the mean time between collisions in iron is 5.2×10−15 what electron current is driven through a 2.4- mm -diameter iron wire by a 0.070 v/m electric field? express your answer as a number of electrons per second.

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11-10-2022
Physics

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the mean time between collisions in iron is 5.2×10−15 what electron current is driven through a 2.4- mm -diameter iron wire by a 0.070 v/m electric field? express your answer as a number of electrons per second.

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AnnuG AnnuG · Answer 1 · 2022-10-19T05:29:10-04:00

The electron current driven through a 2.4 mm diameter iron wire by a 0.070 v/m electric field is 24.34 * [tex]10^{18}[/tex] electrons / s

I = n A e [tex]v_{d}[/tex]

[tex]v_{d}[/tex] = т e E / m

I = Current

n = Free electron density

A = Area

e = Charge

[tex]v_{d}[/tex] = Drift velocity

т = Mean time

E = Electric field

m = Mass of charge

т = 5.2 * [tex]10^{-15}[/tex] s

d = 2.4 mm

r = 1.2 mm = 1.2 * [tex]10^{-3}[/tex] m

E = 0.07 V / m = 0.07 N / C

e = 1.6 * [tex]10^{-19}[/tex] C

m = 9.1 * [tex]10^{-31}[/tex] kg

A = π r²

A = 3.14 * ( 1.2 * [tex]10^{-3}[/tex] )²

A = 4.52 * [tex]10^{-6}[/tex] m²

[tex]v_{d}[/tex] = ( 5.2 * [tex]10^{-15}[/tex] * 1.6 * [tex]10^{-19}[/tex] * 0.07 ) / ( 9.1 * [tex]10^{-31}[/tex] )

[tex]v_{d}[/tex] = ( 0.58 * [tex]10^{-34}[/tex] ) / ( 9.1 * [tex]10^{-31}[/tex] )

[tex]v_{d}[/tex] = 63.7 * [tex]10^{-6}[/tex] m / s

n = ( Avogadro's number * Density ) / Atomic mass

Avogadro's number = 6.023 * [tex]10^{23}[/tex] atoms / mole

Density of copper = 8.92 * 10³ kg / m³

Atomic mass = 63.5 * [tex]10^{-3}[/tex] kg / mole

n = 8.46 * [tex]10^{28}[/tex] / m³

I = n A e [tex]v_{d}[/tex]

I = 8.46 * [tex]10^{28}[/tex] * 4.52 * [tex]10^{-6}[/tex] * 1.6 * [tex]10^{-19}[/tex] * 63.7 * [tex]10^{-6}[/tex]

I = 3897.34 * [tex]10^{-3}[/tex]

I = 3.9 A

I = 3.9 * 6.242 * [tex]10^{18}[/tex] electrons / s

I = 24.34 * [tex]10^{18}[/tex] electrons / s

Therefore, the electron current driven through the iron wire is 24.34 * [tex]10^{18}[/tex] electrons / s

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