With $L\gg R$, the two loops act like two magnetic dipoles with magnetic moment ${\mu }_{0}IA$, where $A$ is the area of the loop. The induction $B$ at loop two due to loop one can be resolved into two components; $B_r$, in the direction of increasing $\mathrm{r}$, and $B_{\theta }$ in the direction of increasing $\theta$. We find

$B_r=\frac{{\mu }_0}{2}I{R}^2\frac{\cos \theta }{L^3}=\frac{{\mu }_0}{2}I{R}^2\frac{x}{{\left(x^2+y^2\right)}^2}$

$B_{\theta }=\frac{{\mu }_0}{2}I{R}^2\frac{\sin \theta }{L^3}=\frac{{\mu }_0}{2}I{R}^2\frac{y}{{\left(x^2+y^2\right)}^2}$

where $x=\mathrm{Lcos}\theta ,y=\mathrm{Lsin}\theta$, and ${\mu }_{0}I\pi R^2$ is the equivalent dipole moment of each loop. The torque on loop two is

$\overrightarrow{\tau }=1\overrightarrow{A}\times \overrightarrow{B}\left(={\mu }_{0}I\overrightarrow{A}\times H\right)$

$\tau =I\pi R^2\left(B_t\sin \theta +B_{\theta }\cos \theta \right)$

  $=\frac{3\pi {\mu }_0{I}^2{R}^4\sin \theta \cos \theta }{4L^3}$

The direction of the torque is pointing into the plane of the paper. The force on loop two is

$\overrightarrow{F}={\overrightarrow{F}}_{\theta }+{\overrightarrow{F}}_r$

where $F_{\theta }=I{R}^2\frac{d{B}_{\theta }}{dx}$

                $=-\frac{{\mu }_0}{4}{I}^2{R}^{2}y\frac{2\times 2x}{{\left(x^2+y^2\right)}^3}=-{\mu }_0{\left(I{R}^2\right)}^2\frac{xy}{{\left(x^2+y^2\right)}^3}=-{\mu }_0{\left(I{R}^2\right)}^2\frac{\sin \theta \cos \theta }{L^4}$

and $F_r=I{R}^2\frac{d{B}_r}{dx}$

           $=\frac{1}{2}{\mu }_0{\left(I{R}^2\right)}^2\frac{y^2-3x^2}{{\left(x^2+y^2\right)}^3}$

           $=\frac{1}{2}{\mu }_0{\left(I{R}^2\right)}^2\frac{{\sin }^2\theta -3{\cos }^2\theta }{{\text{L}}^4}$