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=r[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=8 A
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\\e=1.6\times 10^{-19} C[/tex]
Density of free electrons=[tex]6.9\times 10^{28}m^{-3}[/tex]
Explanation:
We will calculate the density as follows.
n = [tex]\frac{i}{v_{d}eA}[/tex]
As the metallic wire is in the shape of a circle. So, its area will be calculated as follows.
Area = [tex]\pi \times r^{2}[/tex]
= [tex]3.14 \times (2.06 \times 10^{-3})^{2}[/tex]
[tex]v_{d}[/tex] = [tex]5.40 \times 10^{-5}[/tex]
i = 8 A
Hence, we will calculate the density of free electrons as follows.
n = [tex]\frac{i}{v_{d}eA}[/tex]
= [tex]\frac{8}{5.40 \times 10^{-5} \times 3.14 \times (2.06 \times 10^{-3})^{2} \times 1.60 \times 10^{-19}}[/tex]
= [tex]6.950 \times 10^{28} m^{-3}[/tex]
Thus, we can conclude that the density of free electrons in the metal is [tex]6.950 \times 10^{28} m^{-3}[/tex].
