1. Dr Robert Loss Room 301-143 Telephone 351 7747 : Fax 351 2377 Email RLOSSRD@CC.CURTIN.EDU.AU Dr Robert Loss Room 301-143 Telephone 351 7747 : Fax 351 2377 Email RLOSSRD@CC.CURTIN.EDU.AU Perth WesternAustralia University ofTechnology School of Physical Sciences Department of Applied Physics

2. 2 Curtin University P114 1995 A changing magnetic field can produce (induce) an electrical potential in a conductor Moving a conductor through a magnetic field can produce (induce) an electrical potential in the conductor The direction of the induced current opposes the change in the magnetic field The advantages of alternating current over direct current THE BIG IDEAS

3. 3 Curtin University P114 1995 Induced EMF Faraday’s expt. iron core Switch Battery AMMETER

4. 4 Curtin University P114 1995 Faraday’s expt While switch being closed –meter twitches and returns to zero While switch being opened –meter twitches and returns to zero While Switch closed: nothing While Switch open: nothing Switch Battery

5. 5 Curtin University P114 1995  B : The # of lines of B passing through area A Units: tesla/m 2 or weber (Wb) Magnetic flux  B Higher  B Lower  B Rectangular loops: Area A wire loop B B

6. 6 Curtin University P114 1995 Magnetic flux  B  B = B A (where B perp to A)or  B = B A cos  for any angle between B and A Rotating a wire loop in a fixed field also changes  B rotating wire loop B

7. 7 Curtin University P114 1995 Faradays Law of induction Induced emf (V)  number of coils (N)   B/  t  Area of wire loop (A) also  B = BA and  B /  t =  BA/  t

8. 8 Curtin University P114 1995 EMF Example 1 A 100 turn coil of wire experiences a change in magnetic field of 0.2 T every 0.01s. If the coil has an area of 0.05 m 2, what emf is induced in the coil? N =100 A = 10 A  B = 0.2 T  t = 0.01s

9. 9 Curtin University P114 1995 Why the negative sign? What is the direction of the induced current? Consider a static conductor in a region where B is as shown below AND increasing. B inducing increasing X oooo XXX (i) ?

10. 10 Curtin University P114 1995 Lenz’s Law “The induced emf always produces a (induced) current whose magnetic field opposes the original change in magnetic flux” B inducing increasing X o o o o o o o X X X X X X B induced i induced

11. 11 Curtin University P114 1995 Consequences of Lenz’s Law If Lenz’s Law was not correct the induced B induced would ADD to the B inducing - producing more i induced - producing more B induced - etc...... - leading to a “runaway” production of current and finally wire meltdown.

12. 12 Curtin University P114 1995 Two ways to apply Faraday’s Law 1. Keep the conductor stationary and change B eg spin a magnet inside a loop of wire 2. Keep the field stationary and change the orientation of the conductor eg spin a loop of wire inside a magnetic field N S N S Magnet Coils Rotation Brushes

13. 13 Curtin University P114 1995 Applying Faraday’s Law move a conductor through a magnetic field eg moving a length of wire inside a magnetic field conductive loop B velocity v o o o o o o o o o o o o o o o o o o o o o o o o area A, covered per unit time  t

14. 14 Curtin University P114 1995 An application of Faraday’s Law when B, and v are all perpendicular area  A, covered per unit time  t =  A/  t = v velocity v o o o o o o o o o o o o o o o o o o o o o o o o

15. 15 Curtin University P114 1995 Application What potential is produced on a 20 km long wire dragged at right angles by a space shuttle through the earth’s magnetic field at a velocity of 10 km/s. = 20000m v = 10000 km/s B = 0.5 x 10 -4 T emf = B v = 0.00005 x 20000 x 10000 = 10000 V o o o o B

16. 16 Curtin University P114 1995 AC signals V time VoVo Any signal (eg voltage) component which changes periodically over time V o = V p =maximum voltage V PP = peak to peak voltage = 2 x V p Period (T) time for one cycleFrequency (f) = 1/T V PP T

17. 17 Curtin University P114 1995 i = I p sin2  ft v = V p sin2  ft AC Current (i) & Voltage (v) AC current and voltage at any point in time are described as follows frequency Maximum or peak value time

18. 18 Curtin University P114 1995 Applications of AC Amplitude Modulated (AM 720 kHz) Frequency Modulated (FM eg 120 MHz) Power transmission (50 or 60 Hz)

19. 19 Curtin University P114 1995 RMS (root mean squared) The effective value used in power calculations. RMS values for V or I produce the same power as for an equivalent DC values. Power = V 2 /R or I 2 R Power  V 2 or I 2 V V p 2 /2 V2V2

20. 20 Curtin University P114 1995 RMS (root mean squared) The voltage that produces V p 2 /2 equivalent. Likewise for current

21. 21 Curtin University P114 1995 Application What are the peak voltages for AC power with V rms or 240V and 120V respectively? V rms = 0.707 V p or V p = 1.41 V rms = 1.4 x 240 = 340 V and for 120 V = 1.41 x 120 = 170 V

22. 22 Curtin University P114 1995 Electrical power generation + current flows from a to b in top of loop and c to d in bottom of loop components bc and ad do not generate a potential as they do not cross any magnetic field lines as loop rotates it continually generates a potential B a b d c

23. 23 Curtin University P114 1995 Electrical power generation Total V = 2 B v sin  Also V = N A B 2  f sin 2  ft where f = number or rotations per second and A = area of coil, N is the number of coils B a b d c V ab = B v sin  V cd = B v sin 

24. 24 Curtin University P114 1995 The alternator N S AC output Electro magnet wire coils commutators brush

25. 25 Curtin University P114 1995 DC generator brush split ring commutator field time V

26. 26 Curtin University P114 1995 Types of DC generators A + N S - Series A + N S - + A N S - Shunt Compound i  V  i  or i  V is const

27. 27 Curtin University P114 1995 Transformers A device for changing AC voltages V in primary V out N p =6N s =12 secondary soft Fe laminated core

28. 28 Curtin University P114 1995 Transformer Voltages The input to output Voltage ratio can be described in terms of the number of turns (N S /N p ) ratio Transformers

29. 29 Curtin University P114 1995 Application 1 A walkman transformer has 2000 turns on the input and 50 on the output. What V is produced when the input is 250V. N p = 2000 turnsV S /V p = N S /N p N S = 50 turns V S =V p x N S /N p V p = 250 = 250 50/2000 = 6.25 V This assumes the transformer is 100 % efficient Transformers

30. 30 Curtin University P114 1995 Transformer (i, V and P) Although V may increase the power cannot. For 100% efficiency Transformers

31. 31 Curtin University P114 1995 Application 2 A step down transformer has turns ratio of 10:1 What is the input voltage and current if it transfers 150 W at 100V input? N p /N S = 10V s =V p x N S /N p V p = 100 = 100 x 1/10 Power = 150W = 10 V from P = V i i = P/V = 150/10 =15 A Transformers

32. 32 Curtin University P114 1995 Adv and disadv of AC ADVANTAGES AC voltages can easily be increased or decreased using transformers easier to generate reduced transmission losses easier to control unwanted signals DISADVANTAGES Most small appliances require DC

33. 33 Curtin University P114 1995 Summary A changing magnetic field can produce an electrical potential in a conductor Moving a conductor through a magnetic field can produce an electrical potential in the conductor The direction of the induced current opposes the change in the magnetic field The advantages of alternating current over direct current P072 Q21.2 and 27.11, P21.3, 27.1 and 27.5

34. 34 Curtin University P114 1995 Challenge Which of the the following do you think is the most useful/important to society? –Generator –Electric motor –Transformer Briefly explain how it works and why you think it is the most important.