Answer:
[tex]I=0.047A[/tex]
Explanation:
Let's use Ohm's law:
[tex]V=IR[/tex]
or
[tex]I=\frac{V}{R}[/tex] (1)
Where:
[tex]V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance[/tex]
We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:
[tex]R=\rho*\frac{l}{A}[/tex] (2)
Where:
[tex]R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}[/tex]
Keep in mind that the electrical resistivity of the gold is a known constant which is [tex]\rho_g_o_l_d=2.35*10^{-8}[/tex] and the cross sectional area of the conductor is calculated as:
[tex]A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}[/tex]
Because we have a wire in this case, so we assume a cylindrical geometry.
Now replacing our data in (2)
[tex]R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega[/tex]
Finally, we know R and V, so replacing these values in (1) we will be able to find the current:
[tex]I=\frac{0.7}{14.96056465}\approx0.047A[/tex]
