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  • 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 v=v hat{k}_{z} Calculate the pointing vectors

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  v=v\hat{k}_{z} Calculate the pointing vectors S=\frac{1}{\mu _{0}}\left ( E \times B \right ).

 

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Explanation:-

The electric field in an infinitely long thin wire is

\overrightarrow{E}=\frac{\lambda \hat{e}_{s}}{2\pi \epsilon _{0}a}\hat{j}

 The magnetic field due to a long, straight current-carrying wire can be found using Ampere's Law. The magnetic field circles the wire, and at a distance r from the wire, the Magnetic field due to the wire is

\overrightarrow{B}=\frac{\mu _{0}\lambda v}{2\pi a}\hat{i}

The Poynting vector, which represents the power per unit area flowing away from the wire, will point in the direction of the energy flow. The energy flow is perpendicular to both E\mathbf{E}E and B\mathbf{B}B, and it will be directed radially outward from the wire. The equivalent current flowing through the wire is

\overrightarrow{S}=\frac{\lambda ^{2}v}{4\pi^{2}\epsilon _{0} a^{2}}\hat{k}

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