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1 month ago

14

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

lit among the humans, how many free electrons (Ne) would each person get?

1 answer:

eduard [2.7K]1 month ago

7 0

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

2. $5.6\cdot 10^{13}$ electrons per individual

Explanation:

1)

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

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

where

I stands for current

n represents the density of free electrons

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

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

$A=\pi r^2$

where r denotes the wire's radius. Thus, the equation transforms to

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

In this scenario, we have:

I = 8.0 A as the current

$8.5\cdot 10^{28} m^{-3}$ indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

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

So, the resulting 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 entire length of the cord is

L = 3.00 m

And the cross-sectional area is

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

Consequently, the volume of the cord is

$V=AL$ (1)

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

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

Overall, the Earth's population rounds to 8 billion individuals, equating to

$N'=8\cdot 10^9$

Hence, the number of electrons distributed to each person is:

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

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