Answer:
[tex]6.9\times 10^{28}m^{-3}[/tex]
Explanation:
We are given that
Diameter of wire=d=4.12 mm
Radius of wire=[tex]r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Current=I=[tex]8 A[/tex]
Drift velocity=[tex]v_d=5.4\times 10^{-5} m/s[/tex]
We have to find the density of free electrons in the metal
We know that
Density of electron=[tex]n=\frac{I}{v_deA}[/tex]
Using the formula
Density of free electrons=[tex]\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}[/tex]
By using Area of wire=[tex]\pi r^2[/tex]
[tex]\pi=3.14[/tex]
[tex]e=1.6\times 10^{-19} C[/tex]
Density of free electrons=[tex]6.9\times 10^{28}m^{-3}[/tex]
