Discussion
The identity for \( \nabla \cdot (\phi\mathbf{A}) \) with \( \phi \) scalar is
Answer: B
The operator differentiates both factors, and the term acting on the scalar has to be contracted with the vector to stay a scalar:
\[ \nabla \cdot (\phi\mathbf{A}) \;=\; \phi\,(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot \nabla\phi. \]
The sign is positive, as in an ordinary product rule; the minus sign belongs to no standard identity of this kind
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