AC circuit and non-conservative field

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Summary:

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
 

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  • #2
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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.
 
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  • #3
vanhees71
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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!
 
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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!
 
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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.
 
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  • #8
rude man
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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.
 
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  • #9
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I don’t know why you bother posting math when you have been clearly shown that your concept is mathematically invalid. You can’t have it both ways. Your idea cannot both be mathematical and yet you be ok with the mathematical contradictions pointed out in your insights.
 
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  • #10
rude man
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@Dale We don't agree so please let me have my say. Thank you.
 
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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
 
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  • #12
vanhees71
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Fortunately math is right or wrong and not a matter of opinion! I couldn't agree more with @Dale !
 
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