Answer:
[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]
Explanation:
r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]
B = Magnetic field = 3 mT
1 mm = Distance from the surface of the wire
V = Voltage
x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]
R = Resistance
L = Length of wire = 4.5 m
Magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]
Voltage is given by
[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]
Resistivity is given by
[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]
The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].
