The point of this problem is to show how slowly electrons travel on average thro

Question

The point of this problem is to show how slowly electrons travel on average through a thin wire, even for large values of current.

A. A wire made from iron with a cross-section of diameter 0.870 mm carries a current of 12.0 A. Calculate the "areal current density"; in other words, how many electrons per square meter per second flow through this wire? (Enter your answer without units.)

B. The density of iron is 7.85 g/cm3, and its atomic mass is 55.9. Assuming each iron atom contributes two mobile electrons to the metal, what is the number density of free charges in the wire, in electrons/m3? (Enter your answer without units.)

C. Use your results to calculate the drift speed (i.e., the average net speed) of the electrons in the wire.

D. Due to thermal motion, the electrons at room temperature are randomly traveling to and fro at 1.16×105 m/s, even without any current. What fraction is the current's drift speed, compared to the random thermal motion?

Explanation / Answer

A) Current density, J = I/A

= 12/(pi*0.00087^2/4)

= 2.0186*10^7 A/m^2

= 2.0186*10^7 C/(s.m^2)

the charge crossing the given area per unit second = 2.0186*10^7 C

so, no fo electrons crossing per second, N = Q/e

= 2.0186*10^7/(1.6*10^-19)

= 1.26*10^26

B) no of moles present in 7.85 grams, n = mass/molar mass

= 7.85/55.9

= 0.1404 mol

no of atoms present in 7.85 grams = 0.1404*6.023*10^23

= 8.458*10^22

so, charge carrier density = 8.458*10^22*2 per cm^3

= 8.458*10^22*2*10^6 per m^3

= 1.69*10^29

C) vd = J/(n*q)

= 2.0186*10^7/(1.69*10^29*1.6*10^-19)

= 7.46*10^-4 m/s or 0.746 mm/s

d) fraction = 7.46*10^-4/(1.16*10^5)

= 6.43*10^-9

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