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Curl of Biot-Savart Law

  1. Sep 7, 2015 #1
    My understanding of the curl of a vector field is the amount of circulation per unit area with a direction normal to the area. For the vector field described as [tex]\textbf{B} =\boldsymbol{\hat\phi} \frac{\mu_{0}I}{2 \pi r} [/tex] I figured the curl would be something more like this, because it points in the vector normal to the rotation and in the direction of the current.

    [tex]\nabla \times \textbf{B} = \boldsymbol{\hat{z}}\frac{\mu_{0}I}{\pi r^2}[/tex]

    But when I go to calculate the curl of it by hand, I get zero. I know this can't be the case because I can do the line integral around a circle of radius r @ z = 0 and get
    [tex] \iint_{S} {\nabla \times \textbf{B}} \cdot d\textbf{S}=\oint_{C}\textbf{B} \cdot d\textbf{l} = \mu_{0}I [/tex]

    So I know the curl cannot be zero. But when I calculate by hand and by calculator, in cylindrical and cartesian coordinates, I get zero. Why is this the case? Am I doing the math wrong?
     
  2. jcsd
  3. Sep 7, 2015 #2

    Dale

    Staff: Mentor

    It is zero everywhere except at r=0. Try doing a line integral around a path that does not enclose the origin.
     
  4. Sep 7, 2015 #3
    Ahh, that makes more sense. If the curl of B is zero everywhere except for r = 0, I'm guessing the magnitude is probably more in line with a unit impulse function, right?
     
  5. Sep 7, 2015 #4

    Dale

    Staff: Mentor

    Yes, which is what you would expect with an impulsive current density located at r=0.
     
  6. Sep 7, 2015 #5

    jtbell

    User Avatar

    Staff: Mentor

    Note that we have an analogous situation in electrostatics, with the divergence of ##\vec E## and the integral of the flux of ##\vec E## over a closed surface. Consider a point charge and calculate ##\nabla \cdot \vec E## at any point where the charge is not located. Also calculate ##\oint {\vec E \cdot d \vec a}## for (a) a surface that encloses the charge, and (b) for a surface that does not enclose the charge.
     
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