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Curl of vectorfield

  1. Apr 28, 2012 #1
    I would very much like a good intuitive understanding of what the curl of a vector field is. I thought it was a measure of the how much the field tries to rotate something, but that must be wrong because an electric field can have field lines that turn and not just go out radially, but still the resulting field has zero curl. What is curl really a measure of?
     
  2. jcsd
  3. Apr 28, 2012 #2

    BruceW

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    you were correct that the curl is intuitively how much the field lines rotate at a point. (Or as wikipedia puts it: the infinitesimal rotation of a 3-d vector field). The field lines can still turn, but that doesn't necessarily mean that there will be an infinitesimal rotation. The way I imagine it is to think of a small paddle wheel. And if the field lines look like they would turn it, then the curl is non-zero.

    So for example, the electric field created by two stationary charged particles, and assume the magnetic field is zero everywhere. Then imagine putting a small paddle wheel at any arbitrary point. Would it turn?
     
  4. Apr 28, 2012 #3

    tiny-tim

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    hi zezima1! :smile:

    curl is a measurement of the workiness of a force field …

    integrate a conservative force (ie, one whose curl = 0) around any closed path, and you get 0 …

    the work done by a conservative force around any closed path is 0​

    integrate a general force F around any closed path, and you get ∫ F.dl …

    which by stokes' theorem is ∫∫ curlF.dA, ie the flux of the curl through the path

    the work done by F around any closed path is the flux of curl F through it …

    the larger the area, the more curl F inside it, and the more the work done in going round it …

    curl F measures the workiness of F ! :smile:

    (standard example …

    if F is B, the magnetic field, then curlB = ∂E/∂t, the rate of change of the electric field,

    and the work done by B, ∫ B.dl, is …

    ∫∫ curlB.dA = ∫∫ (∂E/∂t).dA = ∂/∂t ∫∫ E.dA = ∂/∂t (electric flux) )​
     
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