Question

A current of I=8.0A is flowing in a typical extension cord of length L=3.00m. The cord is made of copper wire with diameter d=1.5mm. The charge of the electron is e=1.6×10−19C. The mass of the electron is m=9.1×10−31kg. The resisitivity of copper is rho=1.7×10−8Ω⋅m. The concentration of free electrons in copper is n=8.5×1028m−3.The population of the Earth is roughly six 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

⁻Answer:

Ne=1.8 *10^37

Explanation:

Current density J=Iₐ /(πr²)

=  8/(π* (1.5*10⁻³/2)²)= 4.53*10⁶ A/m²

Drift velocity =  J/ne =4.53*10⁶ /(8.5×1028 *1.6×10−19 ) = 3.33 ×10−4 m/s

volume = πr²h= π * (1.5*10⁻³/2)² * 3 = 5.3 ×10−6 m³

$\rho =\frac{m}{ne^{2}} \\\\n=\frac{m}{\rho e^{2}}$

Ne=1.8 *10^37

Answer 2

If all free electrons contained in this extension cord are evenly split among the humans, each person will get 1.416×10¹⁹ electrons

Free electrons:

The current in terms of charge carriers is given by:

I = nev$__d$A

where v$__d$ is the drift velocity

A is the cross sectional area

e = 1.6×10⁻¹⁹C

n = 8.5×10²⁸m⁻³

Thus,

v$__d$ = I/neA

given that I = 8A

diameter d = 1.5mm = 1.5×10⁻³m

$v_d = \frac{8}{(8.5\times10^{28})(1.6\times10^{-19})(\pi\times(1.5\times10^{-3})^{2}/4)}$

v$__d$ = 3.33 ×10⁻⁴m/s

If all free electrons contained in this extension cord are evenly split among the human, then each one will get:

Ne = n/6×10⁹

Ne = 8.5×10²⁸/6×10⁹

Ne = 1.416×10¹⁹ electrons

Learn more about drift velocity:

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