Which, if either, of the two force fields

$$\begin{gathered} {\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathit{y}\mathbf{i}\mathbf{+}\mathit{x}\mathbf{j}\mathbf{+}\mathit{z}\mathbf{k}\mathbf{,}\mathbf{}\mathbf{}\mathbf{} \\ {\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathit{y}\mathbf{i}\mathbf{+}\mathit{x}\mathbf{j}\mathbf{+}\mathit{z}\mathbf{k} \end{gathered}$$

is conservative? Calculate for each field the work done in moving a particle around the circle$\mathbf{x}\mathbf{=}\mathbf{cos}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{sin}\mathbf{t}$in the (x, y) plane.

Two forces field are given as mentioned below.

$$\begin{gathered} F_1=-yi+xj+z{k}^{\text{'}} \\ {F}_2=yi+xj+zk \end{gathered}$$

A force is said to be conservative if $\mathbf{\nabla }\mathbf{\times }\mathit{F}\mathbf{=}\mathbf{0}$ .

The scalar potential is independent of the path. The scalar potential is the sum of potential in all the 3 dimensions calculated separately.

The formula for the scalar potential is.$\mathit{W}\mathbf{=}\mathbf{\int }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$

The force is said to be conservative if $\nabla \times F=0$.

Putthe values given below in the above equation.

$\begin{array}{rcl}\nabla \times {\mathrm{F}}_1& =& \left|\begin{array}{ccc}i& j& k\\ \partial /\partial x& \partial /\partial y& \partial /\partial z\\ -y& x& z\end{array}\right|\\ \nabla \times {\mathrm{F}}_1& =& (0-0)\mathrm{i}-(0-0)\mathrm{j}+(-1-1)\mathrm{k}\\ & \ne & 0\end{array}$

$F_1$is not conservative.

$\begin{array}{rcl}\nabla \times {\mathrm{F}}_2& =& \left|\begin{array}{ccc}i& j& k\\ \partial /\partial x& \partial /\partial y& \partial /\partial z\\ y& x& z\end{array}\right|\\ \nabla \times {\mathrm{F}}_2& =& (0-0)\mathrm{i}-(0-0)\mathrm{j}+(1-1)\mathrm{k}\\ & =& 0\end{array}$

$F_2$is conservative.

For$F_1$ the values are mentioned below.

$$\begin{gathered} r=1, \\ x=\cos \theta , \\ y=\sin \theta \end{gathered}$$

Write the other values.

$$\begin{gathered} dx=-\sin \theta d\theta , \\ dy=\cos \theta d\theta \end{gathered}$$

Put the values in the formula.

$$\begin{gathered} W_1=\int {\mathrm{F}}_1·d\mathrm{r} \\ {W}_1={\int }_0^{2\pi }\left(si{n}^2\theta +co{s}^2\theta \right)d\theta \\ {W}_1=2\pi \end{gathered}$$

Now find for$F_2$ .

$F_2$is conservative so the work it does around a closed path equals zero.

role="math" localid="1664273134330" $$\begin{gathered} W_1=\int {\mathrm{F}}_1·d\mathrm{r} \\ {W}_2=0 \end{gathered}$$

Hence, the force field$F_2$ is conservative.

The work done for both the force is mentioned below

$$\begin{gathered} W_1=2\pi \\ {W}_2=0 \end{gathered}$$