Understanding Non-Conservative Fields in Faraday's Law - Explained

[John94N]

This has been bothering me for a while. Consider a varying magnetic flux through a closed loop. Faraday’s law says the integral of E around the loop is V. This seems to be OK if the loop itself is split and joined with some resistance and with a voltmeter we measure across the resistor. However, if the loop is now closed we are measuring across effectively a short circuit and the voltage must be zero or very low but in any case not the emf developed by the varying magnetic flux. It also seems to work if we measure halfway around the loop and then double the measurement but again doesn’t work going 359.99 degrees aroung the loop. Is there something I am missing here? Probably.

 

[Jolb]

Good question. For most ordinary conducting loops, there actually is some nonzero resistance, so a voltage difference around the loop is permissible. However you're asking about a "perfect" conductor. How can can a perfect conductor, which is an equipotential, have a nonzero voltage difference when we go around the loop?

The answer is that truly perfect conductors, like superconductors, will not permit a magnetic flux to go through them. Strange as it may seem, it is experimentally verified. This is called the Meissner effect, and it explains why a permanent magnet cannot come in contact with a superconductor and instead floats above it.

One way to envision how this actually happens is that when a magnet is brought near the superconducting loop, the loop gets a current induced in it (for free) which causes a magnetic field exactly opposite to the one passing through it, canceling any flux.

 

[John94N]

Jolb, thanks for your response. i guess I was not considering the total B field in Faraday's law, that is the B field generated by the source and that generated by the ring itself.
Thanks John

 

[harrylin]

Jolb, thanks for your response. i guess I was not considering the total B field in Faraday's law, that is the B field generated by the source and that generated by the ring itself.
Thanks John

Right. And as you note, it's not that either field is zero, only that the sum of the two fields is zero. It is similar to electrostatic shielding in which the external electric field is countered by the internal field, but only after and thanks to the displacement of charges ("equipotential" is simply the result).

 

[DrZoidberg]

Maybe this video will be helpful.

 

[John94N]

Thanks for the responses. The videos from Dr Lewin's lecture really hurts my head, but they explain a lot. Thanks, John.