HW 14 - terry(ect328 homework 14 Turner(59130 This...

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terry (ect328) – homework 14 – Turner – (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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 4 . 1 mm 2 can carry a 1 × 10 5 A supercurrent, what is the average (drift) velocity of the superconducting electrons? Correct answer: 30 . 4497 m / s. Explanation: Let : n = 5 × 10 27 m 3 , A = 4 . 1 mm 2 = 4 . 1 × 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 ) (4 . 1 × 10 6 m 2 ) = 30 . 4497 m / s . 002 10.0 points A length of metal wire has a radius of 0 . 004 m and a resistance of 0 . 08 Ω. When the potential difference across the wire is 19 V, the electron drift speed is found to be 0 . 000331 m / s. Based on these data, calculate the density of free electron in the wire. Correct answer: 8 . 92166 × 10 28 m 3 . Explanation: Let : r = 0 . 004 m , R = 0 . 08 Ω , Δ V = 19 V , and v d = 0 . 000331 m / s . The current in the wire is I = Δ V R = 19 V 0 . 08 Ω = 237 . 5 A . From v d = I n q A , the density of free elec- tron is n = I v d q e ( π r 2 ) = 237 . 5 A (0 . 000331 m / s) (1 . 6 × 10 19 C) × 1 π (0 . 004 m) 2 = 8 . 92166 × 10 28 m 3 . 003 10.0 points At 50 C, the resistance of a segment of gold wire is 84 Ω. When the wire is placed in a liquid bath, the resistance increases to 167 Ω. The temperature coefficient is 0 . 0034 ( C) 1 at 20 C. What is the temperature of the bath? Correct answer: 370 . 259 C. Explanation: Let : R 1 = 84 Ω , R 2 = 167 Ω , T 0 = 20 C , T 1 = 50 C , and α = 0 . 0034( C) 1 . Neglecting change in the shape of the wire, we have R 1 = R 0 [1 + α ( T 1 - T 0 )] R 0 = R 1 1 + α ( T 1 - T 0 ) and R 2 = R 0 [1 + α ( T 2 - T 0 )] , where T 0 = 20 C .
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terry (ect328) – homework 14 – Turner – (59130) 2 Thus R 2 = R 1 [1 + α ( T 2 - T 0 )] 1 + α ( T 1 - T 0 ) R 2 + α R 2 ( T 1 - T 0 ) = R 1 + α R 1 ( T 2 -
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  • Fall '09
  • Turner
  • SEPTA Regional Rail, Correct Answer, Jaguar Racing, Series and parallel circuits, Highways in Slovakia, R1 R2 R3

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