PHY 91880 Midterm 3 Exam _ 2020 | PHY91880 Midterm 3 Exam _ Graded A

PHY 91880 Midterm 3 Exam _ 2020 Version 046/AACDC – midterm 03 – turner – (93000) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An infinitely long straight wire carrying a current I1 = 37 A is partially surrounded by a loop as in figure. The loop has a length L = 31.5 cm, a radius R = 11.1 cm, and carries a current I2 = 29.7 A. The axis of the loop coincides with the wire. R L I1 I2 Calculate the force exerted on the loop. 1. 311.146 2. 1247.4 3. 593.321 4. 566.24 5. 1691.62 6. 261.6 7. 509.09 8. 274.081 9. 1129.34 10. 3132.88 Correct answer: 1247.4 µN. Explanation: Let : I1 = 37 A, I2 = 29.7 A, L = 31.5 cm = 0.315 m, and R = 11.1 cm = 0.111 m. The central wire creates field B~ = µ0 I1 2 π R counterclockwise . The curved portions of the loop feels zero force since ~l × B~ = 0 there. The straight portions both feel I2~l × B~ forces to the right, amounting to F~ = I2 2 L µ0 I1 2 π R = µ0 I1 I2 L π R to the right kF~ k = µ0 (37 A) (29.7 A) (0.315 m) π (0.111 m) = 0. N = 1247.4 µN . keywords: 002 10.0 points A bar of negligible resistance and mass of 94 kg in the figure is pulled horizontally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 490 g. The uniform magnetic field has a magnitude of 650 mT, and the distance between the rails is 39 cm. The rails are connected at one end by a load resistor of 46 mΩ. 650 mT 650 mT 650 mT 39 cm 490 g a 46 mΩ 94 kg What is the magnitude of the terminal velocity (i.e., the eventual steady-state speed v∞) reached by the bar? The acceleration of gravity is 9.8 m/s2 . 1. 1.77778 2. 144.023 3. 0. 4. 4.6486 5. 161.262 6. 3.43735Version 046/AACDC – midterm 03 – turner – (93000) 2 7. 1.02823 8. 0. 9. 30.1658 10. 5.657 Correct answer: 3.43735 m/s. Explanation: Let : m = 94 kg , M = 490 g = 0.49 kg , ` = 39 cm = 0.39 m , B = 650 mT , and R = 46 mΩ = 0.046 Ω . B B B B B ` T T M Fg a R a m Fm ~Fg = M ~g ~Fm = I~` × B~ F~net = (M + m)~a = F~g − F~m E = I R = −d ΦB dt ΦB = B~ · A~ E = B ` v . It follows from Lenz’s law that the magnetic force opposes the motion of the bar. When the wire acquires steady-state speed, the gravitational force Fg is counter-balanced by the magnetic force Fm. Fg = M g = Fm = ` I B (1) I = M g ` B . (2) To find the induced current, we use Ohm’s law and substitute in the induced emf, E = −d Φ dt I = |E| R = 1 R d Φ dt . (3) Note: We have ignored the minus sign from the induced emf E because we will eventually evaluate the magnitude of the terminal velocity. The flux is Φ = B A , so |E| = d Φ dt = B dA dt = B ` v , and (4) I = B ` v R . (5) Using Eqs. 2 and 5 and noting that v is the terminal velocity v∞ M g ` B = B ` v∞ R . (6) Solving for the magnitude of the terminal velocity v∞ v∞ = M g R `2 B2 (7) = (0.49 kg) (9.8 m/s2) (0.046 Ω) (0.39 m)2(650 mT)2 = 3.43735 m/s . keywords: 003 10.0 points A uniform non-conducting ring of radius 2.37 cm and total charge 9.5 µC rotates with a constant angular speed of 3.82 rad/s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring? 1. 2.91697e-09 2. 1.10459e-08 3. 2.26777e-09 4. 1.27863e-09 5. 1.32498e-08 6. 1.95952e-09 7. 1.47807e-08 8. 1.01919e-08 9. 2.00442e-09 10. 1.39673e-09Version 046/AACDC – midterm 03 – turner – (93000) 3 Correct answer: 1.01919 × 10−8 A m2. Explanation: The period is T = 2 π ω . The spinning produces a current I = Q f = Q T = Q ω 2 π . Then, the magnetic moment µ is given by µ = I A = Q ω 2 π π R2 . Hence µ = Q R2 ω 2 = (9.5 × 10−6 C) (0.0237 m)2 (3.82 rad/s) 2 = 1.01919 × 10−8 A m2 . 004 10.0 points Consider the setup of a velocity selector for the case where the electric field E~ is pointing downward along y axis. E v − x y z O Choose the direction of the magnetic field B~ such that a negatively charged particle, moving at an appropriate speed in the positive x-direction (ˆı), passes through the E~ and B~ region undeflected. 1. Bb = √12(ˆ + kˆ) 2. Bb = +ˆı 3. Bb = √12(ˆ − kˆ) 4. Bb = √12(−ˆ + kˆ) 5. Bb = +ˆ 6. Bb = −kˆ correct 7. Bb = +kˆ 8. Bb = −ˆı 9. Bb = −ˆ 10. Bb = √12(−ˆ − kˆ) Explanation: The negatively charged particle passes through the selector undeflected when the electric force is equal and opposite to the magnetic force. Since the electric field is pointing in the negative y direction, and the particle’s charge is negative, the particle feels an electric force in the positive y direction. F~E = |q| E ˆ where E is the magnitude of the electric field. We must choose the magnetic field such that the magnetic force is in the negative y direction. The general equation for the magnetic force is given by F~B = q (~v × B~ ) We want F~B to point opposite to F~E, i.e., in the (−ˆ) direction to have a chance of cancelling the net force. Considering only the unit vectors along which the above vectors point, we must satisfy −ˆ = −(ˆı × ˆr) (1) where ˆr denotes the unknown direction in which B~ must point. Since ˆı × (−kˆ) = ˆ we see that equation (1) for ˆr is satisfied by ˆr = −kˆ so the magnetic field must point into the page. This can be verified by using the right hand rule. 005 10.0 pointsVersion 046/AACDC – midterm 03 – turner – (93000) 4 In the circuits labeled A,B, and C, all the thin-filament bulbs, capacitors and batteries are identical. The capacitors are initially uncharged. In each circuit, the batteries are connected for a short time T and then discharged, where T is only 10% of the total charging time through a single thin-filament bulb. Select the choice that correctly identifies the circuits that have the (most charge,least charge) in the correct order. 1. B,C 2. A,C 3. C,A 4. C,B correct 5. B,A 6. A,B Explanation: The charging time is characterized by the time constant τ = RC. Comparing the circuits, we can see that the least resistance is in circuit C, where the bulbs are in parallel; the most resistance is in circuit B, where the bulbs are in series. Therefore, circuit C has the shortest charging time, while B has the longest. So (C,B) is the correct choice - - - - - - - - Continued