superg33k
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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
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