Question

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 6.0-m-long, 4.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.5 mT magnetic field 1.0 mm from the surface of the wire. What is the material’s resistivity?

Answer 1

Answer:

$1.8\cdot 10^{-7} \Omega m$

Explanation:

The magnetic field produced by the wire is given by:

$B=\frac{\mu_0 I}{2 \pi r}$

where

B = 3.5 mT = 0.0035 T is the magnetic field

$\mu_0$ is the vacuum permeability

$I$ is the current

r = 1.0 mm = 0.001 m is the distance from the wire

Solving for I, we find the current in the wire

$I=\frac{2 \pi r B}{\mu_0}=\frac{2 \pi (0.0035 T)(0.001 m)}{4 \pi \cdot 10^{-7} H/m}=17.5 A$

Now we can find the resistance of the wire, R, by using Ohm's law:

$R=\frac{V}{I}=\frac{1.5 V}{17.5 A}=0.086 \Omega$

And the resistance is related to the material's resistivity ($\rho$) by

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

where

L = 6.0 m is the length of the wire

$A= \pi r^2 = \pi (d/2)^2= \pi (0.004 m/2)^2=1.26\cdot 10^{-5}m^2$ is the cross-sectional area of the wire

Solving for $\rho$, we find

$\rho = \frac{RA}{L}=\frac{(0.086 \Omega)(1.26\cdot 10^{-5} m^2)}{6.0 m}=1.8\cdot 10^{-7} \Omega m$