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# 22 - Shie Gary Homework 22 Due 4:00 am Inst Turner This...

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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 ω B r 2 where r is the length of the rod. The E in a disk is therefore the 1 2 ω B r 2 and we can solve for omega to get ω = 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 Ω? Correct answer: 0 . 982443 C. Explanation: From Faraday’s Law for Solenoids E = - N d Φ B dt and Ohm’s Law I = V R , the current through R is I = V R = E R = N R d Φ B dt = N d B A dt 1 R = N d B dt A R . Integrating both sides of the equation above yields Z t t 0 I dt = Z t t 0 N A R d B dt dt = Z B - B N A R dB = N A R Δ B = N A R 2 B . The left hand side of the above equation is just the charge flowing through the R during

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Shie, Gary – Homework 22 – Due: Oct 25 2004, 4:00 am – Inst: Turner 2 this period of time! So, Q = Z t t 0 I dt = N A R 2 B = 0 . 982443 C .
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