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**Unformatted text preview: **Platt, David – Homework 18 – Due: Oct 21 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 5) 10 points The current loop ABCDA carries current I in the direction indicated in the figure, where the segments AB and CD are concentric arcs with radii a and b respectively. a C B b D A O 60 ◦ x y What is the magnitude of the resultant magnetic field k ~ B k at the center of curva- ture O due to the current segment of arc CD ? 1. k ~ B CD k = μ I 12 1 b correct 2. k ~ B CD k = μ I 6 1 b 3. k ~ B CD k = μ I 24 1 b 2 4. k ~ B CD k = μ I 6 1 b 2 5. k ~ B CD k = 0 6. k ~ B CD k = μ I 12 1 b 2 7. k ~ B CD k = μ I 3 1 b 8. k ~ B CD k = μ I 3 1 b 2 9. k ~ B CD k = μ I 24 1 b Explanation: Basic Concepts: Biot-Savart law: d ~ B = μ 4 π I d~s × ˆ r r 2 Conversion from degrees to radians: 60 ◦ = π 3 radians Solution: For current along arc CD , d~s is perpendicular to ˆ r . Therefore, if we are not concerned with the field’s direction, the Biot- Savart law gives dB = μ I 4 π b 2 ds for current along arc CD . Integrating over the segment yields B CD = μ I 4 π b 2 Z D C ds = μ I 4 π b 2 ( bα ) = μ I 4 π b α = μ I 4 π b π 3 = μ I 12 b , where the pseudo unit of α should be radians. Radian denotes a pure number, and is not a unit. 002 (part 2 of 5) 10 points What is the direction of the resultant mag- netic field ~ B at the center of curvature O due to the current segment of arc CD ? 1. pointing into the paper 2. positive y direction 3. undetermined 4. negative x direction 5. positive x direction 6. pointing out of the paper correct 7. negative y direction Explanation: Using the right-hand rule, for d~s × ˆ r , the resultant magnetic field vector ~ B points out of the paper. Platt, David – Homework 18 – Due: Oct 21 2005, 4:00 am – Inst: Ken Shih...

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