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Unformatted text preview: Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih 1 This printout should have 24 questions. Multiplechoice 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 variable resistor is connected across a con stant voltage source. Which of the following graphs represents the power P dissipated by the resistor as a function of its resistance R ? 1. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 2. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 3. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 4. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 5. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 6. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 7. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) cor rect Explanation: The power dissipated in the resistor has several expressions P = E I = E 2 R = I 2 R , where the last two are simply derived from the first equation together with the application of the Ohm’s law. Since the resistor is connected to a constant voltage source E = constant P = E 2 R = constant R , tells us that the power is inversely propor tional to the resistance µ P ∝ 1 R ¶ . 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance (Ω) Power(W) 002 (part 1 of 3) 10 points A rectangular loop consists of 333 closely wrapped turns of wire and has dimensions . 17 m by 0 . 25 m. The loop is hinged along the yaxis, and its plane makes an angle of θ = 12 ◦ with the xaxis. A uniform magnetic field of 0 . 85 T is di rected along the xaxis and the current in the loop is 3 . 2 A in the direction shown. Platt, David – Quiz 2 – Due: Oct 18 2005, 10:00 pm – Inst: Ken Shih 2 x y z 1 2 ◦ B = 0 . 85 T B = 0 . 85 T . 17m . 2 5 m i = 3 . 2 A What is the magnitude of the torque ex erted on the loop? Correct answer: 37 . 6536 Nm. Explanation: Let : n = 333 , ‘ = 0 . 17 m , w = 0 . 25 m , θ = 12 ◦ , B = 0 . 85 T , and I = 3 . 2 A . The field makes an angle of α = 90 ◦ θ with a line perpendicular to the plane of the loop, so the torque acting on the loop is τ = n B I A sin α = n B I ‘ w sin α (1) = (333)(0 . 85 T)(3 . 2 A)(0 . 17 m)(0 . 25 m) × sin(90 ◦ 12 ◦ ) = 37 . 6536 Nm . 003 (part 2 of 3) 10 points Assume that θ is measured in the positive direction; i.e. , clockwise (looking down from above) as shown in the figure. If the loop is at rest, the magnetic torque will 1. try to lift the loop along the yaxis. 2. try to make θ larger. correct 3. try to make θ smaller. 4. not affect θ . 5. None of these 6. try to lower the loop along the yaxis. Explanation: The right hand rule shows that torque tends to rotate the loop clockwise as viewed from above....
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This note was uploaded on 04/07/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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