Faraday's & Ampere's Laws: Conditions of Independence

[atomicpedals]

Under what conditions is Faraday’s Law independent of Ampere’s Law?

I want to say that this is so only in the static case, however this isn't right (or at least there's more to it). What am I missing?

 

[cabraham]

If the filamentary loop subject to Faraday induction has a resistance much greater than its inductive reactance, then the induced voltage is simply $-N d\phi / dt$. Otherwise, Ampere's Law comes into play, and the terminal voltage changes as the load current increases, should resistance fall below inductive reactance value.

Claude

 

[rude man]

emf = -N dΦ/dt is always true, irrespective of the loop current. Φ is the net flux thru the loop.
The thing is, if the loop current is not negligibly small then Φ is not just the externally applied flux but includes the flux set up by the loop current.

In the latter case, emf = externally applied -dΦ/dt + L di/dt
where i is the loop current
L is the loop self-inductance.
The direction of the self-induced flux is such as to oppose the rate of change of the externally applied flux.

 

[cabraham]

emf = -N dΦ/dt is always true, irrespective of the loop current. Φ is the net flux thru the loop.
The thing is, if the loop current is not negligibly small then Φ is not just the externally applied flux but includes the flux set up by the loop current.

In the latter case, emf = externally applied -dΦ/dt + L di/dt
where i is the loop current
L is the loop self-inductance.
The direction of the self-induced flux is such as to oppose the rate of change of the externally applied flux.

Yes of course. I was treating $"\phi"$ as the external flux only. An external time varying flux is incident on a loop resulting in induction. If R is on the same order or less than inductive reactance $L\omega$, then the loop internal flux due to its own current opposes the external flux per Law of Lenz. Then the emf measured at the terminals would be reduced by L di/dt as you stated.

I believe we agree. As long as $\phi$ is understood as the NET flux, superposition of external plus internal, then that value is fixed. I originally treated $\phi$ as the external flux only, which was my interpretation of the OP, which could have been wrong. But I agree with you that in the general case.

Claude

 

[rude man]

We do agree, and it's a bit of a fine point sometimes. I've seen a lot of stated problems that specify an external field, then ask for the current induced in a single loop. They don't state ignoring the self-inductance of the loop. Unfortunately, computing the self-inductance of a single loop is almost prohibitively difficult.

Of course, if you want to get really picky you also have to consider the effect of the loop's mag field on the source of the external field, resulting in a bona fide transformer problem. The stated problems usually avoid this by defining an external field.unaffected by the loop field.