We need to use the Ohms law to give a proper solution,
[tex]J=\sigma E\\J= \frac{E}{\rho}\\\frac{I}{A}=\frac{E}{\rho}\\E=\frac{I}{A}\rho[/tex]
The temperature is 120° c, so calculate [tex]\rho[/tex]
[tex]\rho = 5.25*10^{-8}+5.25*10^{-8}(0.0045)(120-20)[/tex]
[tex]\rho = 7.61*10^{-8} \Omega .m[/tex]
We can get the Are of the cylinder,
[tex]A= \pi \frac{d^2}{4}\\A= \pi (0.6*10^{-3})^2\\A=1.13*10^{-6}m^2[/tex]
To get the electric field replace the values,
[tex]E=\frac{\rho I}{A}\\E= \frac{7.61*10^{-8}12.5}{1.13*10^{-6}}\\E=0.84N/c[/tex]
[tex]Solving for the resistance, we haveR=\frac{\rho L}{A}\\R= \frac{7.61*10^{-8}0.17}{1.13*10^-6}\\R=0.01\Omega[/tex]
At the end we can get the Potential
[tex]V_{max}=IR\\V_{max}=(12.5)(0.01)\\V_{max}=0.12V[/tex]
