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Unformatted text preview: Husain, Zeena – Homework 7 – Due: Mar 8 2004, 4:00 am – Inst: Sonia Paban 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points A 14 m long piece of wire of density 8 . 34 g / m 3 has a diameter of 12 . 208 mm. The resistiv ity of the wire is 1 . 7 × 10 8 Ω · m at 20 ◦ C. The temperature coefficient for the wire is . 0038 ( ◦ C) 1 . Calculate the resistance of the wire at 20 ◦ C. Correct answer: 0 . 00203328 Ω. Explanation: Given: L = 14 m , r = 6 . 104 mm = 0 . 006104 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) · 14 m π (0 . 006104 m) 2 = . 00203328 Ω . 002 (part 2 of 2) 10 points Calculate the difference in the resistance of the wire between 42 ◦ C and 71 ◦ C. Correct answer: 0 . 000224068 Ω. Explanation: Given : T 1 = 42 ◦ C , T 2 = 71 ◦ C , r = 6 . 104 mm = 0 . 006104 m , ρ 20 = 1 . 7 × 10 8 Ω · m , α = 0 . 0038 ( ◦ C) 1 , and L = 14 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 ] · [(42 ◦ C) (20 ◦ C)] } = 1 . 84212 × 10 8 Ω · m . R 1 = ρ 1 L A = ρ 1 L π r 2 = (1 . 84212 × 10 8 Ω · m) · 14 m π (0 . 006104 m) 2 = 0 . 00220327 Ω . ρ 2 = ρ 20 { 1 . 0 + α [ T 2 T 20 ] } = (1 . 7 × 10 8 Ω · m) · { 1 . 0 + [0 . 0038 ( ◦ C) 1 ] · [(71 ◦ C) (20 ◦ C)] } = 2 . 02946 × 10 8 Ω · m . R 2 = ρ 2 L A = ρ 2 L π r 2 = (2 . 02946 × 10 8 Ω · m) · 14 m π (0 . 006104 m) 2 = 0 . 00242733 Ω . So the difference in the resistance is Δ R =  R 2 R 1  =  . 00242733 Ω . 00220327 Ω  = . 000224068 Ω . 003 (part 1 of 1) 10 points Given: A copper bar has a constant velocity in the plane of the paper and perpendicular to a magnetic field pointed into the plane of the paper. Husain, Zeena – 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 right to left ( ⇐ ) correct 2. from left to right ( ⇒ ) 3. from top to bottom ( ⇓ ) 4. from bottom to top ( ⇑ ) 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 figure....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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