11 - Version 003 – Midterm3 – li –(59050 1 This...

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Unformatted text preview: Version 003 – Midterm3 – li – (59050) 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 The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some electrons removed) to the east, what is the direction of the magnetic force on the object? 1. upward correct 2. toward the east 3. downward 4. toward the west Explanation: Use the right-hand rule: point your index finger east and your middle finger north. Your thumb points upward. 002 10.0 points A singly charged positive ion has a mass of 1 . 88 × 10 − 26 kg. After being accelerated through a potential difference of 215 V, the ion enters a magnetic field of 0 . 395 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field. 1. 4.288 2. 0.95032 3. 1.79829 4. 0.8751 5. 2.17081 6. 2.5402 7. 1.25478 8. 2.85525 9. 1.46868 10. 1.05448 Correct answer: 1 . 79829 cm. Explanation: Let : m = 1 . 88 × 10 − 26 kg , q = 1 . 60218 × 10 − 19 C , V = 215 V , and B = 0 . 395 T . We use conservation of energy to find the velocity of the ion upon entering the field, 1 2 m v 2 = q V , v = radicalbigg 2 q V m = radicalBigg 2 (1 . 60218 × 10 − 19 C) (215 V) 1 . 88 × 10 − 26 kg = 60535 . 6 m / s . Therefore, the radius is r = m v q B = (1 . 88 × 10 − 26 kg) (60535 . 6 m / s) (1 . 60218 × 10 − 19 C) (0 . 395 T) = 0 . 0179829 m = 1 . 79829 cm . 003 10.0 points A straight rod moves along parallel conduct- ing rails, as shown below. The rails are con- nected at the left side through a resistor so that the rod and rails form a closed rectangu- lar loop. A uniform field perpendicular to the movement of the rod exists throughout the region. Assume the rod remains in contact with the rails as it moves. The rod experiences no friction or air drag. The rails and rod have negligible resistance. 8 . 2g 7 . 1Ω 1 . 5 T 1 . 5 T . 78A 1 . 1m Version 003 – Midterm3 – li – (59050) 2 At what speed should the rod be moving to produce the downward current in the resistor? 1. 1.50471 2. 0.182543 3. 0.941538 4. 0.643452 5. 5.44923 6. 1.33333 7. 3.35636 8. 1.23333 9. 1.01111 10. 0.819349 Correct answer: 3 . 35636 m / s. Explanation: Let : I = 0 . 78 A , ℓ = 1 . 1 m , m = 8 . 2 g , R = 7 . 1 Ω , and B = 1 . 5 T . m R B B I ℓ From Ohm’s Law, the emf inside the loop is E = I R . and the motional emf is E = B ℓ v . Thus the velocity of the rod is v = E B ℓ = I R B ℓ = (0 . 78 A) (7 . 1 Ω) (1 . 5 T) (1 . 1 m) = 3 . 35636 m / s . 004 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....
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This note was uploaded on 11/02/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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11 - Version 003 – Midterm3 – li –(59050 1 This...

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