This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hussain (wh4892) – hw 11 – Opyrchal – (121104) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A coil is wrapped with 173 turns of wire on the perimeter of a square frame of sides 11 . 8 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 2 . 46 Ω. A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to . 0419 Wb / m 2 in a time of 1 . 17 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 0 . 0862657 V. Explanation: Basic Concept: Faraday’s Law is E = d Φ B dt . Solution: The magnetic flux through the loop at t = 0 is zero since B = 0. At t = 1 . 17 s , the magnetic flux through the loop is Φ B = B A = . 000583416 Wb . Therefore the magnitude of the induced emf is E = N · ΔΦ B Δ t = (173 turns) [( . 000583416 Wb) 0] (1 . 17 s) = . 0862657 V E = 0 . 0862657 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 79 . 3 turns and a total resistance of 15 . 3 Ω. At what rate must the magnitude of B change in order to induce a current of 0 . 122 A in the windings of the coil? Correct answer: 5 . 88462 T / s. Explanation: Basic Concepts: Faraday’s Law of Induc tion E = d Φ B dt Ohm’s law I = V R Solution: With Ohm’s law, the emf induced in one turn of coil is E 1 = I R n = (0 . 122 A) (15 . 3 Ω) (79 . 3 turns) = 0 . 0235385 V , while with Faraday’s law, we get E 1 = d Φ 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 . 0235385 V) (0 . 004 m 2 ) = 5 . 88462 T / s . 003 (part 1 of 2) 10.0 points The clockwise circulating current in a solenoid is increasing at a rate of 9 . 08 A / s. The cross sectional area of the solenoid is 3 . 14159 cm 2 , and there are 338 turns on its 18 . 3 cm length. What is the magnitude of the induced E produced by the increasing current? Correct answer: 2 . 23783 mV. Explanation: Faraday’s Law for solenoid: E = N d Φ dt = N A d B dt . Magnetic field produced by the changing current is B = μ N I L hussain (wh4892) – hw 11 – Opyrchal – (121104) 2 d B d t = μ N L d I d t . Faraday’s Law for solenoid: E = N d Φ dt = N d ( BA ) dt = N 2 A L μ d I d t . Magnetic field induced by current: B = μ N I L . Thus, the induced E is E = μ N 2 L A d I dt = (1 . 25664 × 10 6 N / A 2 ) (338) 2 18 . 3 cm × (3 . 14159 cm 2 ) (9 . 08 A / s) × parenleftbigg 10 3 mV V parenrightbiggparenleftbigg 1 10 2 m cm parenrightbigg = 2 . 23783 mV ....
View
Full
Document
 Spring '08
 moro
 Magnetic Field

Click to edit the document details