Quick QuizzesAnswers to Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 20 Induced Voltages and Inductance Quick Quizzes 1. b, c, a. At each instant, the magnitude of the induced emf is proportional to the rate of change of the magnetic field (hence, proportional to the slope of the curve shown on the graph). 2. (a). All charged particles within the metal bar move straight downward with the bar. According to right-hand rule #1, positive changes moving downward through a magnetic field that is directed northward will experience magnetic forces toward the east. This means that the free electrons (negative charges) within the metal will experience westward forces and will drift toward the west end of the bar, leaving the east end with a net positive charge. 3. (b). According to Equation 20.3, because B and v are constant, the emf depends only on the length of the wire moving in the magnetic field. Thus, you want the long dimension moving through the magnetic field lines so that it is perpendicular to the velocity vector. In this case, the short dimension is parallel to the velocity vector. From a more conceptual point of view, you want the rate of change of area in the magnetic field to be the largest, which you do by thrusting the long dimension into the field. 4. (c). In order to oppose the approach of the north pole, the magnetic field generated by the induced current must be directed upward. An induced current directed counterclockwise around the loop will produce a field with this orientation along the axis of the loop. 5. (b). When the iron rod is inserted into the solenoid, the inductance of the coil increases. As a result, more potential difference appears across the coil than before. Consequently, less potential difference appears across the bulb, and its brightness decreases. 175176 CHAPTER 20 Answers to Conceptual Questions 2. Consider the copper tube to be a large set of rings stacked one on top of the other. As the magnet falls toward or falls away from each ring, a current is induced in the ring. Thus, there is a current in the copper tube around its circumference. 4. The flux is calculated as cosBBA B Aθ⊥Φ= = . The flux is therefore maximum when the magnetic field vector is perpendicular to the plane of the loop. We may also deduce that the flux is zero when there is no component of the magnetic field that is perpendicular to the loop. 6. No. Once the bar is in motion and the charges are separated, no external force is necessary to maintain the motion. An applied force in the x direction will cause the bar to accelerate in that direction. 8. As water falls, it gains velocity and kinetic energy. It then pushes against the blades of a turbine transferring this energy to the rotor or coil of a large alternating current generator. The rotor moves in a strong external magnetic field and a voltage is induced in the coil. This induced emf is the voltage source for the current in our electric power lines. 10. The magnetic field lines around the transmission cable will be circular. If you place your loop around the cable, there will be no field lines passing through the loop, so no emf will be induced. The loop needs to be placed next to the cable, with the plane of the loop containing the cable, to maximize the flux through its area. 12. Let us assume the north pole of the magnet faces the ring. As the bar magnet falls toward the conducting ring, a magnetic field is induced in the ring pointing upward. This upward directed field will oppose the motion of the magnet preventing it from moving as a freely-falling body. Try it for yourself to show that an upward force also acts on the falling magnet if the south end faces the ring. 14. A constant induced emf requires a magnetic field that is changing at a constant rate in one direction — for example, always increasing or always decreasing. It is impossible for a magnetic field to increase forever, both in terms of energy considerations and technological concerns. In the case of a decreasing field, once it reaches zero and then reverses direction, we again face the problem with the field increasing without bounds in the opposite direction. 16. As the magnet moves at high speed past the fixed coil, the magnetic flux through the coil changes very rapidly, increasing as the magnet approaches the coil and decreasing as the magnet moves away. The rapid change in flux through the coil induces a large emf, large enough to cause a spark across the gap in the spark plug. 18. (a) Clockwise. As the south end of the magnet approaches the loop from above, the flux through the loop is directed upward and is increasing in magnitude. The induced current will flow clockwise to produce a downward flux through the loop and oppose the increasing flux due to the magnet. (b) Counterclockwise. When the north end of the magnet is below the loop and is moving away, the flux through the loop is directed upward and is decreasing in magnitude. The induced current will flow counterclockwise to produce upward flux through the loop and oppose the decreasing flux due to the magnet.Induced Voltages and Inductance 177 Answers to Even Numbered Problems 2. (a) (b) 71.00 10 T m−×⋅2 288.66 10 T m−×⋅ (c) 0 4. zero 6. 522.96 10 T m−×⋅ 8. 0.10 mV 10. 34 mV 12. 52.0 Tµ 14. (a) (b) 71.88 10 T m−×⋅2 86.28 10 V−×16. 8.8 A 18. 1.00 m s 20. 2.87 mV 22. 2.8 mV 24. (b) Larger R makes current smaller, so the loop must travel faster to maintain equality of magnetic force and weight. (c) The magnetic force is proportional to the product of field and current, while the current is itself proportional to the field. If B becomes two times smaller, the speed must become four times larger to compensate. 26. left to right 28. (a) left to right (b) no induced current (c) right to left 30. 13 mV 32. (a) 8.0 A (b) 3.2 A (c) 60 V 34. (a) 7.5 kV (b) when the plane of the coil is parallel to the field 36. 1.36 Hµ 40. 521.92 10 T m−×⋅42. (a) (b) 3.00 ms 1.00 kΩ44. (a) 0 (b) 3.8 V (c) 6.0 V (d) 2.2 V178 CHAPTER 20 46. (a) 2.00 ms (b) 0.176 A (c) 1.50 A (d) 3.22 ms 48. (a) 4.44 (b) 5 310 H−×4.55 10 J−×50. (a) 2LRτ= (b) 2LRτ= 52. 3100 A, the galvanometer will definitely show the induced currentµ 54. (a) Amplitude doubles, period unchanged. (b) Amplitude doubles, period cut in half. (c) Amplitude