Chapter 25 Problem 30

Problem

A hollow cylindrical resistor with inner radius $r_{1}$ and outer radius $r_{2}$ , and length $l$ , is made of a material whose resistivity is $\rho$

(a) Show that the resistance is given by

$R={\frac {\rho }{2\pi l}}\ln {\frac {r_{2}}{r_{1}}}$

for current that flows radially outward. [Hint: Divide the resistor into concentric cylindrical shells and integrate.]

(b) Evaluate the resistance $R$ for such a resistor made of carbon whose inner and outer radii are 1.0 mm and 1.8 mm and whose length is 2.4 cm (Choose $\rho =15\times 10^{-5}\Omega \cdot m$.)

(c) What is the resistance in part (b) for current flowing parallel to the axis?

Solution

(a)

$dR=\rho {\frac {dr}{2\pi rl}}$ $R=\int dR=\int _{r_{1}}^{r_{2}}\rho {\frac {dr}{2\pi rl}}={\frac {\rho }{2\pi l}}\ln {\frac {r_{2}}{r_{1}}}$

(b)

$R={\frac {15\times 10^{-5}\Omega \cdot {\textrm {m}}}{2\pi (0.024{\textrm {m}})}}\ln \left({\frac {1.8{\textrm {mm}}}{1.0{\textrm {mm}}}}\right)=5.8\times 10^{-4}\Omega$

(c)

$R={\frac {\rho l}{A}}={\frac {\rho l}{\pi (r_{2}^{2}-r_{1}^{2})}}={\frac {(15\times 10^{-5}\Omega \cdot {\textrm {m}})(0.024{\textrm {m}})}{\pi \left[(1.8\times 10^{-3}{\textrm {m}})^{2}-(1.0\times 10^{-3}{\textrm {m}})^{2}\right]}}=0.51\Omega$