1
$\begingroup$

A long round conductor of cross sectional area $S$ is made of material whose resistivity depends only on a distance $r$ from axis of the conductor $\rho=\frac{\alpha}{r^2}$, where $\alpha $ is constant .Find the resistance per unit length of such conductor .

My work enter image description here

$$dR=\frac{\rho l}{2\pi r dr} \space \text{or}\space dR=\frac{\rho dr}{\pi r^2} $$ $$dR=\frac{\alpha l}{2\pi r^3 dr} \space \text{or}\space dR=\frac{\alpha dr}{\pi r^4} $$ $$\int\frac{1}{dR}=\int_0^{a}\frac{2\pi r^3dr}{\alpha l} \space \text{or} \space \frac{1}{dR}=\frac{\pi r^4}{\alpha dr}$$ $$\frac{1}{R}=\frac{\pi a^2 . a^2}{2\alpha l} \space \text{or} \space \frac{1}{R}=\frac{4\pi a^3}{\alpha}$$ Now, question arises that how should I take length of conductor ? Should I take length of cylinder as length of conductor or $dr$ as length of conductor ? This all depends on how current move but question didn't specify anything about how current move .

$\endgroup$
2
$\begingroup$

I think you have set out the problem incorrectly from the start: Instead of working with the resistance of the shell between $r$ and $r+dr$ and length of material $l$, first work with the conductance, $G(r)$, which is the reciprocal of the resistance. Because the shells are "in parallel" with each other, conductance's add: The incremental conductance due to the shell of radius $r$ and $r+dr$ is $dG(r) = \frac{2 \pi r dr}{\rho l}$, where $\rho$ is the resistivity and given as $\rho = \frac{\alpha}{r^2}$. This gives $dG(r) = \frac{2 \pi r dr}{l \alpha/r^2}$. You can then determine the total conductance as $G = \int_0^R dr \frac{2 \pi r}{l \alpha/r^2}$, where $R$ is the radius of the wire. From this the resistance, $R_{eff} = 1/G$ follows. Then write $R_{eff} = \frac{\rho_{eff} l}{\pi R^2}$.

$\endgroup$
  • $\begingroup$ How you predicted length of conductor ? $\endgroup$ – Aakash Kumar Aug 18 '16 at 2:58
  • $\begingroup$ You are asked to find the resistance per unit length, so with the value of $R_{eff}$ found you don't need to know the explicit length of the conductor. $\endgroup$ – jim Aug 18 '16 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.