A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a $6.5m$long, $2.0mm$-diameter wire of the material, connects it to a $1.5V$battery, and measures a $3.0mT$ magnetic field $1.0mm$ from the surface of the wire. What is the material’s resistivity?

We need to find the material’s resistivity.

Resistance $R$can be determined by finding the wire's resistivity $p$.With the Biot-Savart law, we can calculate the magnitude and direction of a current-carrying wire's magnetic field. At any distance the magnetic field is given by equation in the form

$B_{wire}=\frac{{\mu }_o}{2\mathrm{\pi }}\frac{I}{r}\left(1\right)$

finding the current $I$, so we solve equation $\left(1\right)$for $I$to be in the form

$I=\frac{2{\mathrm{\pi B}}_{\mathrm{wire}}\mathrm{r}}{{\mu }_o}\left(2\right)$

Diameter of $2mm$wire's radius is$1mm$. distance is the distance between the center of the wire and the outside of the wire. therefore, in our problem, we add the value of the radius to the distance from the wire to the probe to arrive at

$r=1mm+1mm=2mm$

Putting the values for $B_{wire},rand{\mu }_o$into eqution $\left(2\right)$to get current

$$\begin{gathered} I=\frac{2{\mathrm{\pi B}}_{\mathrm{wire}}\mathrm{r}}{{\mu }_o} \\ =\frac{2\mathrm{\pi }\left(3\times {10}^{-3}\mathrm{T}\right)\left(2\times {10}^{-3}\mathrm{m}\right)}{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}\right)\left(m/A\right)} \\ =30A \end{gathered}$$

A battery with emf $1.5V$is connected to a wire. To find resistance $R$using Ohm's law

$R=\frac{\epsilon }{I}=\frac{1.5V}{30A}=0.05\Omega$

Depending on the geometry of the material, resistivity $p$is determined by

$p=\frac{RA}{L}\left(3\right)$

length $L$of the wire and $A$is the area of the wire. The wire has $2mm$of diameter , so we get its area by

$A=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2=\mathrm{\pi }{\left(\frac{2\times {10}^{-3}\mathrm{m}}{2}\right)}^2=3.14\times {10}^{-6}{\mathrm{m}}^2$

Putting the value for $A,RandL$into equation $\left(3\right)$to get $p$

role="math" localid="1649581700737" $$\begin{gathered} p=\frac{RA}{L} \\ =\frac{(0.05\Omega )(3.14\times {10}^{-6}{m}^2)}{6.5m} \\ =\left(2.4\times {10}^{-8}\Omega \right)\times \left(m\right) \end{gathered}$$