Answer :
Answer:
Number of electrons are flowing per second is 2.42 x 10¹⁹
Explanation:
The electric current flows through a wire is given by the relation :
[tex]I=envA[/tex] ....(1)
Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.
But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,
[tex]I=Ne[/tex] ....(2)
Here N is the rate of electrons passing through junction.
From equation (1) and (2).
[tex]eN = envA[/tex]
[tex]N=nvA[/tex]
But area of wire, [tex]A=\pi \frac{d^{2} }{4}[/tex]
Here d is diameter of wire.
So, [tex]N = nv\pi \frac{d^{2} }{4}[/tex]
Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.
[tex]N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}[/tex]
N = 2.42 x 10¹⁹ s⁻¹
There are [tex]2.42\times10^{19}[/tex] electrons flowing past a given point each second.
Conduction current:
The current I in a wire with cross-sectional area A is given by:
[tex]I=nev_dA[/tex]
where n is the conduction electron density = 6×10²⁸m⁻³
[tex]v_d[/tex] is the drift velocity = 0.000191 m/s
A is the area = [tex]\frac{\pi d^2}{4}[/tex] ( here d is the diameter = 0.000291m)
In terms of rate of flow of charge the current can be expressed as:
[tex]I=Ne[/tex]
where N is the number of electrons passing through a point per unit time.
Equating both the expressions we get:
[tex]N=nv_dA\\\\N=nv_d(\frac{\pi d^2}{4})\\\\N=6\times10^{28}\times0.000191\times(\frac{3.14\times(0.00291)^2}{4})\\\\N=2.42\times10^{19}\;per\;second[/tex]
Learn more about drift velocity:
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