The work done on a particle of mass m by a force

$$ \vec{F}=K\left[\frac{x}{\left(x^2+y^2\right)^{3/2}} \hat{i}+\frac{y}{\left(x^2+y^2\right)^{3/2}} \hat{j}\right] $$

(with K being a constant with appropriate dimensions), when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is:

Let $ \vec{r}=x \hat{i}+y \hat{j} $

Therefore, $ d \vec{r}=d x \hat{i}+d y \hat{j} $ and $ r=\sqrt{x^2+y^2} $

Work done by the force in moving a particle from A to B is

$$\begin{aligned} W & =\int_{r_A}^{r_B} \vec{F} \cdot d \vec{r} \\\\ & =\int_{r_A}^{r_B} K \left[\frac{x}{(x^2+y^2)^{3/2}} \hat{i}+\frac{y}{(x^2+y^2)^{3/2}} \hat{j}\right] \cdot (d x \hat{i}+d y \hat{j}) \\\\ & =K \int_{r_A}^{r_B} \frac{x d x}{(x^2+y^2)^{3/2}} + \frac{y d y}{(x^2+y^2)^{3/2}} \\\\ & =K \int_{r_A}^{r_B} \frac{1}{(x^2+y^2)^{3/2}} \left[d\left(\frac{x^2}{2}\right) + d\left(\frac{y^2}{2}\right)\right] \\\\ & =K \int_{r_A}^{r_B} \frac{1}{2(x^2 + y^2)^{3/2}} d(x^2 + y^2) \\\\ & =K \int_{r_A}^{r_B} \frac{1}{2 r^3} d(r^2) = K \int_{r_A}^{r_B} \frac{2 r d r}{2 r^3} = K \int_{r_A}^{r_B} \frac{d r}{r^2} \quad (\because r=\sqrt{x^2 + y^2}) \\\\ & =K\left[-\frac{1}{r}\right]_{r_A}^{r_B} = K\left[\frac{1}{r_A} - \frac{1}{r_B}\right] \end{aligned}$$

Here, $ r_A = a $ and $ r_B = a $

Therefore, $ W = 0 $