solution11_pdf

# solution11_pdf - hussain(wh4892 – hw 11 – Opyrchal...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hussain (wh4892) – hw 11 – Opyrchal – (121104) 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 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

{[ snackBarMessage ]}

### Page1 / 8

solution11_pdf - hussain(wh4892 – hw 11 – Opyrchal...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online