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Unformatted text preview: bhakta (nnb232) – HW #6 – Antoniewicz – (56490) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 5.0 points A horizontal circular wire loop of radius 0 . 6 m lies in a plane perpendicular to a uniform magnetic field pointing from above into the plane of the loop, has a magnitude of 0 . 2 T. If in 0 . 15 s the wire is reshaped from a circle into a square, but remains in the same plane, what is the magnitude of the average induced emf in the wire during this time? Correct answer: 0 . 323612 V. Explanation: Let : r = 0 . 6 m , b = 0 . 2 T , and Δ t = 0 . 15 s . The average induced emf E is given by (E) = N parenleftbigg ΔΦ Δ t parenrightbigg = ΔΦ Δ t since N = 1, and ΔΦ = B ( A circle − A square ) = B ( π r 2 − A square ) . Also, the circumference of the circle is 2 π r , so each side of the square has a length L = 2 π r 4 = π r 2 , so A square = L 2 = parenleftBig π r 2 parenrightBig 2 . Thus ΔΦ = B bracketleftbigg π r 2 − parenleftBig π r 2 parenrightBig 2 bracketrightbigg = 0 . 2 T bracketleftBigg π (0 . 6 m) 2 − parenleftBigg π (0 . 6 m) 2 2 parenrightBiggbracketrightBigg = − . 0485418 T · m 2 . and the average induced emf is (E) = − − . 0485418 T · m 2 . 15 s = . 323612 V . 002 (part 2 of 2) 5.0 points The current in the loop during the deforma tion 1. flows in a direction that cannot be deter mined from given information. 2. flows clockwise when viewed from above. correct 3. does not arise. 4. flows counterclockwise when viewed from above. Explanation: The deformation causes the flux through the loop to decrease since the area of the loop is reduced. By Lenz’s law, the induced emf will cause the current to flow in the loop so as to induce a magnetic field that attempts to resist the change of magnetic flux through the loop. A clockwise flow of current, when viewed from above tends to increase the ex isting downward magnetic field through the loop, thereby resisting the decrease of mag netic flux through the loop. 003 (part 1 of 2) 5.0 points A straight rod moves along parallel conduct ing rails, as shown below. The rails are con nected at the left side through a resistor so that the rod and rails form a closed rectangu lar loop. A uniform field perpendicular to the movement of the rod exists throughout the region. Assume the rod remains in contact with the rails as it moves. The rod experiences no friction or air drag. The rails and rod have negligible resistance. bhakta (nnb232) – HW #6 – Antoniewicz – (56490) 2 7 . 8g 4 . 6Ω 2 . 4 T 2 . 4 T . 47A 1 . 4m At what speed should the rod be moving to produce the downward current in the resistor?...
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This note was uploaded on 02/01/2011 for the course PHYSICS 302L taught by Professor Antanioweitz during the Fall '10 term at University of Texas.
 Fall '10
 Antanioweitz
 Physics, Work

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