Test3 (1) - Husain Zeena Homework 7 Due Mar 8 2004 4:00 am...

This preview shows page 1 - 3 out of 70 pages.

Husain, Zeena – 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 – 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 0 . 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 = 0 . 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 | = | 0 . 00242733 Ω - 0 . 00220327 Ω | = 0 . 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. Using the right-hand rule with ~ F = q~v × ~ B , to produce this force on positive charges, the velocity ~v must be directed from right to left ( ). F F v B B - - + + Note: There are four different presentations of this problem and this is the third. 004 (part 1 of 2) 10 points A metal ball having net charge Q = 6 . 3 × 10 - 6 C is thrown out of a window hori- zontally at a speed v = 28 m / s . The window is at a height h = 47 m above the ground. A uniform horizontal magnetic field of mag- nitude B = 0 . 072 T is perpendicular to the plane of the ball’s trajectory.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture