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Unformatted text preview: Niu (qn269) Homework #7 antoniewicz (58855) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The primary of a step-down transformer has 350 turns turns and is connected to a 120 V rms line. The secondary is to supply 24 A at 9 V. Find the rms current in the primary. Correct answer: 1 . 8 A. Explanation: Let : V p = 120 V , V s = 9 V , and I p = 24 A . Because of 100 percent efficiency, we have I p V p = I s V s , so the current in the primary is I p = V s V p I s = parenleftbigg 9 V 120 V parenrightbigg (24 A) = 1 . 8 A . 002 (part 2 of 2) 10.0 points Find the whole number of turns in the sec- ondary, assuming 100 percent efficiency. Correct answer: 26 turns. Explanation: Let : V p = 120 V , V s = 9 V , and N p = 350 turns . Because of 100 percent efficiency, we have V s N p = V p N s , so the number of turns in the secondary is N s = V s V p N p = parenleftbigg 9 V 120 V parenrightbigg 350 turns 26 turns . 003 10.0 points A superconducting solenoid is to be designed to have an interior field of B = 4 . 1 T with a maximum current of I = 1000 A. How many windings are required on a 1- meter solenoid length? Correct answer: 3262 . 68 m 1 . Explanation: 004 (part 1 of 2) 10.0 points Two concentric circular loops of radii b and 2 b , made of the same type of wire, lie in the plane of the page, as shown. The total resistance of the wire loop of radius b is R b = R . b 2 b What is the resistance of the wire loop of radius 2 b ? 1. R 2 b = R 2 2. R 2 b = 4 R 3. R 2 b = 2 R correct 4. R 2 b = R 4 5. R 2 b = R Explanation: Since the two loops are made of the same wire, the total resistances are proportional to Niu (qn269) Homework #7 antoniewicz (58855) 2 the length of the wires (the circumference of the loops). b = 2 b, and 2 b = 2 (2 b ) = 4 b = 2 b . Thus, the total resistance of the wire loop of radius 2 b is R 2 b = 2 R . 005 (part 2 of 2) 10.0 points A uniform magnetic field vector B that is perpen- dicular to the plane of the page now passes through the loops, as shown. B B B B b 2 b a The field is confined to a region of radius a , where a &lt; b , and is changing at a constant rate. The induced emf in the wire loop of radius b is E b = E . What is the induced emf in the wire loop of radius R 2 b = 2 b ? 1. E 2 b = 0 2. E 2 b = E 2 3. E 2 b = 4 E 4. E 2 b = 2 E 5. E 2 b = E correct Explanation: E =- d B dt , where B is the magnetic flux through the wire loop. Since b &gt; a , the magnetic flux through both wire loops is exactly the same. Thus, the induced emf in the wire loop of radius 2 b is also E 2 b = E ....
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This note was uploaded on 11/18/2009 for the course PHY 58855 taught by Professor Antoniewicz during the Spring '09 term at University of Texas.
- Spring '09