How can a vector field with circular components have a zero curl?

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A vector field with circular components can have zero curl if it is defined correctly, as demonstrated by the example of E = (A/r) êφ, which results in a zero curl despite its circular nature. The curl calculation shows that it equals zero, indicating no rotational behavior in the field. However, when evaluating a line integral around a circular path, the result is non-zero due to a singularity at the origin. This singularity invalidates the application of the theorem that would typically ensure the curl's implications hold true. Therefore, the presence of singularities must be considered when analyzing vector fields for curl and circulation.
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If a vector field has any component in a circular direction how can its curl be zero? If I imagine a vortex of water, it makes sense that it will be easier to go with the water in a circle than it would be to go against the water in a circle. Or more mathsy:

A vector field in cylindrical coords is defined by:
\mathbf{E} = \frac{A}{r} \mathbf{ \hat{e}_{\phi} }
where A is a constant. Therefore the curl is zero.
\mathbf{curlE} = \frac{1}{r} \frac{\partial}{\partial r} r \frac{A}{r} \mathbf{ \hat{e}_{z} } = \mathbf{0}
So taking a line integral in a circle, taking it back to the same point should also be zero.
\int^{2 \pi}_{0} \mathbf{E \cdot} r \mathbf{\hat{e_{\phi}}} \partial \phi = \int^{2 \pi}_{0} \frac{A}{r} \mathbf{ \hat{e}_{\phi} } \mathbf{\cdot} r \mathbf{\hat{e_{\phi}}} \partial \phi =\int^{2 \pi}_{0} A \partial \phi = 2 \pi A \neq 0

Thanks in advance
 
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Quite correctly calculated!

The reason is that your function E has a singularity in the region bounded by your curve (i.e, at the origin).

The theorem you had in mind specifically requires, in order to be valid, that such singularities do not exist, so you have misapplied it.
 
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