Answer:
Explanation:
the same question was in my book hehehe so i am telling this from my book
Drift velocity: 3.32\cdot 10^{-4}m/s
2) 5.6\cdot 10^{13} electrons per person
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
In this problem, we have:
I = 8.0 A is the current
is the concentration of free electrons
d = 1.5 mm is the diameter, so the radius is
r = 1.5/2 = 0.75 mm =0.75.10^-3m
Therefore, the drift velocity is:
vd=8.0/(8.5.10^28)(1.6.10^19)π(0.75.10^-3)^2=3.32.10^-4m/s
2)The total length of the cord in this problem is
L = 3.00 m
While the cross-sectional area is
A=π
^2=π
(0.75.10^-3)^2=1.77.10^-6m^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=(8.5.10^28)(1.77.10^-6)(3.0)=4.5.10^23
In total, the Earth population consists of 8 billion people, which is
N^1=8.10^9
Therefore, the number of electrons that each person would get is:
N^e=N/N^1=4.5.10^23/8.10^9=5.6.10^13
hope it helps
