2. 1 Things to know (a)deduce from Faraday’s experiments on electromagnetic induction or other appropriate experiments: (i) that a changing magnetic field can induce an e.m.f. in a circuit (ii) that the direction of the induced e.m.f. opposes the change producing it (iii) the factors affecting the magnitude of the induced e.m.f. (b) describe a simple form of a.c. generator (rotating coil or rotating magnet) and the use of slip rings (where needed)

3. 2 (c) sketch a graph of voltage output against time for a simple a.c. generator (d) describe the structure and principle of operation of a simple iron-cored transformer as used for voltage transformations (e) recall and apply the equations VP / Vs = NP / Ns and VPIP = VsIs to new situations or to solve related problems (for an ideal transformer) (f) describe the energy loss in cables and deduce the advantages of high voltage transmission

4. 3 Electromagnetic Induction Definition: Electromagnetic induction is the production of electricity using magnetism. Need to know: Describe an experiment which shows that a changing magnetic field can induce an e.m.f. in a circuit

5. 4 N S In this experiment, no battery is connected to the coil. Hence no e.m.f. is found in the coil. Sensitive Galvanometer 0 -22 1 Hollow paper or plastic tube A stationary magnet is near the coil

6. 5 N S motion When the magnet is moving towards the coil, an electric current is induced simultaneously. G Induced I 0 -22 1 Hollow paper or plastic tube

7. 6 Faster motion When the magnet is moving faster, the induced current is more. N S G Induced I more 0 -22 1 Hollow paper or plastic tube

8. 7 Not moving When the magnet is not moving, no current is induced even though the magnetic flux is linked with the coil closely. N S G No current 0 -22 1 Hollow paper or plastic tube

9. 8 Faraday’s Law of Electromagnetic Induction: The magnitude (how strong) of the induced emf (or induced current) is directly proportional to the rate of change of the magnetic flux linked with the coil or the rate at which the magnetic flux and wire are cutting each other. This means that when the magnetic field is not moving in relation to the coil, there will be NO induced emf at all.

10. 9 Not moving There is plenty of magnetic flux linkage with the coil, but there is no motion. Is there any induced current in the coil now? Answer: _________ N S G 0 -22 1 Self Test Question Please draw the needle of the galvanometer.

11. 10 What law did you apply when you answer the question in the previous slide?

12. 11 G 0 -22 1 Self Test Question B E C D A Moving constantly N S B E C D A Deflection of G or emf induced

13. 12 G 0 -22 1 Self Test Question: Sketch the graph as the magnet moves from A to E B E C D A Constant speed moving N S B E C D A Deflection of G or emf induced ` Playing back of the graph

14. 13 By now, you have learned that the size or strength of the induced current (or induced e.m.f.) is determined by the speed of change of the magnetic flux linkage with the coil. There is still one more thing about electromagnetic induction you need to investigate. Look at the next slide.

15. 14 When a current is induced in a coil, it has to flow in the certain direction. What factor determines the direction of the induced current?

16. 15 Lenz’s Law of electromagnetic induction: The direction of the induced current is such that its own magnetic effect always opposes the change producing it. This law is actually related to the Law of Conservation of Energy. The coil needs to oppose something in order to obtain energy from it. The coil itself cannot CREATE energy!

17. 16 Beware of a different way the coil can be wound: The paper tube can be taken away to test you

18. 17 Can you spot the difference of winding? Note: The dotted parts are at the back. The solid lines are at the front.

19. 18 N S Motion Please mark one arrow on the left end of the coil and one arrow through the bulb to show how the induced current should flow: Induced CURRENT

20. 19 N S Motion Please mark + or – signs at the points X and Y to show the induced e.m.f. : X Y +

21. 20 N S Motion Please mark + or – signs at the points X and Y to show the presence of induced e.m.f. : X Y This induced emf is still there as long as the magnet is moving, even though the circuit is broken and the induced current cannot flow. Induced emf +

22. 21 G N S Coil is stationary

23. 22 G N S Induced current Coil is in motion, approaching the magnet coil motion N Induced current

24. 23 G N S coil motion N Coil is in motion, approaching the magnet Induced current

25. 24 G N S coil motion N Coil is approaching the magnet Can you see the Right Hand Rule ? This is also called the Dynamo Rule Induced current Existing flux Motion of wire

26. 25 G N S Coil is stationary again Existing flux No more induced current here

27. 26 Fleming’s Right Hand Rule is also called the Dynamo Rule. thuMb -- the Motion of the wire seCond finger -- induced Current First finger -- magnetic Flux (Field) This is actually the result of Lenz’s Law So, sometimes you use the right hand rule instead of Lenz’s Law.

28. 27 This straight wire is moving along the magnetic field No current is induced Wire moving vertically to the flux Current is induced towards you Wire stops moving No current is induced at all Straight wire moving vertically to magnetic field. Current is induced away from you The straight wire stops moving. No current is induced at all. This straight wire is not moving No current is induced

29. 28

30. 29

31. 30 wire cutting flux obliquely wire cutting flux vertically wire moving alongside flux No current is induced in the wire No current is induced in wire Strong induced current Weak induced current

32. 31 N S Magnetic flux

33. 32 N S

34. 33 N S A B C D Motion

35. 34 Motion N S A B C D Induced current Magnetic Flux

36. 35 N S A B C D motion No induced current No pole is needed No induced current

37. 36 N S A B C D motion

38. 37 N S D A B C Motion

39. 38 N S A B C D Motion Induced current Magnetic Flux

40. 39 N S A B C D motion No induced current Magnetic flux

41. 40 N S A B C D motion

42. 41 N S A B C D Motion

43. 42 N S A B C D Induced current Time A D Red arrows represents magnetic flux

44. 43 N S A B C D Induced current Time A D Red arrows represents magnetic flux

45. 44 N S A B C D motion Induced current Time A D

46. 45 N S D A B C Induced current Time A D Flux motion

47. 46 N S A B C D motion Induced current Time A D

48. 47 N S A B C D Induced current Time A D

49. 48 Induced current / emf Time

50. 49 Induced emf Time 2.5V -2.5V

51. 50 The iron core G How would you demonstrate electromagnetic induction here?

52. 51 The iron core AC Source N p turns N s turns output emf Input emf Insulated copper wire Primary windings Insulated copper wire Secondary windings

53. 52 The iron core AC Source N p turns N s turns Secondary emf Primary emf emf primary emf secondary NpNp NsNs =

54. 53 AC Source N p turns N s turns Secondary emf Primary emf emf Time

55. 54 emf Time Primary Alternating emf Peak 12V, 50 Hz Secondary Alternating emf Peak 18V 50 Hz

56. 55 The iron core AC Source N p turns N s turns Secondary emf Primary emf emf primary emf secondary NpNp NsNs = 1 Step-up Transformer

57. 56 The iron core AC Source N p turns N s turns Secondary emf Primary emf emf primary emf secondary NpNp NsNs = 1 Step-down Transformer

58. 57 AC Source N p 200 turns N s 600 turns output emf Input emf 150V Calculate (i) the output emf (ii) the induced current 60  (iii) the primary current

59. 58 AC Source N p 200 turns N s 600 turns output emf Input emf 150V Calculate (i) the output emf 60  VpVp VsVs NsNs = NpNp 150 VsVs 600 = 200 V s = 450 V peak

60. 59 AC Source N p 200 turns N s 600 turns output emf Input emf 150V Calculate (i) the output emf 60  V s = 450 V peak (ii) the induced current I = V R 450 60 = 7.5 A Peak

61. 60 emf, current Time Secondary Alternating emf Peak 450 V, freq 50 Hz 7.5A 450V

62. 61 AC Source N p 200 turns N s 600 turns output emf Input emf 150V Calculate (i) the output emf 60  V s = 450 V peak (iii) the primary current Power= VI, Output power = Input power V s x I s = V p x I p 450x 7.5 = 150x I p (assuming that the transformer is 100% effecient) (ii) I s = 7.5 A,

63. 62 AC Source N p 200 turns N s 600 turns output emf Input emf 150V Calculate (i) the output emf 60  V s = 450 V peak (iii) the primary current IpIp IsIs NsNs = NpNp provided that the transformer has an efficiency of 100%

64. 63 emf, current Time Secondary Alternating emf Peak 450V, freq 50 Hz 7.5A 450V 150V 22.5A

65. 64 The iron core AC Source N p turns N s turns Secondary emf Primary emf Power lost in a transformer, Textbook Page 350 Power losses are : (i) due to the electrical resistance in both the windings Heat is generated in the wires unnecessarily (ii) due to the production of the eddy currents in the iron core. eddy current

66. 65 The iron core AC Source N p turns N s turns Secondary emf Primary emf Such power losses are minimized: (i) by using thicker copper wires in the windings. Less heat is generated in the wires. (ii) by using a laminated iron core.

67. 66 Power Loss in Cables

68. 67 P.S. L N 10KV R = 20  2MW 0V Heating power in the transmitting cables = ? R = 20  P = V I2,000000 = 10000 x I I = 200 A Heating power in the transmitting cables = I 2 R Heating power in cables = 200 2 x 40 = 1,600000 W = 1.6 MW A huge loss 0.4MW Power transmission

69. 68 P.S. L N 20KV R = 20  2MW 0V Heating power in the transmitting cables = ? R = 20  P = V I2,000000 = 20000 x I I = 100 A Heating power in the transmitting cables = I 2 R Heating power in cables = 100 2 x 40 = 400,000 W = 0.4 MW A smaller loss 1.6MW Power transmission