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Homework 09-solutions - sanne(as42476 Homework 09 Yao(59110...

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sanne (as42476) – Homework 09 – Yao – (59110) 1 This print-out should have 26 questions. Multiple-choice 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. force. 2. magnetic flux. 3. pressure. 4. magnetic field. correct 5. area. 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 33 turns, each with an area of 20 cm 2 , is positioned perpen- dicularly to a uniform magnetic field that in- creases its magnitude at a constant rate from 2 . 4 T to 8 . 1 T in 1 . 3 s. If the coil has a total resistance of 0 . 91 Ω, what is the magnitude of the induced current? Correct answer: 0 . 318005 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 = 0 . 318005 A . keywords: 003 10.0 points A two-turn circular wire loop of radius 0 . 431 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0 . 474 T. Now the wire is reshaped from a two-turn circle to a one-turn circle in 0 . 0526 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: 15 . 7768 V. Explanation: Basic Concept: Faraday’s Law is E = N d Φ B dt . Let : r 2 = 0 . 431 m , r 1 = 2 r 2 = 2 (0 . 431 m) = 0 . 862 m , A 2 = π r 2 2 = 0 . 583585 m 2 , A 1 = π r 2 1 = π (2 r 2 ) 2 = 2 . 33434 m 2 , Δ t = 0 . 0526 s , and B = 0 . 474 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 one-turn loop. r 1 is the radius of the new one-turn loop, and r 2 is the radius of the two-turn 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 . 431 m) 2 (0 . 474 T) = 0 . 829858 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
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sanne (as42476) – Homework 09 – Yao – (59110) 2 = vextendsingle vextendsingle vextendsingle vextendsingle ΔΦ B Δ t vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle (0 . 829858 Wb) (0 . 0526 s) vextendsingle vextendsingle vextendsingle vextendsingle = 15 . 7768 V . 004 10.0 points A circular conducting loop is held fixed in a uniform magnetic field that varies in time according to B ( t ) = B 0 exp( a t ) where t is in s, a is in s - 1 and B is the field strength in T at t = 0. At t = 0, the emf induced in the loop is 0 . 084 V . At t = 1 . 97 s, the emf is 0 . 0304 V , .
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