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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 = I 12 1 b correct 2. k ~ B CD k = I 6 1 b 3. k ~ B CD k = I 24 1 b 2 4. k ~ B CD k = I 6 1 b 2 5. k ~ B CD k = 0 6. k ~ B CD k = I 12 1 b 2 7. k ~ B CD k = I 3 1 b 8. k ~ B CD k = I 3 1 b 2 9. k ~ B CD k = I 24 1 b Explanation: Basic Concepts: Biot-Savart law: d ~ B = 4 I d~s r r 2 Conversion from degrees to radians: 60 = 3 radians Solution: For current along arc CD , d~s is perpendicular to r . Therefore, if we are not concerned with the fields direction, the Biot- Savart law gives dB = I 4 b 2 ds for current along arc CD . Integrating over the segment yields B CD = I 4 b 2 Z D C ds = I 4 b 2 ( b ) = I 4 b = I 4 b 3 = I 12 b , where the pseudo unit of 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 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... View Full Document