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Unformatted text preview: soto (rrs946) – Homework 10 – Hoffmann – (56605) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A long, straight wire carries a current and lies in the plane of a rectangular loops of wire, as shown in the figure. 6 . 6cm 26 cm 14cm (33A)sin bracketleftBig (159rad / s) t + δ bracketrightBig → 130 loops (turns) Determine the maximum emf , E , induced in the loop by the magnetic field created by the current in the straight wire. Correct answer: 30 . 5059 mV. Explanation: Let : N = 130 , ω = 159 rad / s , ℓ = 14 cm , a = 6 . 6 cm , b = 26 cm , and I = 33 A . a b ℓ r dr I = I sin( ωt + δ ) Magnetic field near a long wire is B = μ I 2 π r . Faraday’s Law is E = d Φ B dt . The magnitude of the magnetic field is B = μ I 2 π r . Thus the flux linkage is Φ B = N Φ B 1 = μ N I ℓ 2 π integraldisplay a + b a dr r = μ N I ℓ 2 π ln bracketleftbigg a + b a bracketrightbigg sin( ω t + δ ) . Finally, the induced emf E is E = d Φ B dt = μ N I ℓ ω 2 π ln bracketleftbigg a + b a bracketrightbigg cos( ω t + δ ) , which is a maximum when the cosine function yields 1. E max = μ N 2 π I ℓ ω ln bracketleftbigg a + b a bracketrightbigg = (1 . 25664 × 10 − 6 N / A 2 )(130) 2(3 . 14159) × (33 A)(14 cm)(159 rad / s) × ln bracketleftbigg (6 . 6 cm) + (26 cm) (6 . 6 cm) bracketrightbigg × parenleftBig 10 − 2 m cm parenrightBig parenleftbigg 10 3 mV V parenrightbigg = 30 . 5059 mV . soto (rrs946) – Homework 10 – Hoffmann – (56605) 2 002 (part 2 of 2) 10.0 points If at t = 0, δ =0, and a positive I denotes an upward moving current in the figure, then which statement below is correct at t = 0. 1. The sense of current is not determined with the information given. 2. The current in the loop is counter clockwise. correct 3. The current in the loop is clockwise. 4. There is zero current in the loop. Explanation: Note: d I dt ( t =0) = ω I . At t = 0, the magnetic field from the long straight wire is increasing into the loop from above which leads to a finite time rate of change of magnetic flux, although B ( t =0) = 0 . The induced current in the loop then at tempts to produce magnetic field through the loop that penetrates the loop from below ( i.e. , from Lenz’s law it is attempting to resist the change of flux). By the right hand rule, this requires counterclockwise induced current in the loop. 003 10.0 points A toroid having a rectangular cross section ( a = 1 . 58 cm by b = 2 . 71 cm) and inner radius 2 . 74 cm consists of N = 410 turns of wire that carries a current I = I sin ω t , with I = 35 A and a frequency f = 44 . 2 Hz. A loop that consists of N ℓ = 18 turns of wire links the toroid, as in the figure....
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.
 Fall '11
 HOFFMAN
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