1. G0MDK 1 The Nature of Electricity By Chuck Hobson BA, BSc(hons)

2. G0MDK 2 INTRODUCTION This “Nature of Electricity” presentation has been modified to include some of the remarks I usually make as I proceed through the slides. If you have any questions, please email them to me at:

3. G0MDK 3 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS

4. G0MDK 4 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS

5. G0MDK 5 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS

6. G0MDK 6 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS

7. G0MDK 7 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS

8. G0MDK 8 Elektrizität! Was ist es? L'électricité! Qu'est-ce que c'est ? Elettricità! Che cosa è esso? Electricity! What is it? Georg Simon Ohm 1787 - 1854 André Marie Ampère 1775 - 1836 Count Alessandro Volta 1745 - 1827 Michael Faraday 1791 - 1867 Joseph Henry 1797 – 1878 Nikola Tesla 1856 - 1943 Charles Augustin Coulomb 1736 - 1806 James Watt 1736-1819 CREDITS Note they all have units of electricity named after them.

9. G0MDK 9 THE ELECTRON FOUND IN ALL MATTER SOME PROPERTIES Radius < 10 -15 metres Rest mass 9.1 × 10 -28 grams Charge neg. 1.6 × 10 -19 Coulombs HOW SMALL? A thousand trillion electrons side by side measure 0.5m HOW HEAVY? 1.2 thousand trillion trillion electrons weigh one gram HOW POTENT? 6.25 million trillion electrons make a 1 Coulomb charge One Coulomb flowing per second = one Ampere. One gram of electrons contains 176,000,000 Coulombs of charge HEART OF ELECTRICITY

10. G0MDK 10 PUTTING THE ELECTRON TO WORK Note: Ideal model used, Wires have zero resistance, light illuminates instantly and resistance is a fixed 100 ohms.

11. G0MDK 11 PUTTING THE ELECTRON TO WORK

12. G0MDK 12 PUTTING THE ELECTRON TO WORK Note: If an oscilloscope and photo cell at the battery/SW end is triggered at SW closure, the photo cell & oscilloscope would see the light 6.67µs later.

13. G0MDK 13 A QUESTION 1.3.3µs was arrived at using a Radar type calculation 2.Velocity x time results in distance travelled (v x t = d) 3.In Radar v = c, the speed of light- 300 million metres per seconds. 4.Radar range (c x t = d) 300 x 10 6 ms -1 x 3.3 x 10 -6 s = 1000m (one way) 5.In our case we know d (1000m) and c, so we calculate t or 3.3µs Now consider current flow again WHERE DID THE 3.3µs COME FROM?

14. G0MDK 14 CURRENT FLOW

15. G0MDK 15 CURRENT FLOW

16. G0MDK 16 CURRENT FLOW Current measures 1A in fig 3. What about the current in figures 1 & 2?

17. G0MDK 17 CURRENT FLOW The battery doesn’t see the 100 ohm load in figures 1 & 2 All it sees is the characteristic impedance of the pair of wires In our example, this impedance is assumed to be 400 ohms.

18. G0MDK 18 ANOTHER QUESTION 1.No way Jose! 2.Why not? Reason: 1.Electrons are particles with mass as previously stated 2.As particles approach c their masses increase enormously 3.This is in accordance with Einstein’s “Special Relativity” 4.This has been demonstrated at Cern and SLAC 5.Cern and SLAC use GeV’s to reach near c velocities 6.No particles including the electron have ever been accelerated to c Can electrons travel through wire at the speed of light ( c ) where c = 300,000,000 metres per second?

19. G0MDK 19 ELECTRON VELOCITY-1 ELECTRON VELOCITY IN A VACUUM TUBE

20. G0MDK 20 ELECTRON VELOCITY-1 ELECTRON VELOCITY IN A VACUUM TUBE Formula from A-Level Physics Calculation = 26.5 million metres per second (final velocity at the anode) Let’s increase voltage on the Anode of the tube and calc. velocities v = velocity electron reaches at A anode

21. G0MDK 21 ELECTRON VELOCITY-1 Anode voltage (kilovolts) Electron Velocity million metres/s Velocity to speed of light ratio 8.0053.0.177 (12.5%) 16.00750.250 (25%) 64.001500.500 (50%) 128.02120.701 (70%) 256.03001.000 (100%) 512.0424*1.410 (141%) SOMETHING WENT WRONG! Electrons CANNOT exceed c! Increase in mass with velocity (relativistic mass) was not taken into account

22. G0MDK 22 ELECTRON VELOCITY-2 When a mass velocity approaches the speed of light its mass increases This is in accordance with Einstein’s theory “Special Relativity” That is to say: relativistic mass (m r ) = gamma ( ) times m o Relativistic mass calculations are done using the following formulas: #1 #3 #2 #4 where eV = electron volt is a unit of energy used in particle physics

23. G0MDK 23 ELECTRON VELOCITY-2 When a mass velocity approaches the speed of light its mass increases This is in accordance with Einstein’s theory “Special Relativity” That is to say: relativistic mass (m r ) = gamma ( ) times m o Relativistic mass calculations are done using the following formulas: #1 #3 #2 #4 where eV = electron volt is a unit of energy used in particle physics

24. G0MDK 24 ELECTRON VELOCITY-2 When a mass velocity approaches the speed of light its mass increases This is in accordance with Einstein’s theory “Special Relativity” That is to say: relativistic mass (m r ) = gamma ( ) times m o Relativistic mass calculations are done using the following formulas: #1 #3 #2 #4 where eV = electron volt is a unit of energy used in particle physics Known as the Lorentz T ransform

25. G0MDK 25 ELECTRON VELOCITY-2 When a mass velocity approaches the speed of light its mass increases This is in accordance with Einstein’s theory “Special Relativity” That is to say: relativistic mass (m r ) = gamma ( ) times m o Relativistic mass calculations are done using the following formulas: #1 #3 #2 #4 where eV = electron volt is a unit of energy used in particle physics Known as the Lorentz T ransform

26. G0MDK 26 ELECTRON VELOCITY-2 On the last entry, notice the significant mass increase (x 17,219) Velocity m/sGamma 100,000,0001.225 200,000,0001.7320 250,000,0002.4490 290,000,0005.4770 299,000,00017.320 299,999,000547.72 299,999,99917,219

27. G0MDK 27 ELECTRON ENERGY VELOCITY CHART KINETIC ENERGY electron volts VELOCITY million m/s GAMMA mass increase 750keV4.51.00011 66MeV106.4 ( ~ 1/3 c)1.0692 145.1MeV149.5 ( ~ 1/2 c)1.1534 200MeV169.81.2132 400MeV213.91.4264 1.214GeV270.02.2942 4GeV294.545.2644 8GeV298.359.5288 18GeV299.6420.1898 120GeV299.9907128.932 800GeV299.99979853.878 900GeV299.99985960.488 1000GeV (1TeV)299.999881067.1 NOTE 299,999,880m/s is just 120m/s short of the speed of light

28. G0MDK 28 A QUESTION REVISITED ELECTRONS DIDN’T TRAVEL AT c BUT THE SOMETHING DID 1. How about the “nudge” theory (cue ball effect etc.)? Electrons out of the negative terminal nudge the next one on etc. The end result could be electrons at the light bulb 3.3µs later (?) There have been many arguments on this issue since the 1920’s A paper on this notion was submitted to the 1997 IEE.

29. G0MDK 29 RECTANGULAR PULSE WHAT IS HAPPENING DURING INTERVALS: (A – B), (B – C), (C – D)? (A – B) and (C – D)? Nothing! There is NO voltage or current B – C? 100V and 0.25A Note: Z of the transmission line pair = 400 Ohms. What would the situation be in 5.66µs?

30. G0MDK 30 RECTANGULAR PULSE Negative pulse moving back to the Pulse Generator

31. G0MDK 31 SINGLE RECTANGULAR PULSE EXAMINATION RECTANGULAR PULSE Time Domain: Viewed on an Oscilloscope RECTANGULAR PULSE Frequency Domain: Viewed on a Spectrum Analyser Pulse reconstruction formulaFast Fourier Transform formula

32. G0MDK 32 PULSE RECONSTRUCTION-1

33. G0MDK 33 PULSE RECONSTRUCTION-2

34. G0MDK 34 PULSE RECONSTRUCTION-3

35. G0MDK 35 PULSE RECONSTRUCTION-4

36. G0MDK 36 RECAP 1.The nudge (cue ball) explanation of conduction unresolved 2.Electrical energy travels ~ speed of light over wires to a load. 3.Likewise, pulses travel ~ the speed of light over wires. 4.Single pulses are made up of wide spectrums of frequencies 5.Pulse (spectrum of frequencies) travel as TEM signals at ~ c 6.In the circuit comprising a 100V battery, switch & light bulb: the leading edge of a pulse occurs at 100V switch on and the trailing edge of a pulse occurs at 100V switch off 7.Very long pulses have same properties as very short pulses 8.AC signals to µ-wave frequencies) travel as TEM modes 9.Note that the wave length of 50Hz = 6 million metres

37. G0MDK 37 CONCLUSION 1.Electrical energy travels from source to load over wires as TRANSVERSE ELECTROMAGNETIC WAVES (TEM mode) 2.Current drift* (Amperes) is a consequence of EM Waves NOT THE OTHER WAY AROUND 3.This may be difficult to visualize in a pair of wires, but if you consider EM microwaves travelling down a wave guide, there will be surface currents in the wave guide walls. These are also drift currents. They are also the consequence of EM energy * A sample calculation of current drift is shown in appendix 1 of this presentation.

38. G0MDK 38 Appendix 1 ELECTRON DRIFT The current drift rate through a conductor is in the order of mm/s. The drift rate of 1A through a 1mm diametre copper wire is worked out as follows: Current density J = amperes per unit area (J = I / A) so J = 1Amp. / (pi x r 2 ) = 1/(3.14 x 0.0005 2 ) = 1.6 x 10 6 J can also be expressed as J = nev d Transposing: vd = J / (ne) Copper has an electron density n of 8.47 x 10 28 m -3 With e = 1.6 x 10 -19 coulombs of charge: ne = 1.4 x 10 9 Thus: v d = J / (ne) = (1.6 x 10 6 ) / (1.4 x 10 9 ) = 1.14mm/s

39. G0MDK 39 That is the nature of Electricity as I perceive it. Thank you for attending Chuck Hobson BA, BSc(hons)