Question : Prove that curl of position vector is zero vector.
Solution :
We shall Find curl of position vector as follows We know position vector is represented as \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\] And curl position vector is defined as \[\nabla\times\left(\vec{r}\right)\] \[\nabla\times\left(\vec{r}\right)=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)\times\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\] Which can be written as the following determinant \[=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x&\ y&\ z\\\end{matrix}\right|\] \[\nabla\times\left(\vec{r}\right)=\hat{i}\left(\frac{\partial\left(\ z\right)}{\partial y}-\frac{\partial\left(y\right)}{\partial z}\right)-\hat{j}\left(\frac{\partial\left(\ z\right)\ }{\partial x}-\frac{\partial\left(x\right)}{\partial z}\right)+\hat{k}\left(\frac{\partial\left(y\right)}{\partial x}-\frac{\partial\left(x\right)}{\partial y}\right)\] By the rule of partial differentiation If we differentiate a variable with respect to another variable then the partial derivative is zero. That is \[\nabla\times\left(\vec{r}\right)=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0-0\right)\] \[\nabla\times\left(\vec{r}\right)=0\hat{i}+0\hat{j}+0\hat{k}\] Thus we proved that curl of position vector is zero vector