Exam 2-solutions

# Exam 2-solutions - Version 012 Exam 2 shih(57480 This...

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Version 012 – Exam 2 – shih – (57480) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points What current is required in the windings of a long solenoid that has 518 turns uniformly distributed over a length of 0 . 309 m in order to produce inside the solenoid a magnetic field of magnitude 9 . 85 × 10 5 T? The permeablity of free space is 1 . 25664 × 10 6 T m / A. 1. 48.888 2. 36.2653 3. 58.2014 4. 16.9401 5. 26.6502 6. 46.7579 7. 19.2122 8. 20.3536 9. 28.8324 10. 24.6681 Correct answer: 46 . 7579 mA. Explanation: Let : N = 518 , = 0 . 309 m , and B = 9 . 85 × 10 5 T . Magnetic field inside the solenoid is B = μ 0 N I , then I = B ℓ μ 0 N = (9 . 85 × 10 5 T) (0 . 309 m) μ 0 (518) = 46 . 7579 mA . keywords: 002(part1of2)10.0points Consider two conductors 1 and 2 made of the same ohmic material; i.e. , ρ 1 = ρ 2 . Denote the length by , the cross sectional area by A . The same voltage V is applied across the ends of both conductors. The field E is inside of the conductor. V 1 vector E 1 I 1 1 r 1 V 2 vector E 2 I 2 2 r 2 If A 2 = 2 A 1 , ℓ 2 = 2 1 and V 2 = V 1 , find the ratio E 2 E 1 of the electric fields. 1. E 2 E 1 = 1 8 2. E 2 E 1 = 2 3. E 2 E 1 = 4 4. E 2 E 1 = 1 4 5. E 2 E 1 = 1 2 correct 6. E 2 E 1 = 1 16 7. E 2 E 1 = 8 8. E 2 E 1 = 1 3 9. E 2 E 1 = 1 12 10. E 2 E 1 = 1 Explanation: E 2 E 1 = V 2 2 V 1 1 = 1 2 = 1 2 1 = 1 2 . 003(part2of2)10.0points Determine the ratio I 2 I 1 of the currents .

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Version 012 – Exam 2 – shih – (57480) 2 1. I 2 I 1 = 1 16 2. I 2 I 1 = 4 3. I 2 I 1 = 2 4. I 2 I 1 = 1 4 5. I 2 I 1 = 1 correct 6. I 2 I 1 = 1 12 7. I 2 I 1 = 8 8. I 2 I 1 = 1 2 9. I 2 I 1 = 1 3 10. I 2 I 1 = 1 8 Explanation: I 2 I 1 = V 2 R 2 V 1 R 1 = R 1 R 2 = ρ 1 parenleftbigg 1 A 1 parenrightbigg ρ 2 parenleftbigg 2 A 2 parenrightbigg = 1 A 1 2 1 2 A 1 = 1 . 004 10.0points Two long parallel wires are a distance 2 a apart, as shown. Point P is in the plane of the wires and a distance a from wire X . When there is a current I in wire X and no current in wire Y , the magnitude of the magnetic field at P is B 0 . 2 a a X Y P When there are equal currents I in the same direction in both wires, the magnitude of the magnetic field at P is 1. B P = 4 3 B 0 correct 2. B P = 5 3 B 0 3. B P = 10 9 B 0 4. B P = 2 B 0 5. B P = 11 9 B 0 6. B P = B 0 7. B P = 2 3 B 0 8. B P = 7 3 B 0 9. B P = 5 9 B 0 10. B P = 7 9 B 0 Explanation: The superposition principle is applicable to the magnetic field. The magnetic field at P is the sum of two component magnetic fields due to wires X and Y respectively. Furthermore, we know the fields due to X and Y are in the same direction. So we can consider the magnitudes only. B X is known to be B 0 . Applying Am- pere’s law to the magnetic field due to a long straight wire, B = μ 0 I 2 πr 1 r . Therefore, B Y = 1 3 B X = 1 3 B 0 . Finally, we obtain B P = B X + B Y = parenleftbigg 1 + 1 3 parenrightbigg B 0 = 4 3 B 0 . 005 10.0points Find the equivalent resistance between points a and b in the figure.
Version 012 – Exam 2 – shih – (57480) 3 4 . 3 Ω 1 . 5 Ω 3 . 4 Ω 4 . 5 Ω 3 . 1 Ω a b 1. 6.62283 2. 7.9051 3. 8.17143 4. 9.95 5. 8.79328 6. 8.60702 7. 7.31186 8. 10.3432 9. 8.95275 10. 10.2167 Correct answer: 8 . 95275 Ω.

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