BEHAVIOR OF CHARGED PARTICLES IN AN ELECTRIC FIELD METAL COATED PING PONG BALL 1 When power supply is switched on, the ping pong does not move as it is neutral. 2 When the ping pong ball is displaced to the negatively charged plate, positive charges of the ball will be discharged. 3 The ping pong ball becomes negatively charged and repelled to the positively charged plate. 4 At the positively charged plate, the electrons will be transferred to the metal plate until the ball becomes positively charged. 5 The ball will be repelled and attracted to the negatively charged plate. 6 The process keeps repeating (oscillating) until the power supply is switched off. CANDLE FLAME 1 When the power supply is switched on, the candle flame starts to spread out in both directions. 2 The heat from the candle flame cause the air to ionise to form positive ions and negative ions. 3 The negative ions will be attracted to the positively charged plate while the positive ions will be attracted to the negatively charged plate. 4 The spread of flame towards the negatively charged plate will be greater than towards the positvely charged plate as positve ions has larger mass and size. EXAM WATCH Diagram shows a metal coated polystyrene ball hung between two metal plates M and N which is connected to E.H.T. supply. A strong electric field between metal plates M and N is produced when the switch is on. What will happen when metal coated polystyrene ball is displaced to touch the positively charged metal plate while the switch is on? Explain your answer. [4] Negeri Sembilan 2021 Question 10 47 PhysicsCHAPTER 3 ELECTRICITY
Electric Current Definition: Rate of charge flow = S.I unit: ampere, A 1 The direction of current flow is opposite to the direction of electrons flow. 2 Electrons carry negative charges. The charge of an electron, e is −1.6 × 10!"# C. Therefore, charges carried by n electrons: = S.I unit: coulomb, C H T TIPS How many electrons make up 1 coulomb? Given that Q = Ne. The charge in an electron is -1.6 × 10-19 C Q = ne 1 = n (-1.6 × 10-19) n = 6.25 × 1018 *The negative sign is ignored because n is representing the number of electrons only. ✏ CHECKPOINT 1 A conductor conducts a current of 6.5 A for 100 ms. What is the total charge that flows through this conductor in 100 ms? 2 20 C of charges flow through a lamp in 40 s. Calculate the electric current flowing through the lamp. 3 1.25 × 1018 electrons flow across a conductor in 5 seconds. Calculate the current that flows in the conductor. [charge on an electron, e = 1.6 × 10-19 C] = total charge = number of electrons = charge of an electron = current = total charge = time 48 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
4 A steady current of 0.2 A flows across a bulb for 5 minutes. Calculate the number of electrons that flow across the bulb. [charge on an electron, e = 1.6 × 10-19 C] 5 When Ali turned on his car light, the current flowing through his car lamps is 3.6 A. a) What is the amount of charges flow through his car lamp in half an hour? b) Given that 1 electron carries 1.6 × 10-19 C of charge, calculate the number of electrons will flow through his car lamps in half an hour? 6 An electric kettle is used to boil water, it is found that the current that flows through the heating filament of the kettle is 5 A and the total charge that flows through the filament is 3000 C. a) What is the time taken to boil the water? b) A new electric kettle is used to boil the same quantity of water. It is found that the time taken is 15 minutes and the total charge that flows through the heating filament is 6000 C. Calculate the current that flows through the heating filament of the new kettle. 49 PhysicsCHAPTER 3 ELECTRICITY
IDEAS ABOUT ELECTRIC POTENTIAL DIFFERENCE Pressure at point P > pressure at point Q § Water will flow from P to Q due to difference in pressure Gravitational potential energy of ball at X > gravitational potential energy of ball at Y § Ball will fall from X to Y due to difference in the gravitational potential energy Electric potential at X (positive terminal) > electric potential at Y (negative terminal) § Electric current flows from X to Y due to the electric potential difference across battery Electric potential difference drives a current through a conductor. Potential Difference, V Definition: The potential difference across two points is defined as work done to transfer one unit of charge across the two points = S.I. unit: volt, V or J C-1 § The potential difference across an electrical appliance can be measured using a voltmeter, which is connected in parallel to the electrical appliance. ✏ CHECKPOINT 1 The potential difference across two metal plates is 120 V. What is the number of electrons that is transferred across the plates if 960 J of energy is dissipated during the process? [charge on an electron, e = 1.6 × 10-19 C] 2 Lightning is an example of electric current. In a typical lightning, the flash takes 0.2 s to transfer 109 J of energy across a voltage of 5 × 107 V. Use the information given, calculate the amount of charge transferred and the current flows in 0.2 s. = potential difference = work done = total charge 50 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
Ohmic Conductor and Non-ohmic Conductor 1 Ohm’s law states that the potential difference, V across a conductor is directly proportional to the current, I that flows through the conductor. ∝ 2 Based on Ohm’s law, ∝ = = , = , Therefore, = EXPERIMENT 3.1 Inference: Potential difference depends on current. Hypothesis: Current increases, potential difference increases. Aim: To investigate the relationship between current and potential difference. A. Ohmic conductor (constantan wire) Variables: a) Manipulated: Current, b) Responding: Potential difference, c) Constant: Length of constantan wire Apparatus and Materials: dry cell, switch, connecting wires, ammeter, voltmeter, metre rule, rheostat, constantan wire Arrangement of Apparatus: Procedure: 1 Length of constantan wire is fixed. The switch is closed. The rheostat is adjusted until the ammeter reading, = 0.2 A. 2 The voltmeter reading, is recorded. 3 Repeat the experiment with = 0.3 A, 0.4 A , 0.5 A and 0.6 A. Results: Current , / A Potential difference, / V 0.2 0.3 0.4 0.5 0.6 B. Non-ohmic conductor (filament bulb) Variables: a) Manipulated: Current, b) Responding: Potential difference, c) Constant: Length of filament Apparatus and Materials: dry cell, switch, connecting wires, ammeter, voltmeter, metre rule, rheostat, filament bulb = potential difference = current = resistance 51 Physics CHAPTER 3 ELECTRICITY
Procedure: 1 The constantan wire is replaced with a filament bulb. The rheostat is adjusted until the ammeter reading, = 0.14 A. 2 The voltmeter reading, is recorded. 3 Repeat the experiment with = 0.16 A, 0.18 A , 0.20 A and 0.22 A. Results: Current , / A Potential difference, / V 0.14 0.16 0.18 0.20 0.22 Data Analysis: Graphs of against are plotted. Conclusion: Current, increases, the potential difference, increases. OHMIC CONDUCTOR AND NON-OHMIC CONDUCTOR TYPE OF CONDUCTOR OHMIC CONDUCTOR NON-OHMIC CONDUCTOR GRAPH OF V AGAINST I RATE OF INCREASE OF VOLTAGE Constant Increases RESISTANCE Constant Increases TYPES OF CIRCUIT SERIES CIRCUIT PARALLEL CIRCUIT ! = " = # = $ ! = " + # + $ ! = " + # + $ ! = " = # = $ / V / A / V / A 52 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
H T TIPS When solving problems involving complex circuit diagrams, we must always simplify the circuit diagrams to facilitate the calculations. Following are some examples. Potential Difference Divider When two or more resistors are connected in series, potential difference across each resistor and be determined as shown below. " =" " + $ × $ =$ " + $ × ✏ CHECKPOINT 1 Find the effective resistance for each of the following arrangement of resistors. a) b) 53 PhysicsCHAPTER 3 ELECTRICITY
2 Diagram shows the arrangement of three resistors in a circuit. a) Calculate the effective resistance in the circuit. b) Determine the reading on ammeter. c) The resistors which are arranged in parallel are then arranged in series. Calculate the new reading on the ammeter. 3 Diagram shows an electrical circuit with two resistors connected in series. Calculate a) the effective resistance in the circuit, b) the reading of the ammeter, c) the reading of the voltmeter. 4 Diagram shows an electrical circuit with two resistors connected in series. Calculate a) the effective resistance in the circuit, b) the reading of the ammeter, c) the reading of the voltmeter. 54 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
5 Diagram shows an electrical circuit with three resistors connected in series. Calculate a) the effective resistance in the circuit, b) the reading of the ammeter, c) the reading of the voltmeter. 6 Diagram shows two resistors connected in parallel in a circuit. Calculate the readings of ammeters A1, A2 and A3. 7 Diagram shows three resistors connected in parallel in a circuit. Calculate the reading of ammeters A1, A2 and A3. 8 Diagram shows three resistors connected in parallel. Calculate the ration of I1 : I2. 55 PhysicsCHAPTER 3 ELECTRICITY
9 Diagram shows an electrical circuit. Calculate a) the effective resistance of the circuit, b) the reading of the voltmeter, c) the ratio of reading of A1 to reading of A2. 10 Diagram shows an electrical circuit. Calculate a) the effective resistance, b) the reading of the voltmeter, c) the ratio of reading of ammeter A1 to the reading of ammeter A2. 11 Diagram shows an electrical circuit. Calculate a) the effective resistance, b) the readings of ammeter A1 and ammeter A2, c) the reading of the voltmeter. 56 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
12 Diagram shows an electrical circuit with three resistors. a) Switch S is opened. Calculate the i. effective resistance in the circuit, ii. reading of ammeter, iii. reading of voltmeter. b) Then, switch S is closed. Calculate the i. new effective resistance in the circuit, ii. new ammeter reading. 13 Figure shows an electrical circuit with three resistors. a) Switch S is remained open. Calculate the reading of i. Ammeter ii. Voltmeter b) Then, switch S is connected. Calculate i. The effective resistance in the circuit ii. The new reading of the ammeter EXAM WATCH Poor lighting in a classroom causes students not be able to study well. You are asked to modify the classroom and the lighting system of the classroom so that it is brighter, safer and able to save electricity. State and explain the modifications based on the following aspects: § Type of connection of lamp circuits § Colour of classroom’s wall § Type of lamp and colour of light § Additional component for safety [10] Type of lamp connection: Reason: Colour of classroom’s wall: Reason: Type of lamps: Reason: Colour of light: Reason: Safety component: Reason: 57 Physics CHAPTER 3 ELECTRICITY
Resistivity of a Wire, Definition: The resistance per unit length of the material when its cross-sectional area is 1 m2. S.I. unit: ohm meter, 1 Resistivity of a conductor depends on its material. 2 Resistivity is directly proportional to the resistance of a wire, but inversely proportional to the conductivity, . ∝ ∝ 1 3 The resistance of a wire is directly proportional to its resistivity, and length, but inversely proportional to the cross-sectional area, . ∝ ∝ 1 By combining the relationship, = FACTORS THAT AFFECT THE RESISTANCE OF A WIRE LENGTH CROSS-SECTIONAL AREA RESISTIVITY TEMPERATURE ∝ ∝ 1 ∝ ∝ 58 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
EXPERIMENT 3.2 Inference: Resistance of wire depends on the length of wire. Hypothesis: Length of the wire increases, resistance of the wire increases. Aim: To investigate the relationship between the length of wire and the resistance of wire. Variables: a) Manipulated: Length of wire, b) Responding: Resistance, c) Constant: Diameter of wire Apparatus and Materials: dry cell, switch, connecting wires, ammeter, voltmeter, metre rule, rheostat, constantan wire, crocodile clips Arrangement of Apparatus: Procedure: 1 Diameter of wire is fixed. The rheostat is adjusted until the current flowing through the circuit, = 0.5 A. The crocodile clips P and Q are adjusted until the length of wire, = 20.0 cm. 2 The voltmeter reading, is recorded. Resistance, is calculated using the formula = % & . 3 Repeat the experiment with = 40.0 cm, 60.0 cm, 80.0 cm and 100.0 cm. Results: Length of wire, / cm Current , / A Potential difference, / V Resistance, / Ω 20.0 0.5 40.0 0.5 60.0 0.5 80.0 0.5 100.0 0.5 Data Analysis: Graph of against is plotted. Conclusion: Resistance of wire, is directly proportional to / increase linearly with the length of wire, . / Ω / cm 59 Physics CHAPTER 3 ELECTRICITY
EXPERIMENT 3.3 Inference: Resistance of wire depends on the cross-sectional area of wire. Hypothesis: Cross-sectional area of the wire increases, resistance of the wire decreases. Aim: To investigate the relationship between the cross-sectional area of wire and the resistance of wire. Variables: a) Manipulated: Cross-sectional area of wire, b) Responding: Resistance, c) Constant: Length of wire Apparatus and Materials: dry cell, switch, connecting wires, ammeter, voltmeter, metre rule, rheostat, constantan wires of s.w.g. 22, s.w.g. 24, s.w.g. 26, s.w.g. 28, s.w.g. 30, crocodile clips Arrangement of Apparatus: Procedure: 1 Length of wire is fixed. The rheostat is adjusted until the current flowing through the circuit, = 0.5 A. A s.w.g. 22 constantan wire is connected. 2 The voltmeter reading, is recorded. Based on the diameters given, cross-sectional area of wire, = # is calculated. Resistance, is calculated using the formula = % & . 3 Repeat the experiment with wires of s.w.g. 24, s.w.g. 26, s.w.g. 28 and s.w.g. 30. Results: s.w.g. Diameter, / mm Cross-sectional area, / mm2 Current, / A Potential difference, / V Resistance, / Ω 22 0.711 0.5 24 0.559 0.5 26 0.457 0.5 28 0.376 0.5 30 0.315 0.5 Data Analysis: Graph of against is plotted. Conclusion: Resistance of wire, is inversely proportional to / decreases non-linearly with the cross-sectional area of the wire, . Physics Bytes s.w.g. refers to standard wire gauge. s.w.g. increases, diameter of wire decreases. / Ω / mm2 60 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
EXPERIMENT 3.4 Inference: Resistance of wire depends on the resistivity of wire. Hypothesis: Resistivity of the wire increases, resistance of the wire increases. Aim: To investigate the relationship between the resistivity of wire and the resistance of wire. Variables: a) Manipulated: Resistivity of wire, b) Responding: Resistance, c) Constant: Length of wire Apparatus and Materials: dry cell, switch, connecting wires, ammeter, voltmeter, metre rule, rheostat, constantan wire, nichrome wire Arrangement of Apparatus: Procedure: 1 Length of wire is fixed. The rheostat is adjusted until the current flowing through the circuit, = 0.5 A. Constantan wire is connected. 2 The voltmeter reading, is recorded. Resistance, is calculated using the formula = % & . 3 Repeat the experiment with nichrome wire. Results: Type of wire Current, / A Potential difference, / V Resistance, / Ω Constantan 0.5 Nichrome 0.5 Conclusion: Resistance of wire, is increases with the resistivity of the wire, . ✏ CHECKPOINT 1 An aluminium cable with the length of 20 m has a diameter of 3.0 mm. If the resistivity of aluminium is 2.65 × 10-8 Ω m, what is the resistance of the cable? 2 What is the resistance of a copper wire with a diameter of 0.5 mm and length of 2 m? Assume that the wire has the shape of a cylinder. Resistivity of copper = 1.60 × 10-8 Ω m. 61 PhysicsCHAPTER 3 ELECTRICITY
3 Wire X with the length of 120 cm and the cross-sectional area of 0.80 mm2 has the resistance of 3.2 Ω. Calculate the resistivity of wire X. 4 A copper wire with the length and diameter d has the resistance of . What is the resistance of another copper wire with length 2 and diameter " $ in terms of . APPLICATION OF RESISTIVITY OF CONDUCTOR IN DAILY LIFE HEATING ELEMENT § Made of nichrome wire § Heating element is used in water heater and electric kettle to heat up water § Heating element is usually made from material with higher resistivity. § The higher resistance enables the heating element to convert electrical energy to heat energy to heat up the water. § Other than that, heating element must also be made of: i. Material with high melting point so that it does not melt when heating up water ii. Material that can withstand oxidation so that the water heated up is not contaminated by its oxide which is probably toxic. 62 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
CONNECTING WIRE § Copper wires are usually used in electrical wiring in homes because copper have low resistivity. § The low resistance of copper wire enables electrical current to flow more efficiently by reducing the energy loss to the surrounding in the form of heat. § Other than that, copper wire can resist oxidation and do not oxidise easily by action of oxygen in air. § Nevertheless, in the National Grid Network, aluminium cables are used instead of copper even though the resistivity of aluminium is higher than copper. This is because the density of aluminium is way lower than copper. This causes the aluminium cables to be lighter than copper cables. Superconductor 1 For a normal conductor, heat loss occurs when electrical energy passes through the conductor to overcome the electrical resistance. This reduces the efficiency of the conductor. 2 A superconductor is an electrical conductor that conducts electricity with extremely high efficiency. This is because a superconductor has zero resistance at low temperature. 3 Resistance of a superconductor will become zero when it is cooled below a certain temperature which is called the critical temperature, Tc. 4 Advantages of superconductor § Current can flow through superconductors without any energy loss at critical temperature. § Can produce a strong magnetic field § Current can flow through at a higher rate 5 When a small permanent magnet is placed on the surface of a superconductor, a current is induced inside the superconductor. Since the resistance in the superconductor is zero, therefore the induced current will flow continuously. 6 The magnetic field produced by the superconductor is opposed to the magnetic field produced by magnet. Therefore, superconductor will repel the permanent magnet and thus the permanent magnet will float above the superconductor. APPLICATION OF SUPERCONDUCTOR LEVITATING TRAINS 1 MAGLEV train works based on the superconducting electromagnets. 2 There is strong repulsion between the base of the train and the railway tracks that can make the train float about 1 cm above the track. 3 Since the train is floating, the train can move at a very high speed because there is no frictional force between the base of the train and the surface of the track. 63 Physics CHAPTER 3 ELECTRICITY
MAGNETIC RESONANCE IMAGING (MRI) 1 Superconductor is used in MRI system to determine the image of the human body system. 2 When human body is exposed to a strong magnetic field produced by the superconductor, hydrogen atoms in human body fluid and fat molecules are forced to receive energy. 3 The molecules will vibrate at a certain frequency that can be detected and displayed as an image. SUPERCONDUCTING CABLES 1 There are low temperature and high temperature superconducting cables. 2 When the superconducting cable is cooled or heated to their respective critical temperature, this cable can conduct electricity or transmit power with zero resistance. 3 This can increase the efficiency in power transmissions at a lower cost with little power and energy lost. EXAM WATCH Diagram shows a waffle maker with the power rating of 240 V, 1000 W. Waffle maker The number of turns of heating element Type of material used as the heating element The rate of oxidation Fuse J 3 Copper High 3.0 A K 3 Nichrome Low 5.0 A L 1 Nichrome High 3.0 A M 1 Copper Low 5.0 A You are required to study the characteristics of the waffle maker as shown in the table above. Explain the suitability of each characteristic and choose the most suitable waffle maker in shorter time. Give reasons for your choice. [10] Aspect Reason SBP 2021 Question 10 64 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
Electromotive force (e.m.f) 1 The electromotive force (e.m.f.), of a cell is defined as the energy supplied/work done by an electrical source to move 1 coulomb of charge in a complete circuit. Electromotive force (e. m. f. ), = Energy supplied by the cell Charge 2 The potential difference, V, across a component in a circuit is defined as the energy required to transfer a unit of charge across a component. Potential difference, = Energy required to transfer charge across a component Charge COMPARISON BETWEEN ELECTROMOTIVE FORCE AND POTENTIAL DIFFERENCE ELECTROMOTIVE FORCE, POTENTIAL DIFFERENCE, No current flows in the open circuit. A current flows in the closed circuit. Voltmeter gives the reading of the electromotive force, of an electrical sorce. Voltmeter gives the reading of potential difference, across electrical components such as bulb and resistor. 3 The electromotive force of an electrical source is greater than the potential difference across an electrical component. 4 This is due to the internal resistance of the dry cell. 5 The difference between the value of e.m.f. and the value of potential difference is known as voltage drop, v. Internal Resistance 1 The internal resistance of the cell is caused by the chemicals(electrolyte) in the cell. 2 A cell can be illustrated as an e.m.f., , connected in series with the internal resistor, r. 3 When a high resistance voltmeter is connected across a cell, the voltmeter gives the electromotive force of the cell. 4 If a resistor, , is connected to the cell, the voltmeter read the potential difference across the resistor, as well as the potential difference across the terminal of the cell. 5 The value of the electromotive force, is greater than the value of potential difference, . 65 Physics CHAPTER 3 ELECTRICITY
✏ CHECKPOINT 1 Figure shows an electrical circuit with a 20 Ω resistor. When the switch is opened, the voltmeter gives a reading of 3.0 V. When the switch is closed, the voltmeter gives a reading of 2.8 V. Calculate a) the reading of ammeter when the switch is closed, b) the internal resistance of the battery. 2 Figure shows a graph of potential difference, against current, . Calculate the electromotive force, and the internal resistance, of the cell. 3 When a dry cell with electromotive force, and internal resistance, is connected to a 13 Ω resistor, the ammeter gives a reading of 0.2 A. When a 13 Ω resistor is replaced with a 28 Ω resistor, ammeter gives a reading of 0.1 A. Calculate the electromotive force, and the internal resistance, of the dry cell. 4 Figure shows an electric circuit supplied by a battery with e.m.f. of 3.0 V. When the switch is closed, the voltmeter and ammeter give a reading of 2.7 V and 0.3 A respectively. Calculate a) The internal resistance of the battery b) The value of 66 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
EXAM WATCH Diagram shows a circuit before and after the switch is closed. A drop of the voltmeter reading is caused by A the energy that is needed to move charges in the circuit B the voltage that is needed to accelerate charges in the circuit C the energy that is needed to accumulate charges in the circuit D The voltage that is lost due to internal resistance of the cell SBP Model Paper Set 2 Question 29 ARRANGEMENT OF DRY CELLS ARRANGEMENT SERIES PARALLEL Diagram Total e.m.f., '(')* Effective internal resistance, +,,+-'./+ ✏ CHECKPOINT 1 Two batteries with each one has an e.m.f. of 4.5 V is connected in parallel as shown in the diagram. A bulb of 5 Ω and an ammeter are connected in series with the batteries. A voltmeter is connected across the bulb. a) What is the effective e.m.f. of the two batteries connected in parallel? b) What is the effective internal resistance of the batteries connected in parallel? c) What is the reading of the ammeter in the circuit? d) What is the reading of the voltmeter in the circuit? e) What is the voltage drop in the circuit? 67 PhysicsCHAPTER 3 ELECTRICITY
Relationship between Electrical Energy, , Voltage, , Current, , and Time, 1 Electrical energy is an important form of energy because it can change its form into other useful form of energy such as light energy, heat energy, heat energy, sound energy and etc. 2 Electrical energy is produced when work is done to transfer a charge in the electrical field. The S.I. unit for electrical energy is joule, J. Relationship between Power, , Voltage, and Current, Power is defined as the rate of work done or rate of energy transferred. = S.I. unit: watt, W H T TIPS = electrical power = electrical energy transferred = time = = = # = # =# =# = = = = = = = = 68 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
✏ CHECKPOINT 1 About 3.125 × 1019 electrons flow across a light bulb in 2.5 seconds when a battery with potential difference of 3 V is connected to the bulb. If the charge of electron is 1.6 × 10-19 C, calculate a) the energy dissipated b) the electrical power 2 An electrical iron is rated 240 V, 2000 W. a) What is the resistance of its heating element? b) Calculate the current when a 240 V supply is connected to the electrical iron. 3 A coil of heating element has resistance of 80 Ω. If a current of 3.0 A passing through it, calculate a) The supplied voltage b) The electrical power generated c) The electrical energy dissipated in 5 minutes 4 A current of 12 A flows through a light bulb when it is connected to a 240 V supply. Calculate the amount of energy dissipated after 5 minutes. 69 Physics CHAPTER 3 ELECTRICITY
5 Diagram shows an electrical circuit. Calculate a) The effective resistance of the circuit b) The reading of ammeter c) The reading of voltmeter d) The power dissipated in 2 Ω resistor. 6 Diagram shows an electric circuit. Calculate a) The effective resistance, b) The reading of ammeter A1 and voltmeter c) The reading of ammeter A2 d) The ratio of the power dissipated in the 6 Ω resistor to the power dissipated in the 12 Ω resistor. 70 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
Power and Energy Consumption Rate in Various Electrical Appliances 1 Most household electrical appliances that work on heating effect of current are marked with volage and power ratings. 2 For example, an electrical iron which is marked 240 V, 2400 W means the power of 2400 W is consumed when the power supply of 240 V is connected to the electrical iron. 3 The cost of electricity consumed is calculated by the number of units of electricity used. One unit is equivalent to one kilowatt-hour. 4 One kilowatt-hour is the amount of energy consumed by an electrical appliance at the rate of one kilowatt for the duration of one hour. 1 kW h = = = ✏ CHECKPOINT A 2 kW electric kettle takes 10 minutes to boil water. Calculate the cost of electrical energy used to boil the water if the cost of electrical energy is 18 sen per unit consumed. Steps in Reducing Household’s Electrical Energy Used Efficiency of Electrical Appliances 1 Efficiency refers to how effective an electrical appliance work. 2 An effective appliance results minimum energy loss in the form of heat or sound energy. The output energy is almost equal to the input energy. 3 An ineffective appliance usually results a lot of energy loss in the form of heat or sound energy. efficiency = output power input power × 100% 4 The efficiency of an electrical appliances is always lesser than 100%. This is because energy loss is required to overcome frictions or obstructions such as air resistance or electrical resistance. ✏ CHECKPOINT A machine is rated 240 V, 500 W. When the machine is connected to 240 V power supply, the machine is able to lift up a 50 kg load to a height of 3.6 m in 20 seconds. Calculate the efficiency of the machine. [ = 10 m s-2] 71 PhysicsCHAPTER 3 ELECTRICITY
Effective Usage of Electricity 1 Electrical consumption is the main source of energy consumption in a household. All the electrical appliances used at home contribute a great part of the electrical bill. 2 Proper steps should be taken when using electricity so that an electrical appliance can be used effectively and wastage can be prevented. 3 The following shows several steps that should be taken so that electricity can be saved and used wisely. a) Switch off the electrical appliances when they are not in use. b) Always compare the power rating of electrical appliances before buying and using them. c) Choose the appliance that uses electrical energy more effectively. For example, always prefer a fluorescent lamp compared to a light bulb. This is because a light bulb has low efficiency and results a lot of energy loss in form of heat energy. d) Switch on the air conditioner whenever necessary. Set the temperature to give you the condition where you are comfortable with and not the coolest level. Choose the correct horsepower for the correct room size. Clean the filter regularly. If possible, use a fan instead of air conditioner. e) Do not let the door of refrigerator opened all the time. Close the door of the refrigerator when it is not in use. f) Use a washing machine to its maximum capacity. Washing one or two pieces of clothes in one washing machine causes wastages of electricity and water as well. g) When using a vacuum cleaner, make sure that the dust bag is not full. Clean the air filter regularly. Use the vacuum cleaner whenever necessary only. EXAM WATCH Diagram shows a bulb with power rating label 12 V, 6W. What is meant by the power rating label? A The light bulb releases 6 J of energy per second when connected to 12 V potential difference B The light bulb releases 6 J energy per second when electric current flow is 0.5 A C The light bulb releases 6 J energy per second when connected 24 Ω electrical resistance D The light bulb releases 6 W of power when connected to 12 V potential difference MRSM 2023 Question 34 72 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
“WHERE INVISIBLE FORCES DANCE, WEAVING THE FABRIC OF OUR ELECTRIFIED UNIVERSE." ELECTROMAGNETISM 4S P M P H Y S I C S
Magnetic Field Definition: Region where an object experiences magnetic force. § Magnetic field lines leave the north pole of the magnet and enter the south pole. § A magnetic field is produced when a current flows through a conductor. § The right-hand grip rule is used to determine the direction of magnetic field for a straight conductor. a) The right hand grips the wire b) The thumb shows the direction of current c) The other four fingers will show the direction of the magnetic field Right-hand grip rule Cross sectional view Direction of current into the plane Direction of current out of the plane § The pattern and direction of the magnetic field of current-carrying coil and current-carrying solenoid. a current-carrying coil a current-carrying solenoid Pattern of Resulting Magnetic Field 1 The force on a current-carrying conductor in a magnetic field is produced by the interaction between two magnetic fields which are the magnetic field from the electric current in the conductor and the magnetic field form the permanent magnet. 2 These two magnetic fields combine to produce a resultant magnetic field that is known as catapult field. 3 The catapult field exerts a resultant force on the conductor. Definition: A catapult field is the resultant magnetic field produced by the interaction between magnetic field of a current-carrying conductor and the magnetic field of a permanent magnet. Magnetic field of permanent magnet Magnetic field of current-carrying conductor Resultant field of two interacting fields Uniform magnetic field lines from north to south Magnetic field lines in the form of concentric circles, in an anticlockwise direction based on the right-hand grip rule. A catapult field is produced. The distortion of the magnetic field lines causes the wire to move from the stronger magnetic field to the weaker field. 74 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
EXAM WATCH On the diagram below, draw the magnetic field pattern produced. SBP Model Paper Set 1 Question 2 4 The direction of the force on a current-carrying conductor can be determined by using Fleming’s left-hand rule. Factors Affecting the Magnitude of the Force Acting on a Current-carrying Conductor in a Magnetic Field The magnitude of current Current increases, force produced increases. How to increase the current? § Increase the magnitude of e.m.f. § Use thicker and shorter wire (lower resistance) The strength of the magnetic field Magnetic field increases, force produced increases, How to increases the strength of the magnetic field? § Use stronger magnet § Place the magnets closer to each other. 1 We can observe the effect of the increase in current and the strength of magnetic field on the force by observing the height of swing of the copper frame. 75 Physics CHAPTER 4 ELECTROMAGNETISM
2 Another way to study the factors that affect the magnitude of the force acting on a current-carrying conductor in a magnetic field is shown in the diagram below. The distance travelled, d by the conductor represent the magnitude of the force. ✏ CHECKPOINT Diagram shows the arrangement of apparatus to study the effect of the force on a current carrying conductor. Explain the motion of the copper PQ and state the direction of the motion. Effects of a Current-carrying Coil in a Magnetic Field 1 A rectangular coil is formed from a piece of copper wire. 2 A pair of forces acting on sides PQ and SR of a current-carrying coil in a magnetic field. 76 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
3 This pair of forces is produced by the interaction between the magnetic field from the current in the coil and the magnetic field from the permanent magnets. 4 The direction of force can be determined by applying Fleming’s left-hand rule on both sides of coil. 5 The pair of forces have equal magnitude and act in opposite directions on the sides of the coil. 6 The current-carrying coil in the magnetic field will rotate about the axis of rotation. Direct Current Motor 1 Small electrical appliances such as blenders, food mixers and children’s toys have a small direct current motor. Large direct current motors are found in machines such as lifts and electric vehicles. 2 The direct current motor changes electrical energy to kinetic energy by using the turning effect of a current-carrying coil in a magnetic field. 3 An important component in a direct current motor is the commutator that rotates with the rectangular coil. The carbon brushes in contact with the commutator are in their fixed position. First Half Rotation Second Half Rotation § Carbon brush X in contact with red half of the commutator § Carbon brush Y in contact with blue half of the commutator § Direction of the current in the coil A→B→C→D § Side AB of the coil: force acts downwards Side CD of the coil: force acts upwards § Coil rotates in one direction § Carbon brush X in contact with blue half of the commutator § Carbon brush Y in contact with red half of the commutator § Direction of the current in the coil D→C→B→A § Side AB of the coil: force acts upwards Side CD of the coil: force acts downwards § Coil rotates in the same direction as the first half rotation FACTORS AFFECTING THE SPEED OF ROTATION OF AN ELECTRIC MOTOR Magnitude of current in the coil Current in the coil increases, speed of rotation of the electric motor increases Strength of magnetic field Strength of magnetic field increases, speed of rotation of the electric motor increases Number of turns of wire in coil Number of turns of wire in the coil increases, speed of rotation of the electric motor increases 77 Physics CHAPTER 4 ELECTROMAGNETISM
1 Electromagnetic induction is the production of an electromotive force (e.m.f.) in a conductor when there is relative motion between the conductor and a magnetic field or when the conductor is in a changing of magnetic field. 2 Electromagnetic induction occurs when a piece of copper wire is moved across magnetic flux and an electromotive force (e.m.f.) is induced in the wire. If the wire is connected to form a complete circuit, the deflection of the pointer of galvanometer shows induced current is produced. 3 When a bar of magnet is moved towards or away from a solenoid, the turns of the solenoid cut the magnetic field lines. Hence, an electromagnetic induction occurs and an e.m.f. is induced across the solenoid. TYPES OF MOTOR BRUSHLESS MOTOR BRUSHED MOTOR § Has a magnet and a coil § Use magnetic force to produce rotation Coil is stationary, magnet rotates Magnet is stationary, coil rotates No carbon brushes, therefore no friction between the brushes and the commutator Friction between the carbon brush and the commutator causes the carbon brush to wear out No sparking at the commutator Sparking at the commutator Soft operational sound Louder operational noise Physics Bytes Magnetic flux refers to magnetic field lines that pass through a surface. A conductor that moves and cuts through magnetic field lines can be said to cut magnetic flux. 78 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
Factors Affecting the Magnitude of the Induced e.m.f. The induced e.m.f. increases when: § The speed of relative motion increases § The number of turns of the solenoid increases § The strength of magnetic field increases Faraday’s law states that the magnitude of induced e.m.f. is directly proportional to the rate of cutting of magnetic flux. Direction of Induced Current in a Straight Wire and Solenoid 1 Fleming’s Right-hand Rule is used to determine the direction of induced current in a straight wire. 2 The induced current is produced in the solenoid by the relative motion between the bar magnet and the solenoid. 3 For a solenoid, Lenz’s law is used to determine the magnetic polarity at the end of the solenoid when a current is induced. 4 Lenz’s law states that the induced current always flows in a direction that opposes the change of magnetic flux that caused it. 79 PhysicsCHAPTER 4 ELECTROMAGNETISM
EXAM WATCH Diagram shows a magnet is moved into a solenoid. Explain how the direction of the galvanometer is determined. Name the physics law that is used to determine the direction induced current that flows in the solenoid. [4] SBP Model Paper Set 3 Question 11 Radial Field § A radial field is a magnetic field with the field lines pointing towards or away from the centre of a circle like spokes of a wheel. Direct Current Generator and Alternating Current Generator 1 Wind turbine electric generators apply electromagnetic induction to produce induced e.m.f. 2 There are two types of generators, the direct current generator and the alternating current generator. TYPES OF GENERATOR DIRECT CURRENT GENERATOR ALTERNATING CURRENT GENERATOR § Applies electromagnetic induction § Coil is rotated by an external force § Coil cuts magnetic flux § e.m.f. is induced in the coil End of the coil is connected to a split ring commutator Ends of the coil are connected to two slip rings The two sections of the commutator exchange contact with the carbon brush every half rotation Slip rings are connected to the same carbon brush Output is direct current Output current is alternating current 80 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
Working Principle of a Direct Current Generator § When the coil rotates from the vertical position to the horizontal position, the rate of change of magnetic flux linked to the coil causes the magnitude of the induced current and induced e.m.f. in the coil to increase from zero to maximum. When the coil rotates from the horizontal position to the vertical position, the magnitude of the induced current and induced e.m.f. in the coil decreases from maximum to zero. § For each turn of the coil passing through the vertical position, the direction of the induced current in the coil is reversed using Fleming’s right-hand rule. However, the action of the split ring commutator keeps the current in the galvanometer in the same direction as the coil rotates. Working Principle of an Alternating Current Generator § When the coil rotates from the vertical position to the horizontal position, the rate of change of magnetic flux linked to the coil causes the magnitude of the induced current and induced e.m.f. in the coil to increase from zero to maximum. When the coil rotates from the horizontal position to the vertical position, the magnitude of the induced current and induced e.m.f. in the coil decreases from maximum to zero. § For each turn of the coil passing through the vertical position, the direction of the induced current in the coil and galvanometer is reversed. 81 Physics CHAPTER 4 ELECTROMAGNETISM
H T TIPS COMPARISON BETWEEN DIRECT CURRENT MOTOR AND DIRECT CURRENT GENERATOR DIRECT CURRENT MOTOR DIRECT CURRENT GENERATOR Current + Coil = Motor (in magnetic field) Rotation + Coil = Generator (in magnetic field) Use a direct current to rotate the coil. Rotate a coil to produce current. Fleming’s left hand rule Determine direction of rotation Fleming’s right hand rule Determine direction of induced current § Switch on the circuit and current flow. § Coil is magnetised § Interaction between two magnetic field produce catapult field § Force is produced § Rotate rectangular coil § Magnetic flux is cut § Induced e.m.f. is produced § Electromagnetic induction EXAM WATCH Diagram shows a structure of a direct current generator. Explain the working principle of the direct current generator. [4] Kedah 2021 Question 11 82 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
Diagram shows an alternating current generator used to supply alternating current. Generator Magnetic strength Magnetic shape Number of turns of coil Diameter of wire on the coil Q Weak Flat Less Small R Weak Concave Less Big S Strong Flat Less Small T Strong Concave More Big U Strong Flat More Big Table shows the characteristics of an alternating current generator. Explain the suitability of each of the characteristics in the table to produce a generator which can provide the highest induced current, and then determine which generator is most suitable to use. Give reasons for your choice. [10] Aspect Reason Melaka 2022 Question 10 Diagram shows a structure of a direct current generator. The generator produces a small direct current. Suggest modifications that can be made so that it can produce high alternating current (a.c.) State and explain the modifications based on the characteristics of the magnet and wire, and the type of ring. [10] Number of magnet: Reason: Shape of magnet: Reason: Material of the wire: Reason: Characteristics of wire: Reason: Type of ring: Reason: SBP Model Paper Set 3 Question 11 83 Physics CHAPTER 4 ELECTROMAGNETISM
Working Principle of a Simple Transformer 1 The transformer is a device that can raise and lower the voltage of an alternating current using the principle of electromagnetic induction. 2 Types of transformer: § Step-up transformers (increase voltage) § Step-down transformers (decrease voltage) Primary Circuit 1 Alternating current in primary coil 2 Alternating current produces magnetic field that changes in magnitude and direction 3 Magnetic flux from the primary coil is linked to the secondary coil through soft iron core Secondary Circuit 4 The changing of magnetic field induced an alternating voltage across secondary coil 5 !! !" = "! "" EXAM WATCH Diagram below shows a model of a simple transformer built by a student at school laboratory. Referring to the diagram, explain the operation’s principle of a transformer. [4] Kedah 2022 Question 11 Ideal Transformer 1 When an electrical equipment is connected to the output terminals of a transformer, the transformer receives input power from the power supply and supplies output power to the electrical equipment. 2 Therefore, electrical energy is transferred from the primary circuit to the secondary circuit. 84 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
3 A transformer that is operation experiences a loss of energy. Therefore, the output power is less than the input power. 4 The efficiency of the transformer, is defined as = × 100% 5 An ideal transformer is a transformer that does not experience any loss of energy, that is the efficiency, is 100%. != " ##= $$ ✏ CHECKPOINT 1 Diagram shows a transformer that is connected to a 240 V power supplying 48 W of power at a voltage of 6 V to an electronic equipment. Assuming the transformer is ideal, calculate a) The number of turns of the primary coil b) The current in the secondary circuit c) The current in the primary circuit 2 Diagram below shows a transformer used to charge a tablet. a) State the type of transformer shown in the diagram above. b) If the number of turns in the secondary coil is 4 turns, calculate the number of turns in the primary coil. c) If the efficiency of the transformer is 80%, calculate the input current. 85 PhysicsCHAPTER 4 ELECTROMAGNETISM
WAYS TO INCREASE THE EFFICIENCY OF A TRANSFORMER FACTORS EFFECTS WAYS TO REDUCE ENERGY LOSS Resistance of coils Heating of wires causes heat energy to be released to the surroundings § Use thicker wire to reduce resistance § Use copper wire Eddy currents § The changing of magnetic field induced eddy currents in the iron core. §Eddy current heat up the iron core Use a laminated iron core Hysteresis § The iron core is magnetised and demagnetised by the changing of magnetic flux § The energy supplied for magnetism is not fully recovered during demagnetism. Use soft iron core (magnetised and demagnetised easily) Leakage of magnetic flux The magnetic flux produced by the primary current is not fully linked to the secondary coil Secondary coil is wound on the primary coil EXAM WATCH Diagram shows a simple transformer. Transformer Material of wire Thickness of wire Type of core Design of core W Copper Thick Soft iron core Laminated X Copper Thin Soft steel core Solid Y Nichrome Thick Soft iron core Laminated Z Nichrome Thin Soft steel core solid The transformer in the diagram is less efficient. Table shows the characteristics of several transformers. Study each characteristics of the transformers and explain the suitability of each characteristics. Determine the most efficient transformer. Give the reason for your choice. [10] Aspect Reason Kelantan 2023 Question 10 1 Eddy current causes energy loss. However, it also can be beneficial to human beings. 2 The magnetic field induces eddy currents at the base of the pan. The eddy currents heat up the base of the pan. 86 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
EXAM WATCH Diagram shows the structure of an induction cooker. Explain how the induction cooker works to produce eddy currents to heat up the base of the pan. [4] Diagram shows an induction cooker takes a long time to heat up the food. Table shows the characteristics of four induction cooker K, L, M and N. K M L N You are required to determine the most suitable induction cooker to heat up food faster from the following aspects: § Material of the stove top § Material of coil § Coil oxidation rate § Power supply Explain the suitability of the aspects and determine the most suitable induction cooker, Give reasons for your choice. [10] Aspect Reason MRSM 2022 Question 10 87 Physics CHAPTER 4 ELECTROMAGNETISM
Electrical Energy Transmission and Distribution System 1 Transformer play an important role in the transmission and distribution of electricity from the power station to the consumers. 2 At the stage of transmission of electrical energy, step-up transformers are used to increase the voltage in the power cable so that the current in power cable becomes small. This reduces the loss of electrical energy from the power cable. 3 During the distribution of electrical energy, step-down transformers are used to decrease the voltage in the power cable in stages to a suitable value for industrial and residential consumers. EXAM WATCH Diagram below shows the National Grid Network System used to supply electricity for residential area. Using your knowledge in electrical, determine the suitable method for power transmission so that the electric supply is sufficient and safe. Your suggestion should include the following aspects such as type of transformer P, material of the cable, transmission voltage and rate of expansion of the cable. [10] Type of transformer P: Reason: Material of the cable: Reason: Material of the cable: Reason: Transmission voltage: Reason: Rate of expansion of the cable: Reason: Kedah 2022 Question 11 88 Physics Force and Mo+on II Pressure Electricity Electromagne-sm Electronics Nuclear Physics Quantum Physics
"WHERE CIRCUITS CONDUCT SYMPHONIES OF INNOVATION, ORCHESTRATING THE RHYTHM OF MODERN LIFE." ELECTRONICS 5S P M P H Y S I C S
Thermionic Emission and Cathode Rays 1 Thermionic emission is the process of emitting electrons from a heated metal surface. 2 Factors that affect the thermionic emission are: a) Temperature of metal b) Surface area of metal c) Type of metal (tungsten, barium oxide, strontium oxide) 1 § Metal wire such as tungsten filament has many free electrons. § When 6 V d.c. power supply is switched on, the temperature of the tungsten filament increases. § Free electrons gain kinetic energy to leave the metal surface. § This process is known as thermionic emission. 2 § The glass tube is in vacuum condition. The electrons can accelerate towards the anode without colliding with air molecules. § Hence, there is no energy loss and electrons move with maximum velocity. 3 § When a vacuum tube is connected to an E.H.T. power supply, the electrons emitted from the cathode will be attracted to the anode at high velocity to form an electron beam. § This high velocity electron beam is known as cathode rays. § The electron beam will complete the E.H.T. power supply circuit and the milliammeter reading will show that a current is flowing. § If the connection to the E.H.T. power supply is reversed, the milliammeter will not show any reading. § The graph shows that thermionic diode is a non-ohmic component. 90 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
Thermionic Emission and Cathode Rays 1 Cathode rays are beams of electrons moving at high speed in a vacuum. 2 The characteristics of cathode rays can be studied using a deflection tube and a Maltese cross tube. Deflection tube Maltese corss tube Observation Explanation Defelction tube S1 and S2 are turned on § Cathode rays travel in a straight line S1, S2 and S3 are turned on § Cathode rays can be deflected by an electric field § Defected towards positive plate in a parabolic path § Cathode rays are negatively charged Maltese cross tube S1 turned on § Light from the hot tungsten filament is blocked by an opaque object (Maltese cross) to form a shadow. § Lights travel in a straight line S1 and S2 are turned on § Cathode rays are blocked by Maltese cross to form a shadow § Cathode rays travel in a straight line § Cathode rays also produce a fluorescent effect on the screen surrounding the shadow § This shows that cathode rays prossess momentum and kinetic energy S1 and S2 are turned on and magnet is placed near the tube § One shadow is due to light from the hot tungsten filament § The other shadow is due to the deflection of the cathode ray by the magnet bar that is placed near the tube § The deflection of cathode rays can be determined by Fleming’s left-hand rule 91 Physics CHAPTER 5 ELECTRONICS
✏ CHECKPOINT Diagram shows a cathode ray beam moving into an electric field produced by two metal plates connected to a E.H.T. supply. a) On the above diagram, complete the path of cathode ray through the electric field. b) Give one reason for your answer. Velocity of an Electron in a Cathode Ray Tube = § When the E.H.T. power supply is turned on, electrons are attracted by the positively charged anodes. § As there are no air molecules in the vacuum tube, electrons will accelerate to the anode without any collision. Thus, electrons will achieve maximum velocity, !"# when they reach the anode. = 1 2 $ = electrical potential energy = charge of an electron, 1.6 × 10%&' C = potential difference = charge of an electron, 1.6 × 10%&' C = potential difference = mass of electron, 9.11 × 10%(& kg !"#= maximum velocity of an electron 92 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
✏ CHECKPOINT 1 Diagram shows an electron beam that is accelerated from the cathode to the anode in a vacuum. The potential difference across the cathode and the anode is 750 V. [mass of electron, m = 9.11 × 10-31 kg, charge of an electron, e = 1.6 × 10-19 C] a) What is the electrical potential energy of an electron? b) What is the kinetic energy of an electron when it reaches at the anode? c) What is the maximum velocity of an electron when it reaches at the anode? 2 Diagram shows a Maltese cross tube to study the properties of cathode rays. What is the voltage of the E.H.T. used, if the maximum velocity of the cathode rays in the Maltese cross tube is 10% of the velocity of light in a vacuum? 3 Diagram shows an electron beam that is accelerated from the cathode to the anode in a vacuum. The potential difference between the cathode and anode is 1.1 kV. [Mass of electron, ) = 9.11 × 10-31, charge of electron, = 1.6 × 10-19] Calculate a) the electrical potential energy of an electron b) the kinetic energy of an electron on hitting the anode c) the maximum velocity of an electron on hitting the anode 93 Physics CHAPTER 5 ELECTRONICS
4 Diagram below shows a Maltese cross tube used to study the characteristics of cathode rays. a) (i) What is the meaning of cathode ray? (ii) Why is extra high tension (E.H.T.) is used? (iii) State the conversion of energy of cathode ray as it moves from anode to hit the fluorescent screen. b) On the diagram, there have a Maltese cross shadow and a green light region seen on the fluorescent screen. Give a reason why (i) the shadow is formed on the screen? (ii) the green light region is formed on the screen? (iii) What will happen to the Maltese cross image when the evacuated bulb is leaking? Explain your answer. c) The E.H.T. value is 3 kV. Calculate the maximum velocity of the charged particle in the cathode rays. 94 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics
5 Diagram below shows a simple cathode rays tube. The cathode emits electrons when its heater circuit is switched on. a) Name the process that enables the emission of electrons at the hot cathode. b) State one reason why the vacuum tube is used. c) When electron beams flow in the cathode rays tube, a current of 0.02 A is produced in 10 seconds. Calculate the total charge of the electrons. d) If another E.H.T. is applied to P and Q so that the upper plate is positive and lower one is negative, (i) draw the new path of the cathode ray in the region between the two plates. (ii) Give one reason for the answer is (d) (i). (iii) What changes will be made to the new path of the cathode ray if a magnetic field acting vertically into the paper? (iv) State the physics rule used to determine the direction of the cathode ray in (d) (iii). (v) What happens to the path of the cathode ray if both electric and magnetic fields are acting simultaneously in the same region? 95 Physics CHAPTER 5 ELECTRONICS
EXAM WATCH Which of the following statements is correct for cathode rays? A It is a radioactive radiation. B It can be deflected by a magnetic field. C It travels slowly at low temperature. D Its kinetic energy can be converted into elastic potential energy. Diagram shows a shadow formed on the screen of a Maltese cross tube. The shadow is deflected by the magnetic fields. What is the rule used to determine the direction of deflection for the shadow of Maltese cross? A The Fleming’s left hand rule B The Fleming’s right hand rule C The right-hand grip rule Terengganu 2023 Question 34 Pahang 2023 Question 36 Properties of Semiconductors § Conductors are materials which allows current to flow through them easily. This is because conductors have free electrons which can drift between their atoms. Hence, conductors have low resistance. (silver, copper, aluminium) § Insulators are materials which do not conduct electrical current. (rubber, polythene, perspex) § A semiconductor is a material whose resistance is between those of good conductors and those of good insulators. § Example of a semiconductor is silicon. A silicon atom has four electrons in its outermost (valence) shell. § Each electron in the outermost shell can form a covalent bond with one electron in the outermost shell of another atom. Hence, each silicon atom forms four covalent bonds with four neighbouring atoms. § When an electron is removed from a covalent bond, it leaves a vacancy. An electron from a neighbouring atom can fill this vacancy, leaving the neighbouring atom now with a vacancy. Therefore, the vacancy (called a hole) can travel through the lattice and act as a charge carrier. 96 Physics Force and Mo+on II Pressure Electricity Electromagne+sm Electronics Nuclear Physics Quantum Physics