If a time varying magnetic field can give value to the curl of an electric field, why not the other way round? That is, why can't an enclosed loop with some emf produced (basically a current carrying loop) produce a changing magnetic field?

It does produce a constant magnetic field, yes. But according to Faraday's law, curl E=-dB/dt if the curl of the electric field has a value, shouldn't the time derivative also have a value? Meaning: a changing magnetic field should be produced.

  • $\begingroup$ Could you describe the physical situation you’re considering in more detail please? A closed loop of current has E slumming up to zero around it: there’s not average curl of E. $\endgroup$ Oct 10 '19 at 13:18
  • $\begingroup$ It's just a closed loop, like you're mentioning. Oh yes, you're right. I thought that there will be field 'in the wire'. Won't there be a field? $\endgroup$ Oct 10 '19 at 13:19
  • $\begingroup$ Are you sure the curl of the electric field is nonzero in this situation? $\endgroup$ Oct 10 '19 at 13:23
  • $\begingroup$ I'm not sure. Maybe that's where I'm wrong. But how to think of it to be equal to 0? $\endgroup$ Oct 10 '19 at 13:29
  • $\begingroup$ Yes..it is directed along the length of the wire. So the curl for the infinitesimally small area would be 0. $\endgroup$ Oct 10 '19 at 13:42

In a wire carrying a steady current, there is no electric field if the resistance of the wire is zero (i.e. a superconductor). If you make a loop of such wire and induce a current in it, then there is no electric field and the closed line integral of the electric field is zero.

In a wire with finite resistance, there is an electric field and on the face of it, there would be a finite line integral going around a closed loop. However, somewhere in that circuit, there must be an EMF source that has an exactly equal and opposite line integral (if the current is steady).

The net result is a closed line integral of zero (the circuital law) and no changing magnetic field due to steady current.

  • $\begingroup$ Kudos for writing "In a wire carrying a steady current, there is no electric field if the resistance of the wire is zero (i.e. a superconductor)." and emphasizing steady. $\endgroup$ Apr 23 at 4:30
  • $\begingroup$ @MathKeepsMeBusy upvoting an answer is the usual route to do that. $\endgroup$
    – ProfRob
    Apr 23 at 6:09

I may be missing your point, but current carrying loops produce varying magnetic fields in all kinds of situations: AC motors, transformers, inductors, demagnifiers, and many others. In an electromagnetic wave, the interaction of varying electric and magnetic fields determines the rate of propagation of the wave.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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