91Ó°ÊÓ

The radius of the copper pipes is a.

The distance between the two pipes is 2d.

The potential of one pipe is V₀ and another pipe is -V₀.

The conductivity of the material is $\mathrm{σ}$.

The expression to calculate the electric field is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{2}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}\mathbf{s}}\hat{\mathbf{s}}$

Here,is the charge density.

The expression to calculate the potential is given as follows.

$\mathit{V}\mathbf{=}\frac{\mathbf{λ}}{\mathbf{2}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left(\frac{s}{a}\right)$

Consider the diagram shown below.

The combination of the potential is given by,

$V=\frac{\mathrm{λ}}{2{\mathrm{Ï€Î\mu }}_0}In\left(\frac{s_{-}}{s_{+}}\right)$

The combine potential of parallel pipes also can be written as,

$V(y,z)=\frac{\mathrm{λ}}{4{\mathrm{Ï€Î\mu }}_0}In\left(\frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}\right)$

Calculate the locus of the point of constant V,

role="math" localid="1658754924526" $\begin{array}{rcl}\frac{V×4{\mathrm{Ï€Î\mu }}_0}{\mathrm{λ}}& =& In\left(\frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}\right)\\ \frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}& =& e^{\frac{V×4{\mathrm{Ï€Î\mu }}_0}{\mathrm{λ}}}\end{array}$

Lest assume $e^{\frac{V×4{\mathrm{Ï€Î\mu }}_0}{\mathrm{λ}}}=\mathrm{μ}$.

$\begin{array}{rcl}\frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}& =& \mathrm{μ}\\ y^2+b^2+2yb+z^2& =& \mathrm{μ}\left(y^2+b^2-2yb+z^2\right)\\ y^2\left(\mathrm{μ}-1\right)+b^2\left(\mathrm{μ}-1\right)+z^2\left(\mathrm{μ}-1\right)-2yb\left(\mathrm{μ}+1\right)& =& 0\end{array}$

Solve further as,

$y^2+b^2+z^2-2yb\frac{\begin{array}{rcl}& & \left(\mathrm{μ}+1\right)\end{array}}{\begin{array}{rcl}& & \left(\mathrm{μ}-1\right)\end{array}}=0$

Substitute $\mathrm{β}$for $\frac{\begin{array}{rcl}& & \left(\mathrm{μ}+1\right)\end{array}}{\begin{array}{rcl}& & \left(\mathrm{μ}-1\right)\end{array}}$into above equation.

$\begin{array}{rcl}{y}^2+b^2+z^2-2yb\mathrm{β}& =& 0\\ {(y-b\mathrm{β})}^2+z^2+b^2-b^2{\mathrm{β}}^2& =& 0\\ {(y-b\mathrm{β})}^2+z^2& =& b^2\left({\mathrm{β}}^2-1\right)\\ {\left(y-b\frac{\mathrm{μ}+1}{\mathrm{μ}-1}\right)}^2+z^2& =& b^2\left({\left(\frac{\mathrm{μ}+1}{\mathrm{μ}-1}\right)}^2-1\right)\end{array}$

The above equation is circle equation with $y_0=b\left(\frac{\mathrm{μ}+1}{\mathrm{μ}-1}\right)$and radius $\begin{array}{rcl}r& =& b\sqrt{{\left(\frac{\mathrm{μ}+1}{\mathrm{μ}-1}\right)}^2-1}\\ & =& b\sqrt{\frac{{\mathrm{μ}}^2+1+2\mathrm{μ}-({\mathrm{μ}}^2+1-2\mathrm{μ})}{{(\mathrm{μ}-1)}^2}}\\ & =& b\sqrt{\frac{4\mathrm{μ}}{{(\mathrm{μ}-1)}^2}}\\ & =& \frac{2b\sqrt{\mathrm{μ}}}{\mathrm{μ}-1}\end{array}$

Now calculates the parameter $b,\mathrm{μ}$and $\mathrm{λ}$of the image solution.

$$\begin{gathered} \frac{d}{a}=\frac{y_0}{radius} \\ \frac{d}{a}=\frac{d\left({\displaystyle \frac{\mathrm{μ}+1}{\mathrm{μ}-1}}\right)}{{\displaystyle \frac{2b\sqrt{\mathrm{μ}}}{\mathrm{μ}-1}}} \\ \frac{d}{a}=\frac{\mathrm{μ}+1}{2\sqrt{\mathrm{μ}}} \end{gathered}$$

Substitute $\mathrm{Î\pm }$for $\frac{d}{a}$into above equation.

$\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{\mathrm{μ}+1}{2\sqrt{\mathrm{μ}}}\\ 2\mathrm{Î\pm }\sqrt{\mathrm{μ}}& =& \mathrm{μ}+1\end{array}$

Square both the sides of the above equation.

$\begin{array}{rcl}{\left(2\mathrm{Î\pm }\sqrt{\mathrm{μ}}\right)}^2& =& {\left(\mathrm{μ}+1\right)}^2\\ 4{\mathrm{Î\pm }}^2\mathrm{μ}& =& {\mathrm{μ}}^2+1+2\mathrm{μ}\\ {\mathrm{μ}}^2+(2-4{\mathrm{Î\pm }}^2)\mathrm{μ}+1& =& 0\end{array}$

By solving the above equation,

$$\begin{gathered} \mathrm{μ}=\frac{4{\mathrm{Î\pm }}^2-2Â\pm \sqrt{4{\left(1-2{\mathrm{Î\pm }}^2\right)}^2-4}}{2} \\ \mathrm{μ}=2{\mathrm{Î\pm }}^2-1Â\pm \sqrt{1-4{\mathrm{Î\pm }}^2+4{\mathrm{Î\pm }}^2-1} \\ \mathrm{μ}=2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1} \end{gathered}$$

Solve further as,

$\frac{4{\mathrm{Ï€Î\mu }}_0{\mathrm{V}}_0}{\mathrm{λ}}=In\mathrm{μ}$

Substitute $2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}$for $\mathrm{μ}$into above equation.

role="math" localid="1658758997932" $\begin{array}{rcl}\frac{4{\mathrm{Ï€Î\mu }}_0{\mathrm{V}}_0}{\mathrm{λ}}& =& In\left(2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}\right)\\ \mathrm{λ}& =& \frac{4{\mathrm{Ï€Î\mu }}_0{\mathrm{V}}_0}{In\left(2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}\right)}\end{array}$

The line charge in the image problem is given by,

$$\begin{gathered} I=∫J·da \\ I=\mathrm{σ}∫E·da \\ I=\mathrm{σ}\frac{1}{{\mathrm{Î\mu }}_0}{Q}_{enc} \\ I=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\mathrm{λ}\mathrm{Ι} \end{gathered}$$

Solve further as,

$\frac{I}{I}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\mathrm{λ}$

Substitute $\begin{array}{rcl}& & \frac{4{\mathrm{Ï€Î\mu }}_0{\mathrm{V}}_0}{In\left(2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}\right)}\end{array}$for $\mathrm{λ}$into above equation.

$\frac{I}{I}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\left(\begin{array}{rcl}& & \frac{4{\mathrm{Ï€Î\mu }}_0{\mathrm{V}}_0}{In\left(2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}\right)}\end{array}\right)$

Let assume the distance between the pipes is very large as compare to the radius.

d > > a so that, a > > 1.

$\begin{array}{rcl}2{\mathrm{Î\pm }}^2-1Â\pm 2{\mathrm{Î\pm }}^2\sqrt{1-1/{\mathrm{Î\pm }}^2}& =& 2{\mathrm{Î\pm }}^2-1Â\pm 2{\mathrm{Î\pm }}^2\left(1-\frac{1}{2{\mathrm{Î\pm }}^2}-\frac{1}{8{\mathrm{Î\pm }}^2}+....\right)\\ 2{\mathrm{Î\pm }}^2-1Â\pm 2{\mathrm{Î\pm }}^2\sqrt{1-1/{\mathrm{Î\pm }}^2}& =& 2{\mathrm{Î\pm }}^2(1Â\pm 1)-(1Â\pm 1)∓\frac{1}{4{\mathrm{Î\pm }}^2}Â\pm .....\\ 2{\mathrm{Î\pm }}^2-1Â\pm 2{\mathrm{Î\pm }}^2\sqrt{1-1/{\mathrm{Î\pm }}^2}& =& \left\{\begin{array}{l}4{\mathrm{Î\pm }}^2-2-1/2{\mathrm{Î\pm }}^2=..........≈4{\mathrm{Î\pm }}^2\left(+sign\right)\\ -1/4{\mathrm{Î\pm }}^2\left(-sign\right)\end{array}\right.\end{array}$

The current must be decreasing with increasing $\mathrm{Î\pm }$, so therefore,

role="math" localid="1658759243647" $\begin{array}{rcl}i& =& \frac{4\mathrm{Ï€}\mathrm{σ}{\mathrm{V}}_0}{In\left(2{\mathrm{Î\pm }}^2-1Â\pm 2\mathrm{Î\pm }\sqrt{{\mathrm{Î\pm }}^2-1}\right)}\end{array}$

Substitute $\frac{d}{a}$for $\mathrm{Î\pm }$into above equation.

role="math" localid="1658759264126" $\begin{array}{rcl}i& =& \frac{4\mathrm{Ï€}\mathrm{σ}{\mathrm{V}}_0}{In\left(2{\left({\displaystyle \frac{d}{a}}\right)}^2-1Â\pm 2{\displaystyle \frac{d}{a}}\sqrt{{\left({\displaystyle \frac{d}{a}}\right)}^2-1}\right)}\end{array}$

Hence the current per unit length that flows from one pipe to the other is

$\begin{array}{rcl}i& =& \frac{4\mathrm{πσ}{V}_0}{In\left(2{\left({\displaystyle \frac{d}{a}}\right)}^2-1Â\pm 2{\displaystyle \frac{d}{a}}\sqrt{{\left({\displaystyle \frac{d}{a}}\right)}^2-1}\right)}\end{array}$