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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 $\mathrm{v}=v \hat{k}_{\mathrm{z}}$. Calculate the pointing vector
$$
S=\frac{1}{\mu_0}(E \times B)
$$
$$
S=\frac{1}{\mu_0}(E \times B)
$$
Solution:
2239 Upvotes
Verified Answer
Let us consider a cylindrical surface in such a way that the axis of cyclinder lies on wire. So electric field intensity due to long straight wire at a distance $a$ and charge density $\lambda \mathrm{c} / \mathrm{m}$.
$$
\begin{aligned}
\mathrm{E} &=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{a}}(\hat{\mathrm{j}}) \\
\mathrm{B}=& \frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}} \hat{\mathrm{i}}=\frac{\mu_0 \lambda \mathrm{v}}{2 \pi \mathrm{a}}(\hat{\mathrm{i}}) \quad\left[\because \mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\lambda \mathrm{l}}{\mathrm{t}}=\lambda \mathrm{v}\right] \\
\therefore \quad \mathrm{S} &=\frac{1}{\mu_0}[\mathrm{E} \times \mathrm{B}]=\frac{1}{\mu_0}\left[\frac{\lambda(\hat{\mathrm{j}})}{2 \pi \varepsilon_0 \mathrm{a}} \times \frac{\mu_0}{2 \pi \mathrm{a}} \lambda \mathrm{v}(\hat{\mathrm{i}})\right] \\
&=\frac{\lambda^2 \mathrm{v}}{4 \pi^2 \varepsilon_0 \mathrm{a}^2}(\mathrm{j} \times \hat{\mathrm{i}}) \\
\mathrm{S} &=-\frac{\lambda^2 \mathrm{v}}{4 \pi^2 \varepsilon_0 \mathrm{a}^2} \hat{\mathrm{k}}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{E} &=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{a}}(\hat{\mathrm{j}}) \\
\mathrm{B}=& \frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}} \hat{\mathrm{i}}=\frac{\mu_0 \lambda \mathrm{v}}{2 \pi \mathrm{a}}(\hat{\mathrm{i}}) \quad\left[\because \mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\lambda \mathrm{l}}{\mathrm{t}}=\lambda \mathrm{v}\right] \\
\therefore \quad \mathrm{S} &=\frac{1}{\mu_0}[\mathrm{E} \times \mathrm{B}]=\frac{1}{\mu_0}\left[\frac{\lambda(\hat{\mathrm{j}})}{2 \pi \varepsilon_0 \mathrm{a}} \times \frac{\mu_0}{2 \pi \mathrm{a}} \lambda \mathrm{v}(\hat{\mathrm{i}})\right] \\
&=\frac{\lambda^2 \mathrm{v}}{4 \pi^2 \varepsilon_0 \mathrm{a}^2}(\mathrm{j} \times \hat{\mathrm{i}}) \\
\mathrm{S} &=-\frac{\lambda^2 \mathrm{v}}{4 \pi^2 \varepsilon_0 \mathrm{a}^2} \hat{\mathrm{k}}
\end{aligned}
$$
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