AC circuit and non-conservative field

[kidsasd987]

TL;DR

AC circuit generator generates EMF which is due to time-varying magnetic flux.
Then EMF is generated because there is a non-conservative electric field within a circuit.

However, R is defined with respect to the conservative field, where we know v=iR generally holds for an AC circuit. How can we use R derived from an assumption that E=-del V if an AC source generates a non-conservative field within a closed-loop?

Clearly, a non-conservative field can't be del of a scalar function.

*Please refer to the attached question file

 

[Dale]

A non-conservative EMF violates the assumptions of circuit theory, so you cannot use the circuit theory form of Ohm’s law.

Since this becomes a problem in Maxwell’s equations you need to use the compatible form of Ohm’s law: $\vec J = \sigma \vec E$

I am not going to open a word document.

 

[vanhees71]

Ohm's Law is of course valid also for "non-conservative" fields. It's just a transport property of a conducting medium. The constitutive relation in non-relativistic approximation is
$$\vec{j}=\sigma \vec{E}.$$
Since $\vec{E}$ is not conservative there is no electric potential but that doesn't invalidate Ohm's law for a resistor. Just integrate Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$
along the circuit, consisting of your generated and the resistance gives you
$$\mathcal{E}=R i,$$
where $R=l/(A \sigma)$ is the resistance. Note that
$$\mathcal{E}=\oint \mathrm{d} \vec{x} \cdot \vec{E}$$
is NOT a voltage!

 

[kidsasd987]

A non-conservative EMF violates the assumptions of circuit theory, so you cannot use the circuit theory form of Ohm’s law.

Since this becomes a problem in Maxwell’s equations you need to use the compatible form of Ohm’s law: $\vec J = \sigma \vec E$

I am not going to open a word document.

Thanks a lot!

 

[kidsasd987]

Ohm's Law is of course valid also for "non-conservative" fields. It's just a transport property of a conducting medium. The constitutive relation in non-relativistic approximation is
$$\vec{j}=\sigma \vec{E}.$$
Since $\vec{E}$ is not conservative there is no electric potential but that doesn't invalidate Ohm's law for a resistor. Just integrate Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$
along the circuit, consisting of your generated and the resistance gives you
$$\mathcal{E}=R i,$$
where $R=l/(A \sigma)$ is the resistance. Note that
$$\mathcal{E}=\oint \mathrm{d} \vec{x} \cdot \vec{E}$$
is NOT a voltage!

Thanks!

so as far as I understand,

1. so emf is not a voltage because the voltage is defined with respect to conservative electric fields and emf is defined with respect to a non-conservative field.

2. and ohm's law holds because J=σE is just a transport property of a conducting medium (Drude model). so E can be non-conservative. in the word doc, I assumed E = -Del V and found R but for the more general case including the non-conservative field, I should define R from J=σE.

Is this right?

but I have one more question tho.. Is it ok to use R that I found from a conservative field in the word doc?

I assume I can do that because J=σE is dependent on the material property of the conductor and geometry. So once I found R with respect to a conservative field, it should not change when the non-conservative fields are applied. (The generalized resistance equation for R is defined with respect to voltage, not EMF)

This is where I got confused. R is the same for conservative fields and non-conservative fields? It sort of makes sense because it's a material property, but I am not 100% sure.

 

[vanhees71]

For not too large frequency you can use $R$ from DC measurements. For more details, see the Feynman lectures:

https://www.feynmanlectures.caltech.edu/II_32.html

 

[kidsasd987]

For not too large frequency you can use $R$ from DC measurements. For more details, see the Feynman lectures:

https://www.feynmanlectures.caltech.edu/II_32.html

Thanks! Have a great day!

 

[rude man]

Consider a generator with zero resistance as in your word figure of sect. 2: there can be no net E field in the generator wire yet we know from Maxwell that a non-conservative E field is generated - again as you say.

In the wire, in order to reconcile this paradox, an equal and opposite E field is generated in the generator wire which is conservative, yielding net zero E field. Further, since the circulation of a conservative field has to be zero it follows immediately that the E field in the external resistor R is 100% conservative and equal but opposite to the conservative E field in the generator..

Call the non-conservative field $\bf E_m$ and the conservative field $\bf E_s$. The total E field is just $\bf E_m + \bf E_s$ vectorially, everywhere around the circuit.

The relations are then
$emf = \int \bf E_m \cdot \bf dl$ (integration thru all the wire);
since $\oint \bf E_s \cdot \bf dl = 0$ it also follows that
$\oint \bf E \cdot \bf dl = emf$ numerically. Again, as you state.

Whereas
voltage V = $\int \bf E_s \cdot \bf dl$. Circulation of this integral follows Kirchhoff's voltage law $\Sigma \Delta V = 0$.

If there is resistance in the generator wire then the $E_s$ field is reduced by the amount $iR_{int}$ where i = current.
$\oint \bf E_s \cdot \bf dl = 0$ still of course. Current i is consequently reduced by the internal wire resistance to $emf/(R + R_{int}).$. emf is unaffected.

I recommend my blog https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/
for further study.

 

[Dale]

It has been clearly shown that your concept is mathematically invalid. Your idea is full of mathematical self-contradictions that were clearly pointed out in your insights conversation. I would not recommend it to students (or anyone) learning this material.

 

[rude man]

@Dale We don't agree so please let me have my say. Thank you.

 

[Dale]

I have let you have your say by not deleting it, but that doesn’t mean that I should be quiet and let a student be confused

 

[vanhees71]

Fortunately math is right or wrong and not a matter of opinion! I couldn't agree more with @Dale !