I Faraday's Law with Zero Resistance Current Loop

1. Dec 6, 2018

Luxucs

Hi there,

I was recently helping a friend of mine with a fairly standard electromagnetic induction problem (a basic sketch of the set-up is attached) where we have a current loop with resistance $R$ moving through a magnetic dipole and had to roughly sketch out the current induced in the loop due to Faraday's law. The reason we are able to do this is because Faraday's law will tell us the EMF induced in the loop, and Ohm's law, $ε = IR$, will let us find the current from that.

However, later that day, I began to think to myself: what if the resistance around the current loop is zero? The current loop is then an ideal conductor, and we are taught that the potential difference across an ideal conductor is zero. However, if we then look back at Faraday's law,

$\oint \vec E \cdot d \vec l = -\frac {d} {dt}\iint \vec B \cdot d \vec A$

We see that no where does the induction of the EMF depend upon the resistance of the loop, and that it only depends upon there being a changing magnetic flux density through the loop. However, if we assume we get an EMF around the loop, then we should have a current in the loop due to the existence of an electric field. In that case, the induced current should create its own magnetic field through the loop. Since there is no resistance in the loop, intuitively I would think that current could flow "freely" and create a magnetic field that would exactly "cancel out" the magnetic field due to the dipole and result in no net flux in the loop. This then means, due to Faraday's law, that the EMF in the loop should then be zero.

Is my line of reasoning correct here within the realm of classical electrodynamics? Or am I missing something?

File size:
360.6 KB
Views:
63
2. Dec 6, 2018

Staff: Mentor

You are missing the inductance of the loop. Even if it has no resistance it has inductance.

Loops like this can be used to partially shield sensitive equipment from changes in the external magnetic field, but it isn’t perfect.

3. Dec 6, 2018

Delta2

@Dale allow me to say that he is not completely missing the inductance of the loop. At some point in his post he says
Which means he has in mind the self induction mechanism.

@Luxucs your main idea is correct. But it is not the two magnetic fields that get cancelled (the external magnetic field is not exactly cancelled out by the magnetic field due to self induction). It is the respective EMFs that get cancelled. The EMF due to movement of the loop will get cancelled by the EMF due to self induction of the loop. That is because if we use Kirchoff's Voltage Law in the loop we will have
$$\mathcal{E_M}+\mathcal{E_L}+IR=0$$
where $\mathcal{E_M}$ is the EMF due to movement and $\mathcal{E_L}$ the EMF due to self induction. Since $R=0$ we get from the above equation that $\mathcal{E_M}=-\mathcal{E_L}$.

BUT (there is a big great "but" here ) I am not so sure about this too because we are entering the regime of superconducting materials (R=0) which is not completely explained using classical physics.