Electric field produced due to infinitely long charged wire,

$a=$ radius of cylindrical Gaussian surface around wire.

Magnetic field at ' $a$ ' distance from current carrying conductor,

$\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi a} \hat{i}$

but $\mathrm{I}=\frac{q}{t}=\frac{\lambda \mathrm{L}}{t}=\lambda v \quad\left[\because \mathrm{Q}=\lambda \mathrm{L}\right.$ and $\left.\frac{\mathrm{L}}{t}=v\right]$

Here $L=$ length

$\therefore \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \lambda v}{2 \pi a} \hat{i} \quad \ldots$ $(2)$

Now pointing vector,

$\mathrm{S}=\frac{1}{\mu_{0}}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}})$

$\therefore \mathrm{S}=\frac{1}{\mu_{0}}\left[\frac{\lambda}{2 \pi a} \hat{j} \times \frac{\mu_{0} \lambda v}{2 \pi a} \hat{i}\right]$

$=\frac{1}{\mu_{0}}\left(\frac{\lambda}{2 \pi a} \times \frac{\mu_{0} \lambda v}{2 \pi a}\right)(\hat{j} \times \hat{i})$

$=\frac{\lambda^{2} v}{4 \pi^{2} \in_{0} a^{2}}(-\hat{k}) \quad[\because \hat{j} \times \hat{i}=-\hat{k}]$

$\therefore \mathrm{S}=-\frac{\lambda^{2} v}{4 \pi^{2} \in_{0} a^{2}} \hat{k}$