Discussion
Consider a vector field given by \(x\hat{\imath} + y\hat{\jmath} + z\hat{k}\). This vector field is
Answer: B
The field is curl-free but not divergence-free.
For \(\vec{F} = x\hat{\imath} + y\hat{\jmath} + z\hat{k}\):
\[ \begin{aligned} \nabla\cdot\vec{F} &= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} \\ &\quad + \frac{\partial z}{\partial z} \\ &= 1 + 1 + 1 = 3 \end{aligned} \]
so it is not divergence-free; it is the field pointing straight outward from the origin, a uniform source. Each component of the curl, such as \(\partial F_{z}/\partial y - \partial F_{y}/\partial z = 0 - 0\), vanishes because each component depends only on its own coordinate, so \(\nabla\times\vec{F} = 0\). Indeed \(\vec{F}\) is the gradient of \((x^{2} + y^{2} + z^{2})/2\), and every gradient field is curl-free.
That rules out 'divergence-free and curl-free', 'divergence-free but not curl-free' and 'neither'.
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