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 $\vec{v} = v \hat{k}$ . Find the poynting vector at $a$ distance a from the wire..

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$\frac{- λ^{2} v}{4 π^{2} ε_{0} a^{2}} \hat{i}$

$\frac{- λ^{2} v}{4 π^{2} ε_{0} a^{2}} \hat{k}$

$\frac{λ^{2} v}{4 π^{2} ε_{0} a^{2}} \hat{i}$

$\frac{λ^{2} v}{4 π^{2} ε_{0} a^{2}} \hat{k}$

answer is C.

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Detailed Solution

We know that

$\vec{B} = - \frac{μ_{0} I}{2 πa} \hat{i} = - \frac{μ_{0}}{2 π} \frac{λv}{a} \hat{i} [λ = \frac{q}{l} and I = \frac{q}{t} = \frac{λ l}{t} = λv]; \vec{E} = \frac{λ \hat{j}}{2 {πε}_{0} a}$

$\vec{S} = \frac{1}{μ_{0}} [\vec{E} \times \vec{B}] \vec{S} = \frac{1}{μ_{0}} [\frac{λ}{2 {πε}_{0} a} \hat{j} \times \frac{μ_{0}}{2 πa} λv (- \hat{i})] = \frac{λ^{2} v}{4 π^{2} ε_{0} a^{2}} (\hat{i} \times \hat{j}) = \frac{λ^{2} v}{4 π^{2} ε_{0} a^{2}} \hat{k}$