Answer:
[tex]\rho=1.8\times 10^{-8}\Omega.m[/tex]
Explanation:
Given:
width of the wire, [tex]w=1.8\ mm=1.8\times 10^{-3}\ m[/tex]
thickness of the flat wire, [tex]d=0.12\ mm=1.2\times 10^{-4}[/tex]
length of the wire, [tex]l=12\ m[/tex]
voltage across the wire, [tex]V=12\ V[/tex]
current through the wire, [tex]I=12\ A[/tex]
Now the net resistance of the wire:
using ohm's law
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{12}{12}[/tex]
[tex]R=1\ \Omega[/tex]
We have the relation between the resistivity and the resistance as:
[tex]R=\rho.\frac{l}{a}[/tex]
where:
a = cross sectional area of the wire
[tex]\rho =[/tex] resistivity of the wire material
[tex]1=\rho\times \frac{12}{1.8\times 10^{-3}\times 1.2\times 10^{-4}}[/tex]
[tex]\rho=1.8\times 10^{-8}\Omega.m[/tex]
