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A current of I = 3.8 A is passing through a conductor with cross sectional area A = 2.5 x 10^-4 m^2. The charge carriers in the conductor, electrons, have a number density n = 1.3 x 10^27 m^-3.
(a) Express the drift velocity of electrons through 1.A. n, and the fundamental charge e.
(b) Calculate the numerical value of v, in m/s.

Respuesta :

Answer:

Explanation:

current, i = 3.8 A

Area of crossection, A = 2.5 x 10^-4 m²

number density, n = 1.3 x 10^27 m^-3

(a) the relation between the current and the drift velocity is given by

i = n e A v

where, e is the electronic charge, A be the area of crossection area of the wire and v be the drift velocity

v = i / neA

(b)

3.8 = 1.3 x 10^27 x 1.6 x 10^-19 x 2.5 x 10^-4 x v

3.8 = 52000 v

v = 7.3 x 10^-5 m/s

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