Question

A vector $\overrightarrow{P}$ is given by $\overrightarrow{P}=x^{3}y\widehat{a_x}-x^2{y}^2\widehat{a_y}-x^{2}yz\widehat{a_z}$. Which of the given below statements(s) is/are correct regarding above vector?

• A. For$\overrightarrow{P}$ to be irrotational, the curl P will be zero
• B. For$\overrightarrow{P}$ to be solenoidal, the curl P will be zero.
• C. $\overrightarrow{P}$ is irrotational vector.
• D. $\overrightarrow{P}$ is a solenoidal vector.

Answer 1

Answer: (A), (D)

$\overrightarrow{P}$ is a solenoidal vector.

Given, $\overrightarrow{P}=x^{3}y\widehat{a_x}-x^2{y}^2\widehat{a_y}-x^{2}yz\widehat{a_z}$

For solenoidal vector,
$\nabla .\overrightarrow{P}=0$

$\nabla .\overrightarrow{P}=\frac{\partial P_x}{\partial x}+\frac{\partial P_y}{\partial y}+\frac{\partial P_z}{\partial z}$

=$\frac{\partial }{\partial x}\left(x^{3}y\right)-\frac{\partial }{\partial y}\left(x^2{y}^2\right)-\frac{\partial }{\partial z}\left(x^{2}yz\right)$

=$3x^{2}y-2x^{2}y-x^{2}y=0$

As $\nabla .\overrightarrow{P}=0$, so given vector is solenoidal
For vector to be irrotational

$\nabla \times P=0$

$\left[\begin{array}{ccc}\widehat{a_x}& \widehat{a_y}& \widehat{a_z}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^{3}y& -x^2{y}^2& -x^{2}yz\end{array}\right]$

$=\widehat{a_x}(-x^{2}z-0)-\widehat{a_y}(-2xyz-0)+\widehat{a_z}(-2x{y}^2-x^3)\ne 0$

$\therefore \overrightarrow{P}$ is not irrotational