91Ó°ÊÓ

Write the expression for the electric field strength.

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{2}\mathbf{λ}}{\mathbf{x}}\stackrel{\mathbf{Áåž}}{\mathbf{x}}$ .....(1)

Here, ${\mathrm{Î\mu }}_o$ is the permittivity of free space, $\mathrm{λ}$ is the charge density, and x is the distance from the wire.

Write the expression for the magnetic field strength.

$$\begin{gathered} \mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\mathbf{2}\mathbf{λν}}{\mathbf{x}}\hat{\mathbf{y}} \\ \mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{2}\mathbf{π}}\frac{\mathbf{λν}}{\mathbf{x}}\hat{\mathbf{y}} \end{gathered}$$

It is known that:

$\frac{1}{{\mathrm{Î\mu }}_0}={\mathrm{c}}^2{\mathrm{μ}}_0$

Substitute$\frac{1}{{\mathrm{Î\mu }}_0}={\mathrm{c}}^2{\mathrm{μ}}_0$ in equation (1).

$$\begin{gathered} \mathrm{E}=\frac{1}{4\mathrm{Ï€}\left({\displaystyle \frac{1}{{\mathrm{c}}^2{\mathrm{μ}}_0}}\right)}\frac{2\mathrm{λ}}{\mathrm{x}}\hat{\mathrm{x}} \\ \mathrm{E}=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\frac{{\mathrm{볦}}^2}{\mathrm{x}}\hat{\mathrm{x}} \end{gathered}$$

Write the equation for filed tensor as an array form.

${\mathrm{F}}^{\mathrm{μν}}=\left.\left\{\begin{array}{cc}0& {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ -{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& 0\\ -{\mathrm{E}}_{\mathrm{y}}/\mathrm{c}& -{\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{E}}_{\mathrm{z}}/\mathrm{c}& {\mathrm{B}}_{\mathrm{y}}\end{array}\right.\begin{array}{cc}{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ {\mathrm{B}}_{\mathrm{z}}& -{\mathrm{B}}_{\mathrm{y}}\\ 0& {\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{B}}_{\mathrm{x}}& 0\end{array}\right\}\begin{array}{}\\ \\ \\ \end{array}$

Substitute ${\mathrm{B}}_{\mathrm{x}}=0,{\mathrm{B}}_{\mathrm{z}}=0,{\mathrm{E}}_{\mathrm{y}}=0,{\mathrm{E}}_{\mathrm{z}}=0{\mathrm{E}}_{\mathrm{x}}=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\frac{{\mathrm{볦}}^2}{\mathrm{x}}\mathrm{and}{\mathrm{B}}_{\mathrm{y}}=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\frac{\mathrm{λ\pm ¹}}{\mathrm{x}}$in equation (2).

$$\begin{gathered} \mathrm{¹óμν}=\left(\begin{array}{cccc}0& \left(\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{{\displaystyle \frac{2\mathrm{Ï€}\mathrm{x}}{\mathrm{c}}}}\right)& 0& 0\\ -\left(\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{{\displaystyle \frac{2\mathrm{Ï€}\mathrm{x}}{\mathrm{c}}}}\right)& 0& 0& 0\\ 0& 0& 0& 0\\ 0& \left(\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{{\displaystyle 2\mathrm{Ï€}\mathrm{x}}}\right)& 0& 0\end{array}\right) \\ {\mathrm{F}}^{\mathrm{μν}}=\frac{{\mathrm{μ}}_0\mathrm{λ}}{{\displaystyle 2\mathrm{Ï€}\mathrm{x}}}\left(\begin{array}{cccc}0& \mathrm{c}& 0& 0\\ -\mathrm{c}& 0& 0& -\mathrm{v}\\ 0& 0& 0& 0\\ 0& \mathrm{v}& 0& 0\end{array}\right) \end{gathered}$$

Write the equation for the dual tensor as an array form.

$G^{\mathrm{μν}}=\left.\left\{\begin{array}{cc}0& {\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{B}}_{\mathrm{x}}& 0\\ -{\mathrm{B}}_{\mathrm{y}}& {\mathrm{E}}_z/\mathrm{c}\\ -{\mathrm{B}}_{\mathrm{z}}& {\mathrm{E}}_{\mathrm{y}}/\mathrm{c}\end{array}\right.\begin{array}{cc}{\mathrm{B}}_{\mathrm{y}}& {\mathrm{B}}_{\mathrm{z}}\\ -{\mathrm{E}}_{\mathrm{z}}/\mathrm{c}& {\mathrm{E}}_{\mathrm{y}}/\mathrm{c}\\ 0& -{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& 0\end{array}\right\}\begin{array}{}\\ \\ \\ \end{array}$..........(3)

Substitute $0\mathrm{for}{\mathrm{B}}_{\mathrm{z}},0\mathrm{for}{\mathrm{B}}_{\mathrm{x}},0\mathrm{for}{\mathrm{E}}_{\mathrm{y}},,0\mathrm{for}{\mathrm{E}}_{\mathrm{z}},\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{2\mathrm{Ï€}\mathrm{x}}\mathrm{for}{\mathrm{E}}_{\mathrm{x}}\mathrm{and}\frac{{\mathrm{μ}}_0\mathrm{볦}}{2\mathrm{Ï€}\mathrm{x}}\mathrm{for}{\mathrm{B}}_{\mathrm{y}}$in equation (3).

role="math" localid="1654682762200" $$\begin{gathered} {\mathrm{G}}^{\mathrm{μν}}=\left\{\begin{array}{cc}0& 0\\ 0& 0\\ -\left(\frac{{\mathrm{μ}}_0\mathrm{λν}}{2\mathrm{Ï€}\mathrm{x}}\right)& 0\\ 0& 0\end{array}\right.\left.\begin{array}{cc}\left(\frac{{\mathrm{μ}}_0\mathrm{λν}}{2\mathrm{Ï€}\mathrm{x}}\right)& 0\\ 0& 0\\ 0& -\left(\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{{\displaystyle \frac{2\mathrm{Ï€}\mathrm{x}}{\mathrm{c}}}}\right)\\ \left(\frac{{\mathrm{μ}}_0{\mathrm{볦}}^2}{{\displaystyle \frac{2\mathrm{Ï€}\mathrm{x}}{\mathrm{c}}}}\right)& 0\end{array}\right\} \\ {\mathrm{G}}^{\mathrm{μν}}=\left\{\begin{array}{cc}0& 0\\ 0& 0\\ -\mathrm{v}& 0\\ 0& \mathrm{v}\end{array}\right.\left.\begin{array}{cc}0& 0\\ 0& 0\\ 0& -\mathrm{c}\\ 0& 0\end{array}\right\} \end{gathered}$$

Therefore, the field tensor and the dual tensor at the point $\left(\mathrm{x},0,0\right)$is

${\mathrm{F}}^{\mathrm{μν}}=\frac{{\mathrm{μ}}_0\mathrm{λ}}{2\mathrm{Ï€³\ae }}\left(\begin{array}{cccc}0& \mathrm{c}& 0& 0\\ -\mathrm{c}& 0& 0& -\mathrm{v}\\ 0& 0& 0& 0\\ 0& \mathrm{v}& 0& 0\end{array}\right)\mathrm{and}{\mathrm{G}}^{\mathrm{μν}}=\frac{{\mathrm{μ}}_0\mathrm{λ}}{2\mathrm{Ï€³\ae }}\left(\begin{array}{cccc}0& 0& \mathrm{v}& 0\\ 0& 0& 0& 0\\ -\mathrm{v}& 0& 0& -\mathrm{c}\\ 0& 0& \mathrm{c}& 0\end{array}\right)\mathrm{respectivety}$