An equilateral triangle frame with side a carrying $I$ is placed at a distance a from an infinitely long straight wire carrying a current $I$ as shown in the figure. One side of  the frame is parallel to the wire. The whole system lies in the xy-plane. Find the magnetic field acting on the frame.

Force on wire 12 will be $\left(Ia\right)$ in positive x-direction.

Force on 2-3 and 3-1:

$$\begin{gathered} r=a+x \sin 60°=a+\sqrt{3}\frac{x}{2} \\ {B}_r=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I}{r}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I}{a+\sqrt{3}\frac{x}{2}} \\ =\frac{{\mu }_0}{\mathrm{\pi }}\frac{I}{2a+\sqrt{3}x} \\ \therefore {\mathrm{F}}_{231}=\left[2{\int }_{\mathrm{x}=0}^{\mathrm{x}=\mathrm{a}}\left(\mathrm{dF} \cos 60°\right)\right]\left(-\hat{i}\right) \\ =\left[2{\int }_0^{\mathrm{a}}\frac{{\mathrm{\mu }}_0}{\mathrm{\pi }}·\frac{\mathrm{I}}{2\mathrm{a}+\sqrt{3}\mathrm{x}}·\mathrm{I}·\left(\mathrm{dx}\right)\frac{1}{2}\right]\left(-\hat{\mathrm{i}}\right) \\ =\left(\frac{{\mu }_0{I}^2}{\mathrm{\pi }}\right)·\frac{1}{\sqrt{3}}{\left[\ln \left(2\mathrm{a}+\sqrt{3}\mathrm{x}\right)\right]}_0^a\left(-\hat{\mathrm{i}}\right) \\ =\frac{{\mu }_0{I}^2}{\sqrt{3}\mathrm{\pi }}\ln \left(\frac{2+\sqrt{3}}{2}\right)\left(-\hat{\mathrm{i}}\right) \\ \mathrm{and} {\mathbf{F}}_{12}=\frac{{\mu }_0{I}^2}{2\mathrm{\pi }}\left(-\hat{\mathrm{i}}\right) \\ \therefore \mathbf{ }\mathbf{ }{\mathbf{F}}_{\mathbf{total}}=\frac{{\mu }_0{I}^2}{\mathrm{\pi }}\left[\frac{1}{2}-\frac{1}{\sqrt{3}}\ln \left(\frac{2+\sqrt{3}}{2}\right)\right]\left(-\hat{\mathrm{i}}\right) \mathbf{Ans}\mathbf{.} \end{gathered}$$