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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.

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  • $\begingroup$ 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. $\endgroup$ – Bob Jacobsen 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? $\endgroup$ – Swaroop Joshi Oct 10 '19 at 13:19
  • $\begingroup$ Are you sure the curl of the electric field is nonzero in this situation? $\endgroup$ – probably_someone 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? $\endgroup$ – Swaroop Joshi 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. $\endgroup$ – Swaroop Joshi Oct 10 '19 at 13:42
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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.

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If you had the hypothetical situation of a constant current travelling around a loop of wire, then any changing magnetic flux through the loop is calculated as $-\oint \vec{E}\cdot d\vec{l}$.

For an ideal conductor then $\vec{J} = \sigma \vec{E}$ and the current and electric field are of constant magnitude and in the same direction, which is also the direction of the path taken when doing the integral. If it is a perfect loop then the infinitesimal line integrals will sum to zero and so there is no changing magnetic flux through the loop (easy to see if you consider a square "loop" in Cartesian coordinates; the line integrals on opposite sides will be equal in magnitude but opposite in sign).

By Stokes' theorem we know that the curl of the E-field and its closed line integral are related. $$ \int (\nabla \times \vec{E})\cdot d\vec{S} = \oint \vec{E}\cdot d\vec{l}$$ So what matters as far as calculating any changing magnetic flux through the whole loop is the integral of the curl of the E-field over the whole surface bounded by the loop. i.e. It depends on the value of $\nabla \times \vec{E}$ over the whole surface, not just at any single point.

In fact, for a simple, time-invariant current in a long wire there {\it will} be an electric field outside the conductor (because the component of E-field parallel to the wire is continuous at the boundary of the wire), but it must be curl-free because the B-field due to a uniform constant current is constant.

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