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Unformatted text preview: gonzales (pag757) – Homework10 – Hoffmann – (56745) 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. 3 . 6cm 32 cm 9 . 5cm (50A)sin bracketleftBig (272rad / s) t + δ bracketrightBig → 47 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: 27 . 8287 mV. Explanation: Let : N = 47 , ω = 272 rad / s , ℓ = 9 . 5 cm , a = 3 . 6 cm , b = 32 cm , and I = 50 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 )(47) 2(3 . 14159) × (50 A)(9 . 5 cm)(272 rad / s) × ln bracketleftbigg (3 . 6 cm) + (32 cm) (3 . 6 cm) bracketrightbigg × parenleftBig 10 − 2 m cm parenrightBig parenleftbigg 10 3 mV V parenrightbigg = 27 . 8287 mV . gonzales (pag757) – Homework10 – Hoffmann – (56745) 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. There is zero current in the loop. 2. The current in the loop is counter clockwise. correct 3. The sense of current is not determined with the information given. 4. The current in the loop is clockwise. 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 = 2 . 36 cm by b = 2 . 7 cm) and inner radius 2 . 77 cm consists of N = 460 turns of wire that carries a current I = I sin ω t , with I = 60 . 5 A and a frequency f = 34 . 3 Hz....
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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