Question

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 \( \nabla \times \vec{V} \)?

• A. It is zero everywhere in the two-dimensional space.
• B. Its magnitude is non-zero and its direction is out of the two-dimensional plane.
• C. Its magnitude is non-zero and its direction is into the two-dimensional plane.
• D. It points in opposite directions above and below the x-axis.

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Look at the picture as a rule for the field: above the x-axis the arrows point to the right (+x), and their length grows the farther up they are. Below the x-axis the arrows point to the left (-x), and their length grows the farther down they are. This is exactly what happens if the x-component of the field grows in proportion to the y-coordinate, changing sign smoothly as y passes through zero.

Step 2: Key Formula or Approach:
Model the field as \(\vec{V} = c\,y\,\hat{x}\) for some positive constant \(c\). For \(y>0\) this gives \(V_x>0\) growing with \(y\) (arrows to the right, longer higher up); for \(y<0\) it gives \(V_x<0\) growing in size with \(|y|\) (arrows to the left, longer lower down). This matches the sketch. The curl of a 2D field confined to the xy-plane, with only an x-component that depends on y, is
\[ \nabla \times \vec{V} = \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right)\hat{z} \]
Since \(V_y = 0\) here, this reduces to \(\nabla \times \vec{V} = -\dfrac{\partial V_x}{\partial y}\,\hat{z}\).

Step 3: Detailed Explanation:
With \(V_x = c y\), we get \(\dfrac{\partial V_x}{\partial y} = c\), a constant. So
\[ \nabla \times \vec{V} = -c\,\hat{z} \]
Because \(c\) is the same positive constant everywhere in the plane, this curl has the same fixed non-zero magnitude \(c\) at every point, and it always points along \(-\hat{z}\), that is, into the page. It does not flip sign above and below the x-axis, because the slope of \(V_x\) with respect to \(y\) is the same everywhere, even though \(V_x\) itself changes sign.

Step 4: Why the other options are wrong.
Option (A) would be true only if \(V_x\) were independent of \(y\) (flat arrows of the same length everywhere); here the length visibly changes with height, so the curl cannot be zero. Option (B) gets the sign backwards; the curl points out of the page only if \(V_x\) decreased with \(y\), the opposite of what the sketch shows. Option (D) confuses the field itself, which does flip direction across the x-axis, with its curl, which does not; the curl depends on the rate of change of \(V_x\) with \(y\), which stays the same constant slope on both sides of the axis.

Final Answer:
The curl has a fixed non-zero size and points into the plane everywhere.\[ \boxed{\text{Option (C)}} \]