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16 - Kapoor(mk9499 homework16 Turner(60230 This print-out...

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Kapoor (mk9499) – homework16 – Turner – (60230) 1 This print-out should have 12 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 A 3 . 3 g wire has a density of 17 . 1 g / cm 3 and a resistivity of 1 . 65 × 10 7 Ω m. The wire has a resistance of 78 Ω. How long is the wire? Correct answer: 9 . 55134 m. Explanation: Let : m = 3 . 3 g = 0 . 0033 kg , ρ = 17 . 1 g / cm 3 = 17100 kg / m 3 , r = 1 . 65 × 10 7 Ω m , and R = 78 Ω . ρ is density, and r is the resistivity of the wire, so m = ρ V V = m ρ = A ℓ A = m ρ ℓ Thus the resistance is R = r A = r ℓ m ρ ℓ = r ρ ℓ 2 m 2 = R m ρ r = radicalBigg R m ρ r = radicalBigg (78 Ω) (0 . 0033 kg) (17100 kg / m 3 ) (1 . 65 × 10 7 Ω m) = 9 . 55134 m . 002 (part 2 of 2) 10.0 points The wire is made up of atoms with valence 1 and mass 30 kg, where u = 1 . 6605 × 10 27 kg. What is the drift speed of the electrons when there is a voltage drop of 140 . 4 V across the wire? Correct answer: 0 . 00017827 m / s. Explanation: Let : n con = 1 , m μ = 30 kg = 30 kg , V = 140 . 4 V , N A = 6 . 02 × 10 23 , ρ = 17 . 1 g / cm 3 = 17100 kg / m 3 , = 9 . 55134 m , and q e = 1 . 602 × 10 19 C , where m μ is atomic mass, n con is number of conduction electron per an atom, and N A is Avogadro’s number. Since the current I is given by I = n q A v d , where n is the density of charge carriers, q the charge on an electron and v d the drift speed, we have v d = I n q e A = V n q e A R . (1) The density of charge carriers is found from knowing how many electrons there are per atom and then the total number of atoms in the wire. This density is n = n con ρ parenleftbigg m μ N A parenrightbigg = n con ρ N A m μ = (1) (17100 kg / m 3 ) (6 . 02 × 10 23 ) (30 kg) = 3 . 4314 × 10 26 m 3 . Here, the mass of one atom is given by parenleftbigg Atomic mass Avogadro s number parenrightbigg .
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Kapoor (mk9499) – homework16 – Turner – (60230) 2 Since the volume is the cross-sectional area multiplied by the length this tells us that A ℓ = m ρ , so A = m ρ ℓ = (30 kg) (17100 kg / m 3 ) (9 . 55134 m) = 0 . 00018368 m 2 , thus v d = V n q e A R (1) = (140 . 4 V) (3 . 4314 × 10 26 m 3 ) × 1 (1 . 602 × 10 19 C) × 1 (0 . 00018368 m 2 ) × 1 (78 Ω) = 0 . 00017827 m / s . 003 10.0 points Current is carried throughout the body of a type-II superconductor. Assume: The density of superconducting electrons is 5 × 10 27 m 3 . If a Nb 3 Sn (type II) superconducting wire of cross sectional area 2 . 73 mm 2 can carry a 1 × 10 5 A supercurrent, what is the average (drift) velocity of the superconducting elec- trons? Correct answer: 45 . 7304 m / s. Explanation: Let : n = 5 × 10 27 m 3 , A = 2 . 73 mm 2 = 2 . 73 × 10 6 m 2 , and I = 1 × 10 5 A . The drift velocity is v = I q e n A = 1 × 10 5 A (1 . 602 × 10 19 C) × 1 (5 × 10 27 m 3 ) (2 . 73 × 10 6 m 2 ) = 45 . 7304 m / s . 004 10.0 points A length of metal wire has a radius of 0 . 001 m and a resistance of 0 . 11 Ω. When the potential difference across the wire is 17 V, the electron drift speed is found to be 0 . 000365 m / s.
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