Exam 2-solutions physics - Version 104 – Exam 2 – chiu...

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Unformatted text preview: Version 104 – Exam 2 – chiu – (57460) 1 This print-out should have 16 questions. Multiple-choice 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 charge-e, 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 current-carrying 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. 6.88e-07 2. 5.54e-07 3. 4.02e-07 4. 3.66e-07 5. 6.43e-07 6. 3.8e-07 7. 5.33e-07 8. 3.39e-07 9. 3.92e-07 10. 5.13e-07 Correct answer: 6 . 88 × 10 − 7 T....
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This note was uploaded on 03/27/2011 for the course PHYSICS 303L-MI taught by Professor Chiu during the Spring '11 term at University of Texas at Austin.

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Exam 2-solutions physics - Version 104 – Exam 2 – chiu...

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