Question

A $0.60$-mm-diameter wire made from an alloy (a combination of different metals) has a conductivity that decreases linearly with distance from the center of the wire: $\sigma \left(r\right)={\sigma }_0-cr$, with ${\sigma }_0=5.0\times {10}^7{\Omega }^{-1}{\mathrm{m}}^{-1}$and $c=1.2\times {10}^{11}{\Omega }^{-1}{\mathrm{m}}^{-2}$. What is the resistance of a $4.0\mathrm{m}$length of this wire?

Short answer

The resistance of a $4.0m$length of this wire is $0.109m\mathrm{\Omega }$.

Step-by-step solution

Given information

A $0.60$-mm-diameter wire has a conductivity that decreases linearly with distance from the center of the wire: $\sigma \left(r\right)={\sigma }_0-cr$, with ${\sigma }_0=5.0\times {10}^7{\Omega }^{-1}{\mathrm{m}}^{-1}$and $c=1.2\times {10}^{11}{\Omega }^{-1}{\mathrm{m}}^{-2}$.

Explanation

The conductance of the small part of the wire is,

$dG=\frac{\sigma dA}{L}$

Here, $\sigma$is the conductivity of the wire, is the length of the wire, and $dA$is the area of the wire.

As the wire moves toward the center of wire, its conductivity decreases linearly as follows:

$\sigma \left(r\right)={\sigma }_0-cr$

Here, ${\sigma }_0$and $c$are constants and $r$is the distance of the wire from the center of the wire.

Substitute${\sigma }_0-cr$for $\sigma \left(r\right)$in the equation $dG=\frac{\sigma dA}{L}$and solve for $dG.$

$dG=\frac{\left({\sigma }_0-cr\right)dA}{L}$

The area of the wire is,

$A=\pi r^2$

Differentiate the equation on both sides,

$dA=2\pi rdr$

Substitute $2\pi rdr$for $dA$in the equation $dG=\frac{\left({\sigma }_0-cr\right)dA}{L}$and solve for $dG,$.

$dG=\frac{\left({\sigma }_0-cr\right)2\pi rdr}{L}$

Integrate the equation on both sides,

$\begin{array}{r}G={\int }_0^r \frac{\left({\sigma }_0-cr\right)2\pi rdr}{L}\end{array}$

$\begin{array}{r}=\frac{2\pi }{L}{\int }_0^r \left({\sigma }_{0}r-c{r}^2\right)dr\end{array}$

role="math" localid="1649160599266" $=\frac{2\pi }{L}{\left(\frac{{\sigma }_0{r}^2}{2}-\frac{c{r}^3}{3}\right)}_0^r$

$=\frac{2\pi }{L}\left(\frac{{\sigma }_0{r}^2}{2}-\frac{c{r}^3}{3}\right)$

Explanation

The relation between the conductance and resistance is,

$R=\frac{1}{G}$

Here, $R$is the resistance of the wire.

Substitute $\frac{2\pi }{L}\left(\frac{{\sigma }_0{r}^2}{2}-\frac{c{r}^3}{3}\right)$for $G,$

localid="1649160822320" $R=\frac{1}{\frac{2\pi }{L}\left(\frac{{\sigma }_0{r}^2}{2}-\frac{c{r}^3}{3}\right)}\to \left(1\right)$

The relation between diameter and resistance of the wire is,

$r=\frac{d}{2}$

Here, $d$is the diameter of the wire.

Substitute $0.60\mathrm{mm}$for $d,$

$r=\frac{0.60\mathrm{mm}}{2}$

$=0.30\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)$

$=0.30\times {10}^{-3}\mathrm{m}$

Substitute the values in equation (1)

$R=\frac{1}{(4.0\mathrm{m})}\left(\frac{\left(5.0\times {10}^7{\Omega }^{-1}{\mathrm{m}}^{-1}\right){\left(0.30\times {10}^{-3}\mathrm{m}\right)}^2}{2}-\frac{\left(1.2\times {10}^{11}{\Omega }^{-1}{\mathrm{m}}^{-2}\right){\left(0.30\times {10}^{-3}\mathrm{m}\right)}^3}{3}\right)$

$=0.54\Omega$

Final Answer

Therefore, the resistance of the $40\mathrm{m}$ length wire is $0.54\Omega$.