The problem reads:
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 $\vec{S}$ = $\frac{1}{\mu_o}$ $(\vec{E} \times \vec{B)}$ at a distance $a$ from the wire. (Figure 8.1, attached)
Here is my attempt:
For the sake of simplicity, let me consider the fields at a point on the y-z plane. The electric field caused by the wire at a point on the axis will be given by :
$$\vec{E} = \frac{\lambda}{2\pi\varepsilon_0a}\hat{j}$$
Also, as the line charge is now translating in the positive z direction, there is also a current. Considering any cross section of the wire, in a time interval $dt$ a charge equivalent to $\lambda vdt$ would have crossed it, hence the current should be $i=\lambda v$ and the magnetic field due to this wire at the same point would be, $$\vec{B} = -\frac{\mu_0 \lambda v}{2\pi a}\hat{i}$$
Hence the Poynting vector would be, $$\vec{S} = \frac{1}{\mu_o}(\vec{E} \times \vec{B)} = +\frac{\mu_0 \lambda v^2}{4\pi^2 a^2 \varepsilon_0}\hat{k}$$
The given answer is however,
$$\vec{S} = -\frac{\mu_0 \lambda v^2}{4\pi^2 a^2 \varepsilon_0}\hat{k}$$
What has caused this error of sign? Have I misrepresented any directions of the electric/magnetic fields? Help would be appreciated.
(Also, I am quite new to poynting vectors so please do consider that when answering!)
