SHOW WORK PLEASE The point of this problem is to show how slowly electrons trave
ID: 1451857 • Letter: S
Question
SHOW WORK PLEASE 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 wire made from iron with a cross-section of diameter 0.810 mm carries a current of 11.0 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.) You need to calculate the cross-sectional area of the wire. Incorrect. Tries 2/8 Previous Tries The density of iron is 7.88 g/cm3, and its atomic mass is 56.0. 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.) Tries 0/8 Use your results to calculate the drift speed (i.e., the average net speed) of the electrons in the wire. Tries 0/8 Due to thermal motion, the electrons at room temperature are randomly traveling to and fro at 1.15×105 m/s, even without any current. What fraction is the current's drift speed, compared to the random thermal motion?
Explanation / Answer
A)
11.0A means 11.0 C of charge per second passing a given point. This corresponds to 11/1.60x10^-19 = 6.875 x 10^19 electrons per second
Or an electron density of 6.875 x 10^19/(*(0.405x10^-3)^2) = 3.542685678 x 10^13 electron per m^2 per second
B) There are 7.88 (g/cm^3) / 56.2 (g/mol) * 6.022 x 10^23 molecule... electron/molecule
= 8.443658363 x 10^22 electron/cm^3 = 8.443658363 x 10^28 electrons/m^3
C) vd = J/(n*q) = 11.0/(*(0.405x10^-3)^2) / (8.443658363 x 10^28 * 1.6 * 10 ^-19) = 1.580091112 x 10^-3m/s
D) we have 1.580091112 x 10^-3 / 1.15x10^5 = 1.373992272 x 10^-8
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