An infinitely long thin wire carrying a uniform linear static charge density $\lambda $ is placed along the $z-$ axis (figure). The wire is set into motion along its length with a uniform velocity $V = v{\hat k_z}$. Calculate the pointing vector $S = \frac{1}{{{\mu _0}}}(\vec E \times \vec B)$ .
An infinitely long thin wire carrying a uniform linear static charge density $\lambda $ is placed along the $z-$ axis (figure). The wire is set into motion along its length with a uniform velocity $V = v{\hat k_z}$. Calculate the pointing vector $S = \frac{1}{{{\mu _0}}}(\vec E \times \vec B)$ .
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,
$\mathrm{S}=\frac{1}{\mu_{0}}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}})$
$\therefore \mathrm{S}=\frac{1}{\mu_{0}}\left[\frac{\lambda}{2 \pi a} \hat{j} \times \frac{\mu_{0} \lambda v}{2 \pi a} \hat{i}\right]$
$=\frac{1}{\mu_{0}}\left(\frac{\lambda}{2 \pi a} \times \frac{\mu_{0} \lambda v}{2 \pi a}\right)(\hat{j} \times \hat{i})$
$=\frac{\lambda^{2} v}{4 \pi^{2} \in_{0} a^{2}}(-\hat{k}) \quad[\because \hat{j} \times \hat{i}=-\hat{k}]$
$\therefore \mathrm{S}=-\frac{\lambda^{2} v}{4 \pi^{2} \in_{0} a^{2}} \hat{k}$
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