Sketch of a two-dimensional vector field \(\vec{V}\) is shown below. Here, length and arrow head of the arrows denote magnitude and direction of the vector field, respectively. Which of the following statements is correct for \( \nabla \times \vec{V} \)?
Show Hint
Model the sketch as V = c*y (x-hat) with c greater than 0: arrows flip sign at y = 0 but grow the same way in both directions, so the curl, equal to minus dVx/dy, stays one constant non-zero value into the page everywhere.
It is zero everywhere in the two-dimensional space.
Its magnitude is non-zero and its direction is out of the two-dimensional plane.
Its magnitude is non-zero and its direction is into the two-dimensional plane.
It points in opposite directions above and below the x-axis.
Show Solution
The Correct Option isC
Solution and Explanation
Step 1: Set up a small rectangular loop straddling the x-axis. Instead of differentiating directly, use the circulation definition of curl: $(\nabla \times \vec{V})_z = \lim_{A \to 0} \frac{1}{A}\oint \vec{V}\cdot d\vec{l}$ around a tiny loop of area $A$. Pick a thin rectangle centered on a point on the x-axis, with its top edge at height $+\epsilon$ and bottom edge at $-\epsilon$, width $w$ along x.
Step 2: Add up the circulation around the loop. On the top edge (at $y=\epsilon$), the field points in $+x$ with some size $V_x(\epsilon)$; walking the loop counter-clockwise means we traverse the top edge in the $-x$ direction, contributing $-V_x(\epsilon)\,w$. On the bottom edge (at $y=-\epsilon$), the field points in $-x$ with size $V_x(-\epsilon)$ (negative in our sign convention); walking counter-clockwise means we traverse the bottom edge in the $+x$ direction, contributing $+V_x(-\epsilon)\,w$. The short vertical edges contribute nothing since the field has no y-component.
Step 3: Use the antisymmetry of the sketch. Because the picture is antisymmetric about the x-axis (same arrow length above and below at equal distance, just flipped in direction), $V_x(-\epsilon) = -V_x(\epsilon)$. So the total circulation is \[ -V_x(\epsilon)\,w + V_x(-\epsilon)\,w = -V_x(\epsilon)\,w - V_x(\epsilon)\,w = -2V_x(\epsilon)\,w \] This is a fixed negative number for any small loop centered anywhere on the x-axis, never zero.
Step 4: Divide by the loop area. The loop area is $A = 2\epsilon w$, so the z-component of the curl is $\frac{-2V_x(\epsilon)w}{2\epsilon w} = -\frac{V_x(\epsilon)}{\epsilon}$, which stays a fixed negative constant as $\epsilon \to 0$ if $V_x$ grows linearly with $y$. A negative $z$-component means the curl points into the page, matching the direct calculation.
Step 5: Check the loop does not vanish or flip sign. Since the circulation never comes out zero for any placement of the loop, option (A) is ruled out. Since the sign of the circulation stays negative for a loop placed above the axis and for one placed below it (the antisymmetry of the field exactly compensates the flip in the loop's position), option (D) is ruled out too, and the fixed negative sign rules out the out-of-plane option (B).
Final Answer: The circulation test agrees with the direct calculation: the curl is non-zero and points into the plane everywhere.\[ \boxed{\text{Option (C)}} \]
