A point charge $q=6\mu C$ is moving in a straight line with a velocity $\overrightarrow{v}=5\times {10}^4\overrightarrow{i}(m/s)$. When the charge is at the location P (3m, 4m,0) choose the correct statements about the electric & magnetic fields produced by the charge at the origin

For a point charge q moving with a velocity $\overrightarrow{v}$,
$$\begin{gathered} B=\frac{{\mu }_{0}q\overrightarrow{v}\times \overrightarrow{r}}{4\pi t^2} \\ \overrightarrow{r}=[-3\overrightarrow{i}-4\overrightarrow{j}]\left(m\right) \\ r=\sqrt{{(-3m)}^2+{(-4m)}^2=5m} \end{gathered}$$
Thus  $r=\frac{\overrightarrow{r}}{r}=\frac{[-3\overrightarrow{i}-4\overrightarrow{j}]\left(m\right)}{5m}=-0.6\overrightarrow{i}-0.8\overrightarrow{j}$
Substituting the above results into the equation for $\overrightarrow{B}$  we obtain
$$\begin{gathered} \overrightarrow{B}=\frac{{\mu }_{0}q\overrightarrow{v}\times \overrightarrow{r}}{4\pi r^2}=\frac{{\mu }_0}{4\pi }\frac{q\left(v\overrightarrow{i}\right)\times [-0.6\overrightarrow{i}-0.8\overrightarrow{j}]}{r^2} \\ =-\frac{{\mu }_0}{4\pi }\frac{qv\left(0.8\overrightarrow{k}\right)}{r^2} \\ =-({10}^{-7}T.m/A)\frac{(6\times {10}^{-6}C)(5\times {10}^{4}m/s)\left(0.8\right)}{{\left(5m\right)}^2}\overrightarrow{k} \\ =-9.6\times {10}^{-10}\overrightarrow{k}\left(T\right) \end{gathered}$$
Both the electric & magnetic fields at origin will decrease as the charge moves away.