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Physics
Classical Physics
Electromagnetism
Decompose the E field into conservative and non-conservative parts
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[QUOTE="SDL, post: 6826491, member: 731110"] Please consider the followng very important fact. A vector calculus theorem says that if ##\oint\mathbf{\vec{E}}\cdot \mathbf{\vec{dl}}=0## along [B][I]every[/I][/B] closed path, then there exists a scalar function (potential) ##\Phi## such that ##\mathbf{\vec{E}}=-\vec{\nabla} \Phi##. This is equivalent to independance of ##\int_A^B\mathbf{\vec{E}}\cdot \mathbf{\vec{dl}}## from a path from A to B, and we define this integral as "voltage drop": ##V_{AB}=\int_A^B\mathbf{\vec{E}}\cdot \mathbf{\vec{dl}}=\Phi_A-\Phi_B##. If this condition is false, then no such function exists, ##\int_A^B\mathbf{\vec{E}}\cdot \mathbf{\vec{dl}}## depends on a path from A to B, and all node potentials you've written just [B][I]loose any meaning[/I][/B]. And be careful, an assumption "variable magnetic field in the external space and wires is zero" means that there are no solenoidal electric field outside. In your case, this is not true. So you cannot use any "potentials" and "voltage drops". [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Decompose the E field into conservative and non-conservative parts
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