Question

Two long, straight copper pipes, each of radius a, are held a distance 2d apart (see Fig. 7.50). One is at potential V₀, the other at -V₀. The space surrounding the pipes is filled with weakly conducting material of conductivity $\mathit{\sigma }$. Find the current per unit length that flows from one pipe to the other. [Hint: Refer to Prob. 3.12.]

Short answer

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

$i=\frac{4{\mathrm{\pi \sigma V}}_0}{In\left(2{\left({\displaystyle \frac{d}{a}}\right)}^2-1+2{\displaystyle \frac{d}{a}}\sqrt{{\left({\displaystyle \frac{d}{a}}\right)}^2-1}\right)}$

Step-by-step solution

Write the given data from the question.

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 $\sigma$.

Determine the formulas to calculate the current per unit length that flows from one pipe to the other.

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

$\mathit{E}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{s}}\hat{\mathbf{s}}$

Here,is the charge density.

The expression to calculate the potential is given as follows.

$\mathit{V}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathit{I}\mathit{n}\left(\frac{s}{a}\right)$

Calculate the current per unit length that flows from one pipe to the other.

Consider the diagram shown below.

The combination of the potential is given by,

$V=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}In\left(\frac{s_{-}}{s_{+}}\right)$

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

$V(y,z)=\frac{\lambda }{4{\mathrm{\pi \epsilon }}_0}In\left(\frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}\right)$

Calculate the locus of the point of constant V,

role="math" localid="1658754924526" $\begin{array}{rcl}\frac{V\times 4{\mathrm{\pi \epsilon }}_0}{\lambda }& =& In\left(\frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}\right)\\ \frac{{\left(y+b\right)}^2+z^2}{{\left(y-b\right)}^2+z^2}& =& e^{\frac{V\times 4{\mathrm{\pi \epsilon }}_0}{\lambda }}\end{array}$

Lest assume $e^{\frac{V\times 4{\mathrm{\pi \epsilon }}_0}{\lambda }}=\mu$.

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

Solve further as,

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

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

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

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

Now calculates the parameter $b,\mu$and $\lambda$of the image solution.

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

Substitute $\alpha$for $\frac{d}{a}$into above equation.

$\begin{array}{rcl}\alpha & =& \frac{\mu +1}{2\sqrt{\mu }}\\ 2\alpha \sqrt{\mu }& =& \mu +1\end{array}$

Square both the sides of the above equation.

$\begin{array}{rcl}{\left(2\alpha \sqrt{\mu }\right)}^2& =& {\left(\mu +1\right)}^2\\ 4{\alpha }^2\mu & =& {\mu }^2+1+2\mu \\ {\mu }^2+(2-4{\alpha }^2)\mu +1& =& 0\end{array}$

By solving the above equation,

$$\begin{gathered} \mu =\frac{4{\alpha }^2-2\pm \sqrt{4{\left(1-2{\alpha }^2\right)}^2-4}}{2} \\ \mu =2{\alpha }^2-1\pm \sqrt{1-4{\alpha }^2+4{\alpha }^2-1} \\ \mu =2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1} \end{gathered}$$

Solve further as,

$\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }=In\mu$

Substitute $2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1}$for $\mu$into above equation.

role="math" localid="1658758997932" $\begin{array}{rcl}\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{\lambda }& =& In\left(2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1}\right)\\ \lambda & =& \frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{In\left(2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1}\right)}\end{array}$

The line charge in the image problem is given by,

$$\begin{gathered} I=\int J·da \\ I=\sigma \int E·da \\ I=\sigma \frac{1}{{\epsilon }_0}{Q}_{enc} \\ I=\frac{\sigma }{{\epsilon }_0}\lambda {\rm I} \end{gathered}$$

Solve further as,

$\frac{I}{I}=\frac{\sigma }{{\epsilon }_0}\lambda$

Substitute $\begin{array}{rcl}& & \frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{In\left(2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1}\right)}\end{array}$for $\lambda$into above equation.

$\frac{I}{I}=\frac{\sigma }{{\epsilon }_0}\left(\begin{array}{rcl}& & \frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{V}}_0}{In\left(2{\alpha }^2-1\pm 2\alpha \sqrt{{\alpha }^2-1}\right)}\end{array}\right)$

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

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

$\begin{array}{rcl}2{\alpha }^2-1\pm 2{\alpha }^2\sqrt{1-1/{\alpha }^2}& =& 2{\alpha }^2-1\pm 2{\alpha }^2\left(1-\frac{1}{2{\alpha }^2}-\frac{1}{8{\alpha }^2}+....\right)\\ 2{\alpha }^2-1\pm 2{\alpha }^2\sqrt{1-1/{\alpha }^2}& =& 2{\alpha }^2(1\pm 1)-(1\pm 1)\mp \frac{1}{4{\alpha }^2}\pm .....\\ 2{\alpha }^2-1\pm 2{\alpha }^2\sqrt{1-1/{\alpha }^2}& =& \left\{\begin{array}{l}4{\alpha }^2-2-1/2{\alpha }^2=..........\approx 4{\alpha }^2\left(+sign\right)\\ -1/4{\alpha }^2\left(-sign\right)\end{array}\right.\end{array}$

The current must be decreasing with increasing $\alpha$, so therefore,

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

Substitute $\frac{d}{a}$for $\alpha$into above equation.

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

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

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