14 - roofner(bar784 – Homework 14 – Weathers –(22202...

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Unformatted text preview: roofner (bar784) – Homework 14 – Weathers – (22202) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points Four long, parallel conductors carry equal cur- rents. A cross-sectional view of the conduc- tors is shown in the figure below. Each side of the square has length of 0 . 4 m. Note: The current direction is out of the page at points indicated by the dots and into the page at points indicated by the crosses. A B C D P x y . 4 m 3 A out 3 A out 3 A out 3 A out Which diagram correctly denotes the di- rections of the components of the mag- netic field from each conductor at the point P ? The premeability of free space is 1 . 25664 × 10 − 6 N / A 2 . 1. P B A B B B C B D 2. P B A B B B C B D correct 3. P B A B B B C B D 4. P B A B B B C B D 5. P B A B B B C B D 6. P B A B B B C B D Explanation: Let : I = 3 A and ℓ = 0 . 4 m . The directions of the magnetic field due to each wire are given by the right hand rule, where the thumb points in the direction of the current and your fingers curl in the direction of the magnetic field’s circular path. Note: The magnetic field from each wire circulates in a circle around that wire. At the point P, you take the magnetic field direction tangent to the circle formed by magnetic field lines. Consider the magnetic field contributions due to the currents in A and B only: roofner (bar784) – Homework 14 – Weathers – (22202) 2 A B P 3 A out 3 A out B A B B Consider the magnetic field contributions due to the currents in C and D only: C D P 3 A out 3 A out B C B D Consider the results of all four magnetic field contributions superimposed on one an- other: A B C D P . 4 m 3 A out 3 A out 3 A out 3 A out B A B B B C B D 002 (part 2 of 4) 10.0 points What is the magnitude of each of the four components B A , B B , B C , and , B D at the point P? 1. bardbl vector B A,B,C, or D bardbl = μ I 2 π ℓ 2. bardbl vector B A,B,C, or D bardbl = μ I 2 2 π ℓ 3. bardbl vector B A,B,C, or D bardbl = μ I 2 √ 2 π ℓ 4. bardbl vector B A,B,C, or D bardbl = μ I √ 2 π ℓ correct 5. bardbl vector B A,B,C, or D bardbl = μ I ℓ 2 π Explanation: By Ampere’s law, the line integral around any closed path is equal to μ I . Here the path is defined by the circle with radius r . contintegraldisplay vector B · dvectors = μ I B · 2 π r = μ I B = μ I 2 π r , where r = ℓ cos 45 ◦ = ℓ √ 2 is the distance from any of the wires to P, from which it follows that B A = B B = B C = B D ....
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This document was uploaded on 10/26/2011 for the course PHYS 2220 at North Texas.

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14 - roofner(bar784 – Homework 14 – Weathers –(22202...

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