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Faraday-Maxwell Equation

  1. Dec 3, 2008 #1

    With regards to the Faraday Maxwell Equation form of Farday's Law.


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

    Then a curled magnetic field (say for instance a loop of wire with direct current passing through it) would produce a time-varying magnetic field? however as we know it does not..
  2. jcsd
  3. Dec 3, 2008 #2
    It seems to me that your conclusion arises from a notion that the curl of the electric field driving the current somehow depends on the geometry of the wire in which the current runs - it doesn't
  4. Dec 3, 2008 #3


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    The equation doesn't say that. It says that the curl of an E-field gives you the time varying form of a B-field. In the case of a DC current, you get a constant B-field, so [tex]\frac{\partial \mathbf{B}}{\partial t} = 0 [/tex]. Note that the curl of a constant E-field is 0, so there's no contradiction here.
  5. Dec 3, 2008 #4

    Ben Niehoff

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    The confusion is that in a DC circuit, the current clearly travels in a loop. Due to Ohm's law,

    [tex]\vec J = \sigma \vec E[/tex]

    the E field clearly must circle in a loop, too.

    The catch is that just because the flux lines of a vector field form closed loops, does NOT mean that the curl of the field is non-zero! For example, consider the vector field (in cylindrical coordinates)

    [tex]\vec E = \frac{1}{\rho} \hat \phi[/tex]

    The flux lines of this field are circles centered around the z-axis. But the curl is zero everywhere!

    If this vector field represented the velocity of a fluid, then a small object co-moving with the velocity field would NOT rotate, but it would maintain its orientation while traveling around the z-axis. Velocity fields such as this can occur in a free vortex in fluid dynamics.
  6. Dec 4, 2008 #5


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    I believe the OP was referring to the wrong version of the equation. His/her original assertions seems more relevant to the one in integral form:

    [tex]\oint \mathbf{E} \cdot d\mathbf{L} = - \int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{S} [/tex]

    My post would make more sense if I were referring to the above one instead. Since curl is evaluated at a point it doesn't seem to make sense to talk about whether E is in a loop or not.
  7. Dec 4, 2008 #6
    Thank you for all the replies. I now see that where I was getting confused was my incomplete understanding of curl.
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