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Exam 2-solutions physics

# Exam 2-solutions physics - Version 104 Exam 2 chiu(57460...

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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 10.0points 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 e q 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 e q s parenleftbigg 1 a 2 - 1 b 2 parenrightbigg 5. - k e q s parenleftbigg 1 a - 1 b parenrightbigg 6. -2 k e q s parenleftbigg 1 a - 1 b parenrightbigg 7. -2 k e q s parenleftbigg 1 a 2 - 1 b 2 parenrightbigg 8. k e q s parenleftbigg 1 a - 1 b parenrightbigg 9. k e q s parenleftbigg 1 a 2 - 1 b 2 parenrightbigg correct 10. 2 k e q 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 e q 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 10.0points 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.

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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. Explanation: let : B earth = 2 × 10 5 T , r compass = 3 mm = 0 . 003 m , r A = 2 . 5 cm = 0 . 025 m , and θ = 16 . The magnetic field due to a wire is B = μ 0 4 π 2 I r Since tan ( θ ) = B compass B earth , B compass = B earth tan ( θ ). By writing the equation for the magnetic field of a wire for the compass and for point A and dividing the two equa- tions, we obtain B A = B compass r compass r A = B earth tan ( θ ) r compass r A = (2 × 10 5 T) tan (16 ) 0 . 003 m 0 . 025 m = 6 . 88 × 10 7 T .
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