21 - Shie, Gary Homework 21 Due: Oct 22 2004, 4:00 am Inst:...

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Shie, Gary – Homework 21 – Due: Oct 22 2004, 4:00 am – Inst: Turner 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 2) 10 points Assume: The rod remains in contact with the rails as it slides down the rails. The rod experiences no Friction or air drag. The rails and rod have negligible resistance. A straight, horizontal rod slides along par- allel conducting rails at an angle with the horizontal, as shown below. The rails are con- nected at the bottom by a horizontal rail so that the rod and rails Forms a closed rect- angular loop. A uniForm vertical feld exists throughout the region. The acceleration oF gravity is 9 . 8 m / s 2 . 6 . 9 Ω 1 . 6m / s 60 g 0 . 56 T sliding rod 0 . 68 m Viewed From above 6 m 28 0 . 56 T Viewed From the side IF the velocity oF the rod is 1 . 6 m / s, what is the current through the resistor? Correct answer: 77 . 9655 mA. Explanation: Let : = 0 . 68 m , m = 60 g , R = 6 . 9 Ω , v = 1 . 6 m / s , and B = 0 . 56 T . R v m B sliding rod Viewed From above v θ B Viewed From the side Basic Concepts: E = - d Φ B dt The movement oF the rod decreases the area oF the loop, so the ±ux through the loop is changing in time, and there is an induced emF E . IF we denote the area by A , this induced emF is E = - d Φ dt = - d ( B A cos θ ) dt = - B cos θ dA dt . since the ±ux is B · A = BA cos θ , where θ is the angle between the magnetic feld B and the normal vector to the area. The magnetic feld and the angle are both constant and were pulled out oF the di²erentiation. Now, iF we call the distance From the rod to the resistor
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Shie, Gary – Homework 21 – Due: Oct 22 2004, 4:00 am – Inst: Turner 2 x , the emf becomes E = - B cos θ d ( ‘x ) dt = - B ‘ cos θ dx dt = - B ‘v cos θ . Thus the current in the resistor is I = |E| R = B ‘v cos θ R = = (0 . 56 T) (0 . 68 m) (1 . 6 m / s) cos(28 ) (6 . 9 Ω) = 0 . 0779655 A = 77 . 9655 mA . 002 (part 2 of 2) 10 points What is the terminal velocity of the rod? Correct answer: 16 . 8489 m / s. Explanation: The terminal velocity is reached when the forces on the rod cancel, so it feels no more acceleration. The force from the induced cur- rent is, since the rod is perpendicular to the magnetic Feld, F B,total = I ‘B . However, this force is directed parallel to the
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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.

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21 - Shie, Gary Homework 21 Due: Oct 22 2004, 4:00 am Inst:...

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