# hw7sol - Flores Michael Homework 7 Due Mar 8 2004 4:00 am...

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Flores, Michael – Homework 7 – Due: Mar 8 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore making your selection. The due time is Central time. 001 (part 1 o± 2) 10 points A 9 m long piece o± wire o± density 8 . 34 g / m 3 has a diameter o± 15 . 077 mm. The resistiv- ity o± the wire is 1 . 7 × 10 - 8 Ω · m at 20 C. The temperature coe²cient ±or the wire is 0 . 0038 ( C) - 1 . Calculate the resistance o± the wire at 20 C. Correct answer: 0 . 000856982 Ω. Explanation: Given: L = 9 m , r = 7 . 5385 mm = 0 . 0075385 m , and ρ 20 = 1 . 7 × 10 - 8 Ω · m . The area is A = π r 2 and the resistance is R = ρ 20 L A = ρ 20 L π r 2 = (1 . 7 × 10 - 8 Ω · m) · 9 m π (0 . 0075385 m) 2 = 0 . 000856982 Ω . 002 (part 2 o± 2) 10 points Calculate the di³erence in the resistance o± the wire between 52 C and 78 C. Correct answer: 8 . 46698 × 10 - 5 Ω. Explanation: Given : T 1 = 52 C , T 2 = 78 C , r = 7 . 5385 mm = 0 . 0075385 m , ρ 20 = 1 . 7 × 10 - 8 Ω · m , α = 0 . 0038 ( C) - 1 , and L = 9 m . A = π r 2 , and ρ 1 = ρ 20 { 1 . 0 + α [ T 1 - T 20 ] } = (1 . 7 × 10 - 8 Ω · m) · { 1 . 0 + [0 . 0038 ( C) - 1 ] · [(52 C) - (20 C)] } = 1 . 90672 × 10 - 8 Ω · m . R 1 = ρ 1 L A = ρ 1 L π r 2 = (1 . 90672 × 10 - 8 Ω · m) · 9 m π (0 . 0075385 m) 2 = 0 . 000961191 Ω . ρ 2 = ρ 20 { 1 . 0 + α [ T 2 - T 20 ] } = (1 . 7 × 10 - 8 Ω · m) · { 1 . 0 + [0 . 0038 ( C) - 1 ] · [(78 C) - (20 C)] } = 2 . 07468 × 10 - 8 Ω · m . R 2 = ρ 2 L A = ρ 2 L π r 2 = (2 . 07468 × 10 - 8 Ω · m) · 9 m π (0 . 0075385 m) 2 = 0 . 00104586 Ω . So the di³erence in the resistance is Δ R = | R 2 - R 1 | = | 0 . 00104586 Ω - 0 . 000961191 Ω | = 8 . 46698 × 10 - 5 Ω . 003 (part 1 o± 1) 10 points Given: A copper bar has a constant velocity in the plane o± the paper and perpendicular to a magnetic feld pointed into the plane o± the paper.

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Flores, Michael – Homework 7 – Due: Mar 8 2004, 4:00 am – Inst: Sonia Paban 2 B B - - + + If the top of the bar becomes negative rel- ative to the bottom of the bar, what is the direction of the velocity ~v of the bar? 1. from left to right ( ) 2. from bottom to top ( ) 3. from top to bottom ( ) 4. from right to left ( ) correct Explanation: Positive charges will move in the direction of the magnetic force, while negative charges move in the opposite direction. To produce the indicated charge separa- tion, the positive charges in the conductor ex- perience downward magnetic forces while the negative charges in the conductor experience upward magnetic forces leaving the charge separation show in the ±gure. Using the right-hand rule with
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## This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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hw7sol - Flores Michael Homework 7 Due Mar 8 2004 4:00 am...

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