Question

he 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?

Answer 1

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2) $5.6\cdot 10^{13}$ electrons per person

Explanation:

1)

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

$v_d=\frac{I}{neA}$

where

I is the current

n is the concentration of free electrons

$e=1.6\cdot 10^{-19}C$ is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

$A=\pi r^2$

where r is the radius of the wire. So the equation becomes

$v_d=\frac{I}{ne\pi r^2}$

In this problem, we have:

I = 8.0 A is the current

$8.5\cdot 10^{28} m^{-3}$ is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

Therefore, the drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

2)

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Therefore, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$, so the total number of electrons in this cord is

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

In total, the Earth population consists of 8 billion people, which is

$N'=8\cdot 10^9$

Therefore, the number of electrons that each person would get is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$