Question

A $2.0\mathrm{m}$long, $1\mathrm{mm}$diameter wire has a variable resistivity given by

localid="1649324855366" $\mathrm{\rho }\left(\mathrm{x}\right)=\left(2.5\times {10}^{-6}\right)\left[1+{\left(\frac{\mathrm{x}}{1.0}\right)}^2\right]\Omega \mathrm{m}$

where x is measured from one end of the wire. What is the current if this wire is connected to the terminals of a $9.0\mathrm{V}$battery?

Short answer

The current in the wire will be$0.6\mathrm{A}.$

Step-by-step solution

Given information

We have given

Length of the wire = $2\mathrm{m}$

Diameter of the wire = $1\mathrm{mm}$

Resistivity $\mathrm{\rho }\left(\mathrm{x}\right)=\left(2.5\times {10}^{-6}\right)\left[1+{\left(\frac{\mathrm{x}}{1.0}\right)}^2\right]\Omega \mathrm{m}$

Voltage = $9\mathrm{V}$

We have to find the current in the wire.

Simplify

We know the resistance

$\mathrm{R}=\mathrm{\rho }\frac{\mathrm{L}}{\mathrm{A}}$

where area will be

$\mathrm{A}=\mathrm{\pi }\frac{{\mathrm{D}}^2}{4}=\mathrm{\pi }\frac{{10}^{-6}}{4}=7.85\times {10}^{-7}{\mathrm{m}}^2$

We can see here the length is $2\mathrm{m}.$

So x can change from $0\mathrm{to}2.$

Then

localid="1649325644831" $$\begin{gathered} \mathrm{R}=\frac{1}{\mathrm{A}}{\int }_0^2\mathrm{\rho }\left(\mathrm{x}\right)\mathrm{dx} \\ \mathrm{R}=\frac{1}{7.85\times {10}^{-7}}{\int }_0^2\left(2.5\times {10}^{-6}\right)\left[1+{\left(\frac{\mathrm{x}}{1.0}\right)}^2\right]\mathrm{dx} \\ \mathrm{R}=0.318{\left[\mathrm{x}+\frac{{\mathrm{x}}^3}{3}\right]}_0^2 \\ \mathrm{R}=15\Omega \end{gathered}$$

Then current will be

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{9}{15} \\ \mathrm{I}=0.6\mathrm{A} \end{gathered}$$