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# Test 3 - Version 118/ABDBC – midterm 03 – Turner...

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Unformatted text preview: Version 118/ABDBC – midterm 03 – Turner – (60230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A magnetic dipole is falling in a conducting metallic tube. Consider the induced current in an imaginary current loop when the magnet is moving away from the upper loop and when the magnet is moving toward the lower loop. N S N pole I above dipole magnet v S pole I below ? ? z x y Determine the directions of the induced currents I above and I below in an imaginary loop shown in the figure, as viewed from above, when the loop is above the falling magnet and when the loop is below the falling magnet. 1. I above = clockwise and I below = clockwise 2. I above = clockwise and I below = counter-clockwise 3. I above = counter-clockwise and I below = counter-clockwise 4. I above = counter-clockwise and I below = clockwise correct 5. no current flow Explanation: When the falling magnet is below the up- per loop, −→ μ ind must be up to attract the falling magnet and slow it down; i.e. , counter- clockwise as viewed from above. Before reaching the lower loop, −→ μ ind must be down to oppose the falling magnet; i.e. , clockwise as viewed from above. N S N pole I above dipole magnet v S pole I below z x y 002 10.0 points A conducting bar moves as shown near a long wire carrying a constant I = 100 A current. I a v L A B If a = 12 mm, L = 78 cm, and v = 9 . 8 m / s, what is the potential difference, Δ V ≡ V A − V B ? 1. 85.8701 2. 12.74 3. 25.6286 4. 45.2308 Version 118/ABDBC – midterm 03 – Turner – (60230) 2 5. 90.5589 6. 37.4026 7. 26.4 8. 113.554 9. 102.358 10. 43.7349 Correct answer: 12 . 74 mV. Explanation: Given; E = − d Φ B dt = − d dt ( B ℓ x ) = − B ℓ dx dt = − B ℓ v . From Ampere’s law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire is B = μ I 2 π a . Hence the potential difference is Δ V = B Lv = parenleftbigg μ I 2 π a parenrightbigg ( Lv ) = 12 . 74 mV . 003 10.0 points The wire is carrying a current I . x y I I I 180 ◦ O r Find the magnitude of the magnetic field vector B at O due to a current-carrying wire shown in the figure, where the semicircle has radius r , and the straight parts to the left and to the right extend to infinity. 1. B = μ I r 2. B = μ I 4 r correct 3. B = μ I π r 4. B = μ I 4 π r 5. B = μ I 2 π r 6. B = μ I 3 π r 7. B = μ I 3 r 8. B = μ I 2 r Explanation: By the Biot-Savart Law, vector B = μ I 4 π integraldisplay dvectors × ˆ r r 2 . Consider the left straight part of the wire. The line element dvectors at this part, if we come in from ∞ , points towards O, i.e., in the x- direction. We need to find dvectors × ˆ r to use the Biot-Savart Law. However, in this part of the wire, ˆ r is pointing towards O as well, so dvectors and ˆ r are parallel meaning dvectors × ˆ r = 0 for this part of the wire. It is now easy to see that thepart of the wire....
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Test 3 - Version 118/ABDBC – midterm 03 – Turner...

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