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Why Is Faraday's Law not wrong?

  1. Apr 13, 2009 #1
    ok so Faraday's Law says that

    [itex]\oint_C \vec{E} \cdot \vec{dr} = -\frac{\partial}{\partial t} \int_S \vec{B} \cdot \vec{dA}[/itex]

    but we know that [itex]\vec{E}=-\nabla \varphi[/itex]

    and so [itex]\oint_C \vec{E} \cdot \vec{dr} =-\oint_C \nabla \times \nabla \varphi \cdot \vec{dA}=0[/itex] by Stokes' Theorem.

    therefore, why isn't the RHS of Farady's Law just 0 all the time?
     
  2. jcsd
  3. Apr 13, 2009 #2

    Matterwave

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    Haha, I had this exact same question the first time I took E&M

    The electric field is only curl free in electro statics, when there are NO moving charges. When there ARE moving charges the electric field is no longer curl free, and obeys Faraday's law.

    In the terms you presented, in electro statics the E field is due to a gradient of a scalar potential; however, in electro-dynamics, where charges can move, the E field is due to a gradient of a scalar as well as the time derivative of the vector potential. Thus, the closed line integral of E is no longer required to be 0.

    In deeper terms, the electric field is not just due to static charges, but also due to changing magnetic fields. This is exactly what Faraday's law is trying to say.
     
    Last edited: Apr 13, 2009
  4. Apr 13, 2009 #3
    i thought it might be that. what do you mean by curl free?
     
  5. Apr 13, 2009 #4

    Matterwave

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    Curl free means that the curl of the field is zero.
     
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