Q7.41P - Introduction to Electrodynamics

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.]

Step-by-Step Solution

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 1: 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 $\sigma$.

Step 2: 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.

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

Step 3: 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,

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

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.

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

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,

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

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

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

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

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