How are magnetic forces used to produce rotation in an electric motor? What is a brushless motor? How can the concept of electromagnetic induction be applied to benefit human beings? Why are transformers used in the transmission and distribution of electricity? - 4&1 Force ona Current-carrying Conductor in a Magnetic Field 4.2 Electromagnetic Induction 43> Transformers vs) KPM.Learning Standards and List Cae ue ey : 1 4 Information, OJTIED The drop tower is a high technology theme park equipment based on the concept of electromagnetic induction. The passengers on the drop tower will drop from a great height and experience free fall at high speed. They are then slowed down by an arrangement of permanent magnets fixed under their seats and copper strips on the lower section of the drop tower, The motion of the permanent magnet passing the copper strips will activate electromagnetic braking. This can hrp:bitiy! be explained by the concept of electromagnetic induction. 310K Importance/of (A ¥ Knowledge about electromagnetism is important because magnetic forces, electromagnetic induction and transformers have wide applications and affect various aspects of our daily life. Magnetic forces are used in various types of motors such as small electric motors in fans and modern electric motors in modern electric vehicles. The principle of electromagnetic induction is applied in electric generators and transformers for the purpose of generation and transmission of electric power from the power station to the consumer. Various new innovations that use the concept of electromagnetism are being developed by scientists and engineers. —— mmera@ ens * The concept of electromagnets is not only used to slow down motion but also to accelerate the motion of objects to very high velocities. For example, hyperloop transportation uses linear electric motors (without rotation) to accelerate vehicles moving in low pressure tubes. Transportation on land at speeds comparable to the speed of aircraft may become a hnp:/bitty! reality in the near future. 2CTNABX KPM,A 1 Force on a Current-carrying Conductor in a - Magnetic Field Do you know that an electric train as shown in Photograph 4.1 uses a large electric motor while a smart phone uses a small motor? The function of most electric motors is based on the effect of a current-carrying conductor in a magnetic field. eT) Se Coa AL Photograph 4.1 Electric train Aim: To study the effect on a current-carrying conductor in a magnetic field Apparatus: Low voltage direct current power supply, U-shaped steel yoke, a pair of Magnadur magnets and retort stand ‘Materials: Two copper rods without insulation and copper wire (s.w.g. 20 or thicker) without insulation Instructions: 1. Set up the arrangement of apparatus as shown in Figure 4.1 Retort stand Magnadur clamp Steel yoke Copper wire Low voltage direct current power supply2. Tur on the power supply so that current flows into the copper wire. Observe the movement of the copper wire. 3. Tum off the power supply. Reverse the connections to the power supply so that the current in the copper wire is reversed. 4. Tum on the power supply again. Observe the movement of copper wire. 5. Tum off the power supply. Remove the steel yoke, reverse the poles of the Magnadur magnets and put back the steel yoke. 6. Turn on the power supply and observe the movement of copper wire. Discussion: 1. Describe the motion of the copper wire when the power supply is turned on. 2. What is the effect on the copper wire when: (a) the direction of the current is reversed? (b) the poles of the magnet are reversed? State two factors that affect the direction of the force acting on the current-carrying conductor. When a current-carrying conductor is placed in a magnetic field, the conductor will experience a force. ‘The direction of the force depends on the direction of the current and the direction of the magnetic field. Pattern on Resultant Magnetic Field a Figure 4.2 shows a current-carrying conductor that is placed in B—> C—> Dor D> CB A, What is the effect on the coil if the coil carries a current in a magnetic field? 4 Boe D tt dt g D c Figure 4.8 A rectangular coil formed with a copper wire Aim: To observe the turning effect on a current-carrying coil in a magnetic field Instructions: 1. Carry out this activity in pairs. 2. Scan the given QR code to watch a video on the turning effect on a current-carrying coil in a magnetic field. Download Figure 4.9 from the given website and complete the diagram by: (a) labelling the direction of the current in sections AB, BC and CD (b) labelling the direction of the force on the coil at sections AB and CD (©) mark the direction of rotation of the coil a ‘The current-carrying coil in a magnetic field will rotate about the axis of rotation, This rotation is due to a pair of forces of equal magnitude but in opposite directions acting on the sides of the coil. This pair of forces is produced by the interaction between the current-carrying coil and the magnetic field from the permanent magnet. (od BD 43 414 KPMFigure 4.10 shows the pair of forces acting on sides AB and CD of a current- carrying coil. ‘The interaction between the ‘magnetic field from the current-carrying coil and the magnetic field from the permanent magnet as shown in Figure 4.11 produces a catapult field as shown in Figure 4.12, The catapult field exerts a force on sides AB and CD of the coil respectively. ‘This pair of forces rotates the coil. The turning effect on a current-carrying coil in ‘a magnetic field is the working principle of the direct current motor. Side AB Figure 4.11 Direction of magnetic field around sides AB and CD Direct Current Motor Side DC oa Axis of rotation Permanent magnet Permanent magnet Current in Curent out Force Figure 4.10.4 pair of forces acting in a magnetic field ‘causes the coil to rotate Force Force Figure 4.12 Catapult fields is produced Small electrical appliances such as children's toys, portable drills and the hard disk of a computer have a small direct current motor. Larger direct current motors are found in machines such as electric vehicles, lifts and rollers in factories. The direct current motor changes electrical energy to kinetic energy by using the turning effect of a current-carrying coil in a magnetic field. What is the working principle of a direct current motor? =a Aim: To gather information on the working principle of a direct current motor Instructions: 1, Carry out this activity in pairs. 2. Scan the QR code and watch the video of the working principle of the direct current motor. 3. Refer to other materials to obtain additional information. 4. Prepare a multimedia presentation entitled “The Working, Principle of Direct Current Motors”. and its working principle Dea ee scr BD aia aisFigure 4.13 shows a direct current motor during the | proin-tocda first half of its rotation and the second half of its rotation. An important component in a direct current motor is the SUL ee cela 7 ‘Ting but a ring split into two halves? commutator that rotates with the rectangular coil. The Se rea aces eae carbon brushes in contact with the commutator are in their ‘mare sections? fixed positions. ba asl) pee D.C. power supply D.C. power supply First-half rotation Second-half rotation Direction of current in the coil: ABCD Direction of current in the coil: DCBA Carbon brush X in contact with red half of Carbon brush X in contact with blue half WF the commutator of the commutator Carbon brush ¥ in contact with blue half of fa sion terach Yai contact with red ball WF the commutator of the commutator Side AB of the coil: force acts upwards ide CD of the coil: force acts downwards | Side AB of the coil: force acts downwards B.Direction of current in the coi , Direction of current in the eo A-B>C>D Dos Side CD of the coil: force acts upwards Coil rotates in the same direction as the Figure 4.13 Working principle of a direct current motor (les BD asFactors Affecting the Speed of Rotation of an Electric Motor Photograph 4.2 shows a portable device that can function as a screwdriver or a drill. The direct current motor in the device rotates at a low speed when turning a screw. A high speed is necessary when the device is used to drill a hole in the wall. What are the factors that affect the speed of rotation of an 2 electric motor? Photograph 42 Serewslriver and drill Aim: To study the factors that affect the speed of rotation of an electric motor ‘Apparatus: Direct current power supply and a pair of Magnadur magnets Materials: Insulated copper wire (s.w.g. 26), two large paper clips, two pieces of thumb tacks and connecting wires Instructions: 1. Set up the arrangement of apparatus as shown in Figure 4.14. D.C. power supply Copper wire coil Large paper clip magnet genie re ee ert 3. Repeat step 2 using a voltage of 6.0 V. + Switch off the power supply. 4. Add one more Magnadur magnet to produce a stronger ae ce ‘magnetic field. Turn on the power supply and observe the Bot ecerakiaer erat al speed of rotation of the motor. shot time so thal the col vill ot 5. Add more turns to the copper coil. Turn on the power supply ‘become too hot. and observe the speed of rotation of the motor. Discussion: 1, What is the relationship between the current in the coil and the voltage supplied? 2, Describe the change in the speed of rotation of the motor when: (a) the voltage supplied is increased (b) the strength of the magnetic field is increased (©) the number of tums of the coil is increased State the factors that affect the speed of rotation of a motor, D ae (bss reaActivity 4.6 shows that the speed of rotation of an electric motor increases when: | Strength of magnetic | field increases ‘Aim: To study the direct current motors found in used devices to identify the arrangement of the coil and the commutator Instructions: 1. Collect a few direct current electric motors from used devices. 2, With the aid and guidance of your teacher, dismantle each motor and identify the coil, the ‘magnet and the commutator. 3. Observe the position of the coil and the magnet in the motor. 4. Observe also the number of sections in the commutator. 5. Prepare a brief report that compares and contrasts the arrangement of the coil, magnet and commutator in direct current motors. While carrying out Activity 4.7, you may have come across electric motors which do not have a commutator and carbon brushes. Photograph 4.3 shows a brushless motor and a brushed motor. What is the advantage of brushless motors? Carbon brush ‘Magnet Commutator Coil Photograph 4.3 Brushless motor and brushed motor (oss D ae KPM; ‘Aim: To study and report on the advantages of brushless motor compared to brushed motor Instructions: 1. Carry out a Three Stray, One Stay activity. 2, Scan the QR code given or refer to other reference materials to: (a) understand the working principle of brushless motor (b) study the advantages of brushless motor compared to brushed motor 3. Report the findings of your study. Dee ASAE ‘Table 4.2 shows the comparison between brushless motor and brushed motor. Table 4.2 Comparison between brushless motor and brushed motor Pere eee Brushed motor ‘Similarities Has a magnet and a coil Uses magnetic force to produce rotation Coil is stationary, magnet rotates Magnet is stationary, coil rotates No carbon brushes, therefore no friction between _ Friction between the carbon brush and the © | the brushes and the commutator ‘commutator causes the carbon brush to wear out © | No sparking at the commutator Sparking at the commutator © | Soft operational sound Louder operational noise ‘Aim: To design a simple and efficient homopolar motor Materials: Neodymium magnet, A dry cell and copper wire (s.w.g. 18 to 22) Instructions: 1. Carry out this activity in groups. 2, Gather information on homopolar motors from the aspect of: (a) the working principle of homopolar motors BELRe pee (b) the shape and size of the neodymium magnet Sioa (©) various designs of the copper wire that can be tried out eee 3. Use the K-W-L Data Strategy Form. cd4, Sketch the design of a homopolar motor. 5. Construct the homopolar motor according to the suggested design 6. Operate the homopolar motor that you have constructed. 7. Observe the rotation produced and identify the aspects of the design that need to be improved. 8. Discuss the steps of improvement that can be carried out. 9. Improve the homopolar motor if necessary and test the rotation. 10. Based on your experience in designing and constructing the homopolar motor, discuss ways to construct a more efficient motor at a low cost. 11. Present the outcome. History Inthe year 1821, Michael Faraday constructed and demonstrated the operation ‘of 2 homopolar motor atthe Royal Insitute, London. 1. With the aid of a labelled diagram, explain the meaning of catapult field. 2. Figure 4.15 shows the arrangement of apparatus to study the effect of a force on a current-carrying conductor. ‘To current supply Magnadur magnet Copper rod Figure 4.15 (a) What is the direction of the current in the copper wire XY when the switch of the direct current power supply is turned on? (b) Explain the motion of the copper wire XY and state the direction of the motion. 3. State three factors that affect the speed of rotation of a motor. 4. (a) Compare and contrast the structure ofa brushed motor with a brushless motor. (b) State two advantages of brushless motor compared to brushed motor. oss BD 116oa Ws Electromagnetic Induction Photograph 4.4 shows a musician plucking an electric bass guitar. The guitar pickup consisting of four magnets and copper coils produces an electric signal by electromagnetic induction. How does electromagnetic induction produce an electric current without the use of dry cells? Permanent magnet Copper coil Aim: To study electromagnetic induction in a straight wire and a solenoid © Straight wire Apparatus: A pair of Magnadur magnets, copper rod and sensitive centre-zero galvanometer or digital multimeter Material: Connecting wires with crocodile clips Instructions: 1, Set up the apparatus as shown in Figure 4.16. Lz ed Ce Rey rea Figure 4.16 2. Hold the copper rod stationary between the poles of the magnet as shown in Figure 4.16. Observe the reading of the galvanometer. De ined3. Move the copper rod quickly in direction A as shown in Figure 4.16, Observe the deflection of the galvanometer pointer. 4. Repeat step 3 in directions B, C and D. 5, Hold the copper rod with your left hand. Lift up the Magnadur magnet with your right hand. Move the Magnadur magnet in direction A and direction B with the copper rod stationary in between the poles of the magnet. Observe the deflection of the galvanometer pointer. 6. Complete Table 4.3 with a tick (/) for the direction of deflection of the pointer of the secant Results: vam an Sy ‘Stationary Stationary ‘Stationary Moves in direction A Stationary Moves in direction B & ‘Stationary ‘Moves in direction oS ‘Stationary ‘Moves in dit ‘Moves in direction A me Moves in direction B = Discussion: 1. What causes the d the galvanometer pointer? 2. State the away ‘magnetic field between the poles of the magnet. 3. What are the OY een of the copper rod that causes the cutting of magnetic field lines? 4, Explain the cendition where a current is produced in the copper rod. © sot Ap ar magnet, solenoid (at least 400 turns) and sensitive centre-zero galvanometer or digital multimeter nal ‘al: Connecting wires with crocodile clipsInstructions: 1, Set up the arrangement of apparatus as shown in Figure 4.17. 2. Hold the bar magnet stationary near the solenoid as shown in Figure 4.17. Observe the reading of the galvanometer. 3. Push the bar magnet into the solenoid. Observe the deflection of the galvanometer pointer. 4, Hold the bar magnet stationary in the solenoid. Observe the reading of the galvanometer. 5. Pull the bar magnet out of the solenoid. Observe Bar magnet. the deflection of the galvanometer pointer. 6. Hold the magnet stationary. Move the solenoid towards and away from the bar magnet. Observe Figure 4.17 the deflection of the galvanometer pointer. 7. Complete Table 4.4 with a tick (7) for the direction, of deflection of the galvanometer pointer. Results: Table 44 Stationary Moves into the solenoid Stationary “Moves out of A rs Stationary ee Moves towards the pasioney bar magnet . “Moves away from the See ‘bar magnet Discussion: 1. What do you observe about the deflection of the galvanometer pointer when: (a) the bar magnet is moved towards the solenoid? (b) the bar magnet is moved away from the solenoid? 2, State the condition where a current is produced in the solenoid.When a piece of copper wire is moved across magnetic flux, an electromotive force (e.m.f.) is induced in the wire. This phenomenon is known as electromagnetic induction. If the wire is connected to form a complete circuit, a deflection of the galvanometer pointer is observed as shown in Figure 4.18, This shows that induced current is produced. An electromotive force is also induced in the wire if the magnet is moved towards the stationary wire as shown in Figure 4.19. Gio CED Magnetic ux refers to magnetic feld ines that passthrough a surface, Magnetic field line ‘Moving conductor ‘Aconductor that moves and cuts magnetic field tines canbe said to cutmagnetc fuk. cent _ _ Figure 4.18 Magnets are stationary while conductor is moved Coed Kod ae Figure 4.19 Magnets are moved while conductor is stationary (sa) B 21 KPMWhen a bar magnet is moved towards or away from a solenoid, the turns of the solenoid cut the magnetic field lines. Electromagnetic induction occurs and an em. is induced across the solenoid as shown in Figure 4.20. Figure 4.21 shows the ends of the solenoid are connected to agalvanometer Anas ccaemg— to form a complete circuit. ‘The induced i: electromotive force will produce an Figure 4.20 Magnetic fleld lines are cut by the induced current in the circuit and the Soe aay galvanometer pointer shows a deflection. Figure 4.21 Induced current in a complete circuit FAAP SA Activity 4.10 shows that an induced em, is produced by the cutting of magnetic field lines when a magnet and a conductor move towards or away from each other. Electromagnetic induction is the production of an induced e.m.. in a conductor when there is relative motion between the conductor and a magnetic field or when the conductor is in a changing magnetic field Relative mation between two objects isthe motion that results inthe two objects becoming closer to each other or further away from each other. A B Relative mation occurs between objects A and B it * object As stationary and objact 8 moves towards or away from A + object Bis stationary and object A moves towards or away from B + both objects A and B move with dflrent velocities No relative motion between objects A and Bit: + both objects A and B are stationary + both objects A and B move with the same speed inthe same direction Do iu KPM reaFactors Affecting the Magnitude of the Induced e.m.f. Figure 4.22 shows an induction lamp made by a pupil. He found that the LED lights up with different brightness when the magnet in the PVC pipe is shaken at different speeds. What are the factors that affect the magnitude of induced em.f2 ( Figure 4.22 A self-made induction lamp Doe Construction ofan induction lamp BEE pup EE y/FOhR3IU o 7, Bian z hap:/ie Es BAY yqqov Aim: To study the factors affecting the magnitude of induced em£, Apparatus: Two solenoids with 400 and 800 turns respectively, two bar magnets and sensitive centre-zero galvanometer ‘Materials: Connecting wires and rubber band Instructio 1. Set up the apparatus as shown in Figure 4.23 2. Push a bar magnet slowly into the solenoid with 400 tums. Record the maximum reading shown on the galvanometer. Push the bar magnet quickly into the solenoid with 400 tums. Record the maximum reading shown on the galvanometer. Push a bar magnet slowly into a solenoid with 800 turns. Record the maximum reading shown on the galvanometer. Use a rubber band to tie two bar magnets together with like poles side by side. Push the two bar magnets slowly into the solenoid with 800 tums. Record the maximum reading shown on the galvanometer in Table 4.5. - ConnectingResults: Table 4.5 INemnbextat] (i Specdiofil Nemiber of tarra ‘Maximum reading of galvanometer magnets magnet of solenoid First attempt Second attempt Average One Slow 400 One Fast 400 One Slow 800 ‘Two Slow 800 Discussion: 1. Why does the galvanometer pointer deflect when a magnet is pushed into the solenoid? 2. Which factor is studied when the bar magnet is pushed at different speeds into the solenoid? 3. Which factor is studied when the number of magnets pushed into the solenoid is different? 4, How is the magnitude of induced e.m.f. affected by: (a) the speed of magnet? (b) the number of turns of solenoid? (6) the strength of magnetic field? ewan ‘The results of Activity 4.11 show that the magnitude of induced e.m4. is affected by the speed of relative motion between the magnet and the conductor, the number of turns of the solenoid and the strength of the magnetic field. History For the relative motion the speed of relative motion ofa straight wire and increases magnet, the induced the strength of the magnetic eam. increases when: field increases Miebel Faraday (1791-1867) discovered eecromagnetic induct inthe year 1831, He the speed of relative motion successtly constructed an For the relative motion increases electric dynamo. The electric. of a solenoid and the number of turns of the dynamo was used to generate magnet, the induced solenoid increases clectical power. em. increases when: the strength of the magnetic field increases ‘The magnitude of the e.m.f. increases if more magnetic field lines are cut in a certain period of time, Faraday's law states that the magnitude of induced e.m.f. is directly proportional to the rate of cutting of magnetic flux. Din ius KPMDirection of Induced Current in a Straight Wire and Solenoid Be You have observed that the direction of the induced current EERE rer changes when there is a change in the direction of the relative preieng motion between the conductor and the magnet in Activity 4.10. Carry out Activity 4.12 and 4.13 to study the direction of the induced current in a straight wire and solenoid. ‘Aim: To study the direction of the induced current in a straight wire Apa, Ger Apparatus: Thick copper wire, a pair of Magnadur magnets, ek OU Un unc sensitive centre-zero galvanometer or digital multimeter, “6a a dry cell with holder, | kQ resistor and switch Material: Connecting wires with crocodile clips Instructions: 1, Set up the apparatus as shown in Figure 4.24. a © re Aer ae Connecting wire Resistor Figure 4.24 2. Connect the dry cell with its positive terminal at Y and negative terminal at X. 3. Tum on the switch. Observe the direction of deflection of the galvanometer pointer (to the left or to the right) and record your observation in Table 4.6. Determine the direction of the current through the dry cell (X to ¥ or ¥ to X). 4, Reverse the dry cell so that the positive terminal is, at X and negative terminal at ¥ and repeat step 3. 5. Remove the dry cell from the circuit and replace with a piece of thick copper wire. Place the thick copper wire between the pair of Magnadur magnets as shown in Figure 4.25. 6. Move the copper wire upwards (direction 4) Observe and record the direction of deflect of the galvanometer pointer and the direction of the current through the copper wire in Table 4.6. 7. Repeat step 6 by moving the copper wire downwards (direction B). Figure 4.25 = oeneuanin Table 4.6 Direction of deflection of yy. i sat eran Se (to the left or to the right) Dry cell (positive terminal at ¥, negative terminal at X) Dry cell reversed (negative terminal at ¥, positive terminal at X) Copper wire moved upwards (direction A) 1 Copper wire moved downwards (direction B) Discussion: 1. Try to relate the direction of the magnetic field lines, direction of motion of the copper wire and direction of the induced current by using the Fleming’s right-hand rule, 2. Suggest other ways to change the direction of the induced current other than the direction of ‘motion of the copper wire. leming's right-hand rule ‘The direction of the induced current in a straight wire can be determined by using Fleming's right- hand rule as shown in Figure 4.26. “The mation of the magnet downwards is equivalent tothe ‘motion ofthe wire upwards Direction of motion of wireAim: To study the direction of the induced current in a solenoid Apparatus: Solenoid, bar magnet and sensitive centre-zero galvanometer or digital multimeter Material: Connecting wires with crocodile clips Instructions: 1. Set up the apparatus as shown in Figure 4.27. Solenoid Figure 4.27 2. Move the north pole of the bar magnet as shown in Figure 4.27(a): (a) towards end A of the solenoid (b) away from end A of the solenoid 3. Observe the deflection of the galvanometer pointer and determine the polarity of the magnetic field at end A of the solenoid, Scan the QR code for the guideline on determination of the polarity at a solenoid. 4, Move the south pole of the bar magnet as shown in Figure 4.27(b): np: /bitdy/3iforoV (a) towards end A of the solenoid (b) away from end A of the solenoid 5. Observe the deflection of the galvanometer pointer and determine the polarity of the magnetic field at end A of the solenoid. 6. Record your observations and results in Table 4.7. Results: Table 4.7 Direction of deflection of | Magnetic polarity at Rings ometemees "elecenepine, aaratemand (to the left or to the right) (north or south) ‘Towards North Away from ‘Towards South Away from KPMewan Discussion: 1. What is the effect of the motion of the bar magnet on the polarity of the magnet at end A of the solenoid? 2, Predict the polarity of the magnet produced at end A: inthe solenoid by the relative (a) when the south pole of the bar magnet is pushed towards it motion between the bar magnet (b) when the south pole of the bar magnet is pulled away from it | and the solenoid. Induced current is produced For a solenoid, Lenz's law is used to determine the magnetic polarity at the end of the solenoid when current is induced. Lenz's law states that the induced current always flows in a direction that opposes the change of magnetic flux that causes it. Figure 4.28 shows that Lenz’s law is used to determine the direction of induced current in a solenoid. {a (mi Mager pushed | BD 11) Magnet pushed towards solenoid towards solenoid F “mm eee) ap 2 Induced current (2! Induced current produces a north produces a south A pole at the end of A Y pole at the end of the solenoid to repel the solenoid to repel eee eee! pera een fe sign pues 1X saga pte CT cms TTT eam solenoid to attract the ‘magnet moving away from the solenoid solenoid to attract the ‘magnet moving away from the solenoid Figure 4.28 Lenz's law used to determine the direction of induced current in a solenotd ct Current Generator and Alternating Current Generator Photograph 4.5 shows wind turbines electric generators that apply electromagnetic induction, to produce induced e.m-f. There are two types of generators - the direct current generator and alternating current generator. What is the working principle of these current generators? DD 123 424 KPMAim: To gather information on the structure and working principle of the direct current generator and alternating current generator Instructions: 1, Carry out this activity in groups. 2. Examine Figure 4.29 that shows the structure of the direct current generator and alternating current generator. Split ing commutator: Slip ring, (a) Direct current generator (&) Alternating current generator Figure 4.29 3. Surf the Internet to gather more detailed information on the structure and working principle of the direct current generator and alternating current generator. 4. Prepare a multimedia presentation entitled ‘Structure and Working Principle of the Direct Current Generator and Alternating Current Generator’. ‘Table 4.8 Working principle ofthe direct current generator and alternating current generator Direct current generator Alternating current generator Similarities Applies electromagnetic induction Coil is rotated by an external force Coil cuts magnetic flux emf. is induced in the coil Differences Ends of the coil are connected to a split ring Ends of the coil are connected to two slip rings commutator ‘The two sections of the commutator exchange _ Slip rings are connected to the same carbon brush contact with the carbon brush every half rotation Output is direct current Output is alternating current BD 424 KPMoa ‘Aim: To construct a functional prototype current generator (dynamo) by modifying an electric motor Instructions: 1. Carry out this activity in groups. 2. Compare and contrast the structures and working principles ‘note of the direct current motor and the direct current generator. Uso fe ICL sttogy data form Gather and study the following information from reading materials or websites: (a) the method of converting a motor to a dynamo & Uys (b) the ways to produce rotation in a dynamo Sai 7 ' See TL Based on the information you studied, suggest a design for the ie Prototype dynamo and the way to operate it. en Construct the dynamo by modifying electric motor according to the suggested design and test the dynamo. From the results of testing the dynamo, discuss the improvements that need to be made. ‘Make the improvements to the prototype dynamo and test the dynamo again. Present the design of your dynamo, Formative Practice -/4.2) 1. What is the meaning of electromagnetic induction? 2. (a) State Faraday's law. (b) Use Faraday’s law to explain the effect of the speed of rotation of the coil on the magnitude of the induced e.m4. in a current generator. 3. Figure 4.30 showsa simple pendulum with a bar magnet as the bob oscillating near a copper ring. (a) Explain the production of current in the copper ring when Copper the bar magnet is moving towards the ring. ving (b) At the position of the observer in front of the ring as shown in Figure 4.30, state whether the current in the copper ring is clockwise or anti-clockwise. + (c) Explain the effect of the current in the copper ring Observer on the motion of the bar magnet. Figure 4.30 sam paTransformer Working Principle of a Simple Transformer Photograph 4.6 shows a step-up transformer and a step-down transformer that are used in electrical devices. How does a transformer change an input voltage to an output voltage with a different value? aT Tad ‘Transformer Photograph 4.6 Types of transformers used in electrical devices a im: To gather information on the working principle of a simple transformer Instructions: 1. Carry out this activity in groups. Sof iton core 2. Examine Figure 4.31 that shows the circuit diagram for a simple transformer. a on ower put 3. Surf websites or refer to the Form 3 supply C) Ye Primary Secondary Ys terminals Science Textbook to search for information on the working principle of the rusian and gather information eT ea on the following: ea (a) type of power supply in the primary cireuit (b) type of current in the primary circuit (©) magnetic field produced by the current in the primary coil (@) function of the soft iron core (e) the phenomenon of electromagnetic induction in the secondary coil (f) the magnitude of the voltage induced across the secondary coil (g) relationship between Vp, V5 Np and Ng, where V,, = voltage across the primary coil or input voltage V, = voltage across the secondary coil or output voltage number of tums of the primary coil number of tums of the secondary coil 4, Present the findings. ag >=Figure 4.32 shows the structure of a simple transformer and ‘(Qoresomt Figure 4.33 shows the flow map for the working principle of a = simple transformer. Wy does the transformer not work witha direct curent power supply? Laminated sot iron core AC. power /— Leakage of ‘Step-down transformer: EC — magnetie tux Nenouen, Primary col ‘Step-uptansformer: Numbers of tums, Ny, * N,> Ne V,> Vy . Output Magnetic field lines <— terminals Secondaty coil Numbers of turns, 7, bo. Figure 4.32 Structure of a simple transformer S AC. power supply ‘The alternating current produces an produces a magnetic field alternating current that changes in magnitude in the primary coil. and direction ~The magnetic flux from the primary coil is linked to the secondary coil through the soft iron core. ‘The changing magnetic field induces VON Lan alternating voltage across the — VN secondary coil Figure 4.33 Working principle of a simple transformer Ideal Transformer igure 4.34 shows an electrical 7 7 equipment connected to the output terminals of a transformer. The transformer yg, receives input power from the supply Input power ‘Output power Electrical equipment Ve = power supply and supplies the ¥, output power to the electrical equipment. Therefore, electrical Primary circuit Secondary creat energy is transferred from the primary circuit to the Figure 4.24 Electrical equipment connected to the ourput terminals secondary circuit, B Gy 432 163] KPM awareA transformer that is in operation experiences a loss of, energy. Therefore, the output power is less than the input power. ‘The efficiency of the transformer, 7) is defined as Nowadays, there are transformers with very high efficiencies, up to 99%. An ideal transformer is a transformer that does not experience any loss of energy, that is the efficiency, 1) is 100%. SHEE EONE x 100% 100% etal Cee eas For an ideal transformer, efficiency of the transformer, 1) ‘Therefore, output power = input power VT = Vil Ways to Increase the Efficiency of a Transformer ‘The working principle of a transformer involves processes such as the flow of current in the copper coils, the change of magnetic field and electromagnetic induction. These processes cause loss of energy and the transformer is unable to operate at an optimum level. Most of the energy is lost in the form of heat energy. Aim: To gather information and discuss the causes of energy loss in a transformer Instructions: 1, Examine Table 4.9 that shows four main causes of energy loss and their effects, Table 4.9 ae Detfectstat enerag et Resistance of coils» The primary and secondary coils consist of wires that are very long. ‘+ When current flows in the coils which have resistance, heating of the wires occurs. * Heating of the wires causes heat energy to be released to the surroundings. Eddy currents ‘+ The changing magnetic field induces eddy currents in the iron core. + The eddy currents heat up the iron core. + The hot iron core releases heat energy to the surroundings. Hysteresis: ‘+ The iron core is magnetised and demagnetised continuously by the changing magnetic field. + The energy supplied for magnetisation is not fully recovered during demagnetisation. The difference in energy is transferred to the iron core to heat it up. Leakage of ‘+ The magnetic flux produced by the primary current is not fully linked to the ‘magnetic flux secondary coil. (Lass B 432 433 KPMrea 2, Scan the QR code and print Table 4.9. 3. Discuss the ways to increase the efficiency of transformers and complete Table 4.9, 4, Present the outeome of your discussion in the form of a suitable thinking map. Teas ‘The efficiency of a transformer can be increased by reducing the loss of energy in the transformer. Table 4.10 shows ways to reduce the loss of energy in a transformer. Table 4.10 Ways 0 reduce energy loss in a transformer Resistance of coils Use thicker copper wire so that the resistance of the coil is smaller. Eddy currents ‘Use a laminated iron core that consists of thin iron sheets glued together with insulation glue, Hysteresis Use soft iron as the core. Soft iron requires a smaller amount of energy to be magnetised. Leakage of magnetic flux ‘The secondary coil is wound on the primary coil so that all the magnetic flux produced by the primary current will pass through the secondary coil. In the operation of a transformer, eddy currents are a cause of energy loss. However, eddy currents can be beneficial to human beings. Photograph 4.7 shows an induction cooker. Figure 4.35 shows a high-frequency alternating current in the coil producing a magnetic field that changes with a high frequency. This magnetic field induces eddy currents at the base of the pan. The eddy currents heat up the base of the pan. Bady currents in the base of a pan x Ceramic surface Coil High frequeney 4 A.C. power supply Changing ‘magnetic field Photograph 4.7 Induction cooker Figure 4.35 Eddy currents in the induction cooker = baUses of Transformers in Daily Life ‘Transformers have a wide range of uses. The following are some examples of machines that use step-down transformers and step-up transformers. See keane Step-up transformer + Notebook computer charger + Microwave oven + Photocopy machine + Defibrillator + Welding machine + X-ray machine Aim: To search for information on the applications of transformers in daily life Instructions: 1. Carry out a Hot Seat activity. 2. Gather information on the applications of transformers in daily life such as (a) electrical appliances (b) electrical energy transmission and distribution systems 3. Information can be obtained from websites or by referring to your Form 3 Science Textbook. 4, Present your findings in the form of a folio. A transformer that is connected to a 240 V. power supply supplies 27 W of power at a voltage of 18 V to an electronic equipment, 18V [) |] Blectronic as shown in Figure 4.36. Assuming the A) aD 27 |_| equipment transformer is ideal: (a) calculate the number of turns of the primary coil. (b) calculate the current in the secondary circuit. (c) calculate the current in the primary circuit. (@) V, =240V, V,=18V,N, = 60 | (6) Output power=27W |() Vl, = Va Vs Vii, = 27 240 x I, = 18 x 1.5. ca 18x, = 27 18 x15 18 fo e240 aa 18 = 0.1125 A zi =15A ZZ KPMElectrical Energy Transmission and Distribution System ‘Transformers play an important role in the transmission and distribution of electricity from the power station to the consumers, Figure 4.37 shows the use of the step-up transformer and step-down transformer in the system. Generation of Transmission of electricity Distribution of electricity jeneration of 275kV A “Main “Main re cecnig sian cape Ean ¢ [Ef seer siniome Ua t Henny esate Figure 4.37 Electrical energy transmission and distribution system from the power station 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 the power cable becomes small, This reduces the loss of electrical energy from the power cable. 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. 1. A step-down transformer is connected to an alternating current power supply. Explain the working principle of the transformer. 2. A pupil collects the following information on a transformer: Secondary voltage = 6 V Secondary current = 4.80 A (a) Calculate the efficiency of the transformer. (b) Explain two factors that cause the transformer to be non-ideal. 3. Explain how an induction cooker can heat up food in a steel pot. 4, Transformers are used in the electrical energy transmission and distribution system. State the type of transformer used: (a) before transmission of electrical energy (b) at the distribution substation Dw a KPMaia ©) Self -Reflection » bs 1. New things I have learnt in the chapter ‘Electromagnetism’ ae A. 2, The most interesting thing I have learnt this chapter is a Zo 3, The things I still do not fully understand are = 2, ea ad 4, My performance in this chapter. oor 3) (ALA SANS GG very gooa 5. Ineedto___@_to improve my performance in this chapter. 1. Figure 1 shows a conductor hanging from a sensitive spring balance in between a pair of Magnadur magnets. ‘Spring balance ‘Conductor Figure 1 (a) Suggest the polarity of dry cell X, Y and the polarity of magnets P, Q such that the reading of the spring balance increases when switch $ is turned on. (b) Explain why the reading of the spring balance can increase in 1(a). (©) Suggest improvements that need to be made to further increase the reading of the spring balance. 2. With the aid of a labelled diagram, explain how Fleming's left-hand rule is used to determine the direction of the force on a current-carrying conductor in a magnetic field.3, Figures 2 and 3 show the induced currents produced when there is relative motion between a bar magnet and a solenoid. Solenoid Bar magnet Solenoid Bar magnet Figure 2 Figure 3 (a) What is the meaning of induced current? (b) Based on the direction of the current given in Figures 2 and 3, state the magnetic polarities at ends X and ¥ of the solenoid. (c) State the direction of motion of the bar magnet in Figure 2 and Figure 3. @® (A) Suggest two ways to increase the magnitude of the induced current in Figure 3. 4. A transformer is used to step down voltage from 240 V to 6 V for an electronic equipment. ‘The current in the primary coil is 0.18 A. What is the current in the secondary coil? State the assumption that needs to be made in your calculation. 5. Figure 4 shows two identical metal balls and a copper tube. One of the balls is a neodymium ‘magnet while the other is a steel ball. °°. —a Figure 4 Design an activity that can identify which ball is the neodymium magnet. Explain the physics principle used in your activity. 6. Figure 5 shows a wooden block with a bar magnet tied to it sliding with an acceleration down a smooth track. Switch Bar magnet ‘Smooth track Figure 5 When the block arrives at mark X on the track, the switch is turned on. (a) What is produced in the solenoid? Explain your answer. (b) Explain the motion of the block after the switch is turned on, (©) Based on your answers in 6(a) and (b), discuss the effectiveness of electromagnetic braking in stopping a moving object. KPM7. Figure 6 shows a transformer with a bulb at its output terminals, (a) Calculate the value of I. @ ay: By (b) What assumption needs to be made in your calculation in 7(a)? Figure 6 8. A pupil investigated the operation of a transformer and gathered data as shown in Figure 7. Calculate the efficiency of the transformer and suggest improvements to the design of the transformer to increase its efficiency. Solid iron block ‘Secondary voltage = 48 V ‘Secondary current = 0.60 A 9. Figure 8 shows the design of a simple direct current motor that can produce a force to rotate a disc connected to the axle of the motor. A pupil who constructed the motor according to the design made the following observations: «+ speed of rotation of the disc is slow Dry cell + speed of rotation of the disc cannot be controlled + rotation of the disc is not smooth Resistor « the dry cell loses its power in a short time ‘Single turn coil Study the design of the motor and suggest improvements to the design that can overcome the weaknesses identified by the pupil. &® Permanent magnet Figure KPM rewania