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Unformatted text preview: Version 104 – Exam 2 – chiu – (57460) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the setup in Figure above. What is the change in potential energy Δ U = U C U D , in moving an electron from D to C? 1. 2 k eq s parenleftbigg 1 a 2 1 b 2 parenrightbigg 2. k q s parenleftbigg 1 a 2 1 b 2 parenrightbigg 3. k q s parenleftbigg 1 a 1 b parenrightbigg 4. k eq s parenleftbigg 1 a 2 1 b 2 parenrightbigg 5. k eq s parenleftbigg 1 a 1 b parenrightbigg 6.2 k eq s parenleftbigg 1 a 1 b parenrightbigg 7.2 k eq s parenleftbigg 1 a 2 1 b 2 parenrightbigg 8. k eq s parenleftbigg 1 a 1 b parenrightbigg 9. k eq s parenleftbigg 1 a 2 1 b 2 parenrightbigg correct 10. 2 k eq s parenleftbigg 1 a 1 b parenrightbigg Explanation: V C V D = integraldisplay parenleftbigg 2 k q s x 3 parenrightbigg dx = 2 k q s integraldisplay d parenleftbigg 1 2 x 2 parenrightbigg = 2 k q s parenleftbigg 1 2 a 2 + 1 2 b 2 parenrightbigg . (1) Multiplying eq(1) by the electronic chargee, we arrive at the potential energy difference from D to C is given by U C U D = e ( V ( a ) V ( b )) = k eq s parenleftbigg 1 a 2 1 b 2 parenrightbigg . Intuitive reasoning on the sign of Δ U : Nat ural tendency of the motion is from high po tential energy to lower potential energy. Since when the electron is released it should move from C to D, so U C > U D . Alternative explanation: Let the center of the dipole be at the origin. At a distance x along the +ˆ x direction V dipole ( x ) = V q parenleftBig x + s 2 parenrightBig + V − q parenleftBig x s 2 parenrightBig V dipole ( x ) = kq x + s/ 2 + k ( q ) x s/ 2 V dipole ( x ) = ( x s/ 2) ( x + s/ 2) x 2 ( s/ 2) 2 ≈  kqs x 2 So, we have V C ( x ) V D ( x ) = k q s parenleftbigg 1 a 2 1 b 2 parenrightbigg Check: E is along the ˆ x direction, V C is expected to be lower than V D . 002 10.0 points A long currentcarrying wire, oriented North South, lies on a table (it is connected to bat teries which are not shown). A compass lies on top of the wire, with the compass needle about 3 mm above the wire. With the cur rent running, the compass deflects 16 ◦ to the West. At this location, the horizontal com ponent of the Earth’s magnetic field is about 2 × 10 − 5 T. Version 104 – Exam 2 – chiu – (57460) 2 What is the magnitude of the magnetic field at location A, on the table top, a distance 2 . 5 cm to the East of the wire, due only to the current in the wire? 1. 6.88e07 2. 5.54e07 3. 4.02e07 4. 3.66e07 5. 6.43e07 6. 3.8e07 7. 5.33e07 8. 3.39e07 9. 3.92e07 10. 5.13e07 Correct answer: 6 . 88 × 10 − 7 T....
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This note was uploaded on 03/27/2011 for the course PHYSICS 303LMI taught by Professor Chiu during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Chiu
 Physics

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