An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the $z-$axis. The wire is set into motion along its length with a uniform velocity $\vec{V} = v\hat{k}$. Calculate the Poynting vector $\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$.

(N/A) The electric field $\vec{E}$ at a radial distance $a$ from an infinitely long charged wire with linear charge density $\lambda$ is given by Gauss's Law as $\vec{E} = \frac{\lambda}{2\pi\epsilon_0 a} \hat{r}$,where $\hat{r}$ is the radial unit vector in the $xy-$plane.
The moving charged wire constitutes a current $I = \lambda v$. The magnetic field $\vec{B}$ at a distance $a$ from the wire is given by Ampere's Law as $\vec{B} = \frac{\mu_0 I}{2\pi a} \hat{\phi} = \frac{\mu_0 \lambda v}{2\pi a} \hat{\phi}$,where $\hat{\phi}$ is the azimuthal unit vector.
The Poynting vector is defined as $\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$.
Substituting the expressions for $\vec{E}$ and $\vec{B}$:
$\vec{S} = \frac{1}{\mu_0} \left( \frac{\lambda}{2\pi\epsilon_0 a} \hat{r} \times \frac{\mu_0 \lambda v}{2\pi a} \hat{\phi} \right)$
$\vec{S} = \frac{\lambda^2 v}{4\pi^2 \epsilon_0 a^2} (\hat{r} \times \hat{\phi})$
Since $\hat{r} \times \hat{\phi} = \hat{k}$ (the unit vector along the $z-$axis),
$\vec{S} = \frac{\lambda^2 v}{4\pi^2 \epsilon_0 a^2} \hat{k}$.