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solution_pdf 12 - malsam(wgm329 hw 12 turner(56705 This...

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malsam (wgm329) – 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. 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 0 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: 12 C. Explanation: Let : Δ t = 3 . 0 s . I = Δ Q Δ t Δ Q = I Δ t = (2 A)(0 . 5 s) + (4 A)(1 s) + (6 A)(0 . 5 s) + (4 A)(1 s) = 12 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 A. Explanation: I = Δ Q Δ t = 12 C 3 s = 4 A . 003 10.0 points An aluminium wire with a cross-sectional area of 4 . 5 × 10 6 m 2 carries a current of 4 . 1 A. Find the drift speed of the electrons in the wire. The density of aluminium is 2 . 7 g / cm 3 . (Assume that two electrons is supplied by each atom). Correct answer: 4 . 7202 × 10 5 m / s. Explanation: Let: A = 4 . 5 × 10 6 m 2 , I = 4 . 1 A = 4 . 1 C / s , ρ = 2 . 7 g / cm 3 = 2 . 7 × 10 6 g / m 3 , and n e = electrons / atom = 2 . 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 ρ M/N A = n e N A ρ M . Thus the drift speed is v d = I n q e A = I M n e N A ρ q e A = (4 . 1 C / s) (26 . 98 g) (2) (6 . 02 × 10 23 ) (2 . 7 × 10 6 g / m 3 ) × 1 (1 . 602 × 10 19 C) (4 . 5 × 10 6 m 2 ) = 4 . 7202 × 10 5 m / s . 004 10.0 points A current of 4 A flows in a copper wire 4 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 . 21049 × 10 5 m / s. Explanation:
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malsam (wgm329) – hw 12 – turner – (56705) 2 Let : I = 4 A , r = 4 mm 2 = 0 . 002 m , and n e = 9 × 10 28 m 3 . I = j ( π r 2 ) = e n e v d ( π r 2 ) v d = I e n e π r 2 = 4 A (1 . 6 × 10 19 C) (9 × 10 28 m 3 ) × 1 π (0 . 002 m) 2 = 2 . 21049 × 10 5 m / s = 7 . 95775 cm / hours .
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