GPII - Homework 6, Induction - solutions

GPII - Homework 6, Induction - solutions - honea (cth632)...

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Unformatted text preview: honea (cth632) – H6Induction – avram – (1956) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circular coil enclosing an area of 107 cm 2 is made of 207 turns of copper wire as shown schematically in the figure. Initially, a 1 . 02 T uniform magnetic field points perpendicularly left-to-right through the plane of the coil. The direction of the field then reverses to right-to- left. The field reversal takes 1.0 ms. R Magnetic Field B ( t ) During the time the field is changing its direction, how much charge flows through the coil if the resistance is 3 . 75 Ω? Correct answer: 1 . 20491 C. Explanation: From Faraday’s Law for Solenoids E =- N d Φ B dt and Ohm’s Law I = V R , the current through R is I = V R = E R = N R d Φ B dt = N d B A dt 1 R = N d B dt A R . Integrating both sides of the equation above yields integraldisplay t t I dt = integraldisplay t t N A R d B dt dt = integraldisplay B − B N A R dB = N A R Δ B = N A R 2 B . The left hand side of the above equation is just the charge flowing through the R during this period of time! So, Q = integraldisplay t t I dt = N A R 2 B = 1 . 20491 C . 002 10.0 points A two-turn circular wire loop of radius . 383 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0 . 19 T. If the entire wire is reshaped from a two- turn circle to a one-turn circle in 0 . 136 s (while remaining in the same plane), what is the magnitude of the average induced emf E in the wire during this time? Use Faraday’s law in the form E =- Δ( N Φ) Δ t . Correct answer: 1 . 28763 V. Explanation: The average induced emf is given by E =- Δ N Φ Δ t =- N ′ Φ ′- N Φ Δ t =- B ( N ′ A ′- N A ) Δ t . In this case, N = 2 turns, N ′ = 1 turn, and A = π r 2 = π (0 . 383 m) 2 = 0 . 460837 m 2 . Since the total length of the wire is constant, the circumference of the 1 turn coil must be honea (cth632) – H6Induction – avram – (1956) 2 twice the circumference of the 2 turns coil: C ′ = 2 C . Thus 2 π r ′ = 2 (2 π r ) , or r ′ = 2 r = 0 . 766 m , and A ′ = π r ′ 2 = π (0 . 766 m) 2 = 1 . 84335 m 2 . Therefore, the magnitude of the induced emf is E = bracketleftbig 1 · (1 . 84335 m 2 )- 2 · (0 . 460837 m 2 ) bracketrightbig × . 19 T . 136 s = 1 . 28763 V . 003 10.0 points The plane of a rectangular coil, 8 cm by 12 . 6 cm, is perpendicular to the direction of a uniform magnetic field B . If the coil has 119 turns and a total resis- tance of 3 . 4 Ω, at what rate must the magni- tude of B change to induce a current of 0 . 14 A in the windings of the coil?...
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This note was uploaded on 04/12/2011 for the course PHYS 1402 taught by Professor Avram during the Spring '10 term at Austin Community College.

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GPII - Homework 6, Induction - solutions - honea (cth632)...

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