit) as no resistive force is offered by the wire.

Of course there is no potential difference needed - you set the resistance to $0$! Your resistor isn't a resistor at all - its just a plain old wire.

But what if the cross sectional area of that $0Ω$ wire increases on moving? Do we still say potential difference across its ends need to be zero?

Yes, when in a circuit, definitely. Its dimensions don't matter. Its material doesn't matter. Its temperature doesn't matter. Once something has $0$ resistance, it cannot develop a potential difference across it with only a constant current passing through it.

Yes from Ohm's law it needs to be zero,

No. Ohm's law isn't valid here as $R=0$. The current going through a $0$ ohm resistor is independent of the potential across it, which as you yourself said, is always $0$. The current is determined by other elements of the circuit.

Real world perspective

but zero potential difference means that there is no electric field inside that wire, therefore no acceleration and constant velocity of charges.

Even though real world wires have some resistance and therefore some electric field inside them, even if they did not, charges won't be accelerated but neither would they be moving with constant velocity. They would thermalize quite quickly and obtain random velocities. If this was not the case, there would exist a spontaneous current (see Drude model)

But if charge continues to move with constant velocity and area increases on moving ahead then current through wire will increase. $I=neAV_d$

In a non-zero resistance with varying cross-section at equilibrium, the current through it is the same everywhere, its uniformity is unaffected by length or cross-section. This is because if the current entering and leaving a point in the resistor was different, there would be charge buildup/depletion at that point which is not something we model in a resistor.

Another reason the current won't increase with cross-ection is that even though there are more charge carriers, the electric field is smaller.

For your case of $0$ ohm resistor the equation isn't valid anyways since $V_d=0$ so current is zero everywhere.

In addition, the currently accepted answer states that

Actually the thing is ,ohm's law is not a fundamental law .Which means only certain conductor follow it...

Yes. Ohm's law isn't a fundamental law. But neither are most linear relationships like Hooke's law, Curie's law or other formulae that try to model phenomenon linearly. This phenomenological approach is both necessary and useful in describing nature. They are often simple and accurate enough for most daily applications.

However, simply because these laws aren't the most bare metal fundamental descriptions, doesn't mean that they are wrong in their regime of applicability. Such laws provide an approximate description, and when higher accuracy in predictions is needed, more refined, though often more complicated, models are used.

Ohm's law is very well followed by almost all conductors in the real world under ordinary conditions and by all in theory. Why its not being used in your question is because when Ohm gave his law, he was talking about the $IV$ relationship of things that do have resistance - not materials without it.

and OHM's law gives sort of average value of current

It does in the sense that the total current through the wire has to be a statistical average of the Avogadro scale charge carriers in a conductor. This doesn't make the $IV$ relation approximate or impart to it any error. Any other model would do the same. This isn't the reason behind your contradiction.

For a conductor of non uniform cross section ,electric field must be present to keep I along the length constant .

Cross-section has nothing to do with whether field must be present or not. An electric field is always needed to drive a current, constant or not. Moreover, presence of electric field isn't why the current stays constant along the length. As remarked earlier, it has to during-steady state to prevent charge-accumulation.


$^1$ there's another degenerate case: when its $\infty$.

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  • $\begingroup$ "No. Ohm's law isn't valid here as R=0. The current going through a 0 ohm resistor is independent of the potential across it, which as you yourself said, is always 0. The current is determined by other elements of the circuit." I meant about the potential difference . R=0 V=IR ,hence V must be 0 if I doesnt approach infinity . Ofc in normal cases I is indeterminate through a 0 ohm wire $\endgroup$ Jun 15, 2021 at 20:53
  • $\begingroup$ Actually the thing is ,ohm's law is not a fundamental law .Which means only certain conductor follow it... $\endgroup$ Jun 15, 2021 at 20:57
  • $\begingroup$ This was not written by me . Its someone else's answer . I dont know how the same name $\endgroup$ Jun 15, 2021 at 20:57
  • $\begingroup$ @RaghavMadan goodness! sry! thats hillarious! I'll edit. $\endgroup$
    – lineage
    Jun 15, 2021 at 20:59
  • $\begingroup$ @RaghavMadan Actually the thing is ,ohm's law is not a fundamental law it isn't strictly speaking, but it is as fundamental as you need in your question's context. Is your question dealing with superconductors, non-linear active devices, anisotropic materials, high frequency or breakdown regimes? $\endgroup$
    – lineage
    Jun 15, 2021 at 21:06
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In order to keep electric charges inside an ideal conductor of varying cross section, you certainly have to apply a force to the charges, because they can't all move on straight lines (inertial movement). And for that you need an electric field, as you have correctly pointed out. However, this accelerating electric field has nothing to do with Ohm's law (nor does it require a finite resistance).

A zero resistance conductor does not preclude electric fields or potential differences inside of it. It just says that if there are electric fields, the charges are not going to move at constant speed/current, but rather will be accelerated very rapidly (increasing current) according to their inertia. While this happens, effects other than resistance will have an influence on the current, like inductance, or mutual repulsion of the charge carriers.

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The electrons don't move indefinitely on a straight line inside a wire, they colide with the nuclei repeatedly at a scale given by their mean free path, which is of the order of dozens on nanometers in metals. A change in the cross section that happens in scales longer than that should not affect resistance then.

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  • $\begingroup$ This answer seems to describe a resistive wire, but the question is about a nonresistive wire. Perhaps the answer is the same, but an explanation for the case at hand would be helpful. $\endgroup$ Jun 15, 2021 at 23:46

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