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I Faraday's Law

  1. Sep 10, 2018 #1
    The potential difference between two points is given ans the negative of integral of E(vector) <dot product> dl(vector) from initial to final points.
    Therefore, integral integral of E(vector) <dot product> dl(vector) from initial to final point should give the negative of potential difference between them.
    In Faraday's law, closed loop integral of E(vector) <dot product> dl(vector) is given as ε- induced. Why is it not the negative of ε-induced. Should ε-induced not be treated like potential difference?
     
  2. jcsd
  3. Sep 11, 2018 #2

    Charles Link

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    The voltage from an EMF with ## V=+\int E_{induced} \cdot ds ## gets computed in just the opposite way of how the voltage from an electrostatic source gets computed as ## V=-\int E_{electrostatic} \cdot ds ##. It should be noted, the induced ## E ## does not give rise to a potential and, in general, ## \nabla \times E_{induced} \neq 0 ##, so that we cannot write ## E_{induced}=-\nabla \Phi ##. The ## V ## from a Faraday EMF is a voltage, at least when it is observed in an inductor, but it is not a potential type function. ## \\ ## In a conductor, since ## E_{total}=0=E_{induced}+E_{electrostatic} ##, we have ## E_{electrostatic}=-E_{induced} ##. The argument can be made that this is why the voltage from an inductor is in fact ## V=\mathcal{E}=+\int E_{induced} \cdot ds ##.
     
    Last edited: Sep 11, 2018
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