This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: johnson (rj6247) hw 11 Opyrchal (121014) 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 10.0 points A coil is wrapped with 224 turns of wire on the perimeter of a square frame of sides 21 . 6 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 2 . 42 . A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to- . 237 Wb / m 2 in a time of 1 . 16 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 2 . 13524 V. Explanation: Basic Concept: Faradays Law is E =- d B dt . Solution: The magnetic flux through the loop at t = 0 is zero since B = 0. At t = 1 . 16 s , the magnetic flux through the loop is B = B A =- . 0110575 Wb . Therefore the magnitude of the induced emf is E = N B t = (224 turns) [(- . 0110575 Wb)- 0] (1 . 16 s) =- 2 . 13524 V |E| = 2 . 13524 V . 002 10.0 points The plane of a rectangular coil of dimension 5 cm by 8 cm is perpendicular to the direction of a magnetic field B . The coil has 49 . 6 turns and a total resistance of 7 . 21 . At what rate must the magnitude of B change in order to induce a current of 0 . 104 A in the windings of the coil? Correct answer: 3 . 77944 T / s. Explanation: Basic Concepts: Faradays Law of Induc- tion E =- d B dt Ohms law I = V R Solution: With Ohms law, the emf induced in one turn of coil is E 1 = I R n = (0 . 104 A) (7 . 21 ) (49 . 6 turns) = 0 . 0151177 V , while with Faradays law, we get E 1 = d B dt = d ( A B ) dt = A d B dt . So, the rate at which magnetic field changes will be d B dt = E 1 A = (0 . 0151177 V) (0 . 004 m 2 ) = 3 . 77944 T / s . 003 (part 1 of 2) 10.0 points The clockwise circulating current in a solenoid is increasing at a rate of 4 . 88 A / s. The cross- sectional area of the solenoid is 3 . 14159 cm 2 , and there are 464 turns on its 12 . 8 cm length. What is the magnitude of the induced E produced by the increasing current? Correct answer: 3 . 24045 mV. Explanation: Faradays Law for solenoid: E =- N d dt =- N A d B dt . Magnetic field produced by the changing current is B = N I L johnson (rj6247) hw 11 Opyrchal (121014) 2 d B d t = N L d I d t . Faradays Law for solenoid: E =- N d dt =- N d ( BA ) dt =- N 2 A L d I d t . Magnetic field induced by current: B = N I L . Thus, the induced E is |E| = N 2 L A d I dt = (1 . 25664 10- 6 N / A 2 ) (464) 2 12 . 8 cm (3 . 14159 cm 2 ) (4 . 88 A / s) parenleftbigg 10 3 mV V parenrightbiggparenleftbigg 1 10- 2 m cm parenrightbigg = 3 . 24045 mV ....
View Full Document
- Spring '08