Answer:
The potential difference across the wire is 0.08V.
Explanation:
The relationship between voltage [tex]V[/tex], resistance [tex]R[/tex], and current [tex]I[/tex] is
[tex]V=IR[/tex],
Now, the resistance [tex]R[/tex] can be expressed as
[tex]R = \rho \dfrac{L}{A},[/tex]
where [tex]\rho[/tex] is the resistivity, [tex]L[/tex] is the length, and [tex]A[/tex] is the cross sectional area of the wire.
The cross sectional area [tex]A[/tex] of 2mm diameter wire is
[tex]A = \pi (\dfrac{2*10^{-3}}{2} )^2[/tex]
[tex]A = \pi *10^{-6}m^2[/tex],
therefore, the resistance is
[tex]R = (1.68*10^{-8}) \dfrac{(10m)}{\pi *10^{-6}},[/tex]
[tex]R = 0.0535\: \Omega[/tex].
Now, we are told that the current through the wire is 1.5A; therefore the potential difference across it will be
[tex]V = IR = (1.5A)*(0.0535 \Omega)[/tex]
[tex]\boxed{V= 0.08\:volts }[/tex]
To conclude, the potential difference across the wire is 0.08V.
