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Eduardwww [97]
3 years ago
13

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.
Physics
1 answer:
bulgar [2K]3 years ago
8 0

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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Explanation:

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Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

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a.

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b.

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F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

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3 years ago
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