# Midterm3 - Version 098 Midterm 03 Yao(59110 This print-out...

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Version 098 – Midterm 03 – Yao – (59110) 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 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 1 g 3 m / s 7 Ω 1 T 1 T I 7 m 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 2 v 2 R 2 R = B 2 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

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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 0 = 4 π × 10 7 T · m / A . The magnetic field inside a long solenoid is B = μ 0 n I = μ 0 parenleftbigg N L parenrightbigg I . Thus the required current is I = B L μ 0 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. Calculate the radius of the path of the ion in the field. 1. 1.39613 2. 3.70546 3. 0.965351 4. 1.23198 5. 1.14143 6. 1.84606 7. 3.96792 8. 2.5402 9. 2.24475 10. 3.21092 Correct answer: 2 . 24475 cm. Explanation:
Version 098 – Midterm 03 – Yao – (59110) 3 Let : m = 3 . 68 × 10 26 kg , q = 1 . 60218 × 10 19 C , V = 369 V , and B = 0 . 58 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) (369 V) 3 . 68 × 10 26 kg = 56683 . 9 m / s .

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