Why is the curl of Biot-Savart Law equal to zero?

[Dorsh]

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 \textbf{B} =\boldsymbol{\hat\phi} \frac{\mu_{0}I}{2 \pi r} 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.

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

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
\iint_{S} {\nabla \times \textbf{B}} \cdot d\textbf{S}=\oint_{C}\textbf{B} \cdot d\textbf{l} = \mu_{0}I

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?

 

[Dale]

It is zero everywhere except at r=0. Try doing a line integral around a path that does not enclose the origin.

 

[Dorsh]

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?

 

[Dale]

Yes, which is what you would expect with an impulsive current density located at r=0.

 

[jtbell]

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.