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Unformatted text preview: Version 098 – Midterm 03 – Yao – (59110) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Given: Assume the bar and rails have neg ligible resistance and friction. In the arrangement shown in the figure, the resistor is 7 Ω and a 1 T magnetic field is directed out of the paper. The separation between the rails is 7 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 3 m / s . m ≪ 1g 3 m / s 7Ω 1 T 1 T I 7m At what rate is energy dissipated in the resistor? 1. 6.25 2. 63.0 3. 3657.14 4. 256.0 5. 384.0 6. 12544.0 7. 7503.12 8. 1152.0 9. 1393.78 10. 6480.0 Correct answer: 63 W. Explanation: Basic Concept: Motional E E = B ℓ v . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B ℓ v = (1 T) (7 m) (3 m / s) = 21 V . From Ohm’s law, the current flowing through the resistor is I = E R = B ℓ v R = (1 T) (7 m) (3 m / s) R = 3 A . The power dissipated in the resistor is P = I 2 R = B 2 ℓ 2 v 2 R 2 R = B 2 ℓ 2 v 2 R = (1 T) 2 (7 m) 2 (3 m / s) 2 (7 Ω) = 63 W . Note: Third of four versions. 002 10.0 points At times prior to t = 0, the switch is open. The switch is closed at t = 0. 12 mH 4 kΩ 240 V S I When I = 27 mA, what is the potential difference across the inductor? 1. 132.0 2. 210.0 3. 92.0 4. 215.6 5. 218.8 Version 098 – Midterm 03 – Yao – (59110) 2 6. 112.0 7. 84.0 8. 128.0 9. 188.0 10. 140.0 Correct answer: 132 V. Explanation: L R E S I Let : R = 4 kΩ = 4000 Ω , L = 12 mH = 0 . 012 H , and E = 240 V . Basic Concepts: LR Circuits. The simplest method of solution is simply to set the sum of the voltages around the circuit to 0. Then V L = E − V R = E − I R = (240 V) − (0 . 027 A) (4000 Ω) = 132 V . 003 10.0 points What current is required in the windings of a long solenoid that has 2810 turns uni formly distributed over a length of 0 . 162 m in order to produce a magnetic field of magnitude 0 . 000115 T at the center of the solenoid? The permeability of free space is 4 π × 10 − 7 T · m / A . 1. 110.286 2. 13.686 3. 48.568 4. 1.09715 5. 171.846 6. 106.462 7. 5.2759 8. 230.872 9. 70.7077 10. 38.5826 Correct answer: 5 . 2759 mA. Explanation: Let : N = 2810 , L = 0 . 162 m , B = 0 . 000115 T , and μ = 4 π × 10 − 7 T · m / A . The magnetic field inside a long solenoid is B = μ n I = μ parenleftbigg N L parenrightbigg I . Thus the required current is I = B L μ N = (0 . 000115 T) (0 . 162 m) (4 π × 10 − 7 T · m / A) (2810) × 1000 mA 1 A = 5 . 2759 mA . 004 10.0 points A singly charged positive ion has a mass of 3 . 68 × 10 − 26 kg. After being accelerated through a potential difference of 369 V, the ion enters a magnetic field of 0 . 58 T, in a direction perpendicular to the field....
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This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.
 Spring '09
 YAO
 Physics, Resistance

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