[tex]Voltage = 1.47\\L= 22.6m[/tex]
[tex]r=\frac{d}{2} = \frac{0.04in}{2} = 0.02in = 0.508mm = 0.508*10^{-3}m[/tex]
So we can now calculate the area of the wire
[tex]A= \pi *r^2 = \pi (0.508*10^{-3})^2 = 8.107*10^{-7}m^2[/tex]
We need also the resistivity specific of cupper, that is:
[tex]p_{cupper}=1,72*10^{-8}[/tex]
a) We have to calculate the current that is given by,
[tex]R=\frac{pL}{A}\\R=\frac{(1.72*10^{8})(22.6)}{(8.107*10^{-7})}\\R=0.4\Omega[/tex]
b)[tex]I=\frac{V}{R}[/tex]
[tex]I=\frac{1.47}{0.4} = 3.675A[/tex]
c) current density
[tex]j= \frac{I}{A}\\j= \frac{3.675}{8.107*10^{-7}}\\j= 4.533*10^5 A/m^2[/tex]
d) Rate of thermal energy
[tex]\alpha = I^2*R\\\alpha = 3.675^2*0.4 = 5.4W[/tex]
