The point of this problem is to show how slowly electrons travel on average thro
ID: 2037229 • Letter: T
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 wire made from copper with a cross-section of diameter 0.740 mm carries a current of 10.0 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.) Submit Answer Tries 0/16 The density of copper is 8.99 g/cm3, and its atomic mass is 63.6. Assuming each copper atom contributes one mobile electron to the metal, what is the number density of free charges in the wire, in electrons/m3? Enter your answer without units.) Submit AnswerTries 0/32 Use your results to calculate the drift speed (i.e., the average net speed) of the electrons in the wire Submit Answer Tries 0/16 Due to thermal motion, the electrons at room temperature are randomly traveling to and fro at 1.13x105 m/s, even without any current. What fraction is the current's drift speed, compared to the random thermal motion? Submit Answer Tries 0/16 Post Discussion Send FeedbackExplanation / Answer
J = I / A
J = (10)/(pi (0.740 x 10^-3/2)^2)
J = 2.33 x 10^7 A/m^2
number of electron, n = (2.33 x 10^7)/(1.6 x 10^-19)
= 1.45 x 10^26 electrons/m^2 s ....Ans
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rho = 8990 kg/m^3
mass of 1 m^3 = 8990 kg
n= (8990 x 10^3)/63.6 = 1.41 x 10^5 mol
atoms = n NA
free electron = n NA = (1.41 x 10^5)(6.022 x 10^23)
= 8.51 x 10^28 atoms
free electron = 8.51 x 10^28 electrons
free electron density = 8.51 x 10^28 electrons/m^3
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Vd = J / e n
= (2.33 x 10^7) / (8.51 x 10^28 x 1.6 x 10^-19)
= 1.71 x 10^-3 m/s
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ratio = (1.13 x 10^5)/(1.71 x 10^-3)
= 6.61 x 10^8
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