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Physics
Classical Physics
Electromagnetism
Decompose the E field into conservative and non-conservative parts
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[QUOTE="vanhees71, post: 6827576, member: 260864"] I don't know, how you come to take two points and put some voltage on in it in a situation, where you have an EMF due to to a time-varying magnetic field. Due to Faraday's Law, $$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$ there's no scalar potential for ##\vec{E}##, and the EMF between two points depends on how you measure it. That's well demonstrated in this (in)famous lecture by Lewin quoted somewhere above, if I remember right. All you can say is that $$\int_C \mathrm{d} \vec{r} \cdot \vec{E}=-\dot{\Phi}=1 \text{V},$$ where ##\Phi## is the magnetic flux through an arbitrary surface with boundary ##C##. Choosing ##C## to run "along the circuit" the conclusion is that there is a current ##i## and Faraday's Law then tells you that ##I=1 \text{V}/[(0.1+0.9) \Omega]=1 \text{A}##. Which voltages you measure between any two points "on the circuit" depends on the wiring of the voltmeter you attach to the two points! [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Decompose the E field into conservative and non-conservative parts
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