22 - Shie, Gary Homework 22 Due: Oct 25 2004, 4:00 am Inst:...

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Unformatted text preview: Shie, Gary Homework 22 Due: Oct 25 2004, 4:00 am Inst: Turner 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A metal disk lies in the xy-plane, centered about the origin, and rotates with a constant angular velocity about the z-axis. There is a uniform 0 . 0749 T magnetic field parallel to the z-axis. The radius of the disk is 1 . 59 m. At what rate is the disk turning if there is an E of 4 . 59 V develops between the center of the disk and a point on its rim? Correct answer: 48 . 4804 rad / s. Explanation: Imagine a bicycle wheel with spokes. Now let the density of the spokes increase until they fill the wheel. From this you should begin to see that a solid disk can be treated as an infinite number of rods. Why do we do this? Because the solution to a rod rotating in a magnetic field is easy to find; the induced E is just E = 1 2 B r 2 where r is the length of the rod. The E in a disk is therefore the 1 2 B r 2 and we can solve for omega to get = 2 E B r 2 = 48 . 4804 rad / s . 002 (part 1 of 1) 10 points A circular coil enclosing an area of 85 . 5 cm 2 is made of 232 turns of copper wire as shown schematically in the figure. Initially, a 1 . 05 T uniform magnetic field points perpendicularly left-to-right through the plane of the coil. The direction of the field then reverses to right-to- left. The field reversal takes 1.0 ms. R Magnetic Field B ( t ) During the time the field is changing its direction, how much charge flows through the coil if the resistance is 4 . 24 ? Correct answer: 0 . 982443 C. Explanation: From Faradays Law for Solenoids E =- N d B dt and Ohms Law I = V R , the current through R is I = V R = E R = N R d B dt = N dB A dt 1 R = N dB dt A R . Integrating both sides of the equation above yields Z t t I dt = Z t t N A R dB dt dt = Z B- B N A R dB = N A R B = N A R 2 B . The left hand side of the above equation is just the charge flowing through the R during Shie, Gary Homework 22 Due: Oct 25 2004, 4:00 am Inst: Turner 2 this period of time! So, Q = Z t t I dt = N A R 2 B = 0 . 982443 C ....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas at Austin.

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22 - Shie, Gary Homework 22 Due: Oct 25 2004, 4:00 am Inst:...

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