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 .


1 Answer 1


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 $R_{eff} = \frac{\rho_{eff} l}{\pi R^2}$.

  • $\begingroup$ How you predicted length of conductor ? $\endgroup$ Commented Aug 18, 2016 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
    Commented Aug 18, 2016 at 8:52

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