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Unformatted text preview: thwing (jrt2329) – Homework 09 – Yao – (59110) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The Wb/m 2 is a unit of 1. magnetic flux. 2. force. 3. pressure. 4. area. 5. magnetic field. correct Explanation: We know that Wb is the unit of magnetic flux, so Wb/m 2 is the unit of magnetic field, in fact 1 Wb/m 2 = 1 T. 002 10.0 points A flat coil of wire consisting of 30 turns, each with an area of 120 cm 2 , is positioned per pendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 3 . 8 T to 16 T in 4 . 5 s. If the coil has a total resistance of 1 Ω, what is the magnitude of the induced current? Correct answer: 0 . 976 A. Explanation: E = − d Φ B dt Φ B = N integraldisplay vector B · d vector A = N B A E = N A ( B 2 − B 1 ) t I = E R = N A ( B 2 − B 1 ) R t = . 976 A . keywords: 003 10.0 points A twoturn circular wire loop of radius . 622 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0 . 594 T. Now the wire is reshaped from a twoturn circle to a oneturn circle in 0 . 0854 s (while remaining in the same plane). What is the magnitude of the average in duced emf in the wire during this time? Correct answer: 25 . 3618 V. Explanation: Basic Concept: Faraday’s Law is E = − N d Φ B dt . Let : r 2 = 0 . 622 m , r 1 = 2 r 2 = 2 (0 . 622 m) = 1 . 244 m , A 2 = π r 2 2 = 1 . 21543 m 2 , A 1 = π r 2 1 = π (2 r 2 ) 2 = 4 . 86173 m 2 , Δ t = 0 . 0854 s , and B = 0 . 594 T . The wire has a constant length, conse quently the circumference (and radius) of the one turn loop will be twice the circumfer ence (and radius) of the two turn loop, since c = 2 π r . When the wire loop’s shape changed, the radius also changed; i.e. , r 1 = 2 r 2 , where subscript 1 denote the new oneturn loop. r 1 is the radius of the new oneturn loop, and r 2 is the radius of the twoturn loop. The the change in area A = π r 2 , lead to the change of the magnetic flux. ΔΦ B = Φ B 1 − Φ B 2 = π (2 r 2 ) 2 · B − π r 2 2 · B = 3 π r 2 2 B = 3 π (0 . 622 m) 2 (0 . 594 T) = 2 . 1659 Wb . From Faraday’s law, the average induced emf in this period is E = vextendsingle vextendsingle vextendsingle vextendsingle − d Φ B dt vextendsingle vextendsingle vextendsingle vextendsingle thwing (jrt2329) – Homework 09 – Yao – (59110) 2 = vextendsingle vextendsingle vextendsingle vextendsingle ΔΦ B Δ t vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle (2 . 1659 Wb) (0 . 0854 s) vextendsingle vextendsingle vextendsingle vextendsingle = 25 . 3618 V ....
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This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.
 Spring '09
 YAO
 Physics, Force, Work

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