1) Drift velocity: [tex]3.32\cdot 10^{-4}m/s[/tex]
2) [tex]5.6\cdot 10^{13}[/tex] electrons per person
Explanation:
1)
For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:
[tex]v_d=\frac{I}{neA}[/tex]
where
I is the current
n is the concentration of free electrons
[tex]e=1.6\cdot 10^{-19}C[/tex] is the electron charge
A is the cross-sectional area of the wire
The cross-sectional area can be written as
[tex]A=\pi r^2[/tex]
where r is the radius of the wire. So the equation becomes
[tex]v_d=\frac{I}{ne\pi r^2}[/tex]
In this problem, we have:
I = 8.0 A is the current
[tex]8.5\cdot 10^{28} m^{-3}[/tex] is the concentration of free electrons
d = 1.5 mm is the diameter, so the radius is
r = 1.5/2 = 0.75 mm = [tex]0.75\cdot 10^{-3}m[/tex]
Therefore, the drift velocity is:
[tex]v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s[/tex]
2)
The total length of the cord in this problem is
L = 3.00 m
While the cross-sectional area is
[tex]A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2[/tex]
Therefore, the volume of the cord is
[tex]V=AL[/tex] (1)
The number of electrons per unit volume is [tex]n[/tex], so the total number of electrons in this cord is
[tex]N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}[/tex]
In total, the Earth population consists of 8 billion people, which is
[tex]N'=8\cdot 10^9[/tex]
Therefore, the number of electrons that each person would get is:
[tex]N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}[/tex]
If the electrons in this extension cord are shared evenly among people on earth, each person will get 5.6 × 10^12 electrons each.
From the question, we have the following information;
I = current = 8.0 A
n = number of free electrons in copper = 8.5 × 10^28
A = cross sectional area = πr^2 = [3.142 × (1.5 × 10^-3/2)^2 = 1.77 × 10^-6 m^2
v = drift velocity = ?
e = Charge on the electron = 1.6 × 10^-19 C
L = Length of the wire = 3.00 m
The formula for drift velocity is;
I = nAve
Making v the subject of the formula and substituting values;
v = I/nAe
v = 8.0 A/8.5 × 10^28 × 1.77 × 10^-6 m^2 × 1.6 × 10^-19 C
v = 3.3 × 10^-4 m/s
The number of electrons per unit volume is given by;
N = nAL
N = 8.5 × 10^28 × 1.77 × 10^-6 m^2 × 3.00 m
N = 4.5 × 10^23 electrons
Given that there are roughly eight billion people on earth, If all free electrons contained in this extension cord are evenly split among the humans, each would get; 4.5 × 10^23 electrons/8 × 10^10 = 5.6 × 10^12 electrons each
Learn more: https://brainly.com/question/13164491
Missing part:
A current of I = 8.0 A is flowing in a typical extension cord of length L = 3.00 m. The cord is made of copper wire with diameter d = 1.5 mm. The charge of the electron is e = 1.6 Times 10^-19 C. The mass of an electron is m_e = 9.1 Times 10^-31 kg. The resistivity of copper is rho = 1.7 Times 10^-8 Ohm m. The concentration of free electrons in copper is n = 8.5 Times 10^28 m^-3.
1) Find the drift velocity v_d of the electrons in the wire. Express your answer in meters per second, to two significant figures. v_d = 3.30 Times 10^-4 m/s
2) The population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly split among the humans, how many free electrons (Ne) would each person get?
