Question

Let $\vec{F}(x, y, z) = x \hat{i} + x y \hat{j} + \hat{k}$, then $\text{curl}(\vec{F})$ is

• A. $\hat{j} + x \hat{k}$
• B. $y \hat{k}$
• C. $\hat{i} + x \hat{j}$
• D. $x \hat{i} + x y \hat{j} + \hat{k}$

Hint:

Notice that $F_x$ and $F_y$ do not depend on $z$, and $F_z$ is constant. Thus all $z$-derivatives are zero, leaving only the $x,y$ derivative in the $\hat{k}$ component!

Answer 1

The Correct Option is B

Step 1: Concept
The curl of a vector field $\vec{F}(x, y, z) = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ measures the infinitesimal rotation or circulation of the field at a given point. It is calculated using the vector cross product $\nabla \times \vec{F}$.

Step 2: Key Formulas and Approach
\[ \text{curl}(\vec{F}) = \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} F_x & F_y & F_z \end{vmatrix} \] Given $F_x = x$, $F_y = x y$, $F_z = 1$.

Step 3: Step-by-step Explanation

• Compute the $\hat{i}$-component: \[ \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) = \frac{\partial}{\partial y}(1) - \frac{\partial}{\partial z}(x y) = 0 - 0 = 0 \]
• Compute the $\hat{j}$-component: \[ \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) = \frac{\partial}{\partial z}(x) - \frac{\partial}{\partial x}(1) = 0 - 0 = 0 \]
• Compute the $\hat{k}$-component: \[ \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) = \frac{\partial}{\partial x}(x y) - \frac{\partial}{\partial y}(x) = y - 0 = y \]
• Combine components: \[ \nabla \times \vec{F} = 0\hat{i} + 0\hat{j} + y\hat{k} = y\hat{k} \]

Step 4: Final Answer
The curl of $\vec{F}$ is $y \hat{k}$. Thus, Option (B) is correct.