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Faraday's and Ampere's circuital law

  1. Apr 2, 2005 #1
    For electric field [tex]E=E_{0}e^{-\alpha}^{z}cos(\omega t)a_{x}[/tex] in free space (J=0), find B that satisfies Faraday’s law in differential form and then determine if the pair of E and B satisfy Ampere’s circuital law in differential form.

    [tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]

    Can someone give me hint for next step. Thanks
     
  2. jcsd
  3. Apr 2, 2005 #2

    dextercioby

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    Compute the curl.You should have written it (for ease and rigor)

    [tex] \vec{E}(z,t)=E_{0}e^{-\alpha z} \cos \omega t \ \vec{i} [/itex]

    After that,u need to integrate wrt time the negative of the curl you had just computed.

    Daniel.
     
  4. Apr 2, 2005 #3
    From computed curl:

    [tex]\frac{\partial E\vec{i}}{\partial z} = -\frac{\partial B\vec{y}}{\partial t}[/tex]

    And for B I got:

    [tex]B =-E_{0}\alpha\omega e^{-\alpha z} \sin \omega t \ \vec{j} [/tex]

    The answer for B in the book is different so I think I'm doing something wrong
     
  5. Apr 4, 2005 #4

    dextercioby

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    You might.Here's the curl

    [tex]\nabla\times\vec{E}=\left|\begin{array}{ccc}\vec{i}& \vec{j} &\vec{k}\\\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ E_{0}e^{-\alpha z}\cos \omega t & 0 & 0 \end{array} \right| [/tex]

    Okay?

    Daniel.
     
  6. Apr 5, 2005 #5
    This is what I got for curl
    [tex]-E_{0}\alpha e^{-\alpha z} \cos \omega t \ \vec{j} [/tex]
    then using Faraday's law
    [tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]

    [tex]B=\frac{E_{0}}{\omega}\alpha e^{-\alpha z} \sin \omega t \ \vec{j} [/tex]

    I fixed what was wrong, but I still have problem with vector j. Book says it should be z.
     
  7. Apr 5, 2005 #6

    dextercioby

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    It can't,unless there was something different to start with...

    Your work is correct.

    Daniel.
     
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