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Simplified Maxwell's Equation Proof

  1. Feb 28, 2013 #1
    Problem:

    I'm trying to crudely prove the following:

    [itex]\frac{{\partial}B}{{\partial}x}[/itex]=-[itex]\mu_{o}[/itex][itex]\epsilon_{o}[/itex][itex]\frac{{\partial}E}{{\partial}t}[/itex]

    Solution (so far):

    I can get the derivation, but the minus sign eludes me somehow...

    Integrating over a thing rectangular loop of length [itex]l[/itex] and width [itex]dx[/itex], start with the following,

    [itex]\oint{B{\cdot}dl}[/itex]=[itex]\mu_{o}\epsilon_{o}[/itex][itex]\frac{\partial{\Phi_{E}}}{\partial{t}}[/itex]

    Then,

    [itex]\oint{B{\cdot}dl}[/itex]=[itex](B+dE)l-Bl=dEl[/itex]

    Also,

    [itex]\Phi_{E}=EA=E(dx)(l)[/itex]

    ∴[itex]\frac{\partial{\Phi_{E}}}{\partial{t}}=[/itex][itex]\frac{\partial{E}}{\partial{t}}dxl[/itex]

    Equating the equations above,

    [itex]dEl=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}dxl[/itex]

    [itex]dE=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}dx[/itex]

    [itex]\frac{\partial{E}}{\partial{x}}=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}[/itex]

    Any advice is greatly appreciated.
     
  2. jcsd
  3. Apr 1, 2013 #2
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