Chapter 28, Problem 82PQ

A conducting material with resistivity ρ is shaped into a wire of length $\ell$( with a tapered cross section that decreases from radius r₁ on the left end to r₂ on its right end (Fig. F28.82). If the current density in the wire is a constant as a function of the horizontal distance x along the wire, what is the resistance of this tapered wire?

To determine

The resistance of the tapered wire.

Answer to Problem 82PQ

The resistance of the tapered wire is $\underset{\_}{R=\frac{\rho }{\pi }\left(\frac{l}{r_1{r}_2}\right)}$.

Explanation of Solution

The diagram of the tapered wire is given in Figure P28.82. The length of the wire is $l$, the radii of the two end cross sections of the wire is $r_1$ and $r_2$ ($r_1>r_2$). The horizontal distance along the wire is represented as $x$. The current density in the wire is constant as a function of $x$.

From the geometry of the longitudinal section of the wire, it can be deduced that,

  $\frac{\left(r_1-r\right)}{x}=\frac{\left(r_1-r_2\right)}{l}$                                                                                                  (I)

Here, $r$ is the radius at a distance $x$.

From equation (I), write the expression for the radius at a distance $x$ from the left end of the wire.

  $r=r_1-\left(r_1-r_2\right)\frac{x}{l}$                                                                                                  (II)

Write the general expression for the resistance of the wire.

  $R=\rho \frac{l}{A}$                                                                                                               (III)

Here, $R$ is the resistance, $\rho$ is the resistivity, and $A$ is the cross-sectional area of the wire.

Since the cross-sectional area of the wire changes as the distance $x$ changes, for a disk shaped element of this wire with radius $r$ and thickness $dx$, equation (II) can be rewritten as,

  $dR=\rho \frac{dx}{\pi r^2}$                                                                                                         (IV)

Use equation (II) in (IV).

  $dR=\frac{\rho }{\pi }\frac{dx}{{\left[r_1-\left(r_1-r_2\right)\left(\frac{x}{l}\right)\right]}^2}$                                                                                  (V)

Integrate equation (V) with the limits $x=0$ to $x=l$ to find the resistance of the wire.

  $\begin{array}{c}R=\frac{\rho }{\pi }{\displaystyle \underset{0}{\overset{l}{\int }}\frac{dx}{{\left[r_1-\left(r_1-r_2\right)\left(\frac{x}{l}\right)\right]}^2}}\\ =\frac{\rho }{\pi }{\displaystyle \underset{0}{\overset{l}{\int }}\frac{dx}{{\left[\left(\frac{r_2-r_1}{l}\right)x+r_1\right]}^2}}\end{array}$                                                                                (VI)

Take $y=\left(\frac{r_2-r_1}{l}\right)x+r_1$, so that $dy=\left(\frac{r_2-r_1}{l}\right)dx$. Thus, $dx=\frac{dy}{\left(\frac{r_2-r_1}{l}\right)}$. Use this in the integral (VI)

  $\begin{array}{c}R=\frac{\rho }{\pi }\frac{1}{\left(\frac{r_2-r_1}{l}\right)}{\displaystyle \int \frac{dy}{y^2}}\\ =\frac{\rho }{\pi }\frac{1}{\left(\frac{r_2-r_1}{l}\right)}\left(\frac{-1}{y}\right)\end{array}$                                                                                          (VII)

Use $y=\left(\frac{r_2-r_1}{l}\right)x+r_1$ in (VII) and apply the limits.

  $\begin{array}{c}R=\left(\frac{\rho }{\pi }\right){\left[\frac{-1}{\left(\frac{r_2-r_1}{l}\right)\left[\left(\frac{r_2-r_1}{l}\right)x+r_1\right]}\right]|}_{x=0}^{x=l}\\ =-\left(\frac{\rho }{\pi }\right){\left[\frac{1}{\left[{\left(\frac{r_2-r_1}{l}\right)}^{2}x+\left(\frac{r_2-r_1}{l}\right)r_1\right]}\right]|}_{x=0}^{x=l}\\ =-\left(\frac{\rho }{\pi }\right)\left[\frac{1}{\left[{\left(\frac{r_2-r_1}{l}\right)}^{2}l+\left(\frac{r_2-r_1}{l}\right)r_1\right]}-\frac{1}{\left(\frac{r_2-r_1}{l}\right)r_1}\right]\\ =-\left(\frac{\rho }{\pi }\right)\left[\frac{1}{\left[\frac{{\left(r_2-r_1\right)}^2}{l}+\left(\frac{r_2-r_1}{l}\right)r_1\right]}-\frac{1}{\left(\frac{r_2-r_1}{l}\right)r_1}\right]\end{array}$                                        (VIII)

Reduce equation (VIII).

  $\begin{array}{c}R=-\left(\frac{\rho }{\pi }\right)\frac{1}{\left(\frac{r_2-r_1}{l}\right)}\left(\frac{1}{r_2-r_1+r_1}-\frac{1}{r_1}\right)\\ =-\left(\frac{\rho }{\pi }\right)\frac{1}{\left(\frac{r_2-r_1}{l}\right)}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ =\left(\frac{\rho }{\pi }\right)\left(\frac{l}{r_2-r_1}\right)\left(\frac{r_2-r_1}{r_1{r}_2}\right)\\ =\frac{\rho }{\pi }\left(\frac{l}{r_1{r}_2}\right)\end{array}$

Conclusion:

Therefore, the resistance of the tapered wire is $\underset{\_}{R=\frac{\rho }{\pi }\left(\frac{l}{r_1{r}_2}\right)}$.