#### Answer

$0$

#### Work Step by Step

As we know that $div F=\dfrac{\partial A}{\partial x}i+\dfrac{\partial B}{\partial y}j$
Now, we have
$div F=\dfrac{\partial}{\partial x}[\dfrac{-y}{(x^2+y^2)^{1/2}}]+\dfrac{\partial}{\partial y} [\dfrac{x}{(x^2+y^2)^{1/2}}]$
This implies that
$\dfrac{2xy}{2}(x^2+y^2)^{-3/2}-\dfrac{2yx}{2}(x^2+y^2)^{-3/2}=0$