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

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Unformatted text preview: 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 – find 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 field exists throughout the region. The acceleration of gravity is 9 . 8 m / s 2 . 6 . 9 Ω 1 . 6m / s 60 g . 56 T sliding rod . 68 m Viewed from above 1 . 6 m / s 28 ◦ . 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 flux 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 flux is B · A = BA cos θ , where θ is the angle between the magnetic field B and the normal vector to the area. The magnetic field and the angle are both constant and were pulled out of the differentiation. Now, if we call the distance from the rod to the resistor 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 field, F B,total = I B ....
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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

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