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Electromagnetic Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Waves3 Marks
Question
An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis. The wire is set into motion along its length with a uniform velocity $\text{v}=\text{v}\hat{\text{k}}_\text{z}$. Calculate the poynting vector $\text{S}=\frac{1}{\mu_0}(\text{E}\times\text{B})$.
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Answer

The electric field due to infinitely long thin wire
$\vec{\text{E}}=\frac{\lambda\hat{\text{e}}_\text{s}}{2\pi\epsilon_0\text{a}}\hat{\text{j}}$
Magnetic field due to the wire, $\vec{\text{B}}=\frac{\mu_0\text{i}}{2\pi\text{a}}\hat{\text{i}}$
Equivalent current flowing through the wire, $\text{i}=\lambda\text{v}$
Hence $\vec{\text{B}}=\frac{\mu_0\lambda\text{v}}{2\pi\text{v}}\hat{\text{i}}$
$\therefore\ \vec{\text{S}}=\frac{1}{\mu_0}\big[\vec{\text{E}}\times\vec{\text{B}}\big]=\frac{1}{\mu_0}\bigg[\frac{\lambda}{2\pi\epsilon_0\text{a}}\hat{\text{j}}\times\frac{\mu_0\lambda\text{v}}{2\pi\text{a}}\hat{\text{i}}\bigg]$
$\Rightarrow\ \vec{\text{S}}=\frac{\lambda^2\text{v}}{4\pi^2\epsilon_0\text{a}^2}(\hat{\text{j}}\times\hat{\text{i}})=-\frac{\lambda^2\text{v}}{4\pi^2\epsilon_0\text{a}^2}\hat{\text{k}}$

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