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Electromagnetism

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  • 1. ELECTROMAGNETISM 1. Define the following terms & phrases (charge, static electricity, charging by friction/contact/induction, conductor, insulator, uniform/non-uniform charge distribution, earthing and electrical discharge) 2. Describe the behaviour of like and unlike charges. 3. Name and give symbols for the following: DC power supply, cell, battery, switch, lamp, resistor, variable resistor, wires joined, wires crossing, ammeter, voltmeter & fuse 4. State the symbol and metric unit for: charge, current, voltage, resistance & power 5. Define the following terms and phrases: Ohm’s law, DC electricity, series, parallel, current rules and voltage rules. 6. Describe the differences and similarities in the way ammeters and voltmeters are used. 7. Draw and interpret DC circuit diagrams 8. Solve problems using V = IR, P = VI, P = E and Rtotal = R1 + R2 t 9. Describe the magnetic field patterns around permanent magnets, the earth, currents and coils. 10. State the symbol and unit for magnetic field. 11. Use the right hand grip rule to determine relative field (B) and current (I) directions. 12. Describe how the magnetic field due to a current in a straight wire varies with the size of current and the distance from the wire. 13. Solve problems using B = µ0I 2πd Thursday, 4 November 2010
  • 2. Thursday, 4 November 2010
  • 3. LANGUAGE Thursday, 4 November 2010
  • 4. THE LANGUAGE OF ELECTRICITY Term Definition Word list Charge an electrical quantity based on an excess or deficiency of electrons static electricity a form of electricity where charge does not flow continuously Thursday, 4 November 2010
  • 5. NOTES Thursday, 4 November 2010
  • 6. ELECTRO STATICS Thursday, 4 November 2010
  • 7. The process 1. Charge transfer - When two objects are in contact with each other, one object can transfer electrons to the other object. Protons are not transferred because they are “locked” into the nucleus. 2. Charge imbalance - When the two objects are moved away from each other the process of charge transfer is unable to be reversed. • Positively charged objects have had electrons removed • Negatively charged objects have gained electrons Oppositely charged objects attract each other. Those with like charges repel. Proton (+ve charge) Neutron Electron (-ve charge) Empty space Based on atomic structure CHARGING OBJECTS Thursday, 4 November 2010
  • 8. + + - - + + + + - - - - Oppositely charged objects attract each other Objects with the same charge repel each other Demo: The Van der Graaf Generator CHARGE INTERACTION Thursday, 4 November 2010
  • 9. Static electricity around us Read p52 and 53 (Y10 Pathfinder) and then offer some examples to the class discussion: The History of Electricity Generation EXAMPLES Thursday, 4 November 2010
  • 10. LIGHTNING Warm air currents ascend. Ice crystals descend removing electrons from cloud particles in this zone. A zone of positively charged cloud particles results At ground level the air becomes ionized (by losing electrons) and these positively charged particles are attracted to the negatively charged base of the cloud to give rise to a lightning bolt (an upstrike) Thursday, 4 November 2010
  • 11. CLOUD TO GROUND Thursday, 4 November 2010
  • 12. CLOUD TO CLOUD Thursday, 4 November 2010
  • 13. BLUE JETS AND SPRITES Thursday, 4 November 2010
  • 14. CHARGING OBJECTS Methods of charging Induction - the object being charged is not in contact with the object doing the charging (usually a rod or a ruler). It involves charge transfer to or from the earth to generate the charge imbalance in the object being charged. A charge imbalance is a non-uniform charge distribution Eg. ++++ - - - - + + + + This symbol represents a connection to the earth - - Contact - the object being charged is contacted by the other object and charge is transferred directly from one object to another. Friction - the object is rubbed by a material that has a greater or lesser affinity for electrons and a transfer takes place. It is more the contact than the friction that is necessary for the charging to take place. The problem with moisture in the air: Moisture prevents objects from holding a charge because it transfers charge to or from the object resulting in a neutral object. Thursday, 4 November 2010
  • 15. The importance of the material Conductors - are materials that allow charges to flow through them. They do not hold a static charge because any charge imbalance is easily conducted away. Insulators - are materials that doe not allow charges to flow through them. They will hold a static charge because the charge imbalance is not easily conducted away. Example - ESA: Ex 15A Q.1 ESA: Ex 15B Q.1, 2 & 3 Thursday, 4 November 2010
  • 16. SHOCKS AND SPARKS Flow of charge through the body causes the shock Ionisation or flow of electrical energy from the charged object causes the spark Examples Climbing out of a car seat at the gas station. The seat becomes negatively charged because the electrons are moving from the person’s clothing to the seat. When the person is out of the car and in contact with the ground the electrons from the car seat are able to be transferred to the air particles causing these particles to become ionised. It is these negative ions that can move across a gap between the seat and the person. This results in the production of a spark. The shock is caused by the ions losing electrons to the person’s body and these electrons flowing through the body and into the earth (which is charge flowing into the person’s body. These charges Explanation of discharging in terms of electron transfer to/from an object to result in neutrality. Thursday, 4 November 2010
  • 17. CURRENT VOLTAGE Thursday, 4 November 2010
  • 18. INTRODUCTION Thursday, 4 November 2010
  • 19. WHAT IS ELECTRICITY? Thursday, 4 November 2010
  • 20. THE ELECTRICAL CIRCUIT - introducing the idea of the electron pump Thursday, 4 November 2010
  • 21. + - A conducting path Power Supply 1. What is an electric current? 2. What are the two requirements necessary for an electric current to exist? Thursday, 4 November 2010
  • 22. http://regentsprep.org/Regents/physics/phys03/bsimplcir/default.htm 1.Which part of the model represents the power supply? 2. Which part of the model represents the component? 3. What type of current is being modelled? A B THE GRAVITY MODEL Thursday, 4 November 2010
  • 23. 1.Which part of the model represents the power supply? 2. Which part of the model represents the conducting path? 3. Which part of the model represents charge? THE WATER MODEL Thursday, 4 November 2010
  • 24. 1.Which part of the model represents the power supply? 2. Which part of the model represents the conducting path? 3. Which part of the model represents charge? A B C THE BIKE MODEL Thursday, 4 November 2010
  • 25. THE BIKE MODEL one link THINK OF A LINK AS REPRESENTING A COULOMB OF CHARGE 1. In terms of this model, what do you think is meant by the term “current” ?? Thursday, 4 November 2010
  • 26. http://regentsprep.org/Regents/ physics/phys03/bsimplcir/default.htm A conducting path Power Supply + - Requirements: a power supply conducting path around which charge (electrons or ions) can flow. components (and sometimes meters) A component Current is a flow of electrons through a circuit Two types: AC - Alternating current (electrons vibrate back and forth in wires) DC - Direct current (electrons flow in wires in one direction only) Conventional current - the direction in which positive charges would flow in wires if they could. Conventional current is from positive to negative in a circuit (see diagram above) ELECTRIC CIRCUITS Thursday, 4 November 2010
  • 27. Current - is the rate of flow of electrical charge - it is the number of coulombs of electrical charge that passes a point in one second. A coulomb is 6.25 x 1018 charges - I = electric current (measured in amps, A) by an ammeter: Connecting an ammeter I + - A V I Red Red Black Black For charge to flow around an electrical circuit there is a need for a voltage source and a conducting path that is continuous and connects the positive to the negative terminal of the power supply. + - Example charge flows through the circuit as indicated by the arrows ELECTRIC CURRENT Thursday, 4 November 2010
  • 28. VOLTAGE Voltage • Voltage (V) is a measure of the energy lost or gained between two points in a circuit. • It is measured in the units volts , (V) Unit of Voltage: Joule per Coulomb or Volt (JC-1 ) or (V) V = ∆Ep q where V = potential difference or voltage (Volts, V) ∆Ep = change in potential energy that a charge experiences when it moves from one side to the other side of a component (Joule, J) q = the unit of charge (Coulomb, C) If V = 6V then a coulomb of charge has 6J more electrical potential energy at point A than it does at point B 1A V A B + -Consider the voltage across a lamp: Example Thursday, 4 November 2010
  • 29. + - V A • Two wires joined Two wires crossing Cell Lamp Battery (two cells in series) Switch Battery (several cells) Diode Voltmeter Ammeter Resistor Power supply fuse variable resistor (rheostat) demo of circuit components -> Notes CIRCUIT SYMBOLS Thursday, 4 November 2010
  • 30. RESISTANCE & OHM ’S LAW Thursday, 4 November 2010
  • 31. INTRODUCTION Thursday, 4 November 2010
  • 32. RESISTANCE & ELECTROCUTION - 1 Thursday, 4 November 2010
  • 33. RESISTANCE & ELECTROCUTION - 2 Thursday, 4 November 2010
  • 34. FACTORS AFFECTING RESISTANCE Thursday, 4 November 2010
  • 35. FACTORS THAT AFFECT RESISTANCE Also, some materials conduct electricity better than others Eg. Copper is better than iron Thursday, 4 November 2010
  • 36. In an insulator, electrons are fixed In a conductor, electrons are free to flow Label the materials that the arrows are pointing to _______________ _______________ CONDUCTORS & INSULATORS Thursday, 4 November 2010
  • 37. THE VOLTAGE-CURRENT RATIO 1. Consider a lamp in an electrical circuit: 12V 2A 12V represents the energy difference across the lamp. This drives electrons through the lamp at the rate (or “speed”) of 2A. The voltage:current ratio is _____ 2. Consider a different lamp in an electrical circuit: 12V 1A This lamp has higher resistance because 12V across this lamp can only drive electrons through the lamp at a rate of 1A. The voltage:current ratio is _____ This example shows that the greater the voltage:current ratio then the greater the resistance is. Resistance is the voltage:current ratio Thursday, 4 November 2010
  • 38. In an insulator, electrons are fixed In a conductor, electrons are free to flow Label the materials that the arrows are pointing to _______________ _______________ STOP OR GO? Thursday, 4 November 2010
  • 39. Definitions 1. Resistance, R is a measure of the “electrical friction” in a conductor. (the opposition to the flow of current) 2. It is the ratio of the voltage across a conductor to the current through it. Resistance = Voltage Current R = V I Unit of resistance is the ohm, Ω Resistance is given by the slope or gradient of a voltage - current graph Example In an experiment, the voltage across a lamp is measured and recorded as the current is increased 1 A at a time. Calculate the resistance of the lamp. V (V) I (A) 0 1 2 3 4 5 6 24 20 16 12 8 4 RESISTANCE V RI Thursday, 4 November 2010
  • 40. 1 2 3 A conductor that retains a constant temperature as the current is increased: A conductor that is allowed to heat up as the current is increased A conductor that is cooled progressively as the current is increased V (V) I (A)0 1 2 3 4 5 6 24 20 16 12 8 4 V (V) I (A) 0 1 2 3 4 5 6 24 20 16 12 8 4 0 1 2 3 4 5 6 V (V) I (A) 24 20 16 12 8 4 WHATS HAPPENING TO THE RESISTANCE AS THE CURRENT INCREASES? Thursday, 4 November 2010
  • 41. SHAKING MODELLING TEMPERATURE INCREASE IN A WIRE Thursday, 4 November 2010
  • 42. Thursday, 4 November 2010
  • 43. For most conductors, as the temperature increases the increased vibration of particles impedes the flow of electrons. Resistance in the conductor will therefore increase. The graph slopes upwards. V (V) I (A) 0 1 2 3 4 5 6 24 20 16 12 8 4 0 1 2 3 4 5 6 V (V) I (A) 24 20 16 12 8 4 When a temperature of a lamp increases its resistance increases The resistance of a thermistor decreases as its temperature decreases LIMITATIONS OF OHM’S LAW Thursday, 4 November 2010
  • 44. BASIC RESISTANCE PROBLEMS 6. What voltage is needed it a current of 5A is to flow through a resistance of 3Ω? 1. What is the resistance of a bulb it a 240 V supply causes a current of 2 A to flow through it? 2. What current flows through a heating element of 40Ω resistance when the element is plugged into a 240 V supply? 5. A current of 2 A flows through a 6Ω resistor. What is the voltage across it? 3. If a current of 3 A is flowing in a resistor across which there is a voltage of 6 V, what is the resistance? 4. What current must be flowing through a lamp of 0.5Ω resistance if there is a voltage of 6V across it? Thursday, 4 November 2010
  • 45. Resistors which are connected end to end are in series with one another The total resistance of the series combination, Rs is the sum of the resistances R1 and R2. For two or more resistors in series: Rs = R1 + R2 + ........... Resistors which are connected side by side are in parallel with each other. The total resistance of the parallel combination, Rp is less than any individual resistor in the combination. For two or more resistors in parallel the total resistance,Rp is given by: 1 1 1 + .... RP = R1 + R2 RESISTANCE CALCULATIONS R1 R2 R1 R2 Thursday, 4 November 2010
  • 46. Thursday, 4 November 2010
  • 47. EXERCISES 1. Read p52 to p53 QUIETLY 2. Answer questions 1 to 8 on p54 and 55 in your exercise books. There is no need to use full sentences Thursday, 4 November 2010
  • 48. Study the pictures of the appliances shown below and in the table record the materials in each appliance that are conductors and insulators. Material Conductor or Insulator? Material Conductor or Insulator? INSULATORS Thursday, 4 November 2010
  • 49. SERIES & PARALLEL Thursday, 4 November 2010
  • 50. Draw the following circuit diagrams in the spaces provided AND when you have finished, assemble them: 1. 2. 3. Voltmeters are connected _________ components, ammeters are connected _____ a circuit CIRCUITS: diagrams & assembly A + - V + - + - Thursday, 4 November 2010
  • 51. CHARACTERISTICS OF SERIES & PARALLEL CIRCUITS + - Series + - Parallel + - + - 1._ 2._ 3._ 1._ 2._ 3._ components connected one other the other. Single pathway No junctions each component has its own connection with the power supply more than one pathway One or more junctions Thursday, 4 November 2010
  • 52. http://phet.colorado.edu/simulations/ sims.php? sim=Circuit_Construction_Kit_DC_Only CIRCUIT CONSTRUCTION + - 1 A + - 3 A + - 2 A + - 1 A + - 2 A + - 3 A 1. Enter the URL (above) into the address bar of your internet browser. 2. Use the simulation tools to construct each of the following 3 circuits (ensure that you use identical lamps and an the same power supply for each circuit). 3. Record the current in each circuit and explain your observation. 4. Repeat this exercise for the second set of 3 circuits. Thursday, 4 November 2010
  • 53. DRAW THE CIRCUITS AND SET THEM UP Thursday, 4 November 2010
  • 54. CIRCUIT RULES Thursday, 4 November 2010
  • 55. Current in series is constant I1 = I2 = I3 Voltage in series is shared VT = V1 + V2 Note Voltage is shared in proportion to the size of the resistance + - VT A1 A3 V1 V2 A2 I1 I3 I2 VOLTAGE & CURRENT IN SERIES CIRCUITS Thursday, 4 November 2010
  • 56. Current in parallel is shared IT = I1 + I2 Voltage in parallel is constant VT = V1 = V2 Note Current is shared in an inverse proportion to the size of the resistance. For example: If R1 = 5 and R2 = 10 and IT = 3 then I1 = 2 and I2 = 1 in other words “charge splits up as it enters a junction in a circuit” + - VT V1 V2 I1 IT I2 IT R1 R2 “Double the resistance then halve the current” PARALLEL CIRCUITS Thursday, 4 November 2010
  • 57. Example 1 A3 = ________ V1 = ________ V2 = ________ 8V+ - A1 A2 A3 R1 R2 2A 3A V1 V2 8V+ - R1 V2V1 Example 2 3V V1 = ________ Rule used ___________________________________________________________ Rules used ____________________________________________________________________ ____________________________________________________________________ SIMPLE CIRCUIT CALCULATIONS Thursday, 4 November 2010
  • 58. Example 3 8V+ - A1 A2 A3 R1 R2 3A 4A V1 V2 A3 = ________ A4 = ________ V2 = ________ V3 = ________ 3V V3 A4 Example 4 8V+ - A1 A2 A3 3A A1 = ________ A4 = ________ A4 4A Rule used ____________________ ____________________ ____________________ Rules used ______________________ ______________________ ______________________ ______________________ ______________________ Thursday, 4 November 2010
  • 59. For the circuit represented by the circuit diagram above, what is the reading on: (a) V1 (b) A2 (c) V2 (d) V3 1 + 9V - A1 A2 A3 V2 V3 V1 5Ω 10Ω Examples ADVANCED CIRCUIT CALCULATIONS Thursday, 4 November 2010
  • 60. 2 For the circuit represented by the circuit diagram above, what is the reading on: (a) V3 if V2 = 10 V (b) A1 (c) A2 (d) A3 (e) What is the value of resistor R? + 15V - A1 A2 A3 V2 V3 V1 5Ω 10ΩR Thursday, 4 November 2010
  • 61. 3 For the circuit represented by the circuit diagram above, what is the reading on: (a) V1 (b) V2 (c) V3 (d) A2 + 12V - A1 A2 V2 V1 2Ω 3Ω V34.8Ω1 A Thursday, 4 November 2010
  • 62. Examples 1 •• 100 Ω 100 Ω 100 Ω Total resistance = 2 Total resistance = 25 Ω 30 Ω 50 Ω + - R1 R2 RT = R1 + R2 RESISTORS IN SERIES As we add resistors in series the resistance increases and therefore the current drawn decreases As we add resistors in parallel the resistance decreases and therefore the current drawn increases Thursday, 4 November 2010
  • 63. OTHER STUFF Thursday, 4 November 2010
  • 64. ESA: 13B Q.5 to 8 Thursday, 4 November 2010
  • 65. ESA: 13B Q.5 to 8 Thursday, 4 November 2010
  • 66. Thursday, 4 November 2010
  • 67. INTRODUCING “EMF” & “ELECTRICAL POTENTIAL Thursday, 4 November 2010
  • 68. METERS Thursday, 4 November 2010
  • 69. + - A V IT 9 V One of these meters has a very high resistance The other meter has a low resistance. Which is which? Explain your answer Thursday, 4 November 2010
  • 70. + - A V IT 9 V METERS Ammeter 1. connected in series with other components 2. has low resistance so that it doesn’t slow the current that it is supposed to be measuring Voltmeter 1. connected in parallel with other components 2. has high resistance so that it doesn’t allow much current to flow through it. This would reduce the current and voltage through the component. It is supposed to be measuring the voltage Thursday, 4 November 2010
  • 71. POWER & ENERGY Thursday, 4 November 2010
  • 72. • Power is the rate at which electrical energy is transferred into other forms of energy. • It is the amount of work done per second P = E t • It can be shown that the electrical power supplied to a device is given by: P = VI P V I E P t P = Power (Watts, W) E = the amount of energy converted or work done (J) t = the time taken (s) P = power (watts, W) (1W = 1 Js-1 ) V = Voltage (volts, V) I = Current (amps, A) POWER Thursday, 4 November 2010
  • 73. POWER & ENERGY Total energy used by a component/appliance can be calculated from the equation: E = P.t When the power value of the component/appliance is known and this value does not change over time. If power changes over time then this change can be graphed. P (W) t (s)0 2 4 6 8 10 2 4 6 8 10 12 14 E = Area under the graph E = 0.5 (9 + 13) = 4.4 J 10 Thursday, 4 November 2010
  • 74. TOTAL POWER USAGE in a parallel circuit PT = P1 + P2 + P3 + P4 = 12 + 12 + 6 + 6 = 36 W (ii) Example The power usage of the 4 lamps in parallel shown in the circuit below can be calculated in two ways: (i) Use the total voltage (supply voltage) and the total current (current drawn from the supply) to calculate power. (ii) Add the power usage of each of the components in parallel. (i) Lamps 3 & 4: I = P V = 12/12 = 1 A lamps 1 & 2: I = P V = 6/12 = 0.5 A 12 V 12 V 12 W 12 V 12 W 12 V 6 W 12 V 6 W P1 P2 P3 P4 IT = I1 + I2 + I3 + I4 = 0.5 + 0.5 + 1 + 1 = 3 A P = VTIT = 12 x 3 = 36 A Thursday, 4 November 2010
  • 75. LAMP BRIGHTNESS IN CIRCUITS Three main points (i) The brightness of a lamp depends on its power output since for a lamp, power is the rate at which electrical energy is converted into light (and heat) (ii) In a circuit which has values of voltage and current, it is both the voltage and current that determine brightness. (iii)The lamp’s resistance will determine that voltage:current ratio that it possesses Example: The series circuit (below), shows 2 identical lamps. A third identical lamp is added to the circuit. Explain how the brightness of the lamps in the circuit changes + - 12 V Thursday, 4 November 2010
  • 76. EXERCISES Thursday, 4 November 2010
  • 77. TV Stereo Jug Heater Stove torch downlight Voltage (V) 240 240 240 6 Current (A) 0.3 0.02 8 0.05 Resistance (Ω) 250 450 Power (W) 1000 1200 1800 0.3 140 Running time (s) 20 min 20 min 10 min 3 h Energy (J) 1000 125000 300000 POWERING THROUGH THE QUESTIONS Thursday, 4 November 2010
  • 78. 9V+ - 40Ω X 0.1 A 0.5 A 10Ω A 9 V battery is connected in the circuit shown. A current of 0.5 A is found to pass through the 10Ω resistor. (e) Determine the heat energy generated per second in the whole circuit. (a) Calculate the voltage across the 10Ω resistor. (b) Show that the voltage across the parallel combination of resistors is 4.0 V. (c) If 0.1 A passes through the 40Ω resistor, determine the current through resistor X. (d) Show that the resistance of resistor X is 10Ω. “PARALLEL WITHIN SERIES” Thursday, 4 November 2010
  • 79. PARALLEL CIRCUIT IN ACTION A car has two tail lights and two brake lights connected as shown in the diagram: (a) Calculate the resistance of: (i) a tail light (ii) a brake light (b) Calculate the current supplied by the battery when both S1 and S2 are closed. (c) When the driver takes her foot off the brake S2 is opened state what happens to the size of the current from the battery and give a reason for your answer. Thursday, 4 November 2010
  • 80. Y11 Sci - W & W Thursday, 4 November 2010
  • 81. Y11 Sci - W & W Thursday, 4 November 2010
  • 82. (a) 5V (b) V in series is shared. The combination of the 40Ω resistor and X are in series with the 10Ω. Therefore V of the combination is 9 - 5 = 4 (c) Current through X = 0.5 - 0.1 = 0.4A because current in parallel is shared. The total current flowing from the power supply is shared out between the 40Ω resistor and X. (d) I through X = 0.4A and V across X = 4V (since voltage in parallel is constant). R = V/I = 4/0.4 = 10Ω (e) “Heat energy per second” is the definition of power. For the whole circuit, I = 0.5A and V = 9V. P = VI = 9 x 0.5 = 4.5W “PARALLEL WITHIN SERIES”answers PARALLEL CIRCUIT IN ACTION (a) The current through a tail light needs to be calculated first: P = VI => I = P/V = 6/12 = 0.5A For a tail light, R = V/I = 12/0.5 = 24 Ω For a brake light, P = VI => I = P/V = 12/12 = 1A R = V/I = 12/1 = 12 Ω (b) I total = 2 x 0.5 + 2 x 1 = 3 A (c) Less current flows through the battery because there is now more resistance in the circuit because of the reduction in the number of pathways available for charge to flow. Thursday, 4 November 2010
  • 83. MAGNETIC FIELDS Thursday, 4 November 2010
  • 84. THE MAGNETIC FIELD AROUND A BAR MAGNET AND THE EARTH’S MAGNETIC FIELD Thursday, 4 November 2010
  • 85. WHAT IS MAGNETISM? • Magnetism is caused by moving electrons. (The smallest magnetic field is produced by the motion of 1 electron) • When electrons move in a common direction, a magnetic field is produced (sometimes called a magnetic force field) • A force will be exerted on an iron object placed in a magnetic field. • A magnetic field is a region in space where a magnetic force can be detected. The magnetic field around a bar magnet S N The compass needle is itself a tiny magnet (the North pole of this magnet points towards the South end of the magnet) Charm compass Magnetic field lines Strong magnetic field (high density of lines) ---> Demo c bar magnet & major magnet Thursday, 4 November 2010
  • 86. • θ changes with time • Angle of Dip - is the angle that the field lines make with the ground. At the equator, the angle of dip is zero. Near the poles the angle of dip is close to 90 degrees. THE EARTH’S MAGNETIC FIELD compass S N θ (angle of declination) = 11o Geographic North Earth’s axis Magnetic South Thursday, 4 November 2010
  • 87. Home to millions of species including humans, Earth is currently the only place in the universe where life is known to exist. The planet formed 4.54 billion years ago, and life appeared on its surface within a billion years. Since then, Earth's biosphere has significantly altered the atmosphere and other abiotic conditions on the planet, enabling the proliferation of aerobic organisms as well as the formation of the ozone layer which, together with Earth's magnetic field, blocks harmful solar radiation, permitting life on land. THE EARTH’S MAGNETIC FIELD IS ESSENTIAL FOR LIFE ON THE PLANET. The physical properties of the Earth, as well as its geological history and orbit, have allowed life to persist during this period. The planet is expected to continue supporting life for at least another 500 million years. Thursday, 4 November 2010
  • 88. MAGNETIC DOMAINS Thursday, 4 November 2010
  • 89. MAGNETIC MAGIC --> MAGNETIC DOMAINS Thursday, 4 November 2010
  • 90. MAGNETIC THEORY Ferromagnetic materials (Iron, Cobalt and Nickel) can be permanently magnetised. Electrons spinning in atoms have magnetic fields around them. They set up tiny North and South poles. Such an arrangement for an electron is called a dipole moment. [Illustrate a mag. dipole and mention Exchange Coupling] For most elements: magnetic fields cancel. N S e N S Iron, Cobalt and Nickel: Electron structure is such that there is a resultant magnetic field produced by each atom. These atoms are sometimes called atomic magnets.) Thursday, 4 November 2010
  • 91. Regions in a metal where the orientation of the magnetic dipoles is the same are called domains. Fully magnetised => the orientation of the domains is the same Unmagnetised Iron => Domains are scrambled A domain Here, a large number of iron atoms (magnetic dipoles) are aligned. Partially magnetised DOMAINS N NS S Thursday, 4 November 2010
  • 92. BREAKING A MAGNET Thursday, 4 November 2010
  • 93. “Lining up” the domains - Magnetising • Stroke the object end to end with a permanent magnet , in the same direction, using the same pole of the magnet. • Hold the object inside an D.C or A.C solenoid (Domains line up in the direction of the magnetic field) “Scrambling” the domains - Demagnetising Heat or hammer the magnet (This disturbs the alignment of the domains) [CAN ALSO BE DEMONSTRATED WITH THE SOLENOID] MAGNETISING AND DEMAGNETISING [Solenoid Demo] “Domains are induced into alignment” - Picking up iron objects Thursday, 4 November 2010
  • 94. EXERCISES Thursday, 4 November 2010
  • 95. Thursday, 4 November 2010
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  • 98. WIRES Thursday, 4 November 2010
  • 99. A circular magnetic field is formed around a straight current - carrying conductor: View from above • 3D View The direction of the magnetic field lines is given by the Right-hand Thumb Rule The right hand thumb rule: Thumb = direction of the electric current Curled fingers = direction of the circular magnetic field ( “•” represents current directed out of the page) I WIRES Thursday, 4 November 2010
  • 100. I Magnetic field lines • “View” from above” Thursday, 4 November 2010
  • 101. PARALLEL WIRES Piece of card field lines bunch and this leads to the wires repelling each other Current in opposite directions I I Thursday, 4 November 2010
  • 102. B = Magnetic field strength (in Tesla,T) I = Electric current in the wire (in Amps,A) µo= the permittivity of free space (the ability of a material to support a magnetic field (TmA-1 ) d = Distance from the wire (in metres, m) B = µ0I 2πd 1. Reversing the direction of the current reverses the direction of the magnetic field. 2. Magnetic field strength (symbol, B) is measured in NA-1m-1 or Tesla,T. 3. As the current in the wire, I increases the strength of the magnetic field increases B α I i.e. B is proportional to I 4. As the distance,d from the wire increases the strength of the magnetic field decreases. B α 1/d i.e. B is inversely proportional to d Note Thursday, 4 November 2010
  • 103. Example A special meter able to measure the magnetic field strength at any given point in the vicinity of a wire is shown below (taking a reading). It measures the magnetic field strength as 8 x 10-4 T at a distance of 0.01m from the centre of the wire. The current through the wire is 5 A. Calculate the value of the constant µo. Exercises B I d µo 5 x 10-5 T 2 A 20 mm 3.14 x 10-6 6 x 10-5 T 3A 0.159 m 2 x 10-5 TmA-1 7.2 x 10-5 T 3.11 A 2.2 cm 3.2 x 10-6 TmA-1 1.05 x 10-3 3A 20 mm 4.4 x 10-5 TmA-1 Thursday, 4 November 2010
  • 104. CURRENT CARRYING CONDUCTOR AND MAGNETIC FIELD -----> COIL Thursday, 4 November 2010
  • 105. COILS Thursday, 4 November 2010
  • 106. The magnetic field of a solenoid is similar in shape to that of a bar magnet: If the current is known, the poles of the solenoid can be determined using the right hand thumb rule applied earlier to the straight wire: Draw the field lines Complete this: Field lines are parallel in the core of the solenoid which --> the magnetic field in the core is uniform. The density of magnetic field lines is greatest in the core --> the magnetic field strength is greatest in the core. THE SOLENOID Thursday, 4 November 2010
  • 107. Predicting North and South poles: Thumb points to North pole of the solenoid from inside the coil Curled fingers indicate the direction of the current Factors affecting the strength of the magnetic field: 1. Increasing the current increases the magnetic field strength. 2. Increasing the number of turns of wire per given length of the electromagnet increases the magnetic field strength STRENGTH RULZ Thursday, 4 November 2010
  • 108. Uses of electromagnets 1. Electromagnets in relays are able to open and close electrical circuits (eg. starter motor circuit in a car). 2. Used in scrap yards to lift car bodies. 3. Create the ringing sound in electric bells. 4. Electromagnets in the recording heads of tape recorders are used to magnetise the audio tape during recording. A solenoid which contains an iron core is called an electromagnet. Adding an iron core increases the strength of the magnetic field because the iron core itself becomes magnetised and adds to the magnetic field of the solenoid. ELECTROMAGNETS Thursday, 4 November 2010
  • 109. Induced magnetism An unmagnetised object will have have its domains aligned and therefore develop a north and south pole. The object can be picked up by the magnet because opposite poles attract. x x x x x x x x - - - - - - - - North end of coil South end S NN Attraction to South of coil Attraction to North of coil The dipoles in the object change along the rod as the rod is drawn into the coil and it is this dipole change which pulls the rod into the coil Cross-section of coil THE COIL GUN Thursday, 4 November 2010
  • 110. ELECTROMAGNETS IN RELAYS Thursday, 4 November 2010
  • 111. Thursday, 4 November 2010
  • 112. Thursday, 4 November 2010
  • 113. PRACTICAL Thursday, 4 November 2010
  • 114. STATIC Thursday, 4 November 2010
  • 115. plastic rod (charged by rubbing with a cloth) small pieces of torn paper Observation: Explanation: 1 2 Balloon rubbed against hair Removed from head and then brought back to hair PLAYING AROUND WITH STATIC ELECTRICITY Observation: Explanation: Thursday, 4 November 2010
  • 116. 3 Balloon rubbed against jersey Release Observation: Explanation: 4 cotton (a) Each balloon charged separately by rubbing against the sleeve of a jersey (b) Holding the balloons by the cotton, release them, allowing them to come close to each other. Observation: Explanation: Thursday, 4 November 2010
  • 117. Charged plastic rod is held near a thin stream of water http://phet.colorado.edu/new/simulations/sims.php? sim=Balloons_and_Static_Electricity Observation: Explanation: 6 5 (a) Straw, charged at both ends (using a woollen cloth) (b) Straw, also charged using a woollen cloth held horizontally and brought close (c) Repeat (b) using a silk cloth. Observation: Explanation: Thursday, 4 November 2010
  • 118. Cap Insulating material Body of electroscope Leaf Base Aim to charge an electroscope by both induction and by contact and to draw charge distribution diagrams Method 1. Follow the instructions below 2. Write observations as you perform each step 3. Complete the diagrams only after recording the observations (you may need some help with these) Part 1 - Charging by induction Equipment dry cloth/jersey perspex rod ebonite rod electroscope 1. Charge the rod by rubbing it with a dry cloth/jersey and hold the rod near the cap of the electroscope Observation: If the rod was positively charged the charge distribution diagram would look like this: ++++ - - - - + + + + CHARGING OBJECTS Thursday, 4 November 2010
  • 119. Complete the diagram to show how charges would distribute on the electroscope should the rod be negatively charged. - - - - 2. With the rod in this position, earth the cap with your finger. Observation: ++++ - - - - + + + + This symbol represents a connection to the earth Complete the diagram to show how charge moves when the cap of the electroscope is earthed Thursday, 4 November 2010
  • 120. 3. Unearth the cap of the electroscope without removing the charged rod Observation: Draw the charge distribution diagram (by adding to the existing diagram on the right) showing the situation once this charge movement has finished. ++++ - - - - + + + + Draw the resultant charge distribution and the new position of the leaf on the diagram (right). 4. Remove the charged rod Observation: Finally, complete the diagram (right). Thursday, 4 November 2010
  • 121. Part 2 - Charging by contact Method 1. Follow the instructions below 2. Write observations as you perform each step 3. Complete the diagrams only after recording the observations (you may need some help with these) 1. A positively charged rod is held near the cap of the electroscope. 2. The rod makes contact with the cap. 3. The rod is removed. ++++ Thursday, 4 November 2010
  • 122. Cap Insulating material Body of electroscope Leaf Base Aim to charge an electroscope by both induction and by contact and to draw charge distribution diagrams Method 1. Follow the instructions below 2. Write observations as you perform each step 3. Complete the diagrams only after recording the observations (you may need some help with these) Part 1 - Charging by induction Equipment dry cloth/jersey perspex rod ebonite rod electroscope ALL CHARGED UPLab 12 1. Charge the rod by rubbing it with a dry cloth/jersey and hold the rod near the cap of the electroscope Observation: If the rod was positively charged the charge distribution diagram would look like this: ++++ - - - - + + + + The leaf of the electroscope springs up. Thursday, 4 November 2010
  • 123. Complete the diagram to show how charges would distribute on the electroscope should the rod be negatively charged. - - - - 2. With the rod in this position, earth the cap with your finger. Observation: ++++ - - - - + + + + This symbol represents a connection to the earth Complete the diagram to show how charge moves when the cap of the electroscope is earthed - - - - ++++ - - Electrons at the cap are repelled by the negatively charged rod. Leaf of the electroscope drops Electrons at the cap are held in position by the positively charged rod. The earth supplies electrons to the positively charged leaf and lower stem. Thursday, 4 November 2010
  • 124. - - - - 3. Unearth the cap of the electroscope without removing the charged rod Observation: Draw the charge distribution diagram (by adding to the existing diagram on the right) showing the situation once this charge movement has finished. ++++ - - - - + + + + Draw the resultant charge distribution and the new position of the leaf on the diagram (right). 4. Remove the charged rod Observation: Finally, complete the diagram (right). The cap and leaf now have no overall charge. Electrons on the cap are still held in position. Leaf of the electroscope remains in the “dropped” position. The charge distribution has not changed - - - - - - - -+ + + ++ + - - - - - - - - Negative charge redistributes itself around the metal parts of the electroscope leaving the stem and leaf with an overall negative charge The leaf of the electroscope springs up. ++++ + + Thursday, 4 November 2010
  • 125. Part 2 - Charging by contact Method 1. Follow the instructions below 2. Write observations as you perform each step 3. Complete the diagrams only after recording the observations (you may need some help with these) 1. A positively charged rod is held near the cap of the electroscope. 2. The rod makes contact with the cap. 3. The rod is removed. ++++ - - - - + + + + + + + + - Electrons migrate up into the rod + + + + + + + The electroscope is now left with an overall positive charge. Charge separation occurs. Positive repels positive at the stem/leaf Thursday, 4 November 2010
  • 126. 12 Physics > resources > electricity > DC electricity > videos SPARKS Thursday, 4 November 2010
  • 127. • When the generator is turned on, the electric motor begins turning the belt. • The belt is made of rubber and the lower roller is covered in silicon tape. Silicon has a greater affinity for electrons than rubber and so it captures electrons from the belt. The belt in turn must capture electrons from the dome, leaving the dome positively charged. Label the picture of the Van der Graaf (left) using the labels in the box below: Lower roller Belt - A piece of surgical tubing Output terminal - an aluminium or steel sphere Upper roller - A piece of nylon Motor Upper brush - A piece of fine metal wire Lower Brush ______________ ______________ ______________ ______________ ______________ ______________ ______________ Reference: http://science.howstuffworks.com/vdg3.htm THE VAN DER GRAAF - HOW IT WORKS Thursday, 4 November 2010
  • 128. THE VAN DER GRAAF - OBSERVATIONS & EXPLANATIONS 1. Small dome held close to generator dome 2. Hair stands on end when contact is made with the generator dome 3. Aluminium foil plates flying of the top of the generator dome Drawn observation Drawn observation Drawn observation Explanation Explanation Explanation Thursday, 4 November 2010
  • 129. 4. Sparking - a result of ionisation --> thorough step by step explanation 5. The shock that is felt ------> Charge travelling from/into the earth, through the body Thursday, 4 November 2010
  • 130. DC Thursday, 4 November 2010
  • 131. CIRCUIT RULES Thursday, 4 November 2010
  • 132. ADDING BULBS IN PARALLEL 1. Set up each of the following circuits, one after the other (making a mental note of the brightness of the lamps in the circuit. 2. For each circuit read the ammeter and record the current in the space provided. + - 1 A + - 2 A + - 3 A Current = ______ A Current = ______ A Current = ______ A Observation Explanation 8 V 8 V 8 V Thursday, 4 November 2010
  • 133. ADDING BULBS IN SERIES 1. Set up each of the following circuits, one after the other (making a mental note of the brightness of the lamps in the circuit. 2. For each circuit read the ammeter and record the current in the space provided. + - 1 A Current = ______ A + - 3 A Current = ______ A Observation + - 2 A Current = ______ A Explanation Thursday, 4 November 2010
  • 134. CURRENT IN THE SERIES CIRCUIT IS CONSTANT Aim to look for a pattern in the current through bulbs and resistors in a series circuit. 1. Use ONE ammeter in the three different places shown in the circuit diagram. 2. Without changing the setting on the power pack or the variable resistor write the current readings in the spaces provided (below): Equation 8V+ - A2 A1 A3 A1 = ______ A A2 = ______ A A3 = ______ A Thursday, 4 November 2010
  • 135. CURRENT IN THE PARALLEL CIRCUIT IS SHARED Aim to look for a pattern in the current through bulbs and resistors in a parallel circuit. 1. Use ONE ammeter in the each of the four places shown in the circuit diagram. 2. Without changing the setting on the power pack record your results below: Conclusion Current in a parallel circuit is ____________ . Equation relating the currents Results A1 = ____ A A2 = ____ A A3 = ____ A A4 = ____ A 8V+ - A1 A2 A3 A4 Thursday, 4 November 2010
  • 136. VOLTAGE IN THE SERIES CIRCUIT IS SHARED Aim to look for a pattern in the voltages across bulbs and resistors in a series circuit. Conclusion The power supply voltage is _____________ between the components in the circuit 8V+ - V1 V2 V4 V3 A Equation relating the voltages 1. Set up the circuit (below) 2. Use ONE voltmeter in the three different places shown in the circuit diagram. 3. Without changing the setting on the power pack record your results below: Results V1 (power supply) = __ V V2 (variable resistor) = __ V V3 (bulb) = __ V V3 (ammeter) = __ V Thursday, 4 November 2010
  • 137. VOLTAGE IN THE PARALLEL CIRCUIT IS CONSTANT Aim to look for a pattern in the voltages across bulbs and resistors in a series circuit. Conclusion The voltage across components connected in parallel is __________ Equation relating the voltages Results V1 = ___ V V2 = ___ V V3 = ___ V 8V+ - V1 V2 V3 1. Set up the circuit (below) 2. Use ONE voltmeter in the three different places shown in the circuit diagram. 3. Without changing the setting on the power pack record your results below: Thursday, 4 November 2010
  • 138. VOLTAGES AND CURRENTS IN SERIES AND PARALLEL Aim to investigate voltage and current in a series circuit that has a parallel portion in it. 1. Set up the circuit (below) 2. Use ONE voltmeter in the four different places shown in the circuit diagram and ONE ammeter in the four different places shown. 3. Without changing the setting on the power pack record your results below: Results A1 = __A A2 = __A A3 = __A A4 = __A V1 = __V V2 = __V V3 = __V V4 = __V 8V+ - V1 V3 V4 V2 A1 A2 A3 A4 V4 Thursday, 4 November 2010
  • 139. OHMIC CONDUCTORS Thursday, 4 November 2010
  • 140. OHM’S LAW Results Voltage setting of Power pack (V) 2 4 6 8 10 12 Voltage, V (V) Current, I (A) + - A V ice beaker water immersion coil Method 1. Set up the following circuit using iced water to cool the immersion coil. 2. Increase the voltage in regular increments through an appropriate range (widest possible range) Lab 14 Thursday, 4 November 2010
  • 141. Notes •The iced water was used to keep the temperature of the coil constant . •If the iced water was forgotten and the coil was allowed to heat up then the graph would curve up. Draw a graph of Voltage against Current on the grid provided Repeat the experiment but this time replace the coil with a lamp (that will increase in temperature as the current through it increases) Conclusion Thursday, 4 November 2010
  • 142. RESISTANCE Thursday, 4 November 2010
  • 143. Resistance specified MeasurementsMeasurements Resistance calculated from measurements Calculated Power output Resistance specified Voltage Current Resistance calculated from measurements Calculated Power output R1 R2 R3 Resistance calculated (formula provided for parallel resistors) Resistance calculated (formula provided for parallel resistors) across combination of resistors drawn from the power supply For any circuit, set power supply voltage to 8V For any circuit, set power supply voltage to 8V R1 & R2 in series R1 & R2 & R3 in series R1 & R2 in parallel R1 & R2 & R3 in parallel RESISTANCE AND POWER in series and parallel Thursday, 4 November 2010
  • 144. POWER Thursday, 4 November 2010
  • 145. CALCULATING POWER OUTPUT OF APPLIANCES WITH AN ELECTRONIC MONITOR Appliance UNDER CONSTRUCTION Thursday, 4 November 2010
  • 146. ELECTROMAG Thursday, 4 November 2010
  • 147. HANGING MAGNETS I NS 1. Cut out the net and fold it at the dotted lines to create a cradle + NS 2. Suspend the magnet in the cradle 3. Use a short length of cotton to suspend the magnet from a retort stand 4. Repeat using a second magnet. Thursday, 4 November 2010
  • 148. HANGING MAGNETS II 1. Position your two magnets in the orientations shown 2. For each orientation, record your observations NS N S NS N S NS NS NS N S A B C D ObservationObservation ObservationObservation Thursday, 4 November 2010
  • 149. STROKING MAGNETS NS A. Try magnetising an iron nail using a magnet B. Once you have finished, check for a magnetic field using a charm compass. Thursday, 4 November 2010
  • 150. PLOTTING MAGNETIC FIELDS A. Place one or more compasses around a bar magnet Charm compasses (moved around in a variety of positions around the magnet B. Use a pencil to mark the North pole of each magnet using a dot C. Connect the dots using a smooth curve D. Plot several field lines and mark the North and South poles of the magnet. E. Wrap your magnet in glad wrap and spring iron filings over it. Follow the instructions (below) and draw your observations Thursday, 4 November 2010
  • 151. EXAMS Thursday, 4 November 2010
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  • 215. 2009 NCEA Topic TEST FOR 2010 Laid out in a form that is ready for PCopying Thursday, 4 November 2010
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  • 235. Main points: 1(b) & 1(c) - need further elaboration. See revised notes. 3(a) & (b)....... confusion re. magnetic and geographic poles being opposite 3(c) .......... Need to appreciate that coil becomes magnetised which causes dipole and domain alignment in the piston 3 (d) ...... problems creeping in with the algebra Thursday, 4 November 2010
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  • 241. 1 2 CombinedCombined Grade 3 3 88 A 3 3 1010 M 1 1 3E 12 E Thursday, 4 November 2010
  • 242. 2010 PRACTICE EXAM Thursday, 4 November 2010
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  • 246. Worthwhile going into some detail over this. Thursday, 4 November 2010
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  • 249. Need to learn how to do the algebra - BIG PROBLEMS HERE - can be v easily addressed Marks thrown EASY Thursday, 4 November 2010
  • 250. EXERCISES Thursday, 4 November 2010
  • 251. Y11 Sci - W & W Thursday, 4 November 2010
  • 252. 1. Resistance is ______________ __________ . 2. It is responsible for slowing down the ___________ ___ __________ . 3. When there is resistance present in a conductor or an appliance ____________ energy is transformed into ________ energy. 4. Three everyday household appliances that have high resistance are: 5. __________________ , ___________________ , _____________________ 6. Complete the circuit diagram which shows what happens when a third identical lamp is added in parallel to a circuit: + - + - 7. If we continued to add lamps in this fashion explain why the power supply would blow a fuse. __________________________________________________________________ __________________________________________________________________ QUESTIONS ON RESISTANCE Thursday, 4 November 2010
  • 253. APPLYING THE WATER MODEL TO Q.6 1. Pool of water fed by a rut a rut channel 2. Dig a channel and water can flow out of the pool 3. Dig another channel. What effect does this have on the flow through the rut? 3. Dig a third channel ........ What effect does this have on the flow through the rut? Thursday, 4 November 2010
  • 254. “ANSWERS TO EXTRA FOR EXPERT QUESTIONS” Thursday, 4 November 2010
  • 255. + - Draw using arrows the direction of charge flow in each circuit For each circuit, highlight the lamps that will glow. 1 2 + - 3 + - 4 + - NEED FOR A VOLTAGE SUPPLY AND A CONDUCTING PATH Thursday, 4 November 2010
  • 256. CONDUCTORS & METERS 1. Which of the following substances are conductors of electricity: wood, tap water, copper, iron, glass, sodium chloride solution, rubber 2. Draw a circuit diagram that you could use to test for conductivity using a lamp, dry cell and wires. 3. Explain how a fuse protects a circuit. 4. Voltage is .......... 5. Current is .......... 6. Starter Thursday, 4 November 2010
  • 257. In an an Ohm’s law experiment a water-cooled resistor was connected in series with a power supply and an ammeter. A voltmeter was connected to measure the voltage drop across the resistor. The readings on the two meters were recorded. Voltage (V) Current (I) 2 0.15 4 0.31 6 0.45 8 0.59 10 0.75 12 0.92 Draw a graph of V vs I. What is the meaning of its slope and what is its unit? OHM’S LAW Explain the shape of the graph that would be produced should the resistor be allowed to heat up. Thursday, 4 November 2010

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