HW 12 - saldana (avs387) hw 12 turner (56705) 1 This...

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Unformatted text preview: saldana (avs387) hw 12 turner (56705) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The current in a conductor varies over time as shown in the figure below. 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 6 7 1 2 3 4 5 6 Current(A) Time(s) a) How much charge passes through a cross section of the conductor in the time interval t = 0 s to t = 3 s? Correct answer: 13 C. Explanation: Let : t = 3 . 0 s . I = Q t Q = I t = (2 A)(0 . 5 s) + (4 A)(1 s) + (6 A)(1 s) + (4 A)(0 . 5 s) = 13 C . 002 (part 2 of 2) 10.0 points b) What constant current would transport the same total charge during the 3 s interval as does the actual current? Correct answer: 4 . 33333 A. Explanation: I = Q t = 13 C 3 s = 4 . 33333 A . 003 10.0 points An aluminium wire with a cross-sectional area of 3 10 6 m 2 carries a current of 8 . 6 A. Find the drift speed of the electrons in the wire. The density of aluminium is 2 . 7 g / cm 3 . (Assume that three electrons is supplied by each atom). Correct answer: 9 . 90091 10 5 m / s. Explanation: Let: A = 3 10 6 m 2 , I = 8 . 6 A = 8 . 6 C / s , = 2 . 7 g / cm 3 = 2 . 7 10 6 g / m 3 , and n e = electrons / atom = 3 . Assuming n e free conduction electrons per atom, the density of charge carriers is n e times the density of aluminium atoms: n = n e density mass / atom = n e M/N A = n e N A M . Thus the drift speed is v d = I n q e A = I M n e N A q e A = (8 . 6 C / s) (26 . 98 g) (3) (6 . 02 10 23 ) (2 . 7 10 6 g / m 3 ) 1 (1 . 602 10 19 C) (3 10 6 m 2 ) = 9 . 90091 10 5 m / s . 004 10.0 points A current of 2 A flows in a copper wire 8 mm in diameter. The density of valence electrons in copper is roughly 9 10 28 m 3 . Find the drift speed of these electrons. Correct answer: 2 . 76311 10 6 m / s....
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HW 12 - saldana (avs387) hw 12 turner (56705) 1 This...

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