This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: roofner (bar784) – Homework 13 – Weathers – (22202) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An infinite wire is bent as shown in the figure. The current is I . It consists of 3 segments AB , BDE , and EF . AB is vertical with A extending to infinity. EF is horizontal with F extending to infinity. BDE is onequarter of a circular arc. We use the Roman numerals I, II, III and IV to denote the direction of a vector in the xy plane, which begins at the origin and is pointing into the quadrants I, II, III and IV respectively. B O D C E F A O y x I II III IV The direction of the magnetic vector at O due to the current segment BD , that due to DE and that due to EF are given by respectively (“in, out” mean “into, or out of”the xy plane) 1. B BD = in, B DE = out, B ED = out 2. I, II, B ED = in 3. IV, IV, B ED = out 4. B BD = in, B DE = out, B ED = in 5. B BD = out, B DE = out, B ED = out correct 6. B BD = out, B DE = out, B ED = in 7. B BD = in, B DE = in, B ED = in 8. B BD = out, B DE = in, B ED = in 9. B BD = in, B DE = in, B ED = out 10. B BD = out, B DE = in, B ED = out Explanation: The BiotSavart Law is given by d vector B = μ 4 π I d vector l × ˆ r r 2 . Since both d vector l and ˆ r lie in the xy plane, their cross product must be perpendicular to this plane. Apply the BiotSavart Law to each line element of the wire, d vector B is found always to direct out the xy plane at O . 002 (part 2 of 3) 10.0 points What is the magnitude of vector B at the center of the arc due to the arc BDE alone? 1. B = μ I 8 π r 2. B = μ 2 π I r 3. B = μ I 8 r correct 4. B = μ I 4 r 5. B = μ I 2 r 6. B = μ I 2 π r 7. B = μ I r 8. B = μ I 4 π r 9. B = μ π I r 10. B = μ I π r Explanation: Note: The distance r from a current el ement of BDE to O is a constant and the current element I d vector l is always perpendicular to ˆ r . Hence, the magnitude of the vector B at the center of the arc due to the arc BDC alone is roofner (bar784) – Homework 13 – Weathers – (22202) 2 B BDE = μ 4 π I r 2 integraldisplay dl = μ 4 π I r 2 π 2 r = μ I 8 r . 003 (part 3 of 3) 10.0 points Let r = 0 . 063 m and I = 1 . 1 A. The permeability of free space is 1 . 25664 × 10 − 6 N / A 2 . Find the magnitude of vector B at O due to the entire wire ABDEF ? Correct answer: 6 . 23472 × 10 − 6 T. Explanation: Let : μ = 1 . 25664 × 10 − 6 N / A 2 , I = 1 . 1 A , and r = 0 . 063 m . First calculate the magnetic field of a straight wire carrying a current(See figure be low). I r s ∆ y y O θ B From the BiotSavart Law, the contribution to the magnetic field at O due to a current element I ∆ y at y is given by ∆ B = μ 4 π I ∆ y s 2 sin θ = μ I 4 π ∆ θ r sin θ ....
View
Full
Document
This document was uploaded on 10/26/2011 for the course PHYS 2220 at North Texas.
 Fall '00
 Littler
 Magnetism, Work

Click to edit the document details