Three semi-infinite mutually perpendicular conductors are joined at the origin $O$ as shown in the figure.

A current 2I enters through the conductor lying along the z-axis towards the origin O and leaves through the other two as shown in the figure. A charged particle having a charge q and mass m is at the point P, ${\overrightarrow{r}}_p=(\hat{i}+\hat{j}+\hat{k})$, moving with a velocity ${\mathcal{v}}_0\hat{i}$. 

Find the acceleration of the particle.

Take wire aligned along the $y-$axis. Consider a current element at a distance $y$ from the origin. The co-ordinates of this current element are $(0,y,0)$ Hence the position vector of point $P$ w.r.t. current element is
$\overrightarrow{R}=\overrightarrow{r}-y\hat{j}=\hat{i}+(1-y)\hat{j}+\hat{k}$

Hence, $\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi }·\frac{dy\hat{j}\times |\hat{i}+(1-y)\hat{j}+\hat{k}|}{{\left[2+(1-y{)}^2\right]}^{3/2}}$
or       $d\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi }·\frac{dy(\hat{i}-\hat{k})}{{\left[2+(1-y{)}^2\right]}^{3/2}}$

Let $1-y=\sqrt{2}\tan \theta$

$-dy=\sqrt{2}{\sec }^2\theta d\theta$

or $d\overrightarrow{B}=-\frac{{\mu }_{0}I}{4\pi }\frac{{\sec }^2\theta \text{d}\theta }{2{\sec }^3\theta }(\hat{i}-\hat{k})$

or ${\overrightarrow{B}}_y=\frac{{\mu }_{0}I}{8\pi }(\hat{i}-\hat{k}){\int }_{\theta ={\tan }^{-1}({}^1/{}_{\sqrt{2}})}^{-\pi /2}\cos \theta d\theta$

         $=-\frac{{\mu }_{0}I}{8\pi }(\hat{i}-\hat{k})\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right]$

Similarly we can find negative field due to the conductor along x-axis as 

${\overrightarrow{B}}_x=\frac{-{\mu }_{0}I}{8\pi }(\hat{k}-\hat{j}){\int }_{\theta ={\tan }^{-1}(1/\sqrt{2})}^{-\pi /2}\cos \theta d\theta$

     $=-\frac{{\mu }_{0}I}{8\pi }(\hat{k}-\hat{j})\sin \left[{\tan }^{-1}(1/\sqrt{2})\right]$

and the magnetic field at $P$ due to the conductor along $z-$axis as, 

${\overrightarrow{B}}_z=\frac{{\mu }_{0}2I}{8\pi }(\hat{j}-\hat{i}){\int }_{\theta ={\tan }^{-1}(1/\sqrt{2})}^{-(\pi /2)}\cos \theta d\theta =\frac{2{\mu }_{0}I}{8\pi }(\hat{j}-\hat{i})\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right]$

Thus the net magnetic field at $P$ is

$\overrightarrow{B}={\overrightarrow{B}}_x+{\overrightarrow{B}}_y+{\overrightarrow{B}}_z,$ and the acceleration is found from

$\overrightarrow{a}=\frac{q\overrightarrow{v}\times \overrightarrow{B}}{m};\text{ where }\overrightarrow{v}=v_0\hat{i}$

$\overrightarrow{B}=\frac{{\mu }_{e}I}{8\pi }\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right][-\hat{i}+\hat{k}-\hat{k}+\hat{j}+2\hat{j}-2\hat{i}]$

    $=\frac{{\mu }_{0}I}{8\pi }\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right][-3\hat{i}+3\hat{j}]$

$\overrightarrow{a}=\frac{q{v}_0}{m}\left[\frac{{\mu }_{0}I}{8\pi }\right]\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right]\left[\right(-3\hat{i}+3\hat{j})\times \hat{j}]$

   $=\frac{{\mu }_{0}q{v}_{0}I}{8\pi m}\sin \left[{\tan }^{-1}\frac{1}{\sqrt{2}}\right][-3\hat{k}]$