Question

The circulation of \(\vec{F}=y\hat{i}+z\hat{j}+x\hat{k}\) around the circle \(x^2+y^2=1,\ z=0\) is ____.

• A. \(\pi\)
• B. \(\dfrac{\pi}{2}\)
• C. \(\dfrac{\pi}{4}\)
• D. \(-\pi\)

Hint:

For circulation around a closed curve, parametrize the curve and calculate \(\oint_C \vec{F}\cdot d\vec{r}\).

Answer 1

The Correct Option is D

Concept:
The circulation of a vector field \(\vec{F}\) around a closed curve \(C\) is given by the line integral: \[ \oint_C \vec{F}\cdot d\vec{r} \] Here, \[ \vec{F}=y\hat{i}+z\hat{j}+x\hat{k} \] and the curve is the circle: \[ x^2+y^2=1,\qquad z=0 \]

Step 1: Parametrize the circle.
For the unit circle in the \(xy\)-plane, take: \[ x=\cos t,\qquad y=\sin t,\qquad z=0 \] where: \[ 0\leq t\leq 2\pi \] Now, \[ d\vec{r}=dx\hat{i}+dy\hat{j}+dz\hat{k} \] Since: \[ dx=-\sin t\,dt \] \[ dy=\cos t\,dt \] \[ dz=0 \] therefore: \[ d\vec{r}=(-\sin t\,dt)\hat{i}+(\cos t\,dt)\hat{j}+0\hat{k} \]

Step 2: Substitute the parametrization in \(\vec{F}\).
Given: \[ \vec{F}=y\hat{i}+z\hat{j}+x\hat{k} \] Using: \[ x=\cos t,\qquad y=\sin t,\qquad z=0 \] we get: \[ \vec{F}=\sin t\hat{i}+0\hat{j}+\cos t\hat{k} \]

Step 3: Find \(\vec{F}\cdot d\vec{r}\).
\[ \vec{F}\cdot d\vec{r} = (\sin t\hat{i}+0\hat{j}+\cos t\hat{k}) \cdot \left[(-\sin t\,dt)\hat{i}+(\cos t\,dt)\hat{j}+0\hat{k}\right] \] \[ \vec{F}\cdot d\vec{r} = -\sin^2 t\,dt+0+0 \] \[ \vec{F}\cdot d\vec{r} = -\sin^2 t\,dt \]

Step 4: Evaluate the circulation.
\[ \oint_C \vec{F}\cdot d\vec{r} = \int_0^{2\pi}-\sin^2 t\,dt \] \[ = -\int_0^{2\pi}\sin^2 t\,dt \] Using: \[ \int_0^{2\pi}\sin^2 t\,dt=\pi \] we get: \[ \oint_C \vec{F}\cdot d\vec{r} = -\pi \]

Step 5: Final answer.
Therefore, the circulation of the vector field around the given circle is: \[ -\pi \] \[ \therefore \text{Correct Answer is (D)} \]