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