"9 2 2 k?? 8.99 10 Nm / CNOTE: Throughout this unit, it is taken that Coulomb’s Constant, ?19 qC ?? 1.602 10 e Also, that the electronic charge, 9 2 2 NOTE: Throughout this unit, it is taken that Coulomb’s Constant, k?? 8.99 10 Nm / C?19 Also, that the electronic charge, qC ?? 1.602 10e 33.Developing an Electrostatic Series Lab Exercises i. The results from the lab exercise were as in the attached raw data sheet. Yes, the observations would be different if the suspended comb were to be replaced with ametal fork or spoon; this is because the latter duo are good conductors of electricity andtherefore, do not favour charge creation by electrostatic means unlike the suspended combwhich is, inherently, an insulator. 33.Developing an Electrostatic Series Lab Exercises ii. The results from the lab exercise were as in the attached raw data sheet. iii. Yes, the observations would be different if the suspended comb were to be replacedwith a metal fork or spoon; this is because the latter duo are good conductors ofelectricity and therefore, do not favour charge creation by electrostatic means unlikethe suspended comb which is, inherently, an insulator. Instead, the metal fork orspoon would conduct away the charges; neutralizing them as soon as they arecreated. Hence, metallic materials always remain electrostatically neutral. 34. Considering the initial charges q and q separated by a distance r :1 2 1By Coulomb’s Law, the force between these two charges, F is quantified as:1 qq 12 Fk ? .........................(1) 1 2 r 1Now, considering the simultaneous changes introduced and evaluating new force, F :2q ' ? 2q (doubled) 11 q ' ?3q (tripled) 22 r ' ? 2r(doubled) 11 Now, qq'' 12 Fk ? 2 2 (r ') 1 Hence, q 'q ' 2q ?3q q q q q 6 1 2 1 2 1 2 1 2 F ? k ? k ? k ?1.5k 2 2 2 2 2 (r ') (2r) 4 r r 1 1 1 1 qq 12 ??Fk 1.5 .................(2) 2 2 r 1FF ?1.5 21By comparing equations (1) and (2) above, it can be seen that:That is, the magnitude of the electrostatic force experienced increases by a factor of 1.5 if bothchanges occurred simultaneously.22 ?3 N ?? 8.3 10 e ?10 m 1 35. Given, F=6.3N, r=0.5cm=5 and number of electrons in charge q , 1Now, qq 12 Fk ? 2 r But, q ? N q x e e x Thus, 2 N q ? N q (N ? N )(q ) qq e e e e e e e 12 12 12 F ? k ? k ? k 2 2 2 r r r Hence, 2 ?3 2 Fr 6.3 (5 ?10 ) N ? ( ){ } ? ( ){ } ?8.225 e 2 9 22 ?19 2 2 k N ?(q ) 8.99 ?10 8.3 ?10 ?(1.602 ?10 ) ee 1 ? N ?8.225 ?8 e 2 Hence, the second object has about 8 excess electrons.36. Given magnitude of charges as:?17 qC ? ?2.5 ?10 1 ?17 qC ? ?3.0 ?10 2 ?17 qC ? ?3.5 ?10 3These charges are located relative to each other as shown below: To determine the force acting on q :1First and foremost, we know from the laws of electrostatics that: Like charges repel. So, weanalyze the repulsion on q due to q and q each acting independently; and then vectorially sum the1 2 3 effects (Superposition principle).Now, ?34 qq (2.5?? 3 10 ) 12 9 ?23 F ? k ? (8.99 ?10 ) ? 2.697 ?10 N 21 22 (r ) (0.5) 21 ?23 ?FN ? 2.697 ?10 21 Similarly, ?34 qq (2.5?? 3.5 10 ) 13 9 ?23 F ? k ? (8.99 ?10 ) ?8.74 ?10 N 31 22 (r ) (0.3) 31 ?23 ?FN ?8.74 ?10 31 Vectorially representing these two forces:? 1 ? 1 F F31 21Representing these as a triangle of forces in equilibrium:F21 (180?? ? ) 1F FR31 ? R From the right-angle triangle formed by the geometry of the three charges, q , q and q :1 2 3 40 ?? 11 ? ?sin ( ) ?sin (0.8) ?53.13 ? 1 50 ? ? ?53.13 ? 1cos(180 ? ??? ) ? ?cos( ) 11RECALL: 1. sin(180 ? ??? ) ?sin( ) 11 2.Now, by applying Cosine Rule in the triangle of forces in equilibrium, we have:F RThe magnitude of the resultant force on q ,given as:1 22 F ? F ? (F ) ?(F ) ?2(F )(F )cos(180 ? ? ? ) RR 21 31 21 31 1 Thus, 22 F ? (F ) ?(F ) ?2(F )(F )cos( ? ) R 21 31 21 31 1 2 2 ?? 23 23 F ?{ (2.697) ?(8.74) ?2(2.697)(8.74)cos(53.13)} ?10 N ?10.581 ?10 N R ?23 ?FN ?10.581 ?10 RNow, by applying Sine Rule in the triangle of forces in equilibrium, we have:? RThe direction of the resultant force on q ,given as:1 F 2.697 ?? 11 21 ?? ?sin {( )sin( )} ?sin {( )sin(53.13)} ?11.77 ? R 1 F 10.581 R ? ? ?11.77 ? R?23 FN ?? 10.581 10 ??? 11.77 R R The resultant force, F , has a magnitude of,at an angle,to theR F force.31 qC ??4.3 ? 37.Given magnitude of the single point charge,a) Representing the electric field, E around the point charge:EThe electric lines of force are uniformly distributed around the point charge and directed towards?4.3 ?C the point charge . This signifies a uniform electric field, E around this charge. b) Electric field strength, E:Fq Ek ?? 2 qr t ?2 r ?3cm ?3 ?10 m ?6 q (4.3 ?10 ) 97 E ? k ? (8.99 ?10 ) ? 4.595 ?10 N / C 2 ?2 2 r (3 ?10 ) Thus, 7 E?? 4.595 10 N / C c) The magnitude of the force, F:qq e Fk ? 2 r r?? 10cm 0.1m ?? 6 19 qq (4.3 ?10 ?1.602 ?10 ) e 9 ?13 F ? k ? (8.99 ?10 ) ? 6.193 ?10 N 22 r (0.1) ?13 ?FN ? 6.193 ?10 This electric force (attraction towards the point charge) is directed towards the point charge in theleft direction.38. The electric field, E: F 44 E ? ? dV / dx ?1.5 ?10 N / C ?1.5 ?10 V / m q t Thus, 4 dV / dx?? 1.5 10 44 V(x) ? (1.5 ?10 )dx ? (1.5 ?10 )x ? A ?With, A being the constant of integration Now, assuming zero boundary conditions:VA (0) ? 0, ? ? 0 4 ?V(x) ? (1.5 ?10 )x ?2 At, x ?1.2cm ?1.2 ?10 m 4 VV (0.012) ? (1.5 ?10 )(0.012) ?180 Hence, VV ?18039. From first principles, the electric field, E:F 44 E ? ? dV / dx ?3.5 ?10 N / C ?3.5 ?10 V / m q t Thus, 4 dV / dx?? 3.5 10 44 V(x) ? (3.5 ?10 )dx ? (3.5 ?10 )x ? A ? Let, VA (0) ? 0, ? ? 0 4 ?V(x) ? (3.5 ?10 )x At, x?? 15cm 0.15m 43 VV (0.15) ? (3.5 ?10 )(0.15) ?5.25 ?10 Hence, V ?5.25kV40. a) Determining the work done, W: W ? ?F ? distance ? ?F ? ?r But, 1 1 1 1 9?? 6 3 ?F ? F ? F ? k q q { ? } ? (8.99 ?10 )(10 )(3.2 ?10 ){ ? } ?179.797N 21 t 2 2 2 2 rr 0.4 100 21 Thus, ??FN 179.797 distance, ?r ? r ? r ?100 ?0.4 ? 99.6m 12 Hence, 3 WJ ?179.797 ?99.6 ?17.908 ?10 ?? W 17.908kJ b) Since, test object is positively charged, N electrons were lost:e?6 q 10 12 t N ? ? ?6.242 ?10 e ?19 q 1.602 ?10 e12 6.242 ?10 Therefore, electrons were lost from the test object in order to create the positive chargeon the test object.41. Electrical magnetization and demagnetization process of magnetic materials. The process of magnetization of a magnetic material basically involves alignment of magneticdipoles in a specific desired direction. This process is reversible (demagnetization-disorientation ofmagnetic dipoles and domains in the magnetic material). Polarizers are devices actuated byapplication of an electric current for instance electromagnets, which by aligning the magnetic dipolesof the material being magnetized in a particular direction facilitate the magnetization of thematerial. This is achieved by application of a unidirectional current (DC); reversing the direction offlow of the DC in the magnetizing circuit reverses the polarity of the magnet. To electricallydemagnetize the material, an alternating current (a.c) is passed through the circuit, therebydisorienting the magnetic dipoles of the magnetic material, hence achieving demagnetization. 42. For the electromagnet circuit, conventional current flows through the circuit from the positive (+)terminal of the battery to the negative (-) terminal. The magnetic polarity is established using Right- Hand-Grip (RHG) rule: With the fingers of the right hand encircling the direction of flow of theconventional current in the solenoid (coil), the thumb points the north pole of the electromagnet.These are marked in the diagram below:Direction of flow of convetional currentNorth pole (N) South Pole (S)43. Operation of the electric bell:When the switch is closed, an electric current flow in the circuit through the solenoid, thusmagnetizing the electromagnet; by so doing, the magnetic armature to which the hammer isattached gets attracted towards the electromagnet thereby making the hammer to hit the bell(gong) that chimes (sounds), meanwhile the circuit gets disrupted as the make-break contact is open.Consequently, the electromagnet loses its magnetism and the armature is pulled back, by the springattached to it, to its initial position once the initial state is restored, the make-break contact is thusclosed and current again flows in the circuit. The above process is repeated as the bell chimesrepeatedly until the switch of the circuit is open and no current flows in the entire circuit. The variousparts of the system mentioned in this description are as labeled in the diagram below. Bell (gong)44. Force experienced by a current-carrying conductor in a magnetic field:a) The force experienced, 0 ? ? ? 90 ?Where, the angle it makes with the magnetic field: ? ?? 0 When the conductor is rotated in the magnetic field, when , the conductor is parallel to thesin( ?) ?sin(0 ?) ?0 magnetic field and so experiences no force since . Rotating the conductor in the??? 90 magnetic field gradually increases the magnitude of the force experienced as , attaining apeak value (maximum possible force) when the conductor is perpendicular the magnetic field; since sin( ?) ?sin(90 ?) ?1 . At this point, the force experienced by the rotating conductor is: o o b) Comparing forces at 90 and 30 : ??? 90 : ? ? ? ? F ? ILBsin ? ? ILBsin90 ? ? ILB 90 ??? 30 : ? ? ? ? 1 F ? ILBsin ? ? ILBsin30 ? ? ILB 30 2 Thus, ? ? ? ? ? ? ? ? FF ? 2 90 30o o Therefore, the force acting on the conductor at 90 is twice stronger than that acting at 30 .45. Of the three types of fields discussed, so far it is the electrostatic and gravitational fields that aresimilar in most respects: First of all both obey the inverse-square law, secondly the magnitude of theassociated forces is proportional to product of charge and mass quantities respectively. In both, thefield intensity is uniform on throughout the spherical surface at same radius from the reference point,all round the source of source of the field. Furthermore, point sources are allowed in both cases.Unlike magnetic field, electric charges can exist in isolation (monopoles), magnetic monopoles neverexist.46. a) When the switch is closed, the copper ring will be partially attracted by the verystrong magnetic field produced by the solenoid. Thus, the copper ring will slidedownwards along the iron core towards the energized coil.b) With a steady current in the circuit, the copper ring remains attached to the iron coreof the electromagnet.c) When the switch is opened, the copper ring remains in place, due to its own weight.d) Reversing the terminals on the power supply does not affect the above results,because the strong magnetic field produced will still induce magnetism in the copperring making it still move towards the solenoid.47. a) You will feel a “pushing force” as you push a magnet into a solenoid, if the two havelike poles facing each other hence repulsion. On the other hand, you feel a “pullingforce” if the two have unlike poles facing each other hence a mutual attraction. b) With more coils in the solenoid, the magnitude of the repulsive or attraction forcesexperienced would increase hence the “push” or “pull” felt would be more intensethan in the above case (a).48.For charged particle in a magnetic field, the force experienced is given by:? Where, defines the radius of circular orbit.a) Thus, the ion moves faster, being that it has less charge quantity than theion; and being accelerated by an equal force; further justified by the fact thatin the above relation, velocity and charge quantity are inversely proportional,holding everything else constant.b) Theion follows the path with the largest radius; because, according to theabove formula, the radius of the circular path is inversely proportional to thecharge quantity; and vice-versa.49. UNIT 3 KEY QUESTION:Research finding on: “Electric Generator as a Technological Application of Electromagnetism” An electric generator is an electro-mechanical device that converts mechanical energy intoelectrical energy by principle of electromagnetic induction; what is profoundly called the Faraday’slaw which states that: an electromotive force (emf) is generated in an electrical conductor whichencircles a time-varying magnetic flux; the magnitude of which is proportional to the rate of changed ? EN ?? ind dt of the magnetic flux linking the conductor(s).That is: This, basically, is the fundamental principle of operation of an electric generator that in practicalapplications comes in various forms and variations: from a simple bicycle dynamo to the complex? polyphase (1, 2, 3- ...) power supply systems. Electric generators can, basically, be classified undertwo broad categories: ac types (also called alternators) and the dc types (dc generators).When operated in the reverse application mode, an electric generator becomes an electricmotor which operates on the inverse principle which involves conversion of electrical energy intomechanical energy. The basic parts of a dc motor are as shown below. It is worth noting that withminor design modifications, an electric motor suffices being used as an electric generator.The use of electricity, which is generated from the electric generators have revolutionizedseveral facets of the economy ranging from the economic benefits derived: use of polyphase power inindustries for automation, domestic lighting just to mention a few; to the environmental benefits:Hydroelectric power (HEP) being a form of “green” energy as a source of energy significantly reducesdependence on fossil-based sources of energy that adversely contribute to “global warming”. Thefigure below shows parts of a typical electric generator: In a nutshell, electric generators have become a fundamental part of the energy sector of anyeconomy. These devices, though usually ignored by many, have made life comfortable by providingan alternative clean and safe means of energy that finds applications in most if not all facets of theeconomy. At the heart of these devices is an implementation of an idea once conceived by thefamous scientist Michael Faraday: The principle of electromagnetic induction, which regardless of thesource of mechanical power (nuclear, water turbines, fossil fuel etc.) remains to be the fundamentalprinciple of operation of the devices."