If magnetic field is increased at a constant rate inside a zero resistance wire loop,

then, according to Faraday's law ,

$$\varepsilon = \frac{dΦ}{dt} $$

Thus, a potential difference is produced.

Also, according to Kirchhoff's law, net $\varepsilon$ in a loop is $0$. So, in short it says the changing flux doesn't produce any emf in the loop.

Both these laws are understandable when it's a simple circuit or even a superconducting circuit with a battery. But in this case these 2 laws seem contradicting. It may happen that either of the laws don't apply in extreme conditions like in a superconductor. Can somebody help?

I have read this post : Superconducting wire in a Magnetic Field? But it asks about a straight wire and not a loop. Also, it is not answered well (or at least i didn't understand much out of it).

  • $\begingroup$ physics.stackexchange.com/questions/159398/… I think this should answer your question, there are some good answers. $\endgroup$ Commented Sep 16, 2015 at 15:58
  • $\begingroup$ Faraday's law in general does not state anything about potential difference. It states that electromotive force due to changing magnetic field for the whole closed path (meaning circulation of induced electric field for that path) is proportional to rate of change of magnetic flux. This electromotive force/integral being non-zero alone does not mean there is potential difference anywhere in the circuit. In a very symmetric situation, all points of the ring circuit may have the same potential, despite non-zero induced EMF. $\endgroup$ Commented Jul 24, 2022 at 23:42
  • $\begingroup$ Also, Kirchhoff's second circuital law (KSCL, the original formulation) does not state that net EMF in a closed path equals zero. It states net EMF equals sum of terms $R_kI_k$ over all members of the closed path, where $R_k$ is ohmic resistance of the element $k$. Then there is the modern Kirchhoff's Voltage Law (KVL), which states that sum of potential drops in a closed path equals zero. This is always true but not applicable here, because $\epsilon$ is not a potential drop, but induced electromotive force in the circuit. $\endgroup$ Commented Jul 24, 2022 at 23:44
  • $\begingroup$ Only in this special case, since all $R_k$'s are assumed to be zero, from KSCL we get the result that net EMF must be zero as well. See Timaeus' answer. $\endgroup$ Commented Jul 24, 2022 at 23:45

2 Answers 2


The first sentence started with an if. When you start with an if and end with a problem a solution you should consider is that your if never happens.

So if the EMF around a zero resistance loop is zero then we don't expect the total magnetic flux through to change. Is that reasonable? Yes. Since it is a zero resistance loop, it can generate any current it needs, including the exact current that cancels the change in magnetic flux that other sources besides the loop would make.

Induced currents also make a contribution to the magnetic field that makes the flux change less than it otherwise would (Lenz's law). A zero resistance wire can do an excellent job.

for that exact current to get produced don't you need an emf produced first ?

An EMF is needed to supply power to overcome resistance, there is no resistance. For wires without resistance you have to figure out what makes there be current. Its energy and momentum balance, so you'd have to track the flow of energy and momentum from the fields to the charges and currents.

It would help if you were more detailed about an experimental setup. Like placed a loop inside a solenoid and make the solenoid increase its current according to your will. This won't be the only source of magnetic field, so you can't make there be the flux you want through that zero resistance loop.


I agree with the above. The instant you attempt to change the flux linking a loop of zero resistance, an opposing induced magnetic field would immediately negate the attempted change of flux (according to Lenz’s Law). In other words, it is not possible to change the flux linking a zero resistance loop. A similar thing is true if you are moving a straight wire of zero resistance through a magnetic field.A current and magnetic field would be induced which would instantly cancel out the local external field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.