I Deriving expression for resistance in terms of current density

AI Thread Summary
The discussion focuses on deriving an expression for resistance in terms of current density, emphasizing the need to account for the electromotive force (emf) in the relationship between current density and electric field. It argues that current density (j) should be expressed as conductivity times the sum of the electric field (E) and a fictitious electric field (E') due to emf. The voltage difference in circuits with emf is clarified to be V = IR - emf, rather than just IR. Participants express frustration over the complexity of splitting the electric field into components. The conversation concludes with a call for clarity and consensus on the topic.
spindecide
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Is there a way to obtain equation 9.42 (I is current, j is current density, and sigma is conductivity) in the following image (from Modern Electrodynamics by Andrew Zangwill, the part on electromotive force) besides using V=IR and substituting the line integral of j/conductivity for V? The aforementioned way requires that j=conductivity*E, but due to the emf, j should be equal to conductivity*[E + E'], where E' is a fictitious electric field representing the effect of any source of EMF.
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I think this equation is reasonable and I don't see any problem with it.

Circuit-30.jpg
 
alan123hk said:
I think this equation is reasonable and I don't see any problem with it.

But how is R written as the first integral?
 
spindecide said:
But how is R written as the first integral?
My personal opinion as follow.

Circuit-31.jpg
 
But j is not just equal to sigma times E as the effect of the emf needs to be considered, so that j equals sigma times (E + E').
 
A very useful equation is shown here
Circuit-33.jpg
$$ V_1-V_2=IR_{12}-\varepsilon _{12}~~~~~\Rightarrow ~~~~~V_1-V_2=IR_{12}-\int_1^2~dl\cdot~E^{'} $$ That is to say, the voltage difference in a circuit with EMF is not equal to IR, but equal to IR minus EMF. This is actually an equation very familiar and commonly used by electrical and electronics engineers when conducting circuit analysis.

It is also worth noting that $$IR_{12}=(V_1-V_2)+\varepsilon _{12}=\int_1^2~dl\cdot~(E+E^{'}),~~~~\text{so}~~R=\frac {1}{I} \int_1^2~dl\cdot~(E+E^{'}) $$
Everything is in perfect harmony, without contradictions. :smile:
 
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spindecide said:
But j is not just equal to sigma times E as the effect of the emf needs to be considered, so that j equals sigma times (E + E').
You are right, I accidentally wrote the equation wrong.
 
alan123hk said:
You are right, I accidentally wrote the equation wrong.
Can we please stop this split electric field nonsense???
Please.
 
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