Question

For a scalar field \(\psi(\vec{r})\) and a vector field \(\vec{A}(\vec{r})\), \(\nabla\times(\vec{A}\,\psi)\) is equivalent to the expression

• A. \(\psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi)\)
• B. \(\psi(\nabla\times\vec{A}) + \vec{A}\times(\nabla\psi)\)
• C. Null vector
• D. \(\vec{A}\times(\nabla\psi) - \psi(\nabla\times\vec{A})\)

Hint:

Use the product rule \(\nabla\times(\psi\vec{A}) = \psi(\nabla\times\vec{A}) + (\nabla\psi)\times\vec{A}\), then flip the order of the second cross product.

Answer 1

The Correct Option is A

Step 1: Expand the curl directly in Cartesian components.
Instead of the index-notation shortcut, write out $\nabla\times(\psi\vec{A})$ using the determinant definition of curl, with $\vec{A} = (A_x,A_y,A_z)$ so $\psi\vec{A} = (\psi A_x, \psi A_y, \psi A_z)$:
\[ [\nabla\times(\psi\vec{A})]_x = \frac{\partial(\psi A_z)}{\partial y} - \frac{\partial(\psi A_y)}{\partial z} \]

Step 2: Apply the ordinary product rule to each derivative.
Each term is a product of two functions of position, $\psi$ and a component of $\vec{A}$, so the ordinary product rule applies:
\[ \frac{\partial(\psi A_z)}{\partial y} = \psi\frac{\partial A_z}{\partial y} + \frac{\partial \psi}{\partial y}A_z, \qquad \frac{\partial(\psi A_y)}{\partial z} = \psi\frac{\partial A_y}{\partial z} + \frac{\partial \psi}{\partial z}A_y \]
Subtract:
\[ [\nabla\times(\psi\vec{A})]_x = \psi\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right) + \left(\frac{\partial \psi}{\partial y}A_z - \frac{\partial \psi}{\partial z}A_y\right) \]

Step 3: Recognize each piece.
The first bracket is exactly $[\nabla\times\vec{A}]_x$. The second bracket is exactly the $x$-component of $(\nabla\psi)\times\vec{A}$, since for two vectors $\vec{u}\times\vec{v}$, the $x$-component is $u_yv_z-u_zv_y$, with $\vec{u}=\nabla\psi$, $\vec{v}=\vec{A}$. The $y$ and $z$ components follow the same pattern by cyclic symmetry, so as full vectors:
\[ \nabla\times(\psi\vec{A}) = \psi(\nabla\times\vec{A}) + (\nabla\psi)\times\vec{A} \]

Step 4: Match to the answer format.
The options are written with $\vec{A}\times(\nabla\psi)$ rather than $(\nabla\psi)\times\vec{A}$, so swap the order and flip the sign, since a cross product changes sign when the two vectors are swapped:
\[ (\nabla\psi)\times\vec{A} = -\vec{A}\times(\nabla\psi) \]
\[ \nabla\times(\vec{A}\psi) = \psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi) \]

Step 5: Rule out the other three.
This is option (A). Option (B) keeps the wrong sign on the second piece. Option (D) is the negative of the correct expression. Option (C), a null vector, would need the two terms to always cancel exactly, which is not true for arbitrary $\psi$ and $\vec{A}$.

Final Answer:
Component by component expansion confirms the same identity as the index-notation route. \[ \boxed{\nabla\times(\vec{A}\psi) = \psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi)} \]