" 1 1 liUttl il.. •■■■1 immM i„„,i,ii !,.; IlillinHIIIHNIIII liHillliilillillllllllillill I f Jt^ara, 'SStto ^mrfe ..C,.U,.Phy»i-ca Department Cornell University Library QC 125.F83 Lessons in mechanics;.a text-book fw 3 1 924 01 2 332 833 The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924012332833 LESSONS IN ELECTRICITY AND MAGNETISM MESSRS. CONSTABLE AND CO., Limited IO-12 Orange Street, Leicester Square LONDON W. C. 2 represent MESSRS. FRANKLIN & CHARLES IN GREAT BRITAIN AND IRE- LAND, ON THE CONTINENT OF EUROPE, AND IN THE BRITISH COLONIES AND DEPENDENCIES AND SELF-GOVERN- ING DOMINIONS. EXCEPTING CANADA LESSONS IN ELECTRICITY AND MAGNETISM A TEXT BOOK FOR COLLEGES AND TECHNICAL SCHOOLS BY WILLIAM S. FRANKLIN and BARRY MACNUTT BETHLEHEM, PENNSYLVANIA FRANKLIN AND CHARLES LONDON: CONSTABLE & CO., Ltd. 1919 All rights reserved CopYRicajT, 19 19 By franklin AND CHARLES Set up and electrotyped. Published September, ZQig Reprinted August, igao. P.E PRESS OF THE NEW ERA PRINTING COMPANY LAN CASTE F^ PA. PREFACE. This book and a companion volume entitled Lessons in Mech- anics have been arranged to meet the needs of the two-year schedule in elementary physics which has been recently adopted in some of our technical schools, and, in our opinion, most of the added emphasis which this two-year schedule involves should be directed towards the mathematical side, or, let us say, the mathematical bottom of the subject. The two-year schedule in elementary physics, beginning with the freshman year, means that teachers of physics cannot base their work upon college courses in mathematics to the extent that might have been possible heretofore; but we believe, never- theless, that teachers of physics can and should use the more powerful mathematical methods from the beginning. Accord- ingly we have used differential and integral calculus throughout these new texts. The central idea underlying these new texts is that mathe- matical training must be accomplished by the combined efforts of teachers of mathematics and teachers of physics, and it is hoped that these texts may help to bring our teachers of physics to what we believe to be their proper function in the important and difficult matter of mathematical training. If our students were to come from their mathematics teachers fully able to use their mathematics there would be nothing left for the rest of us to do ! The central purpose of these new texts is to facilitate class- room work. Descriptive and explanatory material has been reduced to a minimum, because students will not read more than is absolutely necessary; the development of every topic leads as directly as possible to illustrative numerical problems; and everything is arranged in strict lesson order. VI PREFACE. The authors are indebted to members of the staflf of the De- partment of Physics of the Massachusetts Institute of Tech- nology for many suggestions that have been helpful in the prepar- ation of this book. The Authors. IMPORTANT BOOKS AND HELPFUL REFERENCES. Important Books for the Beginner. Hadley's Electricity and Magnetism, London, Charles Griffen & Co., 1904. J. J. Thomson's Elements of Electricity and Magnetism, Cambridge University Press, 1909. Poynting and Thomson's Electricity and Magnetism, London, Charles Griffen & Co, 19 14 Franklin and MacNatt's Advanced Electricity and Magnetism, Bethlehem, Pa. Franklin and Charles, igiS- Lodge's Modern Views of Electricity, London, Macmillan & Co., 1907. Helpful References for the Beginner. Unusually good discussions of the following topics are given in Poynting and Thom- son's Electricity and Magnetism. Properties of dielectrics and measurement of inductivity, pages 120-133. Pyro-electricity and piezo-electricity, pages 148-163. Para-magnetisih and dia-magnetism, pages 203-207 and 282-300. Magnetism and light (a group of interesting effects), pages 320-340. An instructive discussion of the mechanical conceptions of electromagnetic action is given on pages 245—263 of Franklin and MacNutt's Advanced Electricity and Magnetism. A good discussion of thermoelectric currents is given in J. J. Thomson's Elements of Electricity and Magnetism, pages 306-518. See also Nichols and Franklin's Elements of Physics, Vol. 2, pages 216-221; New York, The Macmillan Co., 1898. A very good discussion of electrolysis is given on pages 176-406 of W. C. D. Whet- ham's Theory of Solution, Cambridge University Press, 1902. Important references for the student of electrical engineering An exhaustive discussion of the magnetic properties of iron is given in J. A. Ewing's Magnetic Induction in Iron and other Metals, The Electrician Publishing Co., London, 1898; and a very good discussion of same by Dr. Saul Dushman in Vol. 19 of the General Electric Review. A good elementary discussion of the mole- cular theory of magnetism is given in Poynting and Thomson's Electricity and Magnetism, pages 192-202. A full discussion of Fortesque's theory of insulator design is given on pages 186- 191 of Franklin and MacNutt's Advanced Electricity and Magnetism. The theory of graded cable insulation and a discussion of the influence of hetero- geneity of dielectric on dielectric stresses is given on pages 156-158 of Franklin and MacNutt's Advanced Electricity and Magnetism. Inductance of a transmission line is discussed on pages 60-62 of Franklin and Mac- Nutt's Advanced Electricity and Magnetism Capacity of coaxial cylinders and parallel cylinders (cables and transmission lines) is discussed on pages 168-185 of Franklin and MacNutt's Advanced Electricity and Magnetism. vii viii IMPORTANT BOOKS AND HELPFUL REFERENCES. A simple introduction to the matliematical theory of electric waves, including a full discussion of transmission line surges and a discussion of wave distortion and line loading is given in Chapter IX of Franklin and MacNutt's Advanced Elec tricity and Magnetism. A good discussion of Ship's Magnetism and the Compensation of the Ship's Compass is to be found on pages 85-100 of Andrew Gray's Treatise on Magnetism and Electricity, Vol. I, London, Macmillan & Co., 1898; and a simpler discussion is given on pages 104-120 of Franklin and MacNutt's A dvanced Electricity and Magnetism. Books for Additional Study. A favorite treatise on Electromagnetic Theory among advanced students is Abra- ham & Foppl's Theorie der Electrizitat, 2 vols. Leipzig, B. G. Teubner, 1904. The classical treatises on Electromagnetic theory are: (o) Faraday's Experi- mental Researches; (b) Maxwell's Electricity and Magnetism, 2 vols., 3d edition, Oxford, 1904; (c) Hertz's Electric Waves, translated by D. E. Jones, London, Macmillan & Co., 1893, and (d) Heaviside's Electromagnetic Theory, 2 vols., Lon- don, The Electrician Pub. Co., 1890. Helps in the study of general theory A very helpful discussion of divergence and curl and of scalar and vector potential is given in chapter IX of Franklin, MacNutt and Charles' Calculus, Bethlehem, Pa., Franklin and Charles, 1909. A good introduction to the study of electric waves is to be found in chapter IX of Franklin and MacNutt's Advanced Electricity and Magnetism; see also Franklin's Electric Waves, Bethlehem, Pa., Franklin and Charles, 1909. Every student should know of the following most important collections of phys- ical and chemical data. I. Tables AnnueUes Internationales de Constants, .i. Physikalisch-chemische Tabellen, LandoIt-BOrnstein. 3. Smithsonian Tables (Fowle); publication No. 2269. THE MORE IMPORTANT BOOKS ON ELECTRONS AND THE ELECTRON THEORY AND ON RADIOACTIVITY. The electron theory is at present the most fruitful branch of theoretical physics, and the following are the most helpful books on this subject. Conduction of Electricity through Gases, J. J. Thomson, Cambridge University Press, 1903 and 1906. Electricity and Matter, J. J. Thomson, New York, Charles Scribner's Sons, 1904. Rays of Positive Electricity, J. J. Thomson, London, Longmans, Green & Co., 1913- Electrons, Oliver Lodge, London, Geo. Bell & Sons, 1909. The Electron, R. A. Millikan, University of Chicago Press, 1917. Theory of Electrons, H. A. Lorentz, Leipsig, B. G. Teubner, 1909. The Emission of Electricity from Hot Bodies, O. W. Richardson, London, Long- mans, Green & Co., 1916. IMPORTANT BOOKS AND HELPFUL REFERENCES. ix Photo-Electricity, H. Stanley Allen, London, Longmans, Green & Co., 1913. Relativity and the Electron Theory, E. Cunningham, London, Longmans, Green & Co., 1915. Radioactivity, Frederick Soddy, London, The Electrician Publishing Co., 1904. The Interpretation of Radium, Frederick Soddy, New York, G. P. Putnam's Sons, 1912. The Chemistry of the Radio Elements, Frederick Soddy, London, Longmans, Green & Co., 1914. Radioactive Substances and their Emanations, Ernest Rutherford, Cambridge University Press, 1913. Practical Measurements in Radioactivity, W. Makower and H. Geiger, London, Longmans Green & Co., 1912. NATIONAL ORGANIZATIONS AND SOCIETIES RELATING TO PHYSICAL SCIENCE AND ENGINEERING. The AMERICA^f Physical Society. Every advanced student of physics and everyone who is interested in physical research of any kind should know about the American Physical Society and its ofGcial organ of publication, The Physical Review. The Secretary of the Society is Professor D. C. Miller of the Case School of Applied Science, Cleveland, Ohio; and the managing editor of the Review is Professor Frederick Bedell of Cornell University, Ithaca, N. Y. The United States Bureau of Standards. Every student of physics and chemistry and every student of engineering should know about the Bureau of Standards. A letter addressed to the Director, Dr. S. W. Stratton, Washington, D. C, will bring full information concerning the activities of the Bureau and a list of its publications. The scientific and tech- nological papers of the Bureau are published chiefly in the Bulletin of the Bureau of Standards, and the Bureaiu Circulars are occasional publications. ENGINEERING SOCIETIES. Name. 1. American Society of Civil Engineers. 2. American Society of Me- chanical Engineers. 3. American Institute of Elec- trical Engineers. 4. American Institute of Min- ing Engineers. 5. American Chemical Society. 6. Illuminating Engineering Society. 7. Institute of Radio Engi- neers. 8. Electrochemical Society. Field. Civil Engineering. Mechanical Engineer- ing. Electrical Engineering. Mining Engineering and Metallurgy. Chemistry and Chemi- cal Industry. Illumination. Radio telegraphy and telephony. Electrochemistry Metallurgy. and Headquarters Office. 29 W. 39th St., New York City. 29 W. 39th St., New York City. 29 W. 39th St., New York City. 29 W. 39th St., New York City. Dr. Chas. L. Parsons, Sec'yi Box 505, Wash- ington, D. C. 29 W. 39th St., New York City. Mr. Davis Sarnofl', Sec'y. Ill Broadway, New York City. Dr. J. W. Richards, Sec'y, South Bethlehem, Pa. * The corresponding establishment in Great Britain is The National Physical Laboratory. xu NATIONAL ORGANIZATIONS AND SOCIETIES. Each of the above societies is broadly concerned with the scientific, engineering, economic and ethical aspects of its particular field. Every engineer should belong to one or more of these societies. They are helpful to their members, and they serve to bring the much needed counsel of the scientist Eind the engineer effectively into municipal, state and national affairs. All of these societies issue important publications, and all, or nearly all, of them make special provision for student members. A letter addressed to the headquar- ters office of any of the societies will bring full information concerning the society. MAGNETIC AND ELECTRIC UNITS AND SYMBOLS. (Used in this text.) Centimeter-gram-second system, m = pole strength. Unit pole is de- iined in Art. 4. H — field intensity. The gauss is defined in Art. lo. Note. H is sometimes used to represent a quantity of heat. Ampere-ohm-volt system. There are no recognized names for the units of m, H and * in the ampere- ohm-volt system. Hence all quan- tities should be expressed in c.g.s. units in any equation containing m or H or *. ^ = magnetic flux. The maxwell or line is defined in Art. 12. i or I = current. The abampere is de- The ampere is defined as one tenth of fined in Art. 21. an abampere. The legal definition of the ampere is given in Art. 24. One ampere =0.1 abampere. r or i? = resistance. A portion of a cir- A portion of a circuit has a resistance of cult has a resistance of one abohm when one erg of heat is developed in it by one abami)ere in one second. e or £ = electromotive force. The ab- volt is the electromotive force between the terminals of a resistance of one abolmi when a current of one ab- ampere is flowing through it. L = inductance. A circuit has an inductance of one abhenry when one abvolt causes the current in the circuit to in- crease at the rate of one ab- ampere per second. one ohm. when one joule of heat is de- veloped in it by one ampere in one second. One ohm =10' abohms. The volt is the electromotive force be- tween the terminals of a resistance of one ohm when a current of one ampere is flowing through it. One volt = 10' abvolts. A circuit has an inductance of one henry when one volt causes the current in the circuit to increase at the rate of one ampere per second. One henry =10' abhenrys. g or Q = electric charge. The abcou- The coulomb is the charge carried in one lomb is the charge carried second by one ampere. in one second by one ab- One coulomb =0.1 abcoulomb. ampere. xiii xiv MAGNETIC AND ELECTRIC UNITS AND SYMBOLS. C = capacity. A conaenser nas a A condenser has a capacity of one farad capacity of one abfarad when one coulomb of charge is drawn when one abcoulomb of out of one plate of the condenser and charge is drawn out of one pushed into the other plate by an plate of the condenser and electromotive force of one volt, pushed into the other plate One farad = lo"* abfarads. by ah electromotive force of one abvolt. THE "ELECTROSTATIC" SYSTEM OF C.G.S. UNITS. (Not used in this text.) The stalcoulomb is the "electrostatic" unit of charge; it is a charge which will repel an equal charge with a force of one dyne at a distance of one centimeter in air. See Art. 69 of Chapter V and see Problem 113 on page 140. One statcoulomb is equal to 3.33 ( X io~") abcoulomb. The statampere is the flow of one statcoulomb per second. The slatvolt is an electromotive force which will do work at the rate of one erg per second when 'propelling'' a current of one statampere. One statvolt is equal to 3 X io'° ab volts. The statohm is a resistzince in which one erg per second of heat is developed by a current of one statampere. The statfarad is the "electrostatic" unit of condenser capacity. A condenser has a capacity of one statfarad when one statvolt will draw one statcoulomb out of one plate and force into it the other plate of the condenser The stathenry is the "electrostatic" unit of inductance. A circuit has an inductance of one stathenry when one statvolt will cause a current in the circuit to in- crease at the rate of one statampere per second. The statgauss is the "electrostatic" unit of magnetic field intensity. A magnetic field has an intensity of one statgauss when it will push sidewise with a force of one dyne on one centimeter of wire carrying a current of one statampere, the wire being at right angles to the field. Etc. Etc. Etc. A very full discussion of the two systems of electric and magnetic units is given on pages 240-253 of Nichols and Franklin's Elements of Physics, Vol. II. CONTENTS. CHAPTER I. PAGE Magnetism and the Magnetic Effect of the Electric Current 1-34 CHAPTER H. Chemical Effect of the Electric Current 35-49 CHAPTER HI. Heating Effect of the Electric Current 50-75 CHAPTER IV. Induced Electromotive Force 76-99 CHAPTER V. Electric Charge and the Condenser 100-12 1 CHAPTER VI. Electric Field and Elementary Theory of Po- tential 122-141 CHAPTER VII. The Atomic Theory of Electricity 142-169 APPENDIX A. Magnetism of Iron 170-184 APPENDIX B. Elementary Theory of Alternating Currents... 185-219 APPENDIX C. Electrical Measurements 220-236 APPENDIX D. Corresponding Equations 237-238 APPENDIX E. Answers to Problems 239-251 INDEX 252-254 XV LEGAL STANDARDS AND PRACTICAL DEFINITIONS OF UNITS. The definitions given or referred to on pages xiii and xiv are the scientifically correct* definitions- of m, H, i, r, e, etc.; but for the engineer everything relating to electrical and magnetic units must be based on the legal standards; see pages 220 and 221. Thus the engineer's definition of the gauss may be based on the legal standard ampere, using equation (6) on page 29; or the engineer may define the unit of magnetic flux (the maxwell) as the amount of flux which must be cut per second to give an induced electromotive force of one abvolt (one hundred-millionth of a volt), in accordance with equation (16) on page 80, and the gauss may be then defined as one maxwell per square centimeter in accordance with equation (4) on page 16. * Some variations in these definitions may be made without departing from the historical and scientific point of view. CHAPTER I. MAGNETISM AND THE MAGNETIC EFFECT OF THE ELECTRIC CURRENT. I. Preliminary statements concerning the electric current and the electric circuit. — Because of the almost universal use of the familiar dry cell or dry battery for operating electric bells and flash lamps, and because of the supply of electric current nearly everywhere for lighting and power, we may, in beginning our study of electricity and magnetism, take the source of electric current for granted and devote our attention to the properties of the electric current; and from the point of view of elementary theory we are concerned almost wholly with three groups of effects, namely, (o) The magnetic effect of the electric current, (b) The chemical effect of the electric current, and (c) The heating effect of the electric current. One aspect of the magnetic effect of the electric current is exem- plified by the electro-magnet which consists of a winding of insulated wire on an iron rod or core.' When the winding of wire is connected to a battery the iron core attracts other pieces of iron and is said to be magnetized. When the winding of wire is disconnected from the battery the iron rod loses its magnetism. A rod of hardened steel may be magnetized in the same way, but a rod of hardened steel retains its magnetism more or less persis- tently when it is removed from the winding of wire or. when the winding or wire is disconnected from the battery. Such a mag- netized rod of hardened steel is called a permanent magnet, or, simply, a magnet. The magnetic effect of the electric current exhibits itself in a variety of ways. Thus the action of the dynamo as an electric generator is one aspect of the magnetic effect of the electric current. One aspect of the chemical effect of the electric current is exem- 2 I LESSONS IN ELECTRICITY AND MAGNETISM. plified by the operation of electro-plating. Thus Fig. i shows a battery connected to two strips of copper C and T both of which dip into a solution of copper sulphate. Under these con- eathon- m_ wire 3 y zine terminal wire ->- cathode- c- - ^anode battery copper sulphate solution Fig. 1. ditions metallic copper is deposited on C and copper is dissolved off T. Another aspect of the chemical effect of the electric current is exemplified by an ordinary battery which is an arrange- ment in which the energy of chemical action maintains a current. See Art. 25. The heating effect of the electric current is exemplified by the familiar flash lamp the essential features of which are shown in lamp/' KM push button dry battery Fig. 2. Fig. 2. The lamp is a piece of very fine tungsten wire mounted in a glass bulb. MAGNETISM AND THE MAGNETIC EFFECT. 3 The electric current and the electric circuit. — When the above described effects are produced an electric current is said to flow through the wire. The production of an electric current always requires an electric generator such as a battery or dynamo. The path of the current is usually a wire, and this path is called the electric circuit. A steady electric current always flows through a complete circuit, that is to say, through a circuit which goes out from one terminal of a battery (or dynamo) and returns to the other terminal of the battery (or dynamo) without a break. Such a circuit is called a closed circuit. When the circuit is not complete it is said to be an open circuit. The electric current ceases to flow through a circuit when the circuit is opened or broken. An electric current which lasts for a very short time, a few thousandths of a second, for example, can flow in an incomplete or open circuit. In such a case very important effects are ob- served at the place where the circuit is broken. These effects, which we may very properly call gap effects, are discussed in chapters V, VI and VII. Conductors and insulators. — The carbon plate of the battery forms a portion of the electric circuit in Figs, i and 2, the wire forms a portion of the circuit in Figs, i and 2, and the solution of copper sulphate forms a portion of the circuit in Fig. i . Any substance which can serve as a portion of an electric circuit (any substance through which the electric current can "flow")i is called an electrical conductor. Thus metals, carbon and salt and acid solutions are electrical conductors. Many substances, suchi as dry wood, rubber, glass and air, cannot* serve as portions of an electric cirucit at ordinary temperatures, that is to say, the electric current cannot "flow" through such substances to any appreciable extent, and such substances are called insulators. * This statement is not strictly true; what is called an insulator is merely a very poor conductor. 4 LESSONS IN ELECTRICITY AND MAGNETISM. PROBLEMS. I. The accompanying diagram Fig. 3 shows a lamp L with connections arranged so that the lamp can be turned on or off at switch A (or B) regardless of how switch B (or .4) stands. Make four diagrams like Fig. 3 showing the four possible com- + supply mains Tt -©- 1 Fig. 3. binations of switch-positions, and indicate the flow of current, if any, by arrows in each case. 2. The six small circles in Fig. 4 represent the contact posts on a double-pole, double-throw "switch, and the dotted lines A C supply mains ^- -o- r — -o- DC supply mains Fig. 4. represent the switch blades. The diagram shows the lamp L taking current from the direct-current mains. Make a diagram showing the lamp taking current from the alternating current mains. 3. The six small circles in Fig. 5 represent the contact posts on a double-pole, double-throw switch with crossed connections AfH-nliM^ AO- AO- ^^^mi .6 b B Fig. 6. MAGNETISM AND THE MAGNETIC EFFECT. 5 adapting it for use as a reversing switch, and the dotted Hnes represent the switch blades. Make a diagram showing a re- versed flow of current through the receiving circuit J?. 4. Figure 6 shows the diagram of connections of an ordinary telegraph relay, a "local" circuit connected to the binding posts BB is opened and closed as the lever L of the relay is moved back and forth by pulses of current coming over the telegraph line which is connected through the binding posts ^^ to ground. M is a. screw with a metal tip, and H is a screw with a hard- rubber insulating tip . Make a diagram showing M and H interchanged, and showing A A and BB connected to each other and to a battery so that the relay will buzz like an ordinary interrupter bell. 5. It is possible to connect any number of bell circuits to one battery, and have any number of push buttons arranged to close each bell circuit independently. Make a diagram showing three bells connected to a single battery with two push buttons arranged to close the circuit of each bell independently. 6. Suppose an electric bell is to ring so as to give a starting signal to the motor-man on a 4-car train, and suppose that the bell is to ring only when every one of the four gate-men pushes a contact button, four contact buttons in all. Make a diagram of the connections. 2. Poles of a magnet. — The familiar property of a magnet, namely, its attraction for iron, is possessed only by certain parts of the magnet. These parts of a magnet are called the poles of the magnet. For example, the poles of a straight bar-magnet are usually at the ends of the bar. Thus Fig. 7 shows the appear- ance of a bar-magnet which has been dipped into iron filings. The filings cling chiefly to the ends of the magnet. When a bar-magnet is susjJended in a horizontal position by a fine thread, it places itself approximately north and south like a compass needle. The north pointing end of the magnet is LESSONS IN ELECTRICITY AND MAGNETISM. called its north pole, and the south pointing end of the magnet is called its south pole. The north poles of two magnets repel each other, the south poles of two magnets repel each other, and the north pole of one magnet attracts the south pole of another magnet ; that is to say, Uke magnetic poles repel each other, and unlike mag- netic poles attract each other. The mutual force action of two magnets is, in general, resolvable into four parts, namely, the forces with which the respective poles of one mag- net attract or repel the respective poles of the other magnet. In the following discussion we con- sider only the force with which one pole of a magnet acts upon one pole of another magnet, not the forces with which one complete magnet acts on another complete magnet. 3. Distributed poles and concentrated poles. — The poles of a bar magnet are always distributed over consider- able portions of the bar. This is especially the case with short thick bars. In the case of a long slim bar magnet, however, the poles are ordinarily approximately concentrated at the ends of the bar. The forces of attraction and repulsion of concentrated magnet poles are easily formulated, therefore the following discussion applies to ideally concentrated poles at the ends of ideally slim bar magnets. 4. Definition of unit pole. — Consider a large number of pairs of magnets a b, c, d, etc., as shown in Fig. 8, the two magnets of each pair being exactly alike.* N N N N N It Jf N S S S S S S S S Fig. 8. Pairs of exactly similar mag- nets. From such a set * That is, the magnets of each pair are made of identically the same kind of steel, subjected to the same kind of heat treatment and magnetized by the same means. MAGNETISM AND THE MAGNETIC EFFECT. ^ it would be possible to select a pair of magnets such that the north pole of one magnet would repel the north pole of the other with a force of one dyne when they (the two north poles) are one centimeter apart ; each pole of such a pair is called a unit pole. That is, a unit pole is a pole which will exert a force of one dyne upon another unit pole at a distance of one centimeter. S. Strength of pole. — Let us choose a slim magnet with unit poles, and let us use one of these unit poles as a "test pole." Viven pole strength m,^ unit test pole _ one centimeter _ /_ (21 dynes / / \ m dyneg Fig. 9. Any given magnet pole is said to have more or less strength ac- cording as it exerts more or less force on our "test pole" at a given distance. And the force m (in dynes) with which the given pole attracts or repels (or is attracted or repelled by) the unit test pole at a distance of one centimeter is taken as the measure of the strength of the given pole. That is, a given pole has m units of strength when it will exert a force of m dynes on a unit pole at a distance of one centimeter, as indicated in Fig. 9. 6. Attraction and repulsion of magnet poles. — Unlike poles attract and like poles repel each other, as stated in Art. 2. When the two attracting or re- pelling poles are unit poles their , *^---^=0 attraction or repulsion is equal ,.^J^^C^_"^----=0 to one dyne when they are one m'=2units '-=0 centimeter apart, and the attrac- pj^ ^^ tion or repulsion of two poles whose respective strengths are m' and m" is equal to m'm" dynes when the poles are one centimeter apart. One may think 8 LESSONS IN ELECTRICITY AND MAGNETISM. of each unit of m' as exerting a force of one dyne on each unit of m". Thus if w' = 3 units and m" = 2 units, then the force of attraction or repulsion will be six dynes, as indicated in Fig. 10, where each dotted line represents one dyne. 7. Coulomb's law. Complete expression for the force of attraction or repulsion of two magnet poles. — Coulomb dis- covered in 1800 that the force of attraction or repulsion of two magnet poles is inversely proportional to the square of the dis- tance between them. But the force of attraction or repulsion of two magnet poles when they are one centimeter apart is m'm" dynes as explained in Art. 6. Therefore, according to Coulomb's law, the force of attraction or repulsion is — -t-^ dynes when the poles are r centimeters apart. That is: „ m'm" , . ^ = —2- w in which m' and m" are the respective strengths of two magnet poles, r is their distance apart in centimeters, and F is the force in dynes with which the poles attract or repel each other. 8. Equation (i) is really a differential equation. — Any physical thing must be to some extent idealized if it is to be formulated mathematically, and, as a rule, this idealization must be excessive if the mathematics is to be simple. The elementary theory of magnetism is a remarkable example of excessive idealization; but without this excessive idealization the mathematics is much too complicated for the beginner.* *Many attempts have been made to develope the elementary theory of mag- netism without excessive idealization, and many of these attempts have been made by men who do not understand divergence and curl. Space distributions, like magnetic field and electric field, are sometimes conditioned by divergence and sometimes conditioned by curl, and the conception of the concentrated magnet pole is artificial not merely because it is a differential but also and. chiefly because the conception of the magnetic pole grows out of the attempt to use the simpler idea of divergence in place of the more difficult idea of curl. The mathematical physicist recognizes the artificial character of the elementary theory of magnetism and he also understands iti and he knows what the alternative is, namely, real downright mathematics. MAGNETISM AND THE MAGNETIC EFFECT. 9 Fig. II represents two indefinitely broad and indefinitely long strips or ribbons of sheet steel MM and M'M'; NN is the edge of one ribbon, S5 is the edge of the other ribbon, and the ribbons are supposed to be magnetized so that a north pole is spread along NN (a units of pole per centimeter of length) and a south pole is I' ,l' ,(', -i/i n(\ III Ill li''iii III I ; 'Ij l'\ aV ^'^m'sa.Ax m"=b.Ay ,||l|l'lVl'lll' Fig. U. spread along SS (b units of pole per centimeter of length) . It is required to find the total force V with which the pole NN pulls straight upwards on 55. Consider the two elements Ax and Ay. The element A* is a concentrated pole m' = a.Ax; the element Ay is a concentrated pole m'' = b.Ay; the force a.Ax X b.Ay AF exerted on Ay by Ax is AF =- . , , _. , according to equation (i); (» —yy + Z>2 n and the vertical component of AF is AF = AF X cos9 = AF. Therefore AV = I(* -yy +D'YiK [(* - yy + m^'^ and the total force V acting on 55 is to be found by integrating this expression. This integration, however, involves too much algebraic manipulation to be given . here. The ideal concentrated pole is really a differential, and the correct definition of dy, for example, is that dy is the amount that y would increase during a finite time dt if the rate of increase of y were to remain constantly at the value it had at the start, at the instant t. It is almost beyond the power of the English language to apply this entirely correct definition of a differential to such a concept as the ideal concentrated pole, and if the correct mathematical definition of an ideal con- centrated pole were to be formulated, the conditional terms "would increase" "if" and "were to remain" would make it entirely useless as a practicable form of thought. It is much better to think of an ideal concentrated pole as an infin- itely small pole I even if such a thing is, strictly speaking, unthinkable! These are the considerations, more or less Irish, to be sure, which lead the engineer and the lO LESSONS IN ELECTRICITY AND MAGNETISM. mathematical physicist to an attitude of contempt for the purist in mathematics who insists on the strictly correct definition of the differential! PROBLEMS. 7. Figure 12 represents two similar bar magnets, the poles being assumed to be concentrated at the extreme ends of the bars and the strength of each pole being 350 units. Find the total force exerted on one magnet by the other. If NC \_ N S :^ ^(Tcm." Fig. 12. 30<;nh S Fig. 13. 8. The two magnets of the previous problem are arranged as shown in Fig. 13. Find the total force action exerted on one magnet by the other. 9. The magnetic field. — ^When iron filings are dusted over a pane of glass which lies fiat on a magnet, the filings arrange themselves in regular filaments as shown in Figs. 14 and 15, if the glass plate is jarred slightly. Figure 14 shows the filaments Fig. 14. MAGNETISM AND THE MAGNETIC EFFECT. II of iron filings in the neighborhood of a single bar magnet, and Fig. 15 shows the filaments of filings between the unlike poles of two large magnets. The region surrounding a magnet is called a magnetic field, and the filaments of iron filings in Figs. 14 and 15 show the trend of what are called the lines of force of the magnetic field. Indeed any region is a magnetic field in which a suspended* magnetic needle {like a compass needle) points in a definite direction, and the Fig. IS. direction in which the north pole of the needle {Joints is called the direction of the magnetic field at the place where the needle is suspended. ID. Intensity of a magnetic field at a point. — ^When a magnet is placed in a magnetic field a force is exerted on each pole of the magnet by the field. Thus the two arrows in Fig. 16 represent the forces which are exerted on the poles of a small magnet which is placed in the magnetic field between two large magnet poles (see Fig. 15). The force H in dynes which a mugnetic field exerts on a " unit test pole" is used as a measure of the strength or intensity of the * The needle is supposed to be suspended at its center of gravity. 12 LESSONS IN ELECTRICITY AND MAGNETISM. field, and this force-per-unit-pole is hereafter spoken of simply as the intensity of the field. The unit of magnetic field intensity (one dyne-per-unit-pole) is called the gauss. That is to say, a N y^ Fig. 16. « The arrows show the forces which act upon the poles of the small magnet. magnetic field has an intensity of one gauss when it will exert a force of one dyne upon a unit pole. Complete expression for the force exerted on a magnet pole by a magnetic field. — A magnetic field of which the intensity is H gausses exerts a force of H dynes upon a unit pole as above explained, and it exerts a force oi mH dynes upon a pole of which the strength is is m units. That is : F = mH (2) in which F is the force in dynes which is exerted on a pole of strength m hy a. field of intensity H. Uniform and non-uniform fields. — A magnetic field is said to be uniform when it has everywhere the same direction and the same intensity, otherwise the field is said to be non-uniform. The earth's magnetic field is sensibly uniform throughout a room. The magnetic field surrounding a magnet is non-uniform. The magnetic field surrounding an electric wire is non-uniform. The oscillating magnet. — A suspended magnet points magnetic north and south. If turned through an angle 6, the suspended magnet is acted upon by a torque T = — ml H sin 6, or, if 6 is MAGNETISM AND THE MAGNETIC EFFECT. 13 small, T = — mlH'd, where 6 is expressed in radians. There- fore the "stiffness coefficient" of the suspended magnet is 6 = mlH, and, as explained in Art. 57 of Lessons in Mechanics, we have ^■kH^K = tnlH where n is the number of oscillations per second of the suspended magnet, K is the moment of inertia of the suspended magnet referred to the axis of suspension, I is the length of the magnet (distance between its poles) , ± w is the strength of the poles of the magnet, and H is the horizontal component of the earth's magnetic field. PROBLEMS. 9. A bar magnet 30 cm. X i cm. X i cm. weighing 235 grams is balanced horizontally on a knife edge at a place where the horizontal component of the earth's magnetic field is 0.18 gauss and the vertical component of the earth's magnet field is 0.48 gauss. The poles of the magnet are assumed to be concentrated at the extreme ends of the bar and the strength of the poles is ± 700 units. Find the horizontal distance of the knife edge from the middle of the bar, taking the acceleration of gravity to be 980 centimeters per second per second. 10. The magnet of the previous problem is suspended so as to stand horizontally and oscillate about a vertical axis. Find the number of complete oscillations per second. II. Direction and intensity of the magnetic field surrounding an "isolated" magnet pole of strength M. — By an "isolated" magnet pole is meant one pole of a very long slim magnet — the other pole being so far away as to be negligible in its action. The magnetic field in the neighborhood of an isolated north pole is everywhere directed away from the pole as shown by the radiating straight lines (lines of force, as they are called) in Fig. 17. The magnetic field in the neighborhood of an isolated 14 LESSONS IN ELECTRICITY AND MAGNETISM. south pole is everywhere directed towards the pole as indicated in Fig. 1 8. Consider two magnet poles M and m which are r centimeters apart as shown in Fig. 19. The force F with which M repels Fig. 17. Fig. 18. m is equal to Mm/r^ according to Art. 7 ; but the force which is exerted on m may also be expressed as mH, where H is the intensity at m of the magnetic field which is due to M. There- r centimeters Fig. 19. fore m H must be equal to Mm/r^, fromwhich we get H = M (3) in which H is the intensity in gausses of the magnetic field produced by ikf at a place which is r centimeters from M. MAGNETISM AND THE MAGNETIC EFFECT. 15 PROBLEMS. 11. Given a bar magnet 30 cm. X i cm. X i cm. with poles of ± 700 units strength, and let us assume that these poles are concentrated at the extreme ends of the bar. Find the intensity of the magnetic field produced by both poles of the magnet at a point which is 18 centimeters from one pole and 24 centimeters from the other pole. Note. — ^When two agencies or causes act together to produce magnetic field the intensity at any point of the field due to both causes is the vector sum of the inten- sities at that point due to the two causes separately. 12. The bar magnet of the previous problem is placed in an east-west position with its middle point one meter east or west of a suspended magnetic needle. The needle points due north before the large magnet is placed in position, find the angle through which the needle is turned by the large magnet, hori- zontal component of the earth's magnetic field being 0.18 gauss. Nole. — The terms north and south, and east and west as used in this problem refer to magnetic north and south and magnetic east and west. 13. A bar magnet i cm. X i cm. X 30 cm. whose mass is 235 grams deflects a suspended magnet needle through an angle of 13I degrees when arranged as described in problem 12; and when suspended and set oscillating as specified in problem 10, the magnet makes 15 complete oscillations in 102 seconds. Calculate the value of the horizontal component of the earth's magnetic field and calculate the strength of each pole of the magnet. Note. — This problem illustrates the essentials of Gauss's method for measuring the horizontal component of the earth's magnetic field and the strength of the poles of a magnet. The distance between the poles is assumed to be 30 centimeters in the solution of the problem whereas it is in fact considerably less than 30 centimeters. Gauss's method as carried out in magnetic surveys or in the laboratory is arranged so as to eliminate this source of error. See Art. 38 of Appendix C. See also Kohrausch's Physical Measurements, pages 240-245; London, J. & A. Churchill, 1894. 12. Definition of magnetic flux. — Figure 20 represents a plane I6 LESSONS IN ELECTRICITY AND MAGNETISM. area of a square centimeters at right angles to a uniform mag- netic field of which the intensity is H gausses. The product aH is called the magnetic flux across the area. That is ^ = aH (4) where $ is the magnetic flux across an area of a square centi- meters at right angles to a magnetic field of which the intensity is H gausses. The unit of magnetic flux is the flux across an area of one square centimeter (a = i) when the area is at right angles to a magnetic field of which the intensity is one gauss {H = i), or it is the flux across n square centimeters of area when the area is at right angles to a mag- netic field of which the intensity is i/wth of a gauss. The unit of mag- netic flux is properly called the max- well; the common usage however is to call the unit of flux the line of force for the following reason : Consider any magnetic field whatever, and imagine a surface or shell BB which is everywhere at right angles to the field as shown in Fig. 2 1 . Imagine the surface BB to be divided up into Fig. 20. Fig. 21. MAGNETISM AND THE MAGNETIC EFFECT. 17 small squares such that one uijit of magnetic flux crosses each square, and imagine lines of force to be drawn so that one line of force passes through each square. Then the magnetic flux across any other surface or shell A A placed anywhere in the field will be equal to the number of these lines of force which cross A A. Proposition. — The magnetic flux $ which emanates from a north pole of strength M is $ = 47rJf (5) Imagine a sphere of radius r to be drawn with its center at the pole M in Fig. 19. The magnetic field due to M is every- where at right angles to this spherical surface and its intensity is ilff/r^ at the spherical surface. Therefore we must multiply the intensity Mjr^ by the area 4irr^ of the sphere to get the flux $, so that $ = 4irM'. In this argument we have neglected the small part of the spherical surface where the steel bar passes through it. An amount of flux * = /^tM emanates from a north pole or comes in to a south pole. 13. Tension of the magnetic field. — The force F in dynes with which the two broad flat magnet poles NN and 55 in Fig. 22 are drawn towards each other is s ^ = r.-^ (0 where s is the area in square centimeters of each pole face in 4 jL steel. steel N N steeL steel B ff Fig. 22. I8 LESSONS IN ELECTRICITY AND MAGNETISM. Fig. 22, and H is the intensity in gausses of the approximately uniform magnetic field in the air gap between the pole faces. The attraction of N N and SS in Fig. 22 shows thai the inter- 1 - — ^ > —~< ^ . *- X Ik \ "k ^ y — » — X < — (dge vieu/i Hat view Fig. 23. vening region is in a state of tension, and the value of this tension in dynes per square centimeter is equal to Ffs or to IP/Sir. As used by the electrical engineer equation (i) is ordinarily somewhat modified as follows : The total magnetic flux $ which crosses from pole face to pole face in Fig. 22 is ^ = sH according to equation (4) of Art. 12 and therefore H = ^/s. Substituting this value of H in equation (i), we get 1 $2 F{m dynes) = — • — OTT 5 (ii) The straightforward proof of equation (i) as given by Maxwell is far beyond the undergraduate, and the following derivation of equation (i) is given chiefly to serve as an example of the sim- plification of the mathematics of a problem by considering ideal- ized conditions, and very certainly the two broad flat magnet poles in Fig. 24 are idealized. MAGNETISM AND THE MAGNETIC EFFECT. 19 Figure 23 shows one end of a very broad flat piece of thin steel magnetized so as to have a north pole spread uniformly over a large portion of one end. Let m be the total strength of this pole. Then 47r»» is the total amount of flux emanating from the pole and this flux emanates as a uniform magnetic field on both sides of the flat pole as indicated in the edge view in Fig. 23. Let Hi be the intensity of this field on either side, then His is the magnetic flux emanating on either side, and 2H1S is the total magnetic flux emanating from the pole. Therefore 2H1S = ^wm so that Hi = (m) Figure 24 shows two poles like Fig. 23 placed side by side, a north pole and a south pole; each pole m being in the field of strength which is due to the other, s and therefore each pole is acted upon by a force m X • other pole according to equation (2). That is to say, the two poles in Fig. 24 are drawn towards each other by a force drawing it towards the 2'jrm^ s (iv) 1st In the region beween the poles in Fig. 24 the field J)f-pqle intensities due to the separate or individual poles are in the sa?ne direction so that the actual intensity jg of the field between the two poles is H — 2H1 so that 4irm ^ ^ ]S H =■ (v) S-pol^ But in each of the regions RR and R'R' the field intensities due to the separate or individual poles are opposite in direction so that no field at all exists in these regions The only magnetic field in Fig. ^ '7' 24 is the field between the poles, and the force in *«■ D equation (iv) is due to the tension of this field. Sub- Fig. 24. stituting the value of m from (v) in (iv) we get F = — .W. 8x Remark. — Equations (i) and (ii) are strictly applicable to Fig. 24, but they give a force which is a little in excess of the true force which draws the two pieces of steel together in Fig. 22, because the entire tension of the field in Fig. 22 does not act on the steel bars. PROBLEMS. 14. The pole face of the field magnet of a dynamo has an area 20 centimeters by 30 centimeters. The magnetic field between the pole faces and the armature core is perpendicular to the pole face at each point and its intensity is .6,000 gausses. Calculate 20 LESSONS IN ELECTRICITY AND MAGNETISM. the number of lines of force which pass from the pole face into the armature core. 15. A steel bar magnet 2 square centimeters in sectional area has poles at its ends and the strength of the poles is + 1500 units and — 1500 units. The bar is cut in two at the middle (before being magnetized, of course) and the two halves are accurately faced so that they fit and thus give practically a continuous bar. How much force would be required to pull the two halves apart? 14. Oersted's experiment. The right-handed screw con- vention as to the direction of current. — Figure 25a represents a top view of a magnetic compass with a wire beneath. When no current flows through the wire the compass needle stands parallel to the wire, let us say; then when the wire is connected to a battery the compass needle turns to the indicated position (or it may turn in the opposite direction). This effect was dis- covered by the Danish physicist Oersted in 1819, the first dis- covery relating to the magnetic effect of the electric current. Fixing the attention on the north pole of the compass needle in Fig. 25a, it may be stated that the north pole starts to circle round the wire (which it cannot actually do because of the sup- wire north end of needle top view Fig. 2So. Fig. 256 MAGNETISM AND THE MAGNETIC EFFECT. 21 porting pivot and because of the force exerted on the south pole of the needle), and the direction of the current in the wire is thought of as the direction a nut on a right-handed screw would travel if turned in the direction in which the north pole of the compass needle starts to circle round the wire. The exact mean- ing of this conventional rule may be understood by comparing Figs. 25c and 2^b. When an iron rod is magnetized by the flow of current round it, the north pole of the rod is at the end towards which a nut would travel (on a right-handed screw) if the nut were turned in the direction in which the current flows round the rod as shown in Fig. 26. When it is desired to , , . r ^direction of now show an end-view 01 a ^ of current wire through which ^ , f ^^^^ ■]/ - % », #. * . 8-pole ( I-tt: _, ."^ - Y— iZ 1 N-poUi current is flowing, the section of the wire is represented by a small circle, current flowing towards the reader is represented by a dot in the circle, as if one were looking endwise at the point of an arrow ; and current flowing away from the reader is represented by a cross in the circle, as if one were looking at the feathered end of an arrow, as shown in Fig. 27. © Fig. 27. Current flowing away from reader. Current flowing towards reader. 15. Another aspect of the magnetic effect of the electric current. Side push of the magnetic field on an electric wire. — One aspect of the magnetic effect of the electric current is de- scribed in Art. i ; another aspect of this effect is indicated in Fig. 25a; and still another aspect of the effect is shown in Fig. 28. A wire AB through which an electric current is flowing is stretched across the end of a magnet, and the wire is pushed sidewise by the magnet as stated in the legend under the figure. 22 LESSONS IN ELECTRICITY AND MAGNETISM. If the current is reversed or if the magnet is turned end for end the side push on the wire is reversed.* The side force on the wire in Fig. 28 is exerted by the magnet, and this force is no doubt transmitted by something which con- nects the magnet and the wire together, namely, the magnetic 7 /A / , • ^ =^-"" N -^ ^ ^ "^^s. \ g. 28. /■ ■ s \ F y/ The wire AB is pushed away from the reader. lines of force which emanate from the magnet. These magnetic lines of force are indicated by the dotted lines in Fig. 28. Figure 29 shows a straight wire AB placed in a narrow air gap between two opposite magnet poles. The fine lines across the gap represent the magnetic lines of force in the air gap, and these lines of force push the wire sidewise (away from the reader in Fig. 29). When an electric wire is placed in a magnetic field at right angles to the lines of force of the field, a force is p. 29. exerted on the wire {a side The wire AB is pushed away from the M«^ On the wire) at right reader. angles to the lines of force and at right angles to the wire. * Let it be clearly understood that the wire in Fig. 28 is neither attracted nor repelled by the magnet. MAGNETISM AND THE MAGNETIC EFFECT. 23 16. The magnetic field surrounding a straight electric wire. — Figure 30 is a photograph of the filaments of iron filings on a horizontal glass plate, the black circle is a hole through the plate, and a straight electric wire passes vertically through this hole. The lines of force of the mag- netic field which is produced by an electric wire encircle the wire, as shown by the fila- ments of iron filings in Fig. !'',!;/■ 30. 17. Explanation of the side push exerted upon an electric wire by a magnetic Figure 15 represei magnetic lines of force tween two opposite magnetic poles, and the attraction of the two opposite poles for each other may be thought of as due to a state of tension in the lines of force. That is, the lines of force may be thought of as stretched, rubber-like filaments leading from pole to pole in Fig. 15, and the attraction of the two opposite Fig. 30. '^y^^J^ ^: Fig. 31. 24 LESSONS IN ELECTRICITY AND MAGNETISM. poles may be thought of as the tendency of these stretched filaments to shorten. Figure 31 shows how the magnetic field between the two oppo- site poles in Fig. 15 is modified by the presence of an electric wire. The glass plate upon which the filings were dusted in Fig. 31 is horizontal, and the black circle represents a hole in the plate through which the vertical electric wire was placed. The lines of force from pole to pole pass mostly to one side of the wire in Fig. 31, and the wire is pushed sidewise by the tension of the lines for force (tendency of the lines of force to shorten) . Rule for determining direction of side push. — ^A wire is at right angles to a magnetic field. The direction of the field* is the direction of the force which would be exerted by the field on a north-pointing magnet pole. When current is started in the wire additional field is produced which circles round the wire in the direction in which a right-handed screw would have to be turned to travel in the direction of the cur- rent. These two fields are in the same direction on one side a of the wire and in opposite directions on the other side b, and the side push on the wire is from a towards h. This rule once it is clearly under- stood is very much better than the well known rule relating to thumb and two fingers of the right hand because the right-handed screw rule is consistently appli- cable to every aspect of the matter under consideration. 18. The direct-current ammeter. — The side push of the mag- netic field on an electric wire as shown in Figs. 28 and 29, and as explained in Art. 17, is made use of in the direct-current am- meter, the essential features of which are shown in Figs. 32 and * This refers, of course, to the field which exists independently of the current in the wire, the field which exists before the current starts to fiow in the wire. Fig. 32. MAGNETISM AND THE MAGNETIC EFFECT. 25 33. The vertical portions, Or limbs, of the pivoted coil play in a narrow air space or gap space between a fixed cylinder of soft iron and the soft iron pole pieces NN and 55 of a steel horse-shoe Fig. 33. magnet. Current is led into and out of the pivoted coil by means of two hair-springs, one at each end of the pivot-axis, and the side push of the magnetic field on the limbs of the coil in the gap spaces turns the coil and moves the pointer over the scale. 19. The magnetic blow-out. — The side push of the magnetic field upon the carrier of an elec- tric current, as shown in Figs. 28 and 29, and as explained in Aft 17, is made use of in the magnetic blow-out. When a electric switch is opened the current continues for a short time to flow across the opening, ^'s- 3i. forming what is called an electric arc, as shown in Fig. 34. This arc melts the contact parts of the switch, and the switch is soon ^witchhldde switch socket. 26 LESSONS IN ELECTRICITY AND MAGNETISM. end 'of ' I- \magitet/ -switch blad^^ -arc -switchsocke^ L3 Fig. 35. The arc is pushed towards or away from the reader. spoiled. This difficulty may be obviated to some extent by always opening the switch quickly and unhesitatingly, but where a switch is to be opened and closed hundreds of times per day, as in the control of a street car motor, it is necessary, to blow out the arc so as to avoid the rapid wear of the switch contacts by fusion. This blowing out of the arc is accomplished by a mag- net placed as shown in Fig. 35. This magnet pushes sidewise on the arc (towards or away from the reader in Fig. 35), and this sidewise push on the arc lengthens it very quickly and breaks the circuit. 20. The electric motor (direct-current type). — The side push of a magnetic field on an electric wire, as shown in Figs. 28 and 29 and as explained in Art. 17, is made use of in the electric motor. The following discussion is intended to explain only the essential principles and to describe only the essential features of the direct-current motor, the actual details of design and construction may be seen by inspecting a com- mercial motor. Figure 36 shows an iron cylinder A A placed between the poles N and 5 of a powerful electro- magnet, the field magnet, as it is called. The air space or gap space between the cylinder A A and each pole face is an intense magnetic field as indicated by the fine lines (lines of force). Fig. 36. MAGNETISM AND THE MAGNETIC EFFECT. 27 Figure 37 shows the cyUnder A A with straight wires laid upon its surface (wires parallel to the axis of the cylinder A A), and the dots and crosses represent electric currents flowing towards the reader and away from the reader, respectively, as as explained in Fig. 27. If we can arrange for the steady flow of electric current as indicated by the dots and crosses in Fig. 37 then Fig. 37. the magnetic field in the gap spaces (the fine lines of force in Fig. 36) will push sidewise on the wires and rotate the armature in the direction, of the curved arrows in Fig. 2>7, and the arrangement whereby the desired flow of current can be realized is most easily imderstood by con- sidering the simplest tf^e of armattxre winding, namely the RING WINDING. An iron ring A A, Fig. 38, is wound uniformly with insulated wire as shown, the ends of the wire being spliced together and sold- ered so as to make the winding endless. Imagine the insulation to he removed from the outward faces of the wire wind- ings on the ring so that two stationary metal or carbon blocks {brushes) a and b can make good electrical contact with the wire winding as Fig. 38. 28 LESSONS IN ELECTRICITY AND MAGNETISM. the ring rotates. Then, if current is led into the winding through brush a and out of the winding through brush b, the current will flow towards the reader in all the wires under the S pole and away from the reader in all the wires under the N pole as indicated by the dots and crosses in Fig. 37. In practice, short lengths of wire are attached to the various turns of wire on the ring and led to copper bars near the axis of rotation, as shown in Fig. 39. These copper bars are insulated from each other, and sliding contact is made with these copper Fig. 39. bars as indicated in Fig. 39, instead of being made as indicated in Fig. 38. The set of insulated copper bars is called the com- mutator. The iron body of the armature (the iron ring in Figs. 38 and 39) is called the armature core. This core is built up of ring-shaped stampings of thin sheet iron. The machine which is here described as the direct-current motor is properly called the direct-current dynamo. It is a motor when it receives electric current from some outside source and is used to drive a pump, or a lathe, or a trolly car. Exactly the same machine, when driven by a steam engine or water MAGNETISM AND THE MAGNETIC EFFECT. 29 wheel, can be used as an electric generator to supply direct cur- rent for driving motors or for operating electric lamps. When so used the machine is called a dynamo electric generator or simply a generator. Remark. — Inasmuch as we are here concerned only with fundamental principles it is not necessary to refer to the mul- tipolar dynamo or to the type of armature winding which is called the drum winding, although most direct current dynamos are nowadays of the multipolar type and have drum windings on their armatures. 21. Strength of electric current magnetically defined. — Consider a straight electric wire stretched across a uniform mag- netic field, the wire being at right angles to the field as shown in Fig, 40. Let us suppose, for a moment, that the field is of 1 1 icire— y N f direction of ^current , s 40. towards reader. Fig Wire is pushed unit intensity. The force in dynes with which this unit field pushes sidewise on one centimeter of the electric wire has been adopted as the fundamental measure of the strength of the cur- rent in the wire. This force-per-unit-length-of-wire-per-unit- field -intensity is called, simply, the strength of the current in the wire, and it is represented by the letter I. The force pushing sidewise on / centimeters of the wire is II dynes ; and if the field intensity is H gausses instead of one gauss, then the force is B. times as great, or IIU dynes. That is F = II H (6) 30 LESSONS IN ELECTRICITY AND MAGNETISM. in which F is the force in dynes pushing sidewise on / centimeters of wire at right angles to a uniform magnetic field of which the intensity is H gausses, and / the wire. is the strength of the current in Definition of the ampere. — The ampere is defined as one tenth of an abampere. Definition of the abampere. — ^A wire is said to carry a cur- rent of one abampere when one centimeter of the wire is pushed sidewise with a force of one dyne, when the wire is stretch- ed across a magnetic field of which the intensity is one gauss, the wire being at right angles to the field. The current / in equation (6) is expressed in abamperes when F is ex- pressed in dynes, I in centi- meters and H in gausses. The abampere is the c.g.s. unit of current. 22. Magnetic field due to a long straight wire carrying I abam- peres of current. — If we can find the force F exerted on the mag- net pole m by the wire in Fig. 41 we will know the intensity H irirel D\ mA. magnet Fig. 41. MAGNETISM AND THE MAGNETIC EFFECT. 31 of the magnetic field at m due to the wire because F = mH according to equation (2). To find the force F let us consider the equal and opposite force F' exerted on the wire by the pole. This force F' may- be calculated as follows. The field intensity at the element of wire Ax due to m is m/r^, the component of m/r^ which is at right angles to the wire is -^ sin 0, and the side force AF' (away from the reader in Fig. 41) exerted on Ax is according to equation (6), where {x^ + D^) is written for r^ and D/ yx? + D^ is written for sin 6. By integrating* equation (i) from :*; = — 00 to a: = + 00 , we get F' = 2ml/ D, whence we get mH = 2ml fD so that H = § (ii) where H is the intensity at m Fig. 41 of the magnetic field due to the indefinitely long wire, I is the current in the wire in abamperes and D is the distance of the point m from the axis of the wire in centimeters. 23. Magnetic field inside of a long uniformly wound coil. — A glass or fiber tube I centimeters long is wound with Z turns of insulated wire and a current of T abamperes flows through the wire. The region inside of the long tube is a uniform mag- netic field, and its intensity H in gausses is H = 4^j-I (7) This equation may be easily established for points along the axis of the coil by straight forward methodsf but the following * This integration involves too mucli algebraic manipulation to be fully ex- plained here. t See Franklin and MacNutt's Advanced Electricity and Magnetism, pages 23-25; published by Franklin and Charles, Bethlehem, Pa. 32 LESSONS IN ELECTRICITY AND MAGNETISM. simple derivation is more intelligible and it furnishes another good example of the simplification of mathematics by the use of an idealized arrangement. Figure 42 represents a long portion (/ centimeters long) of a very slender steel rod which is supposed to be so magnetized that a north pole of total strength m is uniformly spread over the portion. The lines of force of the magnetic field radiate cylindrically from such a pole, and it is easy to show from # = I centimeters steel I irffllflfflt rod< n mMmi ji ,-i aidejeiem .end view Fig. 42. 4Tm that the intensity of this cylindrically radiating field is H = 2m/lr at a point distant r centimeters from the axis of the rod. Imagine the linear "pole" above described to be placed axially in our long coil of wire. The cylindrically radiating magnetic field will engage, as it were, all of the turns in I centimeters of the coil (or Z turns), and, using r for the radius of the turns of wire, it is evident that a total length 27rrZ of wire will be pushed sidewise (end wise with respect to the coil) by the cylindrically radiating magnetic field. Therefore, using equation (6), con- sidering that the force with which the linear pole acts on the wire is equal and opposite to the force with which the coil acts on the pole, and remembering that the force acting on the pole is equal to mH where H is the field due to the coil, we get equation (7). PROBLEMS. 16. Figure 430 shows a magnet placed near a long straight wire carrying current away from the reader. Draw an arrow ■MAGNETISM AND THE MAGNETIC EFFECT. 33 showing force exerted by the wire on each pole of the magnet, and draw a diagram showing the resultant single force exerted on the magnet by the wire. 17. Figure 436 shows a magnet near a long straight wire m ®iwVe I I I I I ! I Fig. 43a. '®wii^ UN. 1, Fig. 436. carrying current towards the reader. The distance of the iV-pole from *the wire is 10 centimeters and the length 'NS is 20 centi- meters. The wire exerts a force of 900 dynes on the iV-pole. Find value and direction of the force exerted by the wire on the 5-pole. What is the value and direction of the force exerted on the wire by the magnet ? Note. The force exerted on either magnet pole by the long straight wire is equal and opposite to the force exerted on the wire by the pole, but the two forces do not have the same line of action I It would seem therefore that the principle of equality of action and reaction (which is not only that action and reaction are equal and opposite but that action and reaction have the same line of action) would not apply to Fig. 436, but if one considers the entire circuit of wire, and no less than an entire circuit ever exists magnetically, this difficulty vanishes. 18. Specify the direction of the side push exerted on the wire ©wire ® hire Fig. 44. Fig. 45. by the magnet pole in Fig. 44. Specify the direction of the side push on the wire in Fig. 45. Note. — The direction of the field due to 5 in Fig. 44 is towards S; the field due to the wire circles round the wire in the direction in which a right-handed screw would have to be turned to travel in the direction of the current; and the wire is pushed away from the side on which these two fields are in the same direc- 34 LESSONS IN ELECTRICITY AND MAGNETISM/ tion. This rule may be understood by a careful study of Figs. 250, 256 and 31. Most mechanically minded men prefer the right-handed screw convention to the rule which makes use of thumb and first two fingers of the right hand. 19. The armature of a direct-current dynamo has a length (under the pole faces) of 30 centimeters, and 250 of the armature wires are under the pole faces at all times. The radius of the armature (measured out to the wires) is 20 centimeters. The magnetic field in the air gaps between pole faces and armature core has an intensity of 6,000 gausses. Each armature wire carries a current of 75 amperes (see Fig. 37). Calculate the side force acting on each wire and calculate the total torque acting on the armature in dyne-centimeters. 20. A thin wooden block 25 X 40 cm. has 10 turns of wire wound around its edge, and the block is balanced in a horizontal position on an east-west axis with its 40 cm. edges north and south. A current of 28 amperes is sent through the winding (using very flexible leads). Find the distance from the balance axis at which a 9800-dyne weight would have to be hung to balance the torque action of the earth's magnetic field. Hori- zontal component of earth's field being 0.21 gauss, and vertical component 0.48 gauss. 21. Two straight parallel wires are 20 centimeters apart. One wire carries a current of 100 amperes and the other carries a current of 250 amperes (both currents in same direction). Do the wires attract or repel each other? How much force is exerted on 100 centimeters of either wire by the other? 22. Imagine an isolated pole of strength m to be placed at the center of a circle of wire in which a current of I abamperes of current is flowing, the radius of the circle being r. Write down an expression for the intensity at the wire of the magnetic field due to the pole, then derive an expression for the force exerted on the wire by the pole and specify the direction of this force, then get an expression for the force exerted on the pole by the wire, and then derive an expression for the intensity at the pole of the magnetic field due to the wire. CHAPTER II. CHEMICAL EFFECT OF THE ELECTRIC CURRENT. 24. The chemical effect of the electric current again con- sidered. — ^A very beautiful experiment showing the deposition of a metal by the electric current is as follows : Two strips of lead are connected to seven or eight dry cells (in series) and dipped into a solution of lead nitrate,* as shown in Fig. 46. The flow of ielre talKoBe'- ■— l^^^-anc; itermihat 4wrc _j'^air6oi» termihat ^ w ■anode —i—i-lead nitrate solution Fig. 46. current decomposes the lead nitrate and deposits beautiful feather-like crystals of metallic lead on the lead strip which is connected to the zinc terminal of the battery. This decomposition of a solution by an electric current is called electrolysis, and the solution which is decomposed is called an electrolyte. Electrolysis is usually carried out in a vessel provided with two plates of metal or carbon. Such an arrange- ment is called an electrolytic cell, and the plates of metal or carbon are called the electrodes. Thus the vessel which contains the lead nitrate solution and the two lead strips in Fig. 46 is an * Ordinary sugar of lead (lead acetate), which can be obtained at any drug store, may be used instead of lead nitrate. 35 36 LESSONS IN ELECTRICITY AND MAGNETISM. electrolytic cell. The electrode at whicn the current enters the solution is called the anode, and the electrode at which the current leaves the solution is called the cathode* The chemical action which is produced by the electric current in the electrolytic cell of Fig. 46 is as follows: The lead nitrate (PbNOs) is separated into two parts, namely, Pb (lead) and NO3 (nitric acid radical) . The lead (Pb) is deposited on the cathode, and the nitric acid radical (NO3) is set free at the anode where it combines with the lead of the anode, forming a fresh supply of lead nitrate which is immediately dissolved in the electrolyte. That is to say, lead is deposited on the cathode and dissolved off the anode. The chemical action produced by the electric current in an electrolytic cell takes place only in the immediate neighborhood of the electrodes. The international standard ampere. — Very careful measure- ments have shown that one ampere deposits 0.001 118 gram of silver per second from a solution of silver nitrate in water; and, inasmuch as it is very difficult to measure a current accurately in terms of its magnetic effect so as to get the value of the current directly in amperes, the ampere has been legal- ly defined as the current which will deposit exactly cathode— -anode —dilute ■ sulphuric acid' Fig. 47. 0.001118 gram of silver in one second from a solution of pure silver nitrate in water. 25. Another aspect of the chemical effect of the electric current. The voltaic cell or electric battery.— When electric * The use of the terms positive and negative is extremely confusing because the positive terminal of a generator (a battery or dynamo) is the terminal out of which current flows, whereas the positive terminal of any receiver is understood to be the terminal at which current enters or flows into the receiver. CHEMICAL EFFECT OF THE ELECTRIC CURRENT. 37 current flows through an electrolytic cell chemical action is produced. For example, Fig. 47 shows a battery forcing current through dilute sulphuric acid, the electrodes being plates of carbon or lead or platinum. The sulphuric acid (H2SO4) is decomposed by the current, being separated into H2 (hydrogen) and SO4 (sulphuric acid radical). The hydrogen appears at the cathode as bubbles of gas and escapes from the cell. The acid radical (SO4) appears at the anode where it breaks up into SO3 and O (oxygen) . The oxygen appears in the form of bubbles and escapes from the cell, and the SO3 combines with the water (H2O) in the cell, forming H2SO4. The net result of the chem- ical action in the cell is therefore to decompose water (H2O) inasmuch as hydrogen gas and oxygen gas are given off by the cell. Now by burning the hydrogen and oxygen heat energy can be obtained, and therefore it is evident that work must be done {hy the battery in Fig. 47) to decompose the H2O in the electrolytic cell. Usually the chemical action which is produced by the current in an electrolytic cell requires the doing of work as above explained, that is to say, an electric generator (battery or dynamo) must be used to force the electric current through the electrolytic cell. In some cases, however, the chemical action which is produced by the flow of ciurent through the electrolytic cell is A SOURCE OF ENERGY. In such a case it is not necessary to use a separate electric generator (battery or dynamo) to force electric current through the electrolytic cell, for such an electro- lytic cell can maintain its own current through the electrolyte from electrode to electrode and through an outside circuit of wire which connects the electrodes. Such an electrolytic cell is called a voltaic cell or an electric battery. That is to say, a voltaic cell is an electrolytic cell in which the chemical action produced by the flow of current is a source of energy. 26. The simple voltaic cell. — The simplest example of an electrolytic cell in which the chemical action produced by the current is a source of energy, is the so-called simple voltaic cell 38 LESSONS IN ELECTRICITY AND MAGNETISM. wtf» which is shown in Fig. 48. It consists of a, carbon or copper electrode C and a clean zinc electrode Z in dilute sulphuric acid. The flow of current through this cell breaks up the H2SO4 into two parts, namely, H2 (hydrogen) and SO4 (sulphuric acid radical). The hydrogen appears at the carbon electrode and escapes as a gas, and the SO4 appears at the zinc electrode where it combines with the zinc to form zinc sul- phate (ZnS04). The combination of the zinc and the SO4 supplies more energy than is required to separate the H2 and SO4, that is to say, the chemical action which is pro- duced by the flow of current through the cell is a source of energy, and the cell '^' ■ itself maintains a flow of current. 27. Voltaic action and local action. — ^Two distinct kinds of chemical action take place in a voltaic cell, namely, (a) the chemical action which depends upon the flow of current and which does not exist when there is no flow of current, and (b) the chemical action which is independent of the flow of current and which takes place whether the current is flowing or not. The chemical action which depends on the flow of current is proportional to the current, that is to say, this chemical action takes place twice as fast if the current which is delivered by the voltaic cell is doubled. This chemical action is essential to the operation of the voltaic cell as a generator of current, its energy is available* for the maintenance of the current which is produced by the cell, and it is called voltaic action. The chemical action which is independent of the flow of current does not help in any way to maintain the current ; it represents a waste of materials, and it is called local action. Local action takes place more or less in every type of voltaic cell. It may be greatly reduced in amount, however, by using pure zinc, and especially by coating the zinc with a thin layer of metallic mercury (amalgamation). * In general not wholly available. CHEMI|AL EFFECT OF THE ELECTRIC CURRENT. 39 Example of local action. — The zinc plate in the simple voltaic cell which is shown in Fig. 48 dissolves in the sulphuric acid even when no current is flowing through the cell, zinc sulphate and hydrogen are formed, and all of the energy of this reaction goes to heat the cell. If the zinc is very pure and if its surface is clean this chemical action takes place very slowly, but if the zinc is impure the action is usually very rapid. The hydrogen which is liberated during this local action appears at the zinc plate. Example of voltaic action. — When the circuit in Fig. 48 is closed, hydrogen bubbles begin to come off the carbon electrode, and zinc sulphate is formed at the zinc electrode. This is voltaic action, and it ceases when the circuit is broken. The essential and important feature of voltaic action is that it is reversed if a current from an outside source is forced back- wards through the voltaic cell, provided no material which has played a part in the previous voltaic action has been allowed to escape from the cell. Thus in the simple voltaic cell, which is described in Art. 26, the sulphuric acid (H2SO4) is decomposed, zinc sulphate (ZnS04) is formed at the zinc electrode, and hydro- gen is liberated at the carbon electrode. If a reversed current is forced through this simple cell, the zinc sulphate previously formed will be decomposed, metallic zinc will be deposited upon the zinc plate, and the sulphuric acid radical (SO4) will be liber- ated at the carbon plate, where it will combine with the trace of hydrogen which is clinging to the carbon plate and form sulphuric acid (H2SO4). In this simple cell, however, the greater part of the liberated hydrogen has, of course, escaped, and the reversed chemical action due to a reversed current cannot long continue. Local action, being independent of the current, is not affected by a reversal of the current. 28. Primary and secondary chemical reactions in the electro- lytic cell. — The decomposition of the electrolyte is the direct or immediate or primary effect of the flow of current, therefore, the decomposition of the electrolyte may be spoken of as the primary chemical action in a electrolytic cell. 40 LESSONS IN ELECTRICITY AND MAGNE||SM When the decomposed parts of the electrolyte rppeav at the electrodes, chemical action usually takes place between these parts and the electrodes or between these parts and the solution, and these chemical actions are called the secondary chemical reactions in the electrolytic cell. For example, the primary chemical reaction in the simple voltaic cell which is shown in Fig. 48 is the decomposition of the sulphuric acid into hydrogen (H2) and sulphuric acid radical (SO4); and the combination of ' the acid radical (SO4) with the zinc of the electrode is a secondary reaction. The primary chemical action in an electrolytic cell usually represents the doing of work ON the cell, and the secondary chemical reactions in an electrolytic cell usually represent the doing of work BY the cell ; therefore secondary reactions are very important in the voltaic cell or electric battery. 29. The use of oxidizing agents in the voltaic cell. — The com- bination of the SO4 with the zinc of the anode is the secondary chemical action in the simple voltaic cell which is shown in Fig. 48. The available energy of the total chemical action which takes place in this cell may be greatly increased, however, by providing an oxidizing agent in the neighborhood of the carbon electrode so that the hydrogen may be oxidized and form water (H2O) at the moment of its liberation by the current. The energy of this oxidation increases the available energy of the chemical action as a whole, and greatly strengthens the cell as a generator of electric current. 30. The chromic acid cell. — ^The Grenet cell is similar to the simple voltaic cell, as shown in Fig. 48, except that the electrode C is of carbon, and chromic acid is added to the electrolyte to furnish oxygen for the oxidation of the hydrogen as it is set free at the carbon electrode. There is, however, a very rapid waste of zinc in this cell by local action even when the zinc is amalga- mated, and the cell is now seldom used. A modified form of the Grenet cell, known as the Fuller cell, is shown in Fig. 49. In this cell the electrolyte e is dilute sulphuric acid, the zinc anode CHEMICAL EFFECT OF THE ELECTRIC CURRENT. 41 wire- Fig. 49. The Fuller cell. Z is contained in a porous earthenware cup, and the chromic acid is dissolved only in that portion of the electrolyte which surrounds the carbon cathode C. In this cell there is not a rapid waste of zinc by local action. 31. Open-circuit cells and clos- ed-circuit cells. — A voltaic cell which can be left standing unused, but in readiness at any time for the delivery of current when its circuit is closed, is called an open- circuit cell. A cell to be suitable for use as an open-circuit cell should above all things be nearly free from local action. The cell most extensively used for open- circuit service is the ordinary dry cell. A voltaic cell which is suitable for delivering a current steadily is called a closed-circuit cell. The gravity Daniell cell which is described in Art. 33 was formerly much used as a closed-circuit cell, but the storage bat- tery is now displacing all other types of vol- taic cells for the delivery of fairly large or steady current. 32. The ordinary dry cell. — A sectional view of this cell is shown in Fig. 50. The containing vessel is a can made of sheet zinc. This can serves as one electrode of the cell, and a binding post is soldered to it. The zinc can is lined with several thicknesses of blotting paper P, and the space between the blotting paper and the carbon rod C is packed with bits of coke and manganese dioxide. The porous contents of the cell are then saturated with a solution of am- Fig. 50. The dry cell. 42 LESSONS IN ELECTRICITY AND MAGNETSISM. monium chloride (sal ammoniac), and the cell is sealed with as- phaltum cement A. The zinc can is usually protected by a covering of pasteboard B. The dry cell has been humorously defined as a voltaic cell which, being hermetically sealed, is always wet; whereas the old-style wet cell was open to the air and frequently became dry. Reputable manufacturers always stamp the date of manu- facture on their dry cells, and a purchaser should not accept a cell which is much more than one or two months old. The condition of a dry cell is most satisfactorily indicated by observing the current delivered when the cell is momentarily short circuited* through an ammeter. When the cell has been exhausted by use or when it has dried out by being kept too long, the short-circuit current is greatly reduced in value. An ordinary dry cell, when fresh, should give about 25 or 30 amperes on a momentary short circuit when the cell is at ordinary room temperature. Fig.TiT 33- The gravity Daniell cell.— This cell The gravity cell. Consists of a copper cathode at the bottom of a glass jar and a zinc anode at the top as shown in Fig. 51. The electrolyte is mainly a solution of zinc sulphate. Crystals of copper sulphate are dropped to the bottom of the cell and a dense solution of copper sulphate sur- rounds the copper cathode. This cell has a considerable amount of local action when it is allowed to stand unused, because of the upward diffusion of the copper sulphate. The cell is used in telegraphy and for operating the "track circuit" relays in automatic railway signalling. PROBLEMS. 23. The anode of an electrolytic cell is a copper rod one inch in diameter, and the cathode is a hollow copper cylinder 6 inches * That is to say, the very low resistance ammeter is connected directly to the terminals of the cell. CHEMICAL EFFECT OF THE ELECTRIC CURRENT. 43 in diameter. The two electrodes are co-axial and they stand on a flat glass plate in an electrolyte which is 8 inches deep over the glass plate. A current of 5 amperes flows through the cell. Find the current density at the cathode and the current density at the anode. Note. — The current density on an electrode is the current per unit of area, and it is generally non-uniform. In the conditions specified in the problem, however, the current density is uniform over the surface of each electrode. Current density at an electrode is important inasmuch as the character of the chemical action at an electrode depends in many cases on current density, and the physical character of a deposited metal depends on the current density. 24. How long a time in seconds would be required for one ampere to deposit one gram-equivalent (107.93 grams) of silver from a solution of pure silver nitrate? 25. A current of / amperes will deposit 0.000338 X / grams of zinc per second from a solution of zinc sulphate, or cause the consumption of 0.000338 X / grams of zinc per second in a grav- ity Daniell cell. The chemical action which takes place in the gravity Daniell cell is identical to what takes place as "local action" when zinc filings are stirred winto a solution of copper sulphate, and experiment shows that 753 calories or 3,160 joules of heat is developed when one gram of zinc filings is stirred into a copper sulphate solution. Assuming that all of the energy of the chemical action in the Daniell cell is available for maintaining current, find joules-per-second-per-ampere available for main- taining current. Note. — ^Joulea-per-second-per-ampere or watts-per-ampere is, of course, a quan- tity which gives watts when multiplied by amperes, and joules-per-second-per- ampere is called electromotive force. The calculated answer to this problem is the electromotive force of the gravity Daniell cell in volts. One volt is one joule-per- oecond-per-ampere. 26. The electromotive force of the chromic acid cell is about 2.25 volts, that is to say, the cell does 2.25 joules of work per second per ampere of delivered current. Assuming that all of the energy of the chemical action in the cell (voltaic action) is represented by the work the cell does in maintaining current, calculate the rise of temperature produced when one gram of 44 LESSONS IN ELECTRICITY AND MAGNETISM. zinc filings is stirred into 500 grams of dilute sulphuric and chro- mic acids, mixed. Take the specific heat capacity of this dilute acid to be the same as water. Note. — Zinc consumed by voltaic action in any form of voltaic cell is 0.000338 gram per second per ampere. 27. A Grenet form of chromic acid cell delivers an average current of 5 amperes for 23 hours. At the beginning the mass of the zinc plate was 150 grams and at the end of the run the mass of the zinc plate was 75 grams. What percentage of the total zinc consumption is due to voltaic action and what per- centage is due to local action? 34. The storage cell. — ^A voltaic cell may be completely re- paired after use by forcing a current backwards through the cell, if there is no local action in the cell, and if all of the materials which take part in the voltaic action remain in the cell. A voltaic cell which meets these two conditions is called a storage cell. The process of repairing the cell by forcing a current through it backwards is called charging, and the use of the cell for the delivery of current is called discharging. The lead storage cell. — The voltaic cell which is most exten- sively used as a storage cell is one in which one electrode is lead peroxide (Pb02), the other electrode is spongy metallic lead (Pb), and the electrolyte is dilute sulphuric acid (H2SO4). This cell is called the lead storage cell. The lead peroxide and the spongy metallic lead are called the active materials of the cell. These active materials are porous and brittle, and they are usually supported in small grooves or pockets in heavy plates or grids of metallic lead. These lead grids serve not only as mechanical supports for the active materials, but they serve also to deliver current to or receive current from the active materials which constitute the real electrodes of the cell. As a lead storage cell is discharged, the active material on both electrodes is reduced to lead sulphate PbSON BaSOt lolution apbtigg lead anode 36 303 98 239 2 2H»0+PbS0t= BtSOt +Pb0^2B* grama' 96 207 303 ampere-houra * '"w* cubic feet (5.074 grams) of hydrogen and one cubic foot of oxygen in an electrolytic cell like Fig. 47. 30. Consider the hydrogen and oxygen which is liberated by 10 amperes in 26.8 hours in an electrolytic cell like Fig. 47, namely, i gram of hydrogen and 8 grams of oxygen. If we should burn this hydrogen and oxygen we would get 34,700 calories or about 146,000 joules of heat. Assuming that the work done in forcing the 10 amperes through the electrolytic cell is all represented by this heat of combustion, calculate the joules- per-second-per-ampere required to force the 10 amperes through the electrolytic cell. Note. — The joules- per-second-per-ampere as here calculated is the electromotive force in volts that would be required to push current through the electroljrtic cell shown in Fig. 47 under the assumption that the work done is equal to the heat- energy that may be obtained by re-combining the hydrogen and oxygen. As a matter of fact an electromotive force of about 2.0 or 2.1 joules-per-second-per- ampere (volts) is required. CHEMICAL EFFECT OF THE ELECTRIC CURRENT. 49 31. The cost of gravity-cell zinc is, say, 6 cents per pound, and the cost of copper sulphate crystals (CUSO4 + 5H2O) is, say, 6.5 cents per pound. Half of the materials consumed in a gravity cell is wasted by local action and about one third of the zinc is left as scrap and is worth about 2 cents per pound. Furthermore the copper which is deposited on the cathode is worth, as scrap about 10 cents per pound. The terminal electromotive force of the cell while it is delivering 0.16 ampere is about 0.72 volt.* What is the cost per kilowatt-hour of the output of the cell making no allowance for labor required to take care of the cell and making no allowance for interest on cost of glass jar. Ans. $1.73. 32. A lead storage cell delivers 10 amperes for 8 hours. Find increase in weight of each electrode. * About 0.72 joule per ampere per second. CHAPTER III. THE HEATING EFFECT OF THE ELECTRIC CURRENT. 36. Heating effect of the electric current again considered. Joule's law. — The heating effect of the electric current is men- tioned in Art. i. The lamp filament in Fig. 2 is heated to a high temperature by the current. Careful observation shows that every portion of an electric circuit is heated move or less by the electric current. Thus the connecting wires in Fig. 2 are heated to some extent. It is important to understand that the heating effect of the electric current is the generation of heat in each portion of a circuit at a definite rate, so many calories or joules per second ; and each portion of an electric circuit grows hotter and hotter until it gives off heat to its surroundings as fast as heat is generated in it by the current. A portion of an electric circuit is said to have a high resistance if a large amount of heat is generated in it per second by a given current, or a low resistance if a small amount of heat is generated in it per second by a given current. Thus a great deal more heat is generated per second in the lamp filament in Fig. 2 than in the connecting wires so that the resistance of the lamp filament is much greater than the resistance of the connecting wires. Joule's law. — An important discovery was made by James Prescott Joule about 1850 concerning the relation between the strength of a current in amperes and the rate at which heat is generated by the current. Joule found that the rate of generation of heat in a given piece of wire is proportional to the square of the current flowing through the wire (Joule's law). To say that the rate of generation of heat in a given wire (or other conductor) is proportional to the square of the current is the same thing as to say that the rate of generation of heat is 50 HEATING EFFECT OF THE ELECTRIC CURRENT. 51 egual to the square of the current multipHed by a constant factor, a factor which has a certain definite value for the given piece of wire. Let us represent this factor (for a given piece of wire) by the letter R. Then RI"^ is the rate of generation of heat in the wire by a current of I amperes, that is RI^ is the amount of heat generated in the wire per second, and RIH is the total amount of heat generated in the wire during t seconds. There- fore we may write : H = RIH (8) where H is the amount of heat generated in a wire during, t seconds by a current of I amperes, and i? is a factor which has a definite value for the given piece of wire. Example. — ^The amount of heat developed in a certain electric lamp in 10 minutes, as found by a calorimeter, is 7143 calories or 30,000 joules, when 0.51 ampere flows through the lamp. Therefore, putting H = 30,000 joules, / = 0.51 ampere, and t = 600 seconds in equation (8), we get R equal to 192.2 joules- per- (ampere) ^-per-second ; but one-joule-per-iampereY-per-second is called an ohm. Therefore the given electric lamp has a resis- tance of 192.2 ohm. Definition of the ohm. — A wire is said to have a resistance of one ohm when one joule of heat is generated in it per second by a current of one ampere. Definition of the abohm. — A wire is said to have a resistance of one abohm when one erg of heat is generated in it per second by a current of one abampere. One ohm is equal to 10' abohms. 37. Dependence of resistance upon the length and size of a wire. Definition of resistivity. — The resistance i? of a wire of given material is directly proportional to the length I of the wire, and inversely proportional to the sectional area j of the wire; that is R = k^- (9) in which fe is a constant for a given material, and it is called 52 LESSONS IN ELECTRICITY AND MAGNETISM. the resistivity* of the material. The exact meaning of the factor k may be made apparent by consiSering a wire of unit length (I = i) and of unit sectional area (a = i). In this case R is numerically equal to k, that is to say, the resistivity of a material is numerically equal to the resistance of a wire of that material of unit length and unit sectional area. Electrical engineers nearly always express lengths of wires in feet and sectional areas in circular mils.f If equation (9) is used TABLE. — Resistivities and Temperature Coefficients. P Aluminum wire (annealed) at 20° C. . ' Copper wire (annealed) at 20° C Iron wire (pure annealed) at 20° C Steel telegraph wire at 20° C Steel rails at 20° C Mercury at 0° C Platinum wire at 0° C German-silver wire at I0° C Manganin wire (Cu 84, Ni 12, Mn 4) at 20° C "la la" metal wire, hard (copper-nickel alloy) at 20° C "Climeix" or "Superior" metal (nickel-steel alloy) at 20° C Arc-lamp carbon at ordinary room temperature Sulphuric acid, 5 per cent, solution at 18° C . . Ordinary glass at 0° C. (density 2.54) Ordinary glass at 60° C Ordinary glass at 200° C 27.4 X10-' 17.24X10-' 95 X10-' ISO Xio~' 120 Xio"' 943.4 X10-' 89.8 X10-' 212 Xio~' 47S X10-' SOD X10-' 800 X10-' 0.005 4.8 ohms 10" ohmst io'2 ohmst 10^ ohmst 16.S 10.4 38.0 9it 72t S40 I27t 286 300t 48ot -t-0.0039 -I-0.0040 -1-0.0045 -|-o.oo43t H-o.o03St -I-0.00088 -fo.oo354 -t-0.O0O25t — o.ooooit -l-o.ooo67t — o.ooost —0.0120* a = resistance in ohms of a bar i centimeter long and i square centimeter sectional area. 6 = resistance in ohms of a wire i foot long and o.ooi inch in diameter. iS = temperature coefBcient of resistance per degree centigrade (mean value between 0° C. and 100° C). Near ordinary room temperature the resistance of an maganinn wire is very nearly independent of temperature. * Between 18° C. and 19° C. t These values differ greatly with different samples. to calculate the resistance of a wire in ohms when the length I of the wire is expressed in feet and the sectional area 5 in circular * Sometimes called specific resistance. The reciprocal of the resistivity of a substance is called its conductivity. t One mil is a thousandth of an inch. One circular mil is the area of a circle of which the diameter is one mil. The area of any circle in circular mils is equsd to the square of the diameter in mils. Thus a wire loo mils in diameter has a sec- tional area of 10,000 circular mils. HEATING EFFECT OF THE ELECTRIC CURRENT. 53 mils, then the value of k must be the resistance of a wire of the given material one foot long and one circular mil in sectional area. The accompanying table gives the resistivities of the more important substances together with their temperature coefficients of resistance. PROBLEMS. 33. An electric heater of the immersion type is placed in a vessel containing 5,000 grams of water, and the temperature of the water is raised from o°C. to 40° C. in five minutes with a current of 25.5 amperes flowing through the heater. What is the resistance of the heater in ohms? All the work done in pushing the current through the heater is represented by the heat that is developed; how many joules-per-second-per-ampere are used in pushing the current through the heater? Note. — One joule-per-second-per-ampere is called a volt. 34. The field coil of a dynamo contains 11,340 grams of copper (specific heat 0.094), weight of cotton insulation negligible. The resistance of the coil is 100 ohms. At what rate does the temperature of the coil begin to rise when a current of 0.5 ampere is started in the coil? Note. — ^At the start, when the coil is at the same temperature as the surrounding air, all of the heat generated in the coil is used to raise the temperature of the coil; later, when the coil has become warmer than the air, a portion of the heat generated in the coil is given off to the air. The problem is simply this: How fast would all the heat generated in the coil cause the temperature of the coil to rise? 35. A given piece of copper wire has a resistance of 5 ohms, another piece of copper wire is i .5 times as long but it has the same weight (and volume) as the first piece. What is its re- sistance? 36. A given spool wound full of copper wire 60 mils in diam- eter has a resistance of 3.2 ohms. An exactly similar spool is wound full of copper wire 120 mils in diameter; what is its re- sistance? Note. — The spool will contain half as many layers and half as many turns in each layer of the larger wire, and the mean length of one turn of wire is the same in each case. 54 LESSONS IN ELECTRICITY AND MAGNETISM. 37. What is the resistance at 20° C. of 2 miles of commercial copper wire 300 mils in diameter? 38. Find the resistance at 20° C. of a copper conductor 100 feet long having a rectangular section 0.5 inch by 0.25 inch. TT ( d y Note. — The area of a circle d mils in diameter is d* circular mils, or — I I 4 \ 1000/ square inches. Therefore the sectional area of a rectangular bar in square inches 4,000,000 , . , ., must be multiplied by to reduce to circular mils. TT 39. What is the resistance at 20° C. of a wrought iron pipe 20 feet long having one inch inside diameter and 1.25 inches outside diameter. Note. — Use resistivity of pure annealed iron. 40. Calculate the resistance in ohms of an arc lamp carbon 0.5 inch in diameter and 12 inches long. 41. Calculate the resistance of a column of 5 per cent solution of sulphuric acid at 18° C, the length of the column being 20 centimeters and the sectional area being 12 square centimeters. 42. Current enters a thin circular disk of platinum at the center of the disk and flows radially outwards towards the edge of the disk. The thickness of the disk is h centimeters, its radius is r centimeters, and its resistance is R ohms. Find , dR an expression for -7- . 38. Variation of resistance with temperature. — The electrical resistance of a conductor which forms a portion of an electrical circuit varies with temperature. Consider, for example, (a) an iron wire, (&) a copper wire, (c) a platinum wire, {d) a German- silver wire, (e) a carbon rod, and (/) a column of dilute sulphuric acid, each of which has a resistance of 100 ohms at 0° C. Then the values of the resistances of {a), (b), (c), (d), (e) and (/) at other temperatures, as determined by experiment, are shown by the ordinates of the curves in Fig. 52. As shown in this figure, iron and copper increase greatly in resistance with rise of tern- HEATING EFFECT OF THE ELECTRIC CURRENT. 55 perature, and German silver increases only slightly in resistance with rise of temperature. But carbon and sulphuric acid decrease in resistance with rise of temperature, carbon to a very slight extent and dilute sulphuric acid to a very great extent. itns / a iiu 6 iloo / // ^ iSo / . / 140 120 ^ c . ^ ^ ^ ^ ^^ '^ 100 An _ \ ^ t>^ i ^ \ e ~" o «a A? Co .80- loo. 120 i.4 Consider the electric lamp in I current I ' pjg g^. Let R be the resis- J — 4 ,.... tance of the lamp in ohms, and t '- - ^=- kattem IE rjfaiup' let I be the current flowing in i „ the circuit in amperes. Then RI"^ is the rate in watts at which / wire -* heat is generated in the lamp. ^'^- ^^- RI {= E) is the electromotive force between the terminals of the lamp. EI {= RP) is the rate at which energy is delivered to the lamp. These statements all refer to the lamp in Fig. 53, not to the entire circuit. To avoid confusion one should always speak of the current IN a circuit; of the resistance OF a circuit (or the resistance of a portion of the circuit) ; and of the electromotive force or potential difference BETWEEN THE TERMINALS OF any portion of a circuit. 44. Definition of the volt. — Consider any portion of an electric circuit, for example consider the lamp in Fig. 53, and let I be the current flowing in the circuit. Then the electromotive force E between the terminals of the lamp is equal to RI as stated in the previous article, and if R is expressed in ohms and I in HEATING EFFECT OF THE ELECTRIC CURRENT. 6 1 amperes, then E (= RT) is expressed in volts. That is, the product ohms X amperes gives volts. One volt is the electromotive force between the terminals of a one-ohm resistance when a current of one ampere is flowing through the resistance. One abvolt is the electromotive force between the terminals of a resistance of one abohm when a current of one abampere is flowing through the resistance. One volt is equal to lo* abvolts. The electromotive force of an ordinary gravity cell is about l.i volts. The electromotive force of an ordinary dry cell is about 1.5 volts. The voltages commonly used for electric lighting and motor service are no volts and 220 volts; that is to say, the voltage between the supply wires in a building is usually either no volts or 220 volts. The usual voltage for electric railway service is 500 volts; that is to say, the voltage between the trolley wire and the rails is generally about 500 volts. 45. The direct-current voltmeter. — Consider an ammeter (see Art. 18) of which the resistance is R ohms. When a current of / amperes flows through the ammeter the electro- motive force across the terminals of the instrument is RI volts, and the scale of the instrument can be numbered so as to give the value of RI in volts instead of giving the value of I in amperes. An ammeter arranged in this way is called a voltmeter. It would seem from the above that the only difference between an ammeter and a voltmeter would be in the numbering of the scale ; but an instrument which is to be used as an ammeter must have a very low resistance in order that it may not obstruct the flow of current in the circuit IN which it is connected, and an •instrument which is to be used as a voltmeter must have a very high resistance in order that it may not take too much current from the supply mains BETWEEN which it is connected. Thus a good ammeter for measuring up to 100 amperes has a resistance of about o.ooi ohm so that one-tenth of a volt would be sufiScient to force the full current of 100 amperSs through the instrument- A good voltmeter for measuring up to 150 volts has a resistance 62 LESSONS IN ELECTRICITY AND MAGNETISM. of about 15,000 ohms, so that about o.oi ampere would flow through the instrument if it were connected across the terminals of a 1 50- volt generator. 46. Measurement of power by ammeter and voltmeter. — The power delivered by a battery or dynamo (direct-current dynamo) is equal to EI watts, where E is the electromotive force be- tween the terminals of the battery or dynamo in volts and I is the current in amperes delivered by the battery or dynamo. The power delivered to a lamp (or in general to any portion of direct-current supply mains Fig. 54. Fig. 55. Ammeter and voltmeter connected for Ammeter and voltmeter connected for measuring power output of generator. measuring power delivered to L. a circuit) is equal to EI watts, where E is the electromotive force between the terminals of the lamp in volts, and / is the current in amperes flowing through the lamp. Note. — Power cannot be measured by an ammeter and a volt- meter arranged as in Figs. 54 and 55 in an alternating-current system. PROBLEMS. 49. A motor takes 79.78 amperes of current from iio-volt mains, and the motor-belt delivers 10 horse-power. What is the efficiency of the motor? Note. — Power output of motor in watts divided by power intake of motor in watts gives the efficiency of the motor. 50. A so-called "25-watt, iio-volt" tungsten lamp takes 25 watts of power when it is connected to iio-volt supply mains. How much current does the lamp take, and what is the resistance of the lamp filament while the lamp is burning? HEATING EFFECT OF THE ELECTRIC CURRENT. 63 51. A motor to deliver 10 horse-power has an efficiency of 89 per cent. The motor is supplied with current at no volts across its terminals. Find the full-load current of the motor. 52. When electrical energy costs 11 cents per kilowatt-hour how much does it cost to operate for 10 hours a lamp which takes 0.227 ampere from no-volt supply mains? 53. Find the cost of energy for operating a 5 horse-power motor at full load for 10 hours, the efficiency of the motor being 85 per cent and the cost of energy being 6 cents per kilowatt- hour. 34. When a certain dynamo electric generator is delivering no current it takes 1.75 horse-power to drive it. When the generator delivers 150 amperes it takes 25 horse-power to drive it. Calculate the electromotive force of the generator on the assump- tion that all of the additional power required to drive it is used to maintain the current of 150 amperes. 55. A coil of wire of which the resistance is to be determined is connected to no- volt direct-current supply mains in series with an ammeter and a suitable rheostat, and a voltmeter is connected across the terminals of the coil. The ammeter reads 13 amperes and the voltmeter reads 80.6 volts. What is the resistance of the coil? 56. Assume the actual cost of electrical energy delivered to a street car to be 0.8 cent per kilowatt-hour. Find the cost of developing 100,000 British thermal units for heating the car first by an electrical heater in which all of the heat generated is available for heating, and second by burning coal costing $8 per ton (2,000 pounds) and giving 14,000 British thermal units per pound of which 30 per cent, say, is lost by incomplete com- bustion and by flue-gas losses. 57. One cubic foot of good illuminating gas costing one tenth of a cent gives about 600 British thermal units when it is burned, and about 20 per cent of the heat of a burner is taken up by the water in a tea kettle. On the other hand about 70 per cent of the healf given off by an ordinary electrical heater is given to a 64 LESSONS IN ELECTRICITY AND MAGNETISM. tea kettle which completely covers the hot disk of the heater, and electrical energy for domestic use costs, say, lo cents per kilowatt-hour. What is the cost of bringing two gallons of water (16.1 pounds or 7570 grams) from 0° C. to 100° C. by a gas burner and what is the cost by electric heater? 47. Voltage drop in a generator. — Let / be the current in amperes which is being delivered by a battery (or dynamo), and let R be the resistance of the battery in ohms. Then a portion of the total electromotive force of the battery is used to force the current through the battery itself. The portion so used is equal to RI according to Ohm's law. If the total electro- motive force of the battery is E voltg, then the electromotive force between the terminals of the battery will be {E — RT) volts. A voltmeter connected to the battery terminals would indicate E volts when the battery is delivering no current, * but the voltmeter would indicate {E — RI) volts at the instant? the battery beginsf to deliver a current of / amperes. The electro- motive force RI which is used to overcome the resistance of a battery (or dynamo) is called the voltage drop in the battery (or dynamo). 48. Voltage drop along a transmission line.t — ^A current of / amperes is delivered to a distant motor or to a distant group of lamps over a pair of wires, the combined resistance of the pair of wires being R ohms. — Let £0 be the voltage across the generator, and let Ei be the voltage across the motor or lamps as shown in Fig. 56. Then £1 is less than Eo, the difference (£0 — -El) is the electromotive force which is used to overcome * Of course the battery delivers current to the voltmeter, but this is a negli- gible current because the resistance of the voltmeter is very large as compared with the resistance of the battery. t Continued flow of current causes a decrease of voltage by polarization as explained in Art. 42. t The electromotive force across the terminals of a battery or between any two given points is frequently spoken of as an electrical "pressure difference" or potential difference, and the voltage drop along a transmission line is frequently called the "pressure drop or potential drop along the line. HEATING EFFECT OF THE ELECTRIC CURRENT. 65 the resistance of both wires, and it is equal to RI volts. This loss of electromotive force along a transmission line is called the voltage drop along the line. For example, the electromotive force across the terminals of a generator is 115 volts. The generator supplies 100 amperes of current to a group of lamps at a distance of 1,000 feet from the generator, and the wire (2,000 feet of it) penenOorlJ J^ Bi (^ motor vr tat^. Fig. 56. which is used for the transmission line has a total resistance of 0.05 ohm. Therefore the voltage drop along the line is 100 amperes X 0.05 ohm, or 5 volts; and the voltage across the terminals of the group of lamps is 1 1 5 volts — 5 volts = 110 volts. PROBLEMS. 58. A gravity Daniell cell of which the electromotive force is 1.07 volts and the resistance is 2.1 ohms is connected to a wire circuit of which the resistance is 5 ohms, (a) What current is produced? (6) What is the electromotive force between the terminals of the cell? (c) What is the electromotive force drop in the cell? 59. A voltmeter connected across the terminals of a set of 60 storage battery cells connected in series reads 120.4 volts when the battery is delivering no current, and the voltmeter reading falls instantly to 112.25 volts when the battery begins to deliver 15 amperes of current. What is the resistance of the battery? Note. — ^When a battery continues to deliver current the voltage falls off because of polarization. The sudden drop of voltage at the instant that current delivery- begins is due almost entirely to the battery resistance. 60. A voltmeter connected across the terminals of a battery reads 15 volts when the battery is not delivering current (except the negligible current which flows through the voltmeter), and the voltmeter reading drops suddenly to 9 volts when a wire 6 66 LESSONS IN ELECTRICITY AND MAGNETISM. circuit having a resistance of 6 ohms is connected to the battery. What is the resistance of the battery? 6i. A dynamo electric generator having an electromotive force of 1 15 volts between its terminals delivers 200 amperes to a group of glow lamps 1,000 feet distant from the generator. Find the diameter in mils of the copper wire required in order that 95 per cent of the power output of the generator may be delivered to the lamps. Note. — If 9S per cent, of the power output of the generator is delivered to the lamps then the electromotive force between the mains at the lamps must be 95 per cent, of 115 volts, or 109.25 volts. 62. What size of copper wire is required to deliver current at no volts to a lo-horse-power motor of 85 per cent efRciency; the motor being 2,000 feet from the generator, and the electro- motive force across the generator terminals being 125 volts? 63. A motor is to receive 100 kilowatts of power from a gen- erator at a distance of 15 miles. A loss of 10 per cent of generator voltage (or 10 per cent of the generator output of power) is to be permitted in the transmission line. Find the generator voltage which must be provided for in order that copper trans- mission wires 200 mils in diameter may be used. 64. If a 10 per cent line loss is allowed as in the previous prob- lem, but if the generator voltage is doubled, what size of copper transmission wires would be used to deliver 100 kilowatts at a distance of 15 miles? I — ^ J ' 1 49. Connections in series. — When I \ two or more portions of an electric cir- ^ \^ cuit are so connected that the entire .=. JL current passes through each portion, TSr then the portions are said to be connected ■ < J in series. Thus Fig. 57 shows two lamps. Fig. 57. ^ ^j^(j i^'^ connected in series. The or- wo amps m series. djnary arc lamps which are used to light city streets are connected in series, and the entire current delivered by the generator flows through each lamp ; but the electromotive HEATING EFFECT OF THE ELECTRIC CURRENT. 67 force of the generator is subdivided. For example, a generator supplies 6.6 amperes at 2,000 volts to a circuit containing 30 arc lamps connected in series. The entire current, 6.6 amperes flows through each lamp, but the electromotive force across the terminals of each lamp is 1/30 of 2,000 volts or 67 volts. The electromotive force of a generator is subdivided among a number of lamps or other units which are connected in series. 50. The voltmeter multiplying coil. — Given a voltmeter which, for example, reads up to 10 volts; one can use such a voltmeter for measuring a higher voltage by connecting an auxiliary resistance in series with it. Thus Fig. 58 shows a voltmeter V supply mains H- — RoKms — ^tf'-j— — SiJ oKms h i © fA/VNAAAAAAAf-" U-RI volts-4*^^ — 9BI vqUs ->j [ I K— ^lORI volts 4 Fig. 58. of which the resistance is R ohms, and which has an auxiliary resistance of gR ohms connected in series with it. Under these conditions the voltage between the mains is found by multiplying the reading of the voltmeter by 10. This may be explained as follows: Let / be the current flowing through the circuit in Fig. 58. Then RI is the electromotive force drop across the terminals of the voltmeter, and gRI is the electromotive force drop across the terminals of the auxiliary resistance. Therefore RI + gRI or loRI is the electromotive force between the mains; but the voltmeter reading is the value of the electromotive force between its terminals, namely RI; therefore the electromotive force between the mains is ten times as great as the voltmeter reading. 68 LESSONS IN ELECTRICITY AND MAGNETISM. t = Fig. 59. Two lamps in parallel. 51. Connections in parallel. — When two or more portions of an electric circuit are so connected that the current divides, part of it flowing through each portion, then the portions are said to be connected in parallel. Thus Fig. 59 shows two lamps L and I ' L', connected in parallel. The ordi- I f 1 nary glow lamps which are used for house lighting are connected in paral- lel between copper mains which lead out from the terminals of the genera- tor ; and (if the resistance of the mains is negligible) the full voltage of the generator acts on each lamp, but the current delivered by the generator is subdivided. For example, a iio-volt generator supplies 1,000 amperes to 2,000 similar lamps connected in parallel with each other between the mains. The full voltage of the generator acts on each lamp, but each lamp takes only 1/2000 of the total current. The current delivered by a generator is subdivided among a number of lamps or other units which are connected in parallel. Remark. — ^When a circuit divides into two branches, the branches are, of course, in parallel with each other, and either branch is called a shunt in its relation to the other branch. 52. The division of current in two branches of a circuit.— Figure 60 shows a battery deliv- ering current to a circuit which branches at the points A and B. Let / be the current delivered by the battery, /' the current in the upper branch, I" the cur- rent in the lower branch, R' the resistance of the upper branch, and R" the resistance of the low- er branch. The product R'l' is the electromotive force be- tween the branch points A and B, also the product R"I" is V — rV or B^i"- j Fig. 60. HEATING EFFECT OF THE ELECTRIC CURRENT. 69 the electromotive force between the branch points A and B. Therefore we have : R'I'.= R"I" (i) The current in the main part of the circuit is equal to the sum of the currents in the various branches into which the circuit divides. Therefore in the present case we have: /■ = /' + /" (ii) By using equations (i) and (ii) the values of /' and I" can both be determined in terms of 7, i?' and i?". It is important to note that a definite fractional part of the total current flows through each branch; and equation (i) shows that the currents I' and I" are inversely proportional to the respective resistances R' and R" . Thus if R' is nine times as large as R" , then /" is nine times as large as /'. 53. The ammeter multiply- ing shunt. — A low-reading volt- meter can be used to measure a higher voltage by connecting an auxiliary resistance (a multi- plying coil) in series with it as explained in Art. 50. A . low- reading ammeter can be used to measure a larger current by con- necting an auxiliary low Iresist- ance (a multiplying shunt) in parallel with it. It is not practicable, however, to use interchangeable shunts with a low resistance instrument (an ammeter). This may be illustrated by an example as follows: The ammeter in Fig. 61 has, let us say, a resistance of o.oi ohm, and let us suppose that a 0.01 ohm shunt 5 is connected across its terminals. Under these conditions one half of the total current flows through the ammeter and one half flows through s. Therefore the value of the total current is twice the ammeter reading. The difiiculty, however, is that if s is detachable there is likely to be an appre- Fig. 61. Impracticable arrangement of amme- ter shunt. 70 LESSONS IN ELECTRICITY AND MAGNETISM. ciable* unknown resistance in the contacts of 5 with the two binding posts p and p' so that j may be in fact lo or 20 per cent greater than it is supposed to be. Any circuit in which binding-post contacts are to he made must be of fairly high resistance if the uncertain resistance at the contacts is to be negligible. Figure 62 shows an ammeter provided with a permanent shunt, j and j being soldered joints. In this case the shunt 5 may be once for all adjusted by the maker of the instru- ment so hat the full deflection of the instrument may corre- spond to any desired number Practicable arrangement of permanent r t r ^ ^ of amperes. In fact a manu- ammeter shunt. "^ facturer usually makes the working part, C, of all of his ammeters alike. The only differ- ence between an ammeter for large current and an ammeter for small current is in the resistance of the shunt s. 54, Combined resistance of a number of branches of a circuit. — (c) The combined resistance of a number of lamps or other units connected in series is equal to the sum of the resistances of the individual lamps, (b) The combined resistance of a number of lamps or other units connected in parallel is equal to the reciprocal of the sum of the reciprocals of the resistances of indivi dual lamps. Proposition (a) is almost self-evident. Prop- osition (b) may be established as follows : Let E be the electro- motive force between the points A and B where the circuit divides into a number of branches (see Fig. 60). Then, accord- ing to Ohm's law, we have: /' =^, (i) £ R' E A R" ♦Appreciable, that is, as compared with o.oi ohm. I'" = ^777 (iii) HEATING EFFECT OF THE ELECTRIC CURRENT. 71 where R', R" and R'" are the resistances of the respective branches, and /', I" and /'" are the currents flowing in the respective branches. Let I be the total current flowing in the circuit (= I' + I" + 1'"). The combined resistance of the branches is defined as the resistance through which the electromotive force E between the branch points would be able to force the total current I. That is, tlje combined resistance is defined by the equation : ^ (iv) / = R in which R is the combined resistance. Adding equations (i), (ii), and (iii), member by member, and substituting EjR for /' + /" + /'", we have E E whence R R''^R"'^R"' R = I ^ I R''^W''^W' (v) (14) 55. Wheatstones's bridge. — Four resistances x, R, a and Fig. 63a. b, Fig. 63, are adjusted so that the galvanometer G gives no deflection (no current through the galvanometer). Then aR a X _=_ or x=-^ (i) 72 LESSONS IN ELECTRICITY AND MAGNETISM. from which x can be calculated if R and the ratio c/6 are known. Proof of equation (t). — There being no current through G in Fig. 630, the same current » flows through x and R, and the some current j flows through a and b. Also since there is no current through G the electromotive force between points 3 and 4 is zero. Therefore the electromotive force between i and 3 (which is equal to xi) must be equal to the electromotive force between i and 4 (which is equal to aj) so that xi = aj (ii) Similarly, the electromotive force Ri between 3 and 2 must be equal to the electro- motive force bj between 4 and 2, so that Ri = bj (iii) Whence, dividing equation (ii) by equation (iii), member by member we get equa- tion (i). The usual arrangement of Wheatstone's bridge, the so-called box bridge, is indicated in Fig. 636, in which the dotted lines show connections outside of the box, x being the unknown resistance which is to be measured. The two resistances a and b are called the ratio arms, and usually each consists of i-ohm, lo-ohm, loo-ohm and i,ooo-ohm coils; and J? is a rheostat, consisting of units, tens, hundreds and thousands of ohms. The unknown resistance x is connected as indicated in Fig. 636, the ratio arms a and b are chosen, the rheostat resistance J? is adjusted, until the galvanometer gives no deflection, and the value of X is then given by equation (i) . 56. Eirchhoff's rules. — The equa- tions which express the current and voltage relations in any complicated network of conductors are most easily formulated by the use of two rules which are known as Kirchhoff's rules. Figure 64 shows a net-work of conductors in which are connected two batteries of which the electro- motive forces E and e act in the directions of the arrows E and e. The currents in the various branches of the net-work are Fig. 64. HEATING EFFECT OF THE ELECTRIC CURRENT. 73 represented by the small letters a, b, c, d, f and g; and the arrows indicate the arbitrarily chosen directions in which the respective currents are to be considered positive. If we should find ultimately that a, for example, is —1.6 amperes it would mean that the current a actually flows in the direction opposite to the arrow a. ■ The resistances of the various branches of the net-work are represented by the capital letters A, B, C, D, F and G. For example A is the resistance of the branch i to 4 including the resistance of the battery E, and B is the resistance of the branch i to 2 including the resistance of the battery e. Kirchhoff's first rule. Current equations. — The algebraic sum of the currents flowing towards* each branch point is zero. Therefore we get: for point i: — a — b-\-c = o (i) for point 2: +b — d— f = o (ii) for point 3: — c + d — g = o (iii) and there is no need for writing down the current equation for point 4 because it is an equation which may be derived from (i), (ii) and (iii). If there are n branch points there are only (n — i) independent current equations. Kirchhoff's second rule. Voltage equations. — ^The algebraic sum of the RI voltage drops around any mesh of the net work is equal to the algebraic sum of the battery voltages in that mesh, due attention being given to directions of arrows a, b, c, d, f, g, e and E with respect to the direction of the curled arrow which shows the chosen direction around the mesh. This matter will be understood by comparing the following equations with Fig. 64. for mesh P we get: + Bb + Dd + Cc = — e (iv) for mesh Q we get : — Dd + Ff — Gg = o (v) for mesh S we get: — Aa — Cc + Gg = + E (vi) Let us suppose that the two electromotive forces E and e, and all of the resistances are given. Then equations (i) — (vi) enable the calculation of the currents. * Or away from. 74 LESSONS IN ELECTRICITY AND MAGNETISM. PROBLEMS. 65. A telegraph line 80 miles long has a ground return (which we will assume to have negligible resistance), and the telegraph wire is, let us say, pure iron 0.100 inch in diameter. The tele- graph instruments which are connected in circuit with the line have a combined resistance of 100 ohms. The line is operated by gravity cells each having an electromotive force of 1.07 volts and an internal resistance of 2 ohms. A current of 10 milli- amperes is required to operate the line. How many gravity cells (connected in series) are required? 66. A millivoltmeter has a resistance of 15.4 ohms. What resistance must be connected in series with the instrument so that the scale reading may give volts instead of millivolts? 67. Three lamps (or other units) are connected in series to iio-volt mains, the resistances of the lamps are 10 ohms, 8 ohms and 4 ohms respectively, find the voltage across the terminals of each lamp. 68. Twenty gravity cells each having an electromotive force of 1.07 volts and an internal resistance of 2 ohms are connected 5 in series X 4 in parallel, and the batteries so connected deliver current to a coil of which the resistance is 6 ohms. What is the current in the coil? 69a. Three resistances of 4, 4 and 2 ohms respectively are connected in parallel ; and two resistances of 6 ohms and 3 ohms respectively are connected in parallel. The first combination is connected in series with the second combination, and to a battery of negligible resistance and of which the electromotive force is 3 volts. What is the current in the 2 ohm resistance and what is the current in the 3 ohm resistance? 69b. A 3-ohm resistance and a 1.5-ohm resistance are con- nected in parallel, and this combination is connected in series with a lo-ohm resistance to a 2-volt battery cell of which the internal resistance is negligible, (a) Find the current in the 3-ohm resistance. (&) How much additional resistance must be connected in the circuit so that the current in the 3-ohm resistance HEATING EFFECT OF THE ELECTRIC CURRENT. 75 may be brought back to its original value after the 1.5-ohin resistance is cut or disconnected? 70. An ammeter has a resistance of 0.05 ohm. The instru- ment is provided with a shunt so that the total current through instrument and shunt is 10 times the current through the am- meter itself. What is the resistance of the shunt? 71. The scale of a direct-reading millivoltmeter has 100 divisions, each division corresponding to one thousandth of a volt between the terminals of the instrument. The instrument is connected to the terminals of a low-resistance shunt, and each division on the instrument scale corresponds to 0.25 ampere in the shunt. What is the resistance of the shunt? 72. A voltmeter which has a resistance of 16,000 ohms is connected in series with an unknown resistance R to iio-volt supply mains, and the reading of the voltmeter is 4.3 volts. What is the value of R? 73. A 40-mile telegraph line is disconnected from ground at both ends, the line is then connected to ground at one end through a 220-volt battery and a direct-reading voltmeter of which the resistance is 16,000 ohms, and the voltmeter reads 2.9 volts. What is the insulation resistance of the 40-mile telegraph line, and what is the insulation resistance of one mile of the line? CHAPTER IV. INDUCED ELECTROMOTIVE FORCE. ANOTHER ASPECT* OF THE MAGNETIC EFFECT OF THE ELECTRIC CURRENT. There is no recognized unit of magnetic flux corresponding to the volt, the ampere, the ohm, etc. Therefore all quantities should be expressed in c.g.s. units (abamperes, abohms, abvolts, abhenrys, etc.) in every equation containing magnetic flux, or a reduction factor must be introduced as in equation (17). 57. Back electromotive force in a motor armature. — Before taking up the mathematical discussion of induced electromotive force it is highly desirable that the reader be brought face to face with induced electromotive force as an unmistakable physical fact, and the best way to do this is to consider the electric motor (direct-current type). A much higher electromotive force is required to push a given current through the motor armature when it is running than when it is standing still. Let the resistance of the armature from brush to brush be one ohm. Then 10 volts will maintain a current of 10 amperes through the armature when it is standing still as indicated in Fig. 65a, but a much higher electromotive force is required to maintain a current of 10 amperes through the running armature. Thus Fig. 65b shows * Side push on the wires of a motor armature when current is flowing through the wires, and back electromotive force (induced electromotive force) in the wires when the motor is running are two aspects of one thing, and this electro-mechan- ical thing is. closely analogous to the following purely mechanical thing: A man walks towards the center of the swinging span of a draw-bridge; the walking man necessarily exerts a side push which helps to turn the span, and the swinging span necessarily creates a back force on the man (a force which the man must overcome as he walks towards the center of the span). The complete parallelism between the equations of electricity and magnetism on the one hand and the equations of mechanics on the other hand is a matter of very great interest, but it cannot be discussed profitably in an elementary text. This parallelism is brought out in any good discussion of generalized coordinates. 76 INDUCED ELECTROMOTIVE FORCE. n lOO volts maintaining a current of lo amperes through the running armature.* It is harder, as it were, to push current through a running motor armature than to push current through the same armature while it is standing still. Something besides resistance must therefore oppose the flow of current through the running arma- supply main amperea flowing through armature T 100 volts I i. 10 amperes Mowing through armature Fig. 65a. Armature not running. supply main Fig. 656. Armature running. ture. In fact a back electromotive force exists in the windings of the running armature. This back electromotive force is pro- duced by the sidewise motion of the armature wires as they cut across the lines of force of the magnetic field in the gap spaces as shown by the fine lines in Fig. 36. An electromotive force produced in this way is called an induced electromotive force. 58. Expressions for induced electromotive force. — Each of the wires on the armature in Fig. 37 is in an intense magnetic field (this statement refers to the wires which are in the gap spaces between the armature core and the pole faces) , each wire is at right angles to the lines of force of this intense field, and each wire is moving sidewise in a direction at right angles to itself and at right angles to fhe lines of force. This state of affairs is represented in Fig. 66 ; the wire be slides on the two metal rails, moving sidewise at velocity v. Let I be the current in abam- peres flowing around the circuit abed in Fig. 66. Then the side push exerted on be by the magnetic field is II H dynes according to equation (6) of Art. 21, where I is the length of the wire be * This effect must be actually shown by an experiment to be believed. 78 LESSONS IN ELECTRICITY AND MAGNETISM. c *^i"v:~.~-~~r b''- '■ :. • nOl S^SyVS':':; *-movi'ng icirel -•■•".".••'js. • -■ "•■->. '•'.•" 1 • -v. ■^P^/^ . ^\ .• • ."•:'•.' ■.■•.'raa . -, c « ■ • . •.;. '.d Fig. 66. Dots represent magnetic lines of force which are perpendicular to the plane of the paper in centimeters and H is the intensity of the magnetic field in gausses. Therefore, neglecting friction, a force equal to F but in the direction of v must be exerted on the wire to make it move, this force will do work on be at the rate Fv or at the rate II Hv ergs per second, and all of the work thus spent in moving the wire reappears as electri- cal work done by the induced electromotive force E in maintaining the current I. Therefore* we must have EI = II Hv, whence E = IHv (15) that is to say, the electromotive force E in abvolts which is induced in the moving wire be in Fig. 66 is equal to the product IHv, where I is the length of the moving wire in centimeters, H is the intensity of the magnetic field in gausses and v is the velocity of the moving wire in centimeters per second. The electromotive force in abvolts which is induced in the moving wire be in Fig. 66 is equal to the rate at which the wire cuts magnetic flux. — Consider the sidewise distance Ax moved by the wire be in Fig. 66 during the short time interval At. Then: Ax = v-At (i) The area swept over by the wire during the time interval At is I- Ax, and the amount of flux crossing this area is A$ = HI- Ax, according to equation (4) of Art. 12, or, using the value of Ax from (i), we have: A^=HlvAt (ii) and, of course, this is the amount of flux which is "cut" by the * This fact was first pointed out by Helmholtz and it includes what has fre- quently been called Lenz's law. INDUCED ELECTROMOTIVE FORCE. 79 wire during the time interval A/. Therefore, dividing A$ by At we have the rate at which the wire cuts flux [in lines of force (or maxwells) per second], and from equation (ii) we have: A$ -=Hlv (iii) That is, the rate at which the wire "cuts" flux is equal to Hlv, but Hlv is the electromotive force in abvolts induced in the moving wire, according to equation (15). Therefore the electro- motive force in abvolts induced in the moving wire be in Fig. 66 is equal to the number of lines of force {maxwells) cut by the wire per second. The electromotive force in abvolts which is induced in the moving wire be in Fig. 66 is equal to the rate of increase of the magnetic flux $ through the circuit abed (across the rectangular area ahcd) . The area of abed is Ix square centimeters, and the magnetic flux $ which passes through the opening abed is found by multi- plying H by the area Ix ; that is : $ = Hlx (iv) Now if X changes, $ must change HI times as fast, that is : d^ ^ dx Tt=^'d^ (-) dx But, 3- is the velocity v at which the wire moves sidewise. at Therefore equation (v) becomes : f=Hfo (vi). Therefore, remembering that Hlv is the electromotive force induced in be, and remembering that $ is the magnetic flux through the opening abed, we have the proposition as stated above. 59. Electromotive force induced in a stationary loop or coil of wire by a changing magnetic field or by a moving magnet. — 80 LESSONS IN ELECTRICITY AND MAGNETISM. Article 58 refers to electromotive force induced in a wire which moves across a permanent or unchanging magnetic field. Let $ be the amount of magnetic flux which passes through a loop of wire ; then if anything whatever causes $ to change, an electro- motive force of E abvolts will be induced in the loop of wire such that ^= "dt or, if there are Z turns of wire in the loop or coil, we will have E=-Zf (.6, No attempt is here made to give a general derivation of this equation. The negative sign is chosen in equation (i6) for the following reason: Let us choose what we are to consider as the positive direction through the opening of a loop of wire, then the positive direction around the loop is conventionally taken as the direction in which a right-handed screw would have to be turned to travel in the chosen direction through the loop. Let us suppose that is positive (the magnetic field which produces * would carry a north magnet pole in the positive d* direction through the loop), and let us suppose that * is increasing. Then — at is positive and experiment shows that the induced electromotive E is in the negative d* direction around the loop, that is, E is negative when -7- is positive. dt 60. The fundamental equation of the direct -current dynamo. — The equation which expresses the electromotive force which is induced in the armature windings of a direct-current dynamo in terms of various data as explained below is called the funda- mental equation of the dynamo. We can derive this equation from equation (15) of Art. 57, but it is instructive to carry out the argument from the beginning. Let us use c.g.s. units through- out, let us consider the dynarno as an electric generator, and let us think of the armature as rotating without energy losses of any kind* such, for example, as friction losses. Then the mech- anical power required to drive the dynamo armature is equal to the * This is not necessary but it avoids tedious qualifying specifications in the discussion. INDUCED ELECTROMOTIVE FORCE. 8 1 electrical power output EI of the armature, where E is the electromotive force induced in the armature windings and I is the current delivered by the armature. Also let us limit the discussion to the simple form of dynamo with a bipolar field magnet and a ring-wound armature, as shown in Figs. 36-39. Let r = radius of armature, measured out to the layer of wires. L = length of armature core parallel to armature shaft. This is also the length of the pole faces parallel to the armature shaft. b = breadth of each pole face measured along the circum- ference of the armature where the wires lie. Z = number of wires on outside of armature. These wires are straight, they lie parallel to the arm- ature shaft, and the length of the portion of each which lies in the gap space is L. n = speed of armature in revolutions per second. H — intensity of magnetic field in gap spaces. We assume this field to have the same intensity every- where in the gap spaces, we assume the lines of force to be radial as shown in Fig. 36, and we ignore the fringe of the magnetic field which spreads out beyond the edges of the pole faces. The total current / is supplied by the coming together at the brushes of 7/2 abamperes flowing through the windings on each side of the armature. That is, the current in each armature wire is 7/2 abamperes. The side push of the magnetic field on each armature wire in the gap spaces opposes the motion of the armature, and is equal to L 777/2 dynes according to equation (6) of Art. 21; and the number of armature wires in the gap spaces at any time is 26 X Z. Therefore the total tangential drag or force on the 2%r . . LHI 2b ^ , _, . „ armature wires is X X Z dynes, and the power F 2 2Trr 82 LESSONS IN ELECTRICITY AND MAGNETISM. required to drive the armature against this dragging force is equal to the product of this force and the velocity (27rm) of the armature wires, that is — — X — X Z X 2irm. But this 2 2Trr power is equal to the power output EI as above explained. Therefore EI = X X Z X 2rrn, or 2 2irr E = {LbH)Zn (i) But the area of a pole face Lb multiplied by H gives the amount of magnetic flux $ which enters the armature core from the N-pole of the field magnet (and leaves the armature core to enter the S-pole of the field magnet), according to equation (4) of Art. 12. Therefore equation (i) becomes E (in abvolts) = $Zw or, since one volt is equal to 10* abvolts, we have E (in volts) = $Zm X IQ-^ (17) Terminal voltage equation of direct-current generator. — Let R be the resistance of the armature of a dynamo, including brush contacts and brushes, and let I be the current delivered. Then RI is voltage drop in the armature, and the voltage Eb be- tween brushes is #Z« X lO"* minus RI. That is Eb = $Zw X io-« - RI (18) Speed equation of the direct -current motor. — Let Eb be the voltage across the brushes of the motor (supply voltage). Then Eb is used in part to overcome the resistance of the armature, the part so used is equal to RI; and Eb is used in part to over- come the back electromotive force Zn X io~' Therefore Eb = #Zw X io~* + RI, and solving for « we get: Eb-RI ^ . "" = $Z X 10- ('9) A very interesting use of this equation is in the discussion of the speed characteristics of the direct-current shunt motor.* * A very simple discussion of this matter is given in W. S. Franklin's Elements of Electrical Engineering, Vol. I, pages 196-200, published by Franklin and Charles, Bethlehem, Pa. INDUCED ELECTROMOTIVE FORCE. 83 PROBLEMS. 74. The armature of a direct-current motor has a resistance of 0.064 ohm and when the motor is running under full load a cur- rent orf 81 amperes is forced through the armature from 1 10- volt supply mains. How much power is delivered to the motor armature? How much power is lost in heating 'the armature wires? What is the back electromotive force in the armature windings? How much power is expended in forcing the current of 81 amperes to flow against the back electromotive force? Where does this power go to? 75. The winding of an electromagnet has a resistance of 22 ohms, and when the winding is connected across i lo-volt supply mains the current in the coil rises from zero at the beginning to 5 amperes ultimate value. The current must therefore pass in succession the values i ampere, 2 amperes, 3 amperes and 4 amperes. Consider the instant when the current is passing the value of 2 amperes, and calculate the values of the following quantities at this instant: (a) The rate at which work is being delivered to the coil, (b) The rate at which work is being spent in overcoming a back electromotive force in the coil due to the increasing magnetism of the rod, and (d) The value of the back electromotive force due to the increasing magnetism of the rod. 76. The winding of the electromagnet in the previous problem has 10,000 turns of wire. Find how fast the magnetic flux through the winding is increasing as the current is passing the value of 2 amperes 77. A long slim magnet of which the strength of each pole is 1,500 units is placed through a coil containing 20,000 turns of wire, and jerked out of the coil so that the magnetic flux through the coil due to the manget drops to zero during 0.004 second. What is the average value during the 0.004 second of the induced electromotive force in the coil? 78. A simple two-pole, direct-current dynamo with a ring winding on its armature has its field winding connected to iio- volt supply mains so that its field excitation is constant. Under 84 LESSONS IN ELECTRICITY AND MAGNETISM. , these conditions the armature flux * is nearly constant (regard- less of current through the armature), and we will assume that # is strictly constant. There are 560 wires (= Z) on the outside of the armature, and when the armature is driven at a speed of 900 revolutions per minute the electromotive force induced in the armaturfr is 1 10 volts as indicated by a voltmeter connected to the brushes (current through armature very small). What is the value of the armature flux $ ? The resistance of the armature between brushes is 0.6 ohms. At what speed will the dynamo run as a motor at full load (10 amperes flowing through armature) when the armature takes current from iio-volt supply mains? 79. A direct-current dynamo (shunt dynamo) operating as a generator is connected as indicated in Fig. 67, and, with no volts ffeld rheostat ihunt Held is f ) armature uinding matA Fig. 67. Shunt dynamo diagram. across its armature terminals it delivers 200 amperes to the dis- tributing mains. The resistance of the shunt field winding and rheostat is 11 ohms. How much current is delivered by the armature? What is the power output of the armature? How much power is expended in field excitation? How much power is delivered to the distributing mains? Note. — After the current in the field winding and the " strength'' of the field magnet become steady in value, all of the power delivered to the field winding reappears as heat in the winding in accordance with Joule's law. Therefore Ohm's law applies to the field winding. No power would be required to maintain the magnetism of the field magnet if a field winding of zero resistance could be obtained, although a zero-resistance field winding could not, of course, be connected as in Fig. 67. When, however, the field magnet is being magnetized (during a small fraction of a second at the beginning) then some of the power delivered to the field winding does not reappear as heat in accordance with Joule's law, but is used to establish the magnetism. INDUCED ELECTROMOTIVE FORCE. 85 80. A thin copper disk r centimeters in radius is mounted on a spindle and driven at a speed of n revolutions per second. The spindle is mounted along the axis of a large and very long coil of wire {I centimeters long) having Z turns of wire distributed uniformly over its length, and a current of J abamperes flows through the coil. The rotating disk has a brush rubbing on its edge and another brush rubs on the tip end of the spindle. These two brushes are connected through a sensitive galvanom- eter to the treminals of a resistance R through which the entire current I flows. The disk is speeded up until the galvanometer gives no deflection. Make a diagram of the connections, and find an expression for R in terms of r, I, Z and n. Note. — This problem refers to the celebrated arrangement which was devised by Lorentz for measuring resistance directly in terms of purely mechanical data. 61. The alternating-current d3mamo. — The alternating-current dynamo is usually called an alternator. The following discussion may be taken as referring to the alternator as a generator. When the alternator is used as a motor it is called a synchronous motor. Consider one of the straight wires on the armature in Fig. 37. The electromotive force which is induced in this particular wire is in one direction while the wire is sweeping across the north pole face and in the opposite direction while the wire is sweeping across the south pole face the field magnet, and if the two ends of this particular wire were to be kept permanently connected to the terminals of an outside circuit by means of sliding contacts, an alternating current would be delivered to the outside circuit. This simple arrangement includes all of the essential features of the simple alternator. The actual arrangement of the simple alternator may be understood with the help of Figs. 68, 69 and 70. These figures show the inwardlj' projecting magnet poles of what are called multipolar field magnets. These field magnets are magnetized or excited by direct current from some outside source, generally a small direct-current generator which is called an exciter. 86 LESSONS IN ELECTRICITY AND MAGNETISM. The large dotted circles in Figs. 68, 69 and 70 represent the ends of the cylindrical armature core, the inner circle represents the front end and the larger circle represents the back end en- U^ UJ rm nn Fig. 68. Possible armature winding diagram for 4-pole alternator. Fig. 69. Possible armature winding diagram for 4-pole alternator. larged so as to be seen. The short heavy radial lines represent ■ the straight wires which lie on the surface of the armature core parallel to the armature shaft, the curved lines BB represent cross-connections on the back end of the armature, the curved lines FF represent cross-con- nections on the front end of the armature, and the lines CC represent connections to the two insulated metal rings which are called collector rings. Two carbon blocks or metal brushes rub on these collector rings thus maintaining permanent con- Fig. 70. nections with the terminals of Possible armature winding diagram for the armature winding, and the 8-poie alternator. ^ outside circuit to which alter- nating current is to be delivered is connected to the two carbon blocks or brushes. INDUCED ELECTROMOTIVE FORCE. 87 The simple alternator above described is called a single-phase alternator. It has a single armature winding and two collec- tor rings. The two-phase alternator has two distinct armature windings, each winding being connected to two collector rings (four collector rings in all). The three phase-alternator has three distinct armature windings, each winding being connected to two collector rings (six collector rings in all). Because of certain relations between the three distinct alternating currents which are delivered by the three armature windings of a three-phase alternator (delivered to three distinct receiving circuits, of course) it is possible to use only three collector rings ; and this is the usual practice. Definition of cycle. Definition of frequency.— The electro- motive force of an alternator (and also the current delivered by an alternator) is subject to rapid reversals in direction, the electromotive force passes through a set of positive values as represented by the ordinates of the portion P of the curve in Fig. 71, and then through a similar set of negative values N, Fig. 71. repeatedly. The two sets P and N together constitute a cycle, and the number of cycles per second is called the frequency of the alternating electromotive force or current. The electromotive force which is induced in the armature winding which is shown in Fig. 68 passes though a set of positive values, let us say, while a given wire is sweeping across the face of a north pole of the field magnet, and through a set of negative values while the given wire is sweeping across a south pole face, so that there are p cycles per revolution of the armature where 88 LESSONS IN ELECTRICITY AND MAGNETISM. p is the number of pairs of field magnet poles. Therefore the frequency f = pn where n is the speed of the armature in revohitions per second. PROBLEMS. 8ia. Make a diagram Hke Fig. 68 and, assuming the armature to rotate in a clockwise direction, indicate by an arrow the direc- tion of the electromotive force which is induced in each of the armature conductors (each short heavy radial line). Show which collector ring is positive and which is negative?* 8ib. Make diagram like that called for in problem 8ia but with the armature in Fig. 68 turned 90° forwards. Show which collector ring is positive and which is negative. 82. A disk of wood of radius r has Z turns of wire wound around its edge, and it is rotated » revolutions per second, about a diameter, in a uniform magnetic field of intensity H, the axis of rotation being at right angles to H. Derive an expression for the electromotive force induced in the coil. 62. The altemating-curreiit transform- er. — The alternating-current transformer consists of two separate windings of wire C on a laminatedf iron core as shown in Fig. 72. One of these windings is connected to an alternator and the rapid reversals of the magnetism of the core induce an alternat- ing electromotive force in the other wind- ing. The winding which receives alternat- ing current from an alternator is called the primary coil and the other coil which deliv- Alternating-current trans- ers alternating current at a higher or lower voltage is called the secondary coil. Step-down transformation. — ^A small alternating current may * The positive terminal of a generator is tlie terminal at which current flows out of the generator to the receiving circuit, t See Art. 64. Fig. 72. INDUCED ELECTROMOTIVE FORCE. 89 be delivered at high voltagfe to the coil-of-many-turns, in which case the coil-of-few-turns will deliver a large alternating current at low voltage. This constitutes what is called step-down trans- formation. Step-up transformation. — A large alternating current may be delivered at low voltage to the coil-of-few-turns, in which case the coil-of-many-turns will deliver a small alternating current at high voltage. This constitutes what is called step-up trans- formation. PROBLEMS. 83. A transformer has 100 turns of wire in one of its coils A and 1 ,000 turns of wire in its other coil B. Coil A is connected for a few thousandths of a second to iio-volt, direct-current, supply mains. Assuming resistance of coil A to be negligible, calculate rate of growth of magnetic flux through the transformer core and calculate electromotive force induced in coil B. 84. Let us assume that a negligibly small current in coil A, alone, or a negligibly small current in coil B, alone, will produce an indefinitely large magnetic flux $ through the core of the transformer of the previous problem, then the magnetizing action (amperes X turns) of the current in one coil must always be equal and opposite to the magnetizing action (amperes X turns) of the current in the other coil. Why? Coil A is connected for a few thousandths of a second to iio-volt, direct-current, supply mains, and coil B is connected to a "non-inductive" receiving circuit of which the resistance is 2,155 ohms. The resistance of coil A is zero and the resistance of coil B is 45 ohms. Find the values of the following quantities during the few thousandths of a second, (a) Current delivered by coil B, (b) Current in coil A , and (c) Voltage across terminals of coil B. Note. — ^A "non-inductive" circuit is one in which the current is always given by Ohm's law even at the instant that an electromotive force begins to act upon it. 8s. Calculate the current which is delivered by coil B and the voltage across the terminals of B when the resistance of coil A is 0.4 ohm, everything else being as specified in the previous problem. 90 LESSONS IN ELECTRICITY AND MAGNETISM. 63. The induction coil. — ^An iron rod or core wound with insu- lated wire can be repeatedly magnetized and demagnetized by connecting a battery to the winding and repeatedly making and breaking the circuit; and the increase and decrease of mag- netism of the core thus produced can be utulized to induce elec- tromotive forces in an auxiliary coil of wire wound on the iron core. Such an arrangement is called an induction coil. The winding through which the magnetizing current from the battery flows is called the primary coil, and the auxiliary winding in which Fig. 73. Induction coil. the desired electromotive forces are induced is called the secondary coil. The iron core is always made of a bundle of fine iron wires or strips of sheet iron. A general view of an induction coil is shown in Fig. 73, and the diagram of connections is shown in Fig. 74. When the iron core is magnetized, the block of iron a is attracted, and the battery circuit is broken at the point p. The iron core then loses its magnetism, and the spring 5 brings the points at p into contact again so that the battery current again flows through the circuit and magnetizes the iron core. The iron block a is then at- tracted again, and the above action is repeated. When the iron core of an induction coil is magnetized, a INDUCED ELECTROMOTIVE FORCE. 91 momentary pulse of electromotive force is induced in the second- ary coil; and when the iron core is demagnetized a reversed momentary pulse of electromotive force is induced in the second- ary coil. Electromotive forces are induced only while the core is being magnetized or demagnetized, and each pulse of electro- motive force may be made very large in value (manv thousands battery Fig. 74. Induction coil diagram. of volts) by using many turns of wire in the secondary coil and by providing for the quickest possible magnetization or demagne- tization of the core. A battery cannot, however, magnetize a core very quickly when connected to a magnetizing coil; in fact a very considerable fraction of a second is required for the core to become magnetized. Therefore during the magnetization of the iron core of an induction coil the electromotive force induced in the secondary coil is a comparatively weak pulse of fairly long duration. On the other hand the use of the condenser CC Fig. 74, causes the iron core of the induction coil to he demagnetized very quickly as explained in Art. 66, and this quick demagnetization induces in the secondary coil an intense pulse of electromotive force of very short duration. 92 LESSONS IN ELECTRICITY AND MAGNETISM. The iron core of the alternating-current transformer should form a complete "magnetic circuit" as shown in Fig. 72, but the induction coil must have its iron core in the form of an open "magnetic circuit" as shown in Fig. 74, because after the core has been magnetized it is the energy of this magnetism which becomes avail- able when the primary circuit is broken, and the greater part of the available energy of a magnet resides in the air field near the poles of the magnet. 64. Eddy currents. Lamination. — Figure 75 shows an end view of an iron rod surrounded by a wire ring. While the iron Hiimenf of iron wire ring Fig. 75. Current Induced in wire ring. -iroa rod Fig. 76. Current induced in filament of solid iron rod. rod is being magnetized or demagnetized an electromotive force is induced in the ring, according to Art. 59, and an electric cur- rent is produced in the ring in the direction of the small arrows or in the opposite direction. Figure 76 shows the end of a larger iron rod. While the rod is being magnetized or demagnetized an electric current is produced in the circular filament of iron. The increasing or decreasing magnetism of the central portion C of the rod in Fig. 76 has the same action on the filament of iron in Fig. 76 as the iner easing or decreasing magnetism of the iron rod in Fig. 75 has on the wire ring in Fig. 75. Every circular filament in an iron rod has more or less current induced in it while the rod is being magnetized or demagnetized. Thus the currents which are induced in a solid iron rod while it is being magnetized or demagnetized are in the directions of the arrows in Fig. 77 or in the opposite directions, and these currents are called eddy currents. One effect of these eddy currents is to INDUCED ELECTROMOTIVE FORCE. 93 make it impossible to magnetize or demagnetize a solid iron rod quickly and another effect is to generate heat in the rod. If the rod is a bundle of thin strips of sheet iron, as shown in Fig. 78, or a bundle of fine iron wires, as shown in Fig. 79, then the eddy currents (as shown in Fig. 77) cannot flow because the strips of iron in Fig. 78 and the iron wires in Fig. 79 are suffi- Fig. 77. Fig. 78. Currents induced in solid iron rod. Fig. 79. ciently insulated from each other by the thin coating of oxide which always covers the iron. Eddy currents are not only produced in a solid iron rod while it is being magnetized or demagnetized, but eddy currents are also generally produced in a piece of solid iron (or in any solid piece of metal) which moves near a magnet. Thus if the cylinder A A, Fig. 36, were solid and if it were set rotating as indicated by the curved arrows in Fig. 37, eddy currents would be produced in it as indicated by the circles with dots and crosses in Fig. 80. solid inn cylindei' rotating Fig. 80. Eddy currents in rotating solid cylinder; away from reader on right side, towards reader on left side. 94 LESSONS IN ELECTRICITY AND MAGNETISM. If the cylinder is built up of thin sheet iron disks or stampings, these eddy currents cannot flow because the disks are sufficiently insulated from each other by films of on oxide. An iron rod or core which is built up of stampings of thin sheet iron or of fine iron wires is said to be laminated. Armature cores of dynamos and transformer cores are always laminated. The damping disk of the watt-hour meter. — ^An interesting example of eddy currents is afforded by the damping disk of the watt-hour meter, a copper or aluminum disk which rotates between the poles of a horse-shoe magnet. 65. Inductance of a coil or circuit. — ^While a boat is being brought up to full speed a part of the propelling force is used to accelerate the boat (to cause its velocity to increase). Simi- larly, when a circuit is connected to supply mains (direct-current supply mains) some time elapses before the current reaches its steady value as given by Ohm's law, and during the whole of this time RI is less than the supply voltage E* Therefore during J, - the short time that is required for the current •■~-~.j^^^jr^ ^ turns to reach its final steady value a portion of iw-- Tra"^ the supply voltage E is used to cause the current in the circuit to grow. Figure 81 represents a coil of Z turns of wire connected to a battery (battery not shown) . The current grows from zero to a final steady value as given by Ohm's law, and while the current is growing Fig. 81. ° ° the magnetic flux * through the coil (the flux that is due to the current in the coil) grows. In the following discussion $ is the average flux per turn of wire,] a.nd $ is at each instant proportional to the current i, that is to say, when i is doubled the value of $ is doubled. Therefore we may write: ^ = ki (i) * This is evident when we consider that RI = E or / = EjR when the current reaches its final steady value. t There is more flux through the outer turns than through the inner turns, and * represents the average at any instant. INDUCED ELECTROMOTIVE FORCE. 95 where ^ is a constant for the given coil. If i is growing it is evident, that must grow k times as fast, that is we must have d^ di But the growing flux induces an electromotive force in the coil and this induced electromotive force E' is £' = - Z~ (iii) according to equation (i6). Therefore to make the current grow an outside electromotive force (a portion of the electromotive force of the battery in Fig. 8i) must be used to overcome E', and the portion so used is equal and opposite to E' or equal to + Z -7-. Therefore the electromotive force which must act on a circuit or d^ coil to make the current in the circuit or coil grow is equal to Z-t- di or to kZ-r,- Let us use the letter L for the quantity kZ, at that is L = kZ (iv) Then we have e = L^^ (20) where e is the part of the electromotive force of the battery in di . , Fig. 81 which is causing the current to grow, -7 is the rate of growth of the current in amperes per second (or in abamperes per second), and L is what is called the inductance of the coil or circuit. The portion, /, of the propelling force which causes the velocity dv . n of a boat to grow (increase) is f = m-j . The inductance L of a circuit is precisely analogous to mass in mechanics. See Appendix D. 96 LESSONS IN ELECTRICITY AND MAGNETISM. Remark. — If there is an iron core in the coil in Fig. 8i, the flux * will be nearly proportional to i for small values of the current, but as the iron core approaches magnetic saturation $ is no longer even approximately proportional to i. A coil with an iron core does not, therefore, have a definite inductance. Flux -turns. — The quatity $ in the above discussion is the average flux per turn of wire as stated, and the product Z is the number of linkages of lines of force and turns of wire, and it is usually called, simply, flux-turns. Putting k = L/Z from (iv) in (i), we get ^Z = Li v2i) Definition of inductance. — The inductance of a circuit is defined by equation (20), and a circuit has unit of inductance when one unit of electromotive force will cause the current in the circuit to increase at the rate of one unit per second. A circuit has an inductance of one henry when one volt will cause the current in the circuit to increase at the rate of one ampere per second. A circuit has an inductance of one abhenry when one abvolt will cause the current in the circuit to increase at the rate of one abampere per second. One henry equals 10' abhenrys. Non-inductive circuit. — ^When the inductance of a circuit is negligibly small the circuit is said to be non-inductive. The di inductance is neglibigly small when the electromotive force L -j dt is very small in comparison with Ri. Kinetic energy of the electric current. — The kinetic energy in joules which is associated with a current of I amperes in a circuit or coil of which the inductance is L henrys is W = hLP (22) The kinetic energy of a current in a coil resides in the surrounding magnetic field. Imagine the circuit to have zero resistance, then if an electro- INDUCED ELECTROMOTIVE FORCE. 97 motive force E acts on the circuit the current will grow in di accordance with equation (20) so that E = L— . Therefore at 17 • -^^ El = Lt -7. But Ei is the rate at which work is done on the . . dW circuit —,—, so that dW_ di 'dt ~ -^* dt Integrating this equation from i = o to i = J we get equa- tion (22). Differential equation of growing current. — A constant electro- motive force E acts on a circuit. Let i be the value of the current at any instant before the final steady value (E/R) of the current is established. Then a portion of E is used to, over- come resistance, and the portion so used is equal to Ri ; and the remainder of E is used to make the current grow, and this di remainder is therefore equal to L —, according to equation (20). Therefore we have E=^Ri+ L% (23) Integrating* this equation subject to the condition * = o when t = o (which means that time is reckoned from the instant that the constant electromotive force E begins to act on the circuit), we get . E E R R -(BlL).t (24) where e is the Naperian base (2.7183). Equations (23) and (24) apply to a circuit such as is shown in Fig. 81. Differential equation of decasring current. — If the electro- motive force [electromotive force E in equation (23)] ceases to act on the circuit, that is, if the battery in Fig. 8r is short *A full discussion of this integration, and of the integration of equation (25), is given in Franklin, MacNutt and Charles' Calculus, pages 176-181; published by Franklin and Charles, Bethlehem, Pa., 1913. 8 98 LESSONS IN ELECTRICITY AND MAGNETISM. circuited by a heavy copper connection between its terminals, then £ = o, and equation (23) becomes = Ri+ Lj^ (25) which is the differential equation of decaying current. Integrating equation (25) ,we get i = /e-(«/i)« (26) where / is the value of the current when / = o, and e is the Naperian base. Calculation of inductance* of a coil which consists of a single layer of wire wound on a long cylindrical wooden rod. — This is the only case in which a simple formula can be established for calculating L. The magnetic field intensity inside of the coil is 4irz7, where 2 is the number of turns of wire per centi- meter of length of coil and I is the current in the coil in ab- amperesf (see Art. 23). Therefore 7rr^X4xz7 is the flux through the opening of the coil where r is the radius of the coil, and zl X 7rr^ X 47rz7 is the flux-turns due to current I where / is the length of the coil and zl is the total number of turns of wire in the coil. Therefore from equation (21) we have ^Z = LI = ^wVzHI whence L = 47rWZ * (27) This equation gives L in abhenrys; to reduce to henrys divide by lo^ PROBLEMS. 86. A coil of wire has a resistance of 20 ohms and an induc- * The calculation of inductance is in general very complicated. See Formulas and Tables for Calculation of Inductance, by E. B. Rosa and F. W. Grover, Bulletin 0} the Bureau of Standards, Vol. 8, No. i, 191 1. An important and interesting discussion of this subject is given by S. Butterworth, in the Philosophical Magazine for April 1915. The absolute measurement of inductance is discussed by Rosa and Grover in the Bulletin of the Bureau of Standards, Vol. 1, No. ^, 1905. t In all equations involving magnetic field intensity or magnetic flux c.g.s. units should be used (abamperes, abvolts etc.), because there are no recognized units of field intensity and flux which correspond to the ampere, volt, ohm, henry, etc. INDUCED ELECTROMOTIVE FORCE. 99 ance of 0.6 henry. When connected to iio-volt direct-current supply mains the current rises to a steady value of 5.5 amperes. How fast does the current start to increase in the coil when it is first connected to the no- volt supply mains and how long would it take for the current to reach its steady value of 5.5 amperes if it continued to increase at this initial rate? Note. — The time which would be required for the current to reach its final steady value at its initial rate of increase is called the time constant of the circuit. 87. From the definition or the time constant of a circuit as given in the note to the previous problem show that the time constant of any circuit is equal to LjR where L is the induc- tance of the circuit and R is the resistance of the circuit. 88. Find the rate of growth of current in the coil of problem 86 at the instant that the current is passing the value of 3 am- peres. 89. The choke coil of a lightning arrester consists of 40 turns of wire in a cylindical one-layer coil 10 centimeters in diameter and 30 centimeters long. Calculate the approximate inductance of the coil in henrys. At the instant of a lightning stroke an electromotive force of 40,000 volts acts across the terminals of this coil. How fast does the current begin to grow in the coil? 90. A current is left to die away in a circuit of which the resistance is 10 ohms and the inductance is 0.05 henry. How fast is the current decreasing as it passes the value 4 amperes? 91. The coil of problem 86 is connected to no-volt direct- current supply mains. What is the value of the growing current after 0.02 second? 91. An alternating current i flows through a circuit of which the inductance is L and of which the resistance is negligible. The expression for i is i = I sin cot where / and oj are constants. Find an expression for the electromotive force which must act on the circuit to cause the current to change in the prescribed manner. CHAPTER V. ELECTRIC CHARGE AND THE CONDENSER. 66. The elimination of the water hammer effect by an air cushion. The elimination of the spark at break by a condenser. jy^^ — The water hammer effect ■- which is produced when a hy- drant is suddenly closed is some- times sufficiently intense to burst the pipe or injure the valve of the k dmht hydrant. In some cases, there- fore, it is desirable to protect the hydrant by an air cushion as indicated in Fig. 82. When the hydrant in Fig. 82 is closed (however quickly) the moving water in the pipe is brought to rest gradually as it slowly compresses the air in the chamber CC. pipe Fig. 82. centrifugal pump check valve pipe Fig. 83a. Figure 83a shows a centrifu- gal pump maintaining a stream of water through a circuit of pipe. If the check valve is sudednly closed, the water will continue for a short time to flow through the circuit into the chamber A and out of the Figure 836 shows a battery maintaining a "current of elec- tricity" through a circuit. If the circuit is broken at p, the electric current will continue for a short time to flow through the circuit into the metal plate A and out of the metal plate ELECTRIC CHARGE AND THE CONDENSER. 1 01 chamber B. This continued flow of the water into chamber A and out of chamber B causes a bending (a mechanical stress) of the elastic diaphragm wire B. This continued flow of the electric current into plate A and out of plate B causes what we may think of as an "electrical bending" (an elec- •— battery P Dz r=io Fig. 836. DD, and this bending soon stops the flow of water; then the diaphragm unbends and produces a reversed surge of water through the circuit of pipe. trical stress) of the layer of insulating material DD, and this "electrical bending" soon stops the flow of current; then the layer of insulating ma- terial "unbends" and produces a reversed surge of electric current through the circuit. The two metal plates A and B, Fig. 836, together with the layer of insulating material between them constitute what is called a condenser. A condenser is usually made of sheets of Fig. 84. tin foil separated by sheets of waxed paper. Thus the heavy horizontal lines in Fig. 84 represent sheets of tin foil, and the fine dots represent insulating material. In order that the follow- ing effects may actually be observed the condenser in Fig. 836 must be made of a large number of sheets of tin foil and waxed paper. I02 LESSONS IN ELECTRICITY AND MAGNETISM. The actual flow of current into the metal plate A and out of the metal plate B when the circuit is broken at p in Fig. 836 is shown by the fact that no spark at break is produced when the condenser AB is connected, whereas a very perceptible spark at break is produced when the condenser AB is not connected- The reversed surge of current which takes place after the original current has been stopped in Fig. 83& may be shown as follows: Disconnect the condenser AB, make and break contact at p, hold a magnetic compass near one end of the core of the inductance* coil, and the core will be found to have retained a large portion of its magnetism; in other words, the core will not have become by any means completely demagnetized when the circuit is broken and the current reduced to zero. Then connect the condenser A B, make and "break the circuit at p as before, and again test the core of the inductance coil with a compass. The core will now be found to have lost nearly the whole of its magnetism because of the reversed surge of current. A reversed surge of current from the condenser in Figs. 73 and 74 is the cause of the very quick demagnetization of the core of an induction coil. 67. The momentary flow of current in an open circuit. Idea of electric charge. — When the metallic contact at p in Fig. 836 is broken the electric circuit remains closed as long as the current continues to flow across the break in the form of a spark. The intensely heated air in the path of a spark is a conductor. When the condenser AB is connected as shown in Fig. 836, there is no spark at break, and the circuit is actually opened at the moment the contact at p is broken. The continued flow of current through the circuit after the contact at p is broken is an example of the momentary flow of current in an open circuit. The current continues for a very short time to flow through the open circuit into plate A and out of plate B after the circuit is broken at p, and the two plates A and B are said to become electrically charged. 68. Definition of the coulomb or ampere-second. — ^An electric * A coil which has inductance. Do not confuse this term with induction coil. ELECTRIC CHARGE AND THE CONDENSER. 1 03 current in a wire may be looked upon as the transfer of "electric- ity" along the wire, and the quantity Q of "electricity" which flows past a point on the wire during t seconds may be defined as the product of the strength of the current and the time, that is we may write : Q = Il (28) in which I is the strength of the current in amperes, and Q is the quantity of electricity which flows past a point on the wire during t seconds. It is evident from equation (28) that the product of amperes and seconds gives quantity of electricity, and therefore the unit of quantity of electricity is most conve- niently taken as one ampere-second,, meaning the amount of elec- tricity which during one second flows past a point on a wire in which a current of one ampere is flowing. The ampere-second usually called the coulomb. One ampere-hour is the quantity of electricity carried in one hour by a current of one ampere. The c.g.s. unit of charge* in the "electromagnetic system" is the amount of charge carried in one second by a current of one abampere, and it is called the ahcoulomh. The abcoulomb is equal to 10 coulombs. 69. Electrostatic attraction. The electrostatic voltmeter. — When a momentary current flows into plate A and out of plate B in Fig. 836, the plates are said to become electrically charged, ■ as stated above, the plate into which the momentary current flows is said to become positively charged and the plate out of which the momentary current flows is said to become negatively charged. Two plates which have thus been oppositely charged attract each other when the intervening insulating material is a fluid like air or oil. This attraction between two oppositely charged metal plates is utilized in the electrostatic voltmeter which consists of a very delicately suspended metal plate and a stationary metal plate, both carefully insulated. The electromotive force to be measured is connected to these plates, a momentary flow of current charges * The word charge as here used means quantity of electricity. 104 LESSONS IN ELECTRICITY AND MAGNETISM. one plate positively and the other plate negatively, the suspended plate is moved by the electrostatic attraction between the plates, and a pointer attached to the movable plate plays over a divided scale. The c.g.s. unit of charge in the "electrostatic system" is that amount of charge which if concentrated on a very small body would repel a similar charge with a force of one dyne at a distance of one centimeter. The "electrostatic system" system of units is not used in this text. 70. Measurement of electric charge. The ballistic galvanom- eter. — A very large charge of electricity may be determined by observing the time during which the charge will maintain a sensibly constant measured current. Thus, a given storage cell can maintain a current, say, of ten amperes for eight hours so that the discharge capacity of the storage battery is equal to eighty ampere-hours. The quantities of charge which are most frequently encountered in the momentary flow of electric current in open circuits are, however, exceedingly small. For example, the terminals of a given condenser are connected to iio-volt direct-current supply mains, and the momentary flow of current represents the transfer of, say, o.oooi of a coulomb which corre- sponds to a flow of one ampere for a ten-thousandth of a second. It is evident that such a small amount of electric charge cannot be measured by the method above suggested. Such small quanti- ties of electric charge are measured by means of the ballistic galvanometer. This galvanometer is an ordinary D'Arsonval galvanometer. When a momentary pulse of current is sent through such a galvanometer, the suspended coil is set in motion, and a certain maximum deflection or throw of the coil is produced. Let d be the measure of this maximum deflection or throw on the galvanometer scale. A certain amount of charge Q is rep- resented by the momentary pulse of current, and this amount of charge is proportional to the throw d. That is, we may write: Q^kd (29) in which ^ is a constant for the given galvanometer, and it is called the reduction factor of the galvanometer. ELECTRIC CHARGE AND THE CONDENSER. 105 The value of the reduction factor k is generally determined in practice by sending through the galvanometer a known amount of charge Q and observing the throw d produced thereby. 71. Definition of condenser capacity. — The amount of charge which is drawn out of one plate and forced into the other plate of a condenser is proportional to the electromotive force which acts upon the condenser. Therefore we may write : Q= CE (30) where Q is the quantity of charge which is drawn out of one plate and forced into the other plate of a condenser when an electromotive force of E volts is connected so as to act upon the condenser, and C is a constant for a given condenser. The factor C is adopted as a measure of what is called the capacity of the condenser. It is evident from the above equation that C, the capacity of a condenser, is expressed in coulombs-per-volt. One coulomb- per-volt is called a farad, that is to say a condenser has a capacity of one farad when an electromotive force of one volt will draw one coulomb out of one plate of the condenser and force one coulomb into the other plate of the condenser. A condenser has a capacity of one abfarad when one abcoulomb of charge would be drawn out of one plate and forced into the other plate by one abvolt. One abfarad is equal to 10' farads. Condenser capacities as usually encountered in practice are very small fractions of a farad. Thus the capacity of a condenser made by coating with tin foil the inside and outside of an ordi- nary one-gallon glass jar would be about one five-hundred- millionth of a farad, or 0.002 of a microfarad. A microfarad is a millionth of a farad, and in practice capacities of condensers are usually expressed in microfarads. The approximate dimensions of a one-microfarad condenser are as follows: 501 sheets of tin foil separated by sheets of paraf- fined paper 0.05 centimeter in thickness, the overlapping portions of the tin foil sheets being 25 centimeters X 25 centimeters, as shown in Fig. 85. I06 LESSONS IN ELECTRICITY AND MAGNETISM. Two pieces of melal of any shape separated by insulating material constitute a condenser; the only reason for using sheets of metal with thin layers of insulating ma- gi V\ terial between is to obtain a large ■^j \ \ capacity. '■§1 \\ Two examples illustrating the A \ \ use of the ballistic galvanometer. I \ \ — (a) Comparison of condenser I — J capacities. — A condenser of which the capacity is C farads is charged by an electromotive I __ 25 ce ntimeters '^' ■ force of E volts and discharged through the ballistic galvanometer. Let d be the observed throw of the galvanometer, then CE = kd (i) This equation is evident when we consider that CE is the charge which has been drawn out of one plate of the condenser and forced into the other plate by the charging electromotive force E, and this amount of charge flows through the galvanometer when the condenser is discharged ; furthermore' the charge which flows through the ballistic galvanometer is equal to kd according to equation (29) of Art. 70. Another condenser of which the capacity is C farads is charged by the same electromotive force E, and discharged through the ballistic galvanometer; and the observed throw of the galvanometer is d' scale divisions. Then: CE = kd' (ii) Dividing equation (i) by equation (ii) member by member, we get C__d_ C'~ d' A condenser of which the capacity has been carefully measured (at the Bureau of Standards, for example) is called a standard ELECTRIC CHARGE AND THE CONDENSER. 107 condenser. If C in equation (i) is the known capacity of a standard condenser and if the value of E is known, the value of k (the reduction factor of the galvanometer) may be calculated, the throw d being observed (6) The measurement of magnetic flux. — ^An iron rod has a winding of wire through which a magnetizing current may be passed, and the magnetism of the rod may be reversed by revers- ing the magnetizing current* this changing the magnetic flux through the rod from — $ to + $ so that the total change of flux through the rod will be 2$. An auxiliary winding of Z turns of wire surrounds the iron rod, this auxiliary winding is connected to a ballistic galvanometer, and d is the observed throw of the ballistic galvanometer when the magnetic flux through the iron rod is suddenly reversed. Then Rkd 2* = -^ where R is the total resistance of the ballistic galvanometer circuit (including, of course, the resistance of the auxiliary winding). If R is expressed in ohms, and if k is in coulombs per division, then Rkd , ..... 2$ = —^ X 10' (ui) Derivation of equation (Hi). Let — be the rate of change of the magnetic flux at any instant during the above mentioned reversal. There —j— is the value of the electromotive force induced in the auxiliary winding (algebraic sign does not concern us) and h R is the current in the ballistic galvanometer circuit.f dt But the current in the ballistic galvanometer circuit is the rate — at which electric * The magnetizing current has to be reversed several times before the reversal of current will change the flux through the rod from + * to — *. t Any delay of current growth due to inductance causes a prolongation of current -ow later, and it can be shown that the inducta nee of the galvanometer circuit has no effect on the total amount of charge which passes through the circuit when * is reversed. Io8 LESSONS IN ELECTRICITY AND MAGNETISM. charge is passing through the galvanometer. Therefore dQ _Z d* dl ~ R dt This equation means that Q changes ZjR times as fast as # so that total change in Q (total Q passing through the galvanometer) is ZjR times total change in *. But the total change of $ is 2$ as above explained so that Q = kd = xZ^IR. PROBLEMS. 93. Imagine the current in a circuit to increase at a constant rate from zero to 50 amperes in 3 seconds. Find the number of coulombs of charge carried through the circuit. 94. An electromotive force which acts on a condenser increases at a constant rate from zero to 1,000 volts during an interval of 0.005 second. The capacity of the condenser is 20 microfarads. Find the value of the current. 95. A 6o-cycle alternating electromotive force of which the the maximum value is A/2 X no volts acts on a condenser Find the average value of the current during the 1/120 of a second during which the electromotive force acting on the condenser changes from — 155 volts to + 155 volts, the capacity of the condenser being 20 microfarads. 96. The alternating electromotive force which is represented by the ordinates of the zigzag line in Fig. 86 acts on a condenser y^ — *' /\ Fig. 86. of which the capacity is 20 microfarads. Plot a curve of which the ordinates represent the successive instantaneous values of the current flowing into and out of the condenser. 97. An alternating electromotive force e = £ sin ut acts on a condenser of which the capacity is C. Find an expression for the current flowing into and out of the condenser. 98. A standard one-microfarad condenser is charged by 14.34 volts and discharged through a ballistic galvanometer producing ELECTRIC CHARGE AND THE CONDENSER. 109 a throw of 20.7 scale divisions. What is the value of the reduc- tion factor of the galvanometer in abcoulombs per division? A very small Coil of 100 turns of wire, mean radius of the turns being 1.2 centimeters, is connected to the above ballistic gal- vanometer, and placed between the poles of a powerful magnet with the plane of the coil at right angles to the lines of force. When the coil is quickly jerked out from between the poles of the magnet the galvanometer throw is observed to be 18.3 divi- sions. The resistance of the galvanometer circuit (total) is 3,580 ohms. Find the intensity of the magnetic field between the poles of the powerful magnet. 72. Inductivity of a dielectric. — The insulating material be- tween the plates of a condenser is called a dielectric. Indeed, the insulating material between any two oppositely charged bodies is called a dielectric. The capacity of a condenser depends upon the size of the plates, upon the thickness of the dielectric Fig. 87o. Fig. 876. and upon the nature of the dielectric. The dependence of the capacity of the condenser upon the nature of the dielectric is a matter which must be determined purely by experiment. Thus Fig. 87a represents two metal plates with air between them, and Fig. 876 represents the same plates immersed in oil. The dis- tance between the plates is understood to be the same in Figs. 87a and 87&. . Let C be the capacity of the condenser in Fig. no LESSONS IN ELECTRICITY AND MAGNETISM. 870 with air as the dielectric, and let C be the capacity of the condenser in Fig. 876 with a given kind of oil as the dielectric. The ratio C'/C is called the *«//./ 77/777////, ^active material Fig. 132. The a-partides are deflected towards the reader, and the j8-particles are deflected away from the reader. in Fig. 133 represents the path of the ball, D being the horizontal distance traveled by the ball and d being the vertical distance fallen by the ball under the deflecting force of gravity. If D Fig. 133. is known and d observed, the velocity of the ball is given by the equation V' = 2d (i) in which g is the acceleration of gravity. Action of the electric field on a moving charged particle. — Consider a charged particle moving upwards through an electrical ATOMIC THEORY OF ELECTRICITY. i6S field as indicated in Fig. 134. Let q be the charge on the par- ticle in abcoulombs, and let / be the intensity of the field in abvolts per centimeter. Then the force F in dynes pulling i ledd'hloei Fig. 134. Electric field from left to right. lead bUtek '. ', lead block Fig. 135. Magnetic field towards reader. sidewise on the particle is /g so that the sidewise acceleration of the particle is /g/wJ, where m is the mass of the particle in grams. It is evident that the particle moves in the same kind of path as the cannon ball in Fig. 133 (resistance of air zero). Therefore using fglm for g in equation (i) , we have or 2 dm q 2ii? (ii) When the velocity v has been determined, as explained below, this equation gives the value of the ratio g/w. Action of the magnetic field on a moving charged particle. — Figure 135 represents a charged particle moving upwards through a magnetic field. The moving particle is equivalent to an electric current and the side-wise force F is equal to gvH where q is the charge on the particle, v is the velocity of the particle and H is the intensity of the magnetic field in gausses. The force F is always at right angles to v and therefore the path of the par- l66 LESSONS IN ELECTRICITY AND MAGNETISM. tide is a circle, and the radial acceleration is Flm or gvH/m where m is the mass of the particle. But the acceleration of a particle moving at velocity »• in a circular path of radius r is v^/r, and therefore gvH v^ m r or q V When measurements have been taken from which r can be calculated, when H is known, and when v has been determined as explained below, this equation also gives the value of the ratio g/w. Calculation of velocity of moving particle. — Equating the two values of g/jw from equations (ii) and (iii) and solving for v, we get m ,. . " = 2drH ('^^ The significance of D and d may be understood by comparing Figs. 133 and 134, and both may be easily measured; also /, r and H are easily determined, and therefore, when the neces- sary measurements have been made, equation (iv) gives the value of V. In this way it is found that the velocity of the a-particles from radium is about 20,000 miles per second, and the velocity of the /3-particles from radium is about 160,000 miles per second. The mass of a moving particle increases with its velocity, and the above equations should be modified to take this increase of mass into account; the purpose of the above discussion, however, is to set forth only the simplest ideas of the subject. A very good discussion of the methods for determining the values of m and g is given in a very recent book, The Electron, Professor R. A. Millikan. 112. Electron theory of conduction in metals, contact poten- tial-differences, and thermo-electromotive forces. — The electron theory has given a wonderfully consistent and accurate inter- ATOMIC THEORY OF ELECTRICITY. 167 pretation o. many of the phenomena of the discharge of electricity through gases, whereas the application of the electron theory to metallic conduction, to contact potential-differences and to thermo-electromotive forces is by no means entirely consistent or accurately interpretative, and the following discussion is justified chiefly by the fact that the phenomena themselves are familiar. The older atomic theory, likewise, has given a wonderfully com- plete and accurate interpretation of the properties of gases, whereas the application of the atomic theory to liquids and solids is by no means complete or accurately interpretative, and yet everyone nowadays accepts the atomic theory as applying to liquids and solids. The fundamental hypothesis on which the following discussion is based is that a block of metal contains a great number of free electrons (negatively charged particles) which wander about in the block of metal very much as the atoms of a gas wander about in a containing vessel. The electric current is carried through a wire by a steady drift of the free electrons, and heat is generated by the impact of the electrons with the atoms or molecules of the metal, an electron being set in motion again after each impact by the electromotive force along the wire. When the metal is heated the irregular to and fro motion of the free electrons is increased, an increased pressure (electron pressure) is produced by this increased motion very much as the pressure of a gas is increased by the increased molecular motion when the gas is heated, and when the temperature is very high some of the electrons escape from the metal as described inArt. loi . Contact difference of potential. — Figure 136 represents two blocks of different kinds of metal A A and BB welded together along cc. Metal A contains rnore free electrons per cubic centimeter than metal B, therefore the electron pressure in A is greater than the electron pressure in B, therefore electrons tend to drift from the high pressure region P to the low pressure region p, the effect of this drift is to leave an excess of positive i68 LESSONS IN ELECTRICITY AND MAGNETISM. ■.■;'/'.:.^.'fi'-;». c -le « B ■P.' charge in A and produce an excess of negative charge in B, and the drift continues until these charges become large enough to balance the difference of electronic pressure. Thus Fig. 136 shows an electron e being pulled back from B towards A by the electric forces. When the balanced condition is reached there is a definite electromotive force or potential difference between A A and BB which is called a contact potential difference. Such contact potential differences have long been known to exist, they are quite small, only a few tenths of a volt, and they cannot be detected by an ordinary voltmeter because they always bal- ance out when different metals are connected in a circuit (temperature being everywhere the same). The thermo-element and thermo-electromotive force.* — When a circuit is made of two metals as indicated in Fig. 137, a Fig. 136. c c represents an actual contact or junction of A and B very greatly magnified and idealized. copper Fig. 137. metal A icmtem^J metal B — f+aP -^p+ip Fig. 138. current flows around the circuit when the two junctions Ji and Ji are at different temperatures. In fact a definite electro- motive force acts around the circuit when the temperatures of Ji and J2 are given. This electromotive force, which can be measured by a galvanometer G is called a thermo-electromotive force, and the circuit of two metals is called a thermo-element. Figure 138 represents a circuit formed of two metals A and * A fairly complete discussion of thermo-electromotive forces is given in Nichols and Franklin's Elements of Physics, Vol. -^, pages 216-221. See also J. J. Thomson's Elements of Electricity and Magnetism, pages 506-518. ATOMIC THEORY OF ELECTRICITY. 169 B, one junction Ji being at temperature T and the other junction Ji being at temperature T + AT. The electronic pressure in metal A increases from P to P + AP, and the electronic pressure in metal B increases from p to p + Ap due to increase of temperature from P to P + A P as indicated AP P in the figure, and as in the case of gases — — = — . That is to say Ap p the increase of electronic pressure AP is larger than the increase of electronic pressure Ap, consequently the contact potential difference at J2 is not equal to the contact potential difference at Ji and, as a result,* current flows round the circuit. * This statement ignores the potential drop along A due to changing electronic pressure along A, and it ignores the potential drop along B due to changing elec- tronic pressures along B. No attempt is here made to give a complete discussion. Indeed a complete discussion does not lead to results which are in accordance with all the known facts. APPENDIX A. THE MAGNETISM OF .IRON. 1. Magnetizing force in iron. — ^When an iron rod is placed in and parallel to a magnetic field the rod is magnetized, and after the rod is magnetized its free poles modify the previously existing magnetic field. The actual magnetic field which acts upon and magnetizes the rod may he very different from the original field. For example an iron rod 20 centimeters long is placed inside of a long coil where the magnetic field, before the rod is put in place, has an intensity of 40 gausses. Let us suppose the rod to be magnetized so that each of its poles has a strength of i ,600 units. The actual magnetic field is now due to the combined action of the coil and the free poles of the rod, and near the middle of the rod the intensity of this actual field is 40 gausses minus 32 gausses, or 8 gausses, because, according to Art. 11 of Chapter I, the poles of the rod produce a field intensity of 32 gausses near the middle of the rod and in a direction opposite to the field due to the coil. The "magnetizing force" or magnetizing field means always the net intensity of field due not only to outside agencies but due also to the free magnet poles on the magnetized iron itself. There are two cases in which free magnet poles are without appreciable influence, namely, (a) When a very long and slender iron rod is magnetized, and (&) When the magnetic circuit is wholly of iron so that the magnetic flux never passes from iron to air (a north pole) or from air to iron (a south pole). In case a the poles are so weak and so far away from the middle portions of the rod that their influence is negligible, and in case b there are no free poles. 2. Flux density in iron. Definition of magnetic permeability. — Consider a long coil of wire (a long tube wound with wire 170 MAGNETISM OF IRON. 171 over its entire length). The intensity of the magnetic field inside of this coil in gausses is Z . R = \TC I (I) where Z is the number of turns of wire on the long tube, / is the length of the tube in centimeters, and / is the current m abamperes. Let 2 be the sectional area of the opening or bore of the tube. Then when the tube is filled with air, or any ordinary non-mag- netic material, the magnetic flux through the tube is g^H, accord- ing to Art. 12 of Chapter I, and the flux per unit area (thej^M^ density) is, of course, H. If the long tube is filled with iron the magnetic flux through the tube is many times as great and the flux per unit area (the flux density) is correspondingly great. Indeed we may write B -= IJ.H (2) where H is the "magnetizing force" or magnetizing field, B is the flux density in the iron rod, and /i is what is called the permeability of the iron. The permeability of a given sample of iron is not constant but it grows less and less as the flux density B increases (as the iron approaches what is called mag- netic saturation). The permeability of air and of all ordinary non-magnetic materials is unity. The accompanying table gives the corresponding values of B and H for wrought iron, for cast iron and for soft cast steel. TABLE. Magnetic Properties of Iron and Steel. Wrought Iron (Hopkinsoo). Cast Iron (M.E.Thompson). Soft Cast Steel. H B /» zr B (« If IS /» 10 12,400 1,240 10 5,000 500 10 9.700 970 20 14.330 716 20 6,600 330 20 13.380 669 30 IS. 100 503 30 7.290 246 30 13.500 483 40 IS.S50 389 40 7.850 195 40 15.250 3S1 so IS.9S0 319 so 8,360 169 SO IS.840 317 60 16,280 271 60 8,800 146 60 16,300 272 70 i6,soo 23S 70 9,200 131 70 i6,7So 239 172 LESSONS IN ELECTRICITY AND MAGNETISM. coif oooooo oooooo oooooo oooooo oooooo 800000 ooooo oooooo 3. The magnetic circuit. Definition of magnetomotive force. — The path of any portion of magnetic flux is always a closed path or circuit, and in many cases in practice this closed path or circuit is nearly all of iron. Therefore as the simplest case let us consider a magnetic circuit, so called, which is wholly of iron. Thus the iron rod which is shown in Fig. I is a magnetic circuit, the mag- netizing action of the coil on the rod depends upon the average value of the magnetizing field, the average value of H, along the rod; and the product IH is called the magnetomotive force around the circuit which is formed by the rod, where I is the length of the rod (distance around the circuit) and H is the average magnetizing field {component of actual field parallel to the rod at each point). This definition of magnetomotive force applies not only to an entire circuit, but to any portion of a circuit. Thus IH is the magnetomotive force acting on any portion of a magnetic circuit where / is the length of the portion and H is the average magnetizing field along and parallel to the portion. 4. Magnetomotive force of a coil. — The magnetomotive force around any circuit or path which links with a coil or winding of Fig. 1. wire is 4irZ7 (3) where Z is the number of turns of wire in the coil and I is the current in the coil in abamperes. This equation is derived in Art. 7 of this Appendix. 5. Calculation of the magnetomotive force required to produce a specified amount of magnetic flux around a given magnetic circuit. (a) Divide the specified amount of flux by the sectional area of each portion of the given magnetic circuit, wrought iron. MAGNETISM OF IRON. 173 cast iron, steel or air as the case may be. This gives the flux density B in each portion of the circuit.* ip) Knowing B for each part of the magnetic circuit, take from the above table the corresponding value of H for each part of the circuit. In an air gap the value of H is equal to B. (c) Multiply the value of H for each portion of the magnetic circuit by the length / of that portion in centimeters. This gives the magnetomotive force required for each portion. {d) Add the magnetomotive forces found under (c) to get the total magnetomotive force. (e) The total required magnetomotive force is equal to ^tZI, where Z is the number of turns of wire in the magnetizing coil and / is the magnetizing current in abamperes. Then if Z is given the value of F may be found, or if I is given the value of Z may be found. Example. — The magnetic circuit of a dynamo consists of wrought iron (in armature core and field magnet coresf), cast iron (in field magnet yoke and pole pieces), and air gap. The dimensions of each portion of the magnet are estimated as fol- lows: Wrought iron portion 50 centimeters long and 120 square centimeters in sectional area, cast iron portion 40 centimeters long and 202 square centimeters in sectional area, and air portion is 2.5 centimeters long and 300 square centimeters in sectional area. How many ampere-turns are required to force 1,600,000 lines of magnetic flux through the circuit, ignoring the leakage of flux through the surrounding air from pole piece to pole piece? (The flux density in the wrought iron portion is B — 13,330 The flux density in the cast iron portion is B = 7,270 The flux density in the air portion is B = 5,330. The value of H in the wrought iron is H = 14.8 gausses. (b) ■ The value of H in the cast iron is H = 30.0 gausses. . The value of H in the air gaps is H = 5330 gausses. * No account is here taken of what is called magnetic leakage, t Portions of the field magnet structure on which the magnetizing coils are wound. (c) 174 LESSONS IN ELECTRICITY AND MAGNETISM xne magnetomotive force required for the wrought iron portion is 740 gauss-centimeters or gilberts. The magnetomotive force required for the cast iron portion is 1,200 gilberts. The magnetomotive force required for the air gaps is 13,320 gilberts. (d) The total required magnetomotive force is 15,260 gilberts. (e) Dividing 15,260 by 47r gives Z/ = 1,215 abampere- turns or 12,150 ampere- turns. 6. To calculate the magnetic flux produced around a given magnetic circuit by a specified magnetomotive force. — When the given magnetic circuit consists of different materials, as in the above example, this problem is solved by calculating the magnetomotive forces required to produce a series of assumed values of flux, these results are plotted as a curve, and the flux corresponding to the given magnetomotive force is taken from this plotted curve. Magnetic reluctance. — The electrical engineer usually carries out a magnetic circuit calculation as follows: After finding the flux density B in any portion of the magnetic circuit, and tak- ing the value of n from the table, he calculates the quantity I I where I is the length in centimeters and a is the sectional M 2 area in square centimeters of the portion of the magnetic cir- cuit. This quantity is called the magnetic reluctance of the por- tion. Having thus determined the magnetic reluctances of the various portions of the circuit, the reluctance of the entire mag- netic circuit is found exactly as the electrical resistance of an entire electrical circuit is found from the resistances of the various portions of the electrical circuit. Then the magnetic flux through the circuit is found by dividing the magnetomotive force by the total reluctance of the magnetic circuit. This method is the exact equivalent of the method which is outlined above. MAGNETISM OF IRON. 175 PROBLEMS. 1. An iron rod 2 centimeters square and 20 centimeters long is magnetized to an intensity of 1,000 units pole per square centimeter section when it is placed in a region which, but for the action of the poles of the rod, would be a uniform field parallel to the rod and of an intensity of 102 gausses. Assuming the poles of the rod to be concentrated at its ends, calculate the net magnetizing field at the center of the rod. 2. A laminated iron core two feet long and J inch in diameter is wound from end to end with a layer of wire containing 50 turns. Assuming that the permeability of the iron is constant and equal to 1,150, and ignoring the demagnetizing action of the free poles of the rod, calculate the inductance of the winding in henrys. 3. The intensity of the magnetic field in the air gap between the pole face and the armature core of a dynamo is 3,500 gausses and the field is at right angles to pole face and armature surface. The distance across the air gap is 3/8 inch. Find the magneto- motive force across the air gap in gilberts and in ampere-turns. 4. A slim rod 25 centimeters long is made into a link which passes through a coil of 50 turns of wire in which a current of 15 amperes is flowing. Find the average value along the rod of the component parallel to the rod of the magnetic field due to the coil. Express the result in gausses. 5. A transformer has a laminated soft-iron core of which the sectional area is 120 square centimeters and the mean length of the magnetic circuit formed by the core is 100 centimeters. How much current must flow through a winding of 500 turns of wire to produce a magnetic flux of 1,767,000 lines? 6. How much flux will be produced through the laminated iron core of the previous problem by a current of 8.2 amperes in the 500 turns of wire? 7. How much magnetic flux would be forced through the mag- netic circuit which is specified in the example of Art. 5 by a magnetomotive force of 13,500 ampere-turns? 176 LESSONS IN ELECTRICITY AND MAGNETISM. 7. Proposition. — The magnetomotive force along a path in a magnetic field is equal to the work per unit pole done by the magnetic field upon a magnet pole while the pole is made to travel along the path; that is, the magnetomotive force along a path is equal to WIm, where W is the work done by the field upon a pole of strength m while the pole is made to travel along the path. That is : F=— (4) m in which F is the magnetomotive force along the path. Equation (4) is usually taken as the definition of magneto- motive force, but it is necessary here to derive it from the defi- nition of magnetomotive force which is given in Art. 3. If H is the average value along a path of the component of a magnetic field parallel to the path at each point, then mH is the average force parallel to the path with which the field acts on a pole of strength m while the pole moves along the path so that lmH=W W is the work done on the pole by the field. Therefore IH = — m where IH is the magnetomotive force along the path. 8. Proposition. — The magnetomotive force of a coil is given by the equation: F = ^irZi (3) in which Z is the number of turns of wire in the coil, and i is the strength of the current in the wire in abamperes. Proof. — Before proceeding to the derivation of equation (3), it is necessary to find an expression for the total work W done in keeping the current i in a coil constant while the magnetic flux through the opening of the coil is increased by a specified amount $ ; W being expressed in ergs, i in abamperes, and $ in maxwells or lines. Now, while the flux is increasing, a back electromotive force equal to Z — abvolts is induced in the coil, and therefore (assuming the coil to have zero resistance for the MAGNETISM OF IRON. 177 sake of simplicity of statement) an electromotive force e = Z ^- dt will have to act on the coil to keep the back electromotive force from decreasing the current. Consequently work will have to d^ be done on the coil at the rate ei or Z -^ X i, to keep the cut^ at rent from decreasing. That is : dW dt = Zi dt (i) But Zi is a constant, because * is being kept from changing. Therefore by integrating (i) from W = o and # = o we get: W = Zi^ . (ii) in which W is the work in ergs done to keep a current of i abamperes constant in a coil (of zero resistance) while the mag- netic flux through the opening of the coil increases by the amount of $ maxwells, and Z is the number of turns of wire in the coil. We shall now proceed to the proof of equation (3). Let ZZ, Fig. 2, represent a coil of Z turns of wire. Im- agine NS to be a flexible magnet. Let the north pole of this flexible magnet be carried through the coil and around to its initial position along the dotted path. The flexible magnet will then link with the coil of wire as shown in Fig. 3. Let F be the magnetomotive force along the dotted path, and let m be the strength of the pole N which has been carried around the path. Then, according to equation (4) , Fm is the work done on the pole by the magnetic field of the coil. This work done on the momng pole by the magnetic field of the coil is the work which 13 iooo ioooz i ooo Fig. 2. 178 LESSONS IN ELECTRICITY AND MAGNETISM. is spent in keeping the current constant in spite of the electromotive force induced in the coil by the moving pole. This is evident when we consider that work done on the pole by the field must be made good that is, energy must be supplied from somewhere : and when we consider that the energjy stored in the system is the same in Figs. 2 and 3. Therefore, the only possible source of the energy for doing work on the moving pole is the work that is done to keep the current in the coil constant. Now the two poles of the flexible magnet are in the same positions before and after the movement; therefore, of the total number of lines of force which radiate from these poles, the same number pass through the coil be- fore and after the move- ment. On the other hand, the flux ^irm (see Art. 12 of chapter I), which passes along the magnet from pole to pole, passes through the coil after the movement, so that the flux through the opening of the coil is increased by the amount ^irm by the movement of the pole. Therefore, according to equation (ii), Zi X ^irm is the work spent in keeping the current constant during the movement of the pole and, since this is equal to the workFm done upon the pole by the magnetic field of the coil, we have Fig. 3. or Fm = /\viZm F = 47rZt 9. Work required to magnetize iron. — ^When an iron rod is magnetized by sending an electric current through a coil of wire surrounding the rod, an opposing electromotive force is induced in the coil by the growing magnetism of the rod, and the work done in forcing the current against this opposing electromotive force is the work expended in magnetizing the rod. MAGNETISM OF IRON. 179 The work W, in ergs, which is done in magnetizing V cubic centimeters of iron from a given initial flux density B' to a given final flux density B", is given by the equation: W = - \ H-dB (5) Proof. — In order to avoid the complications which arise on account of the perceptible demagnetizing action of- the poles of a short iron rod, let us consider a very long slim rod I centi- meters in length and g square centimeters in sectional area. Suppose this rod to be placed in a long coil having z turns of wire per centimeter of length or Iz total turns. When the coil of wire is first connected to the battery or other source of current, the current in the coil (beginning at zero) rises in value during the time that the rod is being magnetized, and during this time the magnetic flux through the rod is growing in value. Let d^/dt be the rate at which the flux is increasing at a given instant, and let i be the value of the current at this instant. Then Iz X d^fdt is the induced electromotive force in the coil which at the given instant is opposing the current i, so that Iz X d^/dt X i is the rate, dW/dt, at which work is being done at the given instant in magnetizing the rod. That is: dW ^ . d^ ~df = ^'''^ so that: dW = Izi ■ d^ ■ (i) in which dW is the amount of work done during the time that the flux has increased by the amount d^ and while the current has the mean value i. Now ^ — Bq or di = q • dB, and zi = H/^t from equa- tion (i) of Art. 2 of this Appenidx, so that equation (i) becomes: la dW = — H ■ dB 47r l8o LESSONS IN ELECTRICITY AND MAGNETISM, or, since Iq = V, we have: dW = — H ■ dB or W = — \ H-dB In magnetizing a short iron rod, more work is done than is accounted for by equation (5). The additional work goes to estabhsh the magnetic field in the neighborhood of the magnetic poles of the rod. Equation (5) expresses the work which is spent within the iron. 10. Graphical representation of work done in magnetizing iron. — Let the curve Opp', Fig. 4, be drawn so that the co- ordinates represent corresponding values of B and H for a given sample of iron. The branch Op represents the values of B and H when the iron is being magnetized for the first time, and the branch pp' represents the values of B and H when, after the iron has been magnetized up to the point p, the value of H is slowly reduced to zero. The curve B-axis gj -Q ^^^ jj jg^ decreasing values of H ^a^^" does not coincide with the curve jor increasing ■ / values of H. Now, as explained later, the ^ total area Opa represents the work done upon the iron in magnetizing it up to the point p, J and the area pp'a represents the work which H-axis is regained from the iron when the magnetiz- ing field drops slowly to zero. The work re- gained is less than the work required to magnetize the iron. The work which is lost is represented by the shaded area in Fig. 4. That area in Fig. 4 represents work may be shown as follows: Abscissas represent values of H to scale, so that we may write: H = ax (i) MAGNETISM OF IRON. l8l Ordinates represent values of B to scale, so that we may write: B = by (ii) or dB = b • dy Substituting these values of H and dB in equation (5), we have abV r W = t X ■ dy 41- J in which a is the number of units of H represented by one unit of abscissa, and b is the number of units of B represented by one unit of ordinate in Fig. 4. Now j x • dy is the area between any portion of the B and H curve and the y-axis. Therefore, ab V/4.T is the number of ergs of work represented by each unit of area between the B and H curve and the y-axis. II. Magnetic hysteresis. The magnetic cycle. — The diver- gence of the B and H curve for increasing values of H from the B and H curve for decreasing values of H is called mag- netic hysteresis ; or, rather, the tendency of iron to retain a previous magnetic state which is the cause of this divergence is called magnetic hysteresis. One effect of magnet hysteresis is that the work regained when iron is demagnetized is less than the work which must be spent to magnetize the iron, as pointed out in Art. 10. The magnetization of a given portion of the iron core of a transformer is reversed with each reversal of the alternating current which flows through the primary coil, the iron is thus repeatedly carried from a certain degree of magnetization in one direction (a certain positive value of B) to the same degree of magnetization in the opposite direction (the same negative value of B), and back to the original degree of magnetization. Such a magnetic double-reversal is called a magnetic cycle. At the end of a cycle the iron comes to precisely the same condition that it had at the beginning of the cycle. The importance of the 1 82 LESSONS IN ELECTRICITY AND MAGNETISM. B-axis Fig. 5. magnetic cycle in the determination of the amount of heat produced by magnetization will be appreciated from tho^foUow- ing two statements : (a) When a mass of iron is magne- tized along the B and H curve Op of Fig. 4 and then partially demagnet- ized along the curve pp', a portion of the work done- upon the iron during the first stage Op is regained during the stage pp', a portion is lost as heat, and a portion remains in the iron as energy of magnetization ; and no experi- mental method has been devised for determining the second and third por- ions of work separately. (b) When, however, a mass of iron is carried through a mag- netic cycle, the algebraic sum of the work spent upon the iron during the cycle is lost as heat, inasmuch as the magnetic energy in the iron is exactly the same at the beginning and at the end of the cycle. Figure 5 shows the relation between B and H during a complete magnetic cycle. The total work spent on the iron is given by the value of the integral -— \ X- dy (i) extended over the whole cycle; but the value of I a; • rfy extended over the whole cycle is the area enclosed by the B and H curve. Therefore, the total energy, in ergs, lost in V cubic centimeters of iron per magnetic cycle, is equal to ab V/^w X area enclosed by the B and H curve. This energy loss is called the hysteresis loss, and it is all converted into heat. The meanings of a and b are explained in Art. lo. The hysteresis loss per cycle increases with the range of flux MAGNETISM OF IRON. 183 density, and it may be expressed with sufficient accuracy for most practical purposes by the empirical equation: W = vVB^-^ (6) which is due to Steinmetz. In this equation W is the loss of energy in ergs per cycle, V is the volume of the iron in cubic centimeters, ±-B is the range of flux density during the cycle, and r; is a coefficient which is nearly constant for a given kind of iron or steel. The following table gives the approximate values of t; for different kinds of iron and steel. TABLE. Values of Hysteretic Coefficient ij. Best quality of sheet iron for transformer cores, annealed 0.0013 Sheet iron for armature cores, annealed 0.003 Ordinary sheet iron, annealed 0.004 Soft annealed czist iron 0.008 Soft machine steel 0.0095 Cast steel 0.12 Hardened steel 0.25 12. The magnetic testing of iron (Rowlands' method). A ring- shaped sample of the iron or steel to be tested is wound with a magnetizing winding of Z turns of wire as indicated in Fig. 6, arrangements are made whereby the magnetizing current in this winding .^<^ \t-^i^ can be increased and decreased by sudden steps, an auxiliary winding of Z' turns is wound on the ring and connected to a ballistic galvanometer, and the throw of the ballistic is ob- p- g served at every step of the magne- tizing current. The object of the test is to get data from which a hysteresis loop like Fig. 5 can be plotted, but we will here ex- plain only how to get the values of B and H for the extreme points of the hysteresis loop. A measured magnetizing current of I abamperes is sent through the magnetizing winding and reversed several times so 1 84 LESSONS IN ELECTRICITY AND MAGNETISM as to wipe out residual effects of previous magnetic conditions of the sample, then a reversal of I changes the magnetic flux (through the sample) from a certain positive value $ to an equal negative value $ thus giving a total change of flux 2$ which may be. calculated from the observed throw of the ballistic galvanometer as explained in Art. 71 of Chapter V. Thus we get the value of #, and dividing $ by the sectional area of the ring-shaped sample we get the value of flux density B. The corresponding value of H is equal to ^irZI/l where / is the magnetizing current in abamperes and / is the mean circumferen- tial length of the ring-shaped sample. Problems. 8. (a) Find the work in ergs spent in magnetizing a wrought iron bar 3 inches square and 20 inches long from a neutral condi- tion to B = 16,000 lines per square centimeter, using the tabular values of B and H given in Art. 2. (b) Find the work in ergs required to magnetize a cast-iron rod of the same size from a neutral condition to B = 9,000 lines per square centimeter. Note. — Plot the B and H curves from the tabular values given in Art. 2. Divide the areas between the curves and the B axis into a number of narrow strips, and calculate the area of each strip. Add these areas to get the total area. 9. A transformer core contains 96 cubic inches of the best quality of transformer iron. The core is carried through 60 magnetic cycles per second between the limits B = d= 3,500 gausses, by means of an alternating current. Find the hysteresis loss in the core in watts. APPENDIX B. ELEMENTARY THEORY OF ALTERNATING CtTRRENTS. 13. Undamped oscillations and damped oscillations. — An os- cillating electrical circuit like Fig. 89 or Fig. 93 of Chapter V always has more or less resistance, and an oscillating circuit always loses energy more or less rapidly by the emission of electro- magnetic waves. Therefore electrical oscillations once started die away rapidly. Such oscillations are called damped oscillations. The frequency equation (36) of Art. 78, Chapter V, applies to an ideal oscillating circuit which has no resistance and which loses no energy by the emission of electromagnetic waves, a circuit which once started would continue to oscillate indefinitely per- forms what would be called undamped oscillations. When an oscillating circuit like Fig. 93 of Art. 78, Chapter V, has resistance R, equation (36) is not applicable, but the frequency is given by _ and each successive maximum positive value of the current is equal to the previous maximum value divided by «"'', where e is the Naperian base, T is i// of a second, and The current is called a decaying oscillatory current, and equations (i) and (ii) come from the solution of the differential equation of the circuit. Consider a circuit like Fig. 93 of Chapter V, the capacity of the condenser being C, the inductance of the circuit being L and the resistance of the circuit being R. Let g be the charge on the condenser at any instant. Then the electromotive force across the condenser is gj C, this electromotive force pushes back 18S l86 LESSONS IN ELECTRICITY AND MAGNETISM. wards on the circuit so that it is to be considered as negative, part of it is used to overcome resistance and part of it is used to make the current charge. The part used to overcome resistance da is Ri or R-;j , and the part used to make the current change is di d^q L-T- or I. TiT . Therefore we have at ar C ~ ^dt'^ ^dt' whence we have da d^q a+CR^+CLj=o This is the differential equation of the circuit, and its general solution is 2 = Qr"^ sin {cot - 6) ■ ^^ u or, since i = 'j: > we have i = It~°* cos (aj< — ff) (iii) where I is written for the value of i when t = o. This is the equation of the decaying oscillatory current. 14. Free and forced oscillations. — When a pendulum is started and left to itself its oscillations are called its free oscil- lations; if, however, a periodic force (a steady series of impulses) acts on a pendulum the pendulum soon settles to a steady state of oscillatory motion in which the frequency is exactly the same as the frequency of the force, regardless of the frequency of the free oscillations of the pendulum. This steady state of oscillation which is produced by a periodic force constitutes what is called forced oscillations. Similarly, an electric circuit like Fig. 93 of Chapter V performs free oscillations of which the frequency is given by equation (i) above when it is started and left to itself, but if such a circuit (or any circuit) is acted upon by an alter- nating electromotive force of any given frequency it soon settles to a steady condition of oscillation {forced oscillation) in which the frequency is that of the applied alternating electromotive force ELEMENTARY THEORY OF ALTERNATING CURRENTS. 187 regardless of the frequency of the free oscillations of the cir- cuit. The whole of the elementary theory of alternating currents has to do with steady states of forced oscillation. 15. The alternating-current ammeter and voltmeter. — In using an ordinary alternating-current ammeter or voltmeter the frequency is so high that the pointer of the instrument stands still at a definite scale reading, so many "amperes" or "volts" as the case may be, and it is important to consider precisely what this constant reading means. It evidently does not mean the average value of the alternating current or electromotive force, because the average value is zero. Indeed a direct-current ammeter or voltmeter of the type described in Art. 18 of Chapter I measures the true average value of a rapidly fluctuating current or electromotive force, and such an instrument gives zero reading for an alternating current or electromotive force. To understand the meaning of an ampere reading or voltage reading on an alternating-current instrument let us consider the electrodynamometer type of ammeter or voltmeter. This instrument consists of a fixed coil and a delicately pivoted coil connected in series with each other, the current which produces the deflection flows through both coils, the fixed coil exerts a force on the pivoted coil which moves the coil and the attached pointer, and this force is at each instant strictly proportional to the square of the current for a given position of the pivoted coil. (a) Let c? be the constant or direct current which produces a certain deflection or scale reading 5 on an electrodynamometer type of ammeter. Then the steady deflecting force exerted on the pivoted coil is equal to k ^, where ^ is a constant. (6) Let i be the varying value of an alternating current which gives the same steady deflection s. Then the instantaneous value of the force exerted on the pivoted coil is ^?, and the average value of this force is ^ X average i^. Now the steady deflecting force under (o) must be equal to 1 88 LESSONS IN ELECTRICITY AND MAGNETISM. the average deflecting force under (b) because the deflection s is the same. Therefore* kif^ = k X average i^ or ^ = average i^ or d = Vaverage i^ (i) But the value of tf is marked on the ammeter scale where the pointer stands under (o) above. Therefore t? is the reading 5 of the ammeter under (b). Therefore, according to eqiiation (i), the reading of an alternating-current ammeter gives the square- root-of-the-average-value-of-the-square of the alternating current. The voltmeter. — A large non-inductivef resistance is connected in series with the fixed and pivoted coils of an electrodynamo- meter instrument so that the instrument may be connected directly across the supply mains. Let R be the value of this resistance including the resistance of the coils of the instrument. (a) Let S be the constant or direct electromotive force which produces a certain deflection or scale reading s. Then the steady deflecting force exerted on the pivoted coil is k (!)■ U (b) Let e be the varying value of an alternating electromotive force which gives the same steady deflection 5. ThenJ the instantaneous value of the force exerted on the pivoted coil is k and the average value of this force is — X average e^. * The reader should use his calculus to establish the following two propositions, namely, (i) Average (ax) = o X average x, where a is a constant, and (2) Average (x + y) = average * + average y. t A circuit is said to be non-inductive when its inductance L is negligibly- small, and the inductance is negligibly small when the electromotive force L — dt which is required to make the current change is very small as compared with the electromotive force Ri which is required to overcome resistance. t The electromotive force e which acts on a circuit of resistance R and in- di dl ductance L is equal to Ri + L — , and if Z, — is negligible we have e = Ri or at dt i = ejR. ELEMENTARY THEORY OF ALTERNATING CURRENTS. 189 Now the steady deflecting force under (a) must be equal to the average deflecting force under (b) because the deflection s is the same. Therefore k — X average e^ or S' = average e^ t — 6) a current leading 6° ahead of e i = I sin {cot + ( (10) Remark. — Consider a rotating clock-diagram vector whose varying projection represents a harmonic alternating electro- motive force or current. We shall hereafter speak of such a vector simply as representing the electromotive force or current. 19. Relation between maximum value and effective value of hannonic alternating electromotive force or current. — The effec- tive value of an alternating electromotive force or current, as measured by a voltmeter or ammeter, is the square-root-of-the- 14 £ 77 _ jjj — Vi I j = H 194 LESSONS IN ELECTRICITY AND MAGNETISM. average-value-of-the-square of the electromotive force or current as explained in Art. 15; and the maximum value of a harmonic alternating electromotive force or current is the quantity E in equation (8) or the quantity I in equation (9), as explained in Art. 16. The effective value of a harmonic alternating electromotive force {or current) is equal to its maximum value E {or I) divided by the square root of 2. That is If (II) (12) in which E and / represent effective values as indicated by a voltmeter or ammeter. These relations may be established by re- I r'" ducing the expression, (effective value)^ = — I E^ sin^ x-dx. 2X Jo Use of effective values instead of maximum values in the clock diagram. — According to Art. 17 the lengths of the vectors or lines in a clock diagram represent maximum values E and I; but the ratio of maximum values to effective values is constant according to equations (11) and (12), therefore it is permissible to think of the clock-diagram vectors as representing effective values. Indeed this is always done when the clock-diagram is used to show the relation between several alternating electromotive forces or currents. PROBLEMS. 11. A 6o-cycle harmonic alternating current is 25° behind an electromotive force in phase. What is the value of the phase difference expressed as a fraction of a cycle? What is the value of the phase difference expressed as a fraction of a second? 12. The voltmeter value of a harmonic electromotive force is no volts. What is its maximum value! This alternating electromotive force produces harmonic alternating current of 20 amperes (ammeter value) through a circuit and the current ELEMENTARY THEORY OF ALTERNATING CURRENTS. 195 is in phase with the electromotive force. What is the maximum rate at which work is being delivered to the circuit? 20. Addition of harmonic alternating electromotive forces of the same frequency. — Consider two alternators A and B con- nected in series with each other as shown in Fig. 130. Let a and b be the values of the electromotive forces of machines A and B respectively at any instant, and let c be the value Fig. 13a. a, b and c represent instantaneous values of electromotive force and c = a +b. Lengths of A, B and C represent voltmeter values of o, 6 and c. at the same instant of the electromotive force between the mains. Then c = a -\- b at each instant. If a and b have the same frequency / (and a constant phase difference 6) , then c will be a simple harmonic electromotive force of fre- quency f as represented by the clock-diagram vector C in Fig. 13&. To make this evident let us think of ^, B and C as representing maximum values so that their projections will represent instantaneous values but the projection of the diagonal C is at each instant equal to the sum of the projections of the sides A and B of the parallelogram. Therefore c = a -\- b, where c, a and b axe the projections of C, A and B. Resolution of a harmonic alternating electromotive force into harmonic parts. — ^Any given harmonic alternating electromotive force c (represented by the clock-diagram vector C in Fig. 13&) may be resolved into two harmonic alternating electro- 196 LESSONS IN ELECTRICITY AND MAGNETISM. motive forces a and h (represented by the clock-diagram vectors A and B in Fig. 13&). 21. Addition of harmonic alternating currents of the same frequency. — Consider two alternators A and B connected in parallel with each other as indicated in Fig. 14a. Let a and b Fig. 14o. Fig. 146. a, b and c represent instantaneous Lengths oi A, B and C represent arame- values of current and c = o + ft. ter values of u, b and t. be the values of the currents delivered by machines A and B respectively at any instant and let c be the value at the same instant of the current in the mains. Then c = a -{- b at each instant. If a and b have the same frequency / (and a con- stant phase difference d), then c will be a simple harmonic current of frequency f as represented by the clock-diagram vector C in Fig. 14&. Resolution of a harmonic alternating current into parts. — ^Any given harmonic alternating current c (represented by the clock- diagram vector C in Fig. 146) may be resolved into two har- monic alternating currents a and b (represented by the clock- diagram vectors A and B in Fig. 14&). supply main B T I « I l6 .! L T aupply main Fig. 15a. a, b and c represent instantaneous values of electromotive force and c = a +b B Fig. 156. Lengtlisof A, B and C represent volt- meter values of a, 6 and c. ELEMENTARY THEORY OF ALTERNATING CURRENTS. 197 22. Examples of resolution into parts. — (o) Dividing of an alternating electromotive force between two coils in series. Figure 150 shows two coils, or units of any kind, connected in series between alternating-current supply mains. The supply voltage c divides into two parts a and b such that c — a -\- b at each instant; but from Fig. 156 it is evident that the voltmeter value of c is not equal to the sum of the voltmeter values of a and b, unless the phase difference d is zero. (&) Dividing of an alternating current between two coils in supply main a i U m supply main Fig. 16a. , b and c represent instantaneous values of current and c = a + b. Fig. 166. Lengths of A, B and C represent am- meter values of a, b and <-. parallel. Figure i6a shows two coils, or units of any kind, connected in parallel with each other between alternating- current supply mains. The total current c divides into two parts a and b such that c = a -j- b at each instant ; but from Fig. 166 it is evident that the ammeter value of c is not equal to the sum of the ammeter values of a and b, unless the phase difference 6 is zero. PROBLEMS. Electromotive forces and currents understood to be harmonic. 13. Two similar alternators running at precisely the same speed are connected in parallel as indicated in Fig. 140. The current A is 60 amperes, the current B is 40 amperes, and the current C is 85 amperes (ammeter values, of course). What is the phase difference of currents A and Bl 14. The voltages A, B and C in Fig. 150, as read by a volt- meter are found to be 75 volts, 80 volts and no volts respectively. What is the phase difference between A and Bf 198 LESSONS IN ELECTRICITY AND MAGNETISM. 15. The currents A, B and C in Fig. 1 6a, as read by an am- meter are found to be 10.3 amperes 9.9 amperes and i.i amperes respectively (current C being very much smaller than either A or B). What is the phase difference between A and Bl Note. — The electromotive force C in Fig. 150 may be much smaller than either A or B. 23. Instantaneous and average values of power. — Let e be the value at a given instant of the electromotive force acting on a receiving circuit, and let i be the value of the current at the same instant ; then ei is the power in watts which is being delivered to the receiving circuit at the given instant, and the average value of ei is the average power delivered to the receiv- ing circuit. This average power is usually called, simply, the power delivered to the receiving circuit, and it is ordinarily not equal to the product EI where E is the voltmeter value of the electromotive force and I is the ammeter value of the current. Some idea of the pulsating character of the instantaneous power ei as delivered by alternating-current supply mains may Fig. 17. Electromotive force e and current i in phase with each other, in this particular case is equal to EI. Average power be obtained from Figs. 17, 18 and 19. Negative values of the power ei mean power returned to the supply mains by the receiving circuit. ELEMENTARY THEORY OF ALTERNATING CURRENTS. 199 The relation between electromotive force e, current i, and power ei in Fig. 19 is completely analogous to the relation be- Fig. 18. Current i is 60° (one sixth of a cycle) behind electromotive force e in phase. Average power in this particular case is JE/. tween the torque e which is exerted by the hair spring on the balance wheel of a watch, the angular velocity i of the wheel, Fig. 19. Current » is 90° (one quarter of a cycle) behind electromotive force e in pha Average power in this particular case is zero. 200 LESSONS IN ELECTRICITY AND MAGNETISM. and the power ei or rate at which work is delivered to the balance wheel by the spring. Consider the instant when the balance wheel is at its extreme position ; during the following quarter of a cycle the torque e is setting the wheel in motion, e and i are both in the same direction, and ei is positive which means that work is done by the spring on the wheel ; during the next quarter of a cycle the wheel is being stopped by the spring, e is opposed to i, and ei is negative which means that work is being done on the spring by the wheel ; and so on. In this statement fric- tion is ignored. 24. The wattmeter. — Power taken from alternating-current supply mains is nearly always measured by means of a watt- meter which consists of a delicately pivoted coil of fine wire A connected directly across the supply mains in series with a large non-inductive re- sistance R, and a stationary coil 0000 °^ course wire B connected in £ series with the receiving circuit L ^. as indicated in Fig. 20. The force Fig. 20. ^ action between the coils A and B moves the pivoted coil A , and the pointer which is attached to the pivoted coil plays over a divided scale. (a) Let 5 be the scale reading of the instrument when the instrument is connected to direct-current supply mains (connec- tions the same as in Fig. 20) with constant voltage S across the mains and with constant current d flowing through coil B and the receiving circuit. The force exerted on the pivoted coil A S is then equal to k • -^- 9, where ^ is a constant. (5) Let the instrument give the same scale reading 5 when it is connected to alternating-current supply mains as indicated in Fig. 20, let e be the value of the alternating electromotive force at a given instant, and let i be the value at the same instant of the alternating current flowing through coil B and ELEMENTARY THEORY OF ALTERNATING CURRENTS. 20I the receiving circuit. Then the value of the current in coil A at the given instant is e/R, the instantaneous value of the force exerted on the pivoted coil is k • r=' i, and the average value k of this force is -^ X average ei. But since the deflection of the K. instrument is the same as under (a) above this average force 8 must be equal to the steady force k • -:=. • S. Therefore K k . , ^/ ->/ Fig. 29. Fig. 30. line and the receiving circuit. The voltage E^ across the termi- nals of the receiving circuit is nearly 90° ahead of I in phase as shown, and the generator voltage Eq is the vector sum of Ei, rl and xl as before. In the present case, however, the line resistance produces a difference in phase between £0 and Ei, and the line reactance produces a difference in value between Ea and Ei-t (c) Voltage drop in a transmission line which delivers current to a condenser or any receiving circuit having negative reactance, but having small resistance. — This case is shown in Fig. 30, and Eo is the vector sum of £1, rl and xl as before ; but, Ei and xl are. nearly opposite in direction, inasmuch as £1 is * This statement is made on tlie assumption that */ is very small as com- parea with £1. t This statement is made on that assumption that rl is very small as com- pated with Ei. ,2l6 LESSONS IN ELECTRICITY AND MAGNETISM. nearly 90° behind I under the assumed conditions. Therefore in this case the line resistance produces a difference in phase between £0 and £1 and the line reactance produces a difference in value between £0 and £1, making £1 greater than £0. {d) Voltage drop in a transmission line which delivers current to a receiving circuit whose power factor (cos &) has any given value. — Let the line 01, Fig. 31, represent the current flowing through the transmission line and the receiving circuit. Let the line £1 represent the voltage across the receiving circuit, being the phase difference between / and £1 as shown. Let r be the resistance and x the reactance of the transmission line, and let £0 be the voltage at the generator. Lay off rl parallel to 01, and xl at right angles to 01. Then the differ- ence between the values of £1 and £0 is the required transmis- sion line drop. The value of Vr^P -f- x^P is called the impedance drop, and the difference between the numerical values of £0 and £1 is called simply the {voltage) drop in the line. 31. Circuits in parallel. — ^The general problem of circuits in parallel is somewhat complicated if one attempts to find algebraic expressions for the combined resistance and reactance of two or more circuits in parallel. Two simple cases of this problem are, however, of considerable interest, namely, (a) the problem of finding the power factor of two receiving circuits in parallel, the power factor of each and the current delivered to each being given, and (&) the problem of compensating for lagging currents. (a) Power factor of receiving circuits in parallel. — Let the line Fig. 32. ELEMENTARY THEORY OF ALTERNATING CURRENTS- 217 OE, Fig. 32, represent the voltage across the two receiving circuits. Let /i be the current delivered to receiv- ^^*^JZT la" ^ ing circuit number one and cos Oi its power factor, and let li be the current deliv- ered to receiving circuit number two and cos d^ its power factor. Lay off the clock diagram in Fig. 32 carefully to scale, and draw the diagonal I as shown. The angle d between E and I is then the angle whose cosine is the required power factor of the two receiving circuits in parallel. (&) Compensation for lagging currents. — When a transmission line delivers current to an inductive receiving circuit, power is delivered over the line to the receiving circuit during the time that the product ei is positive, and power is delivered back over the line from the receiving circuit to the generator during the time that the product ei is negative (compare Art. 23). This back- ward flow of energy from receiving circuit to generator represents an essentially unnecessary service of the transmission line, and it is desirable, and in some cases feasible, to reduce this backward flow of energy by connecting a condenser or something which is equivalent to a condenser (such as an over-excited synchronous motor) in parallel with the receiving circuit. Let OE, Fig. 33, represent the voltage at the terminals of the receiving circuit, and let 01 represent the lagging current delivered to the receiving circuit. Let R be the resistance of the receiving circuit and X its reactance. Then X tan e = -^ and X sm 6 = , =^ Vi?2 -I- X2 2l8 LESSONS IN ELECTRICITY AND MAGNETISM. The component Ob of the current / is equal to / sin 6 or IXIMB? + X^. But / is equal to E/^1r^ + X\ so that the component Ob is equal to EX/{R^ + X^). If a condenser of capacity C is connected across the transmission line at the receiver end, then the current flowing into the condenser will be 90° ahead of E, or parallel to Oc, Fig. 33, and its value will be >£ Fig. 33. equal to E divided by the condenser reactance i/ojC, or to EcaC; and if this current is numerically equal to Ob, the sum of / and Oc will be in phase with E. That is, the receiving circuit and the condenser together will take current in phase with E, and the instantaneous value of power delivered over the trans- mission line will never be negative. The capacity of the con- denser required to produce this result is given by the relation above mentioned, namely, EuC = EX/{R' + X^), which gives X C = a)(i?2 + X^) PROBLEMS. 32. A transmission line of which the resistance is 4 ohms and the reactance is 3 ohms delivers 100 amperes to a non-inductive receiving circuit, the electromotive force across the terminals of the receiving circuit being 10,000 volts, (a) What is the value of the generator voltage, and (b) what is the phase diff'erence between generator voltage and the voltage across the receiving circuit? Ans. (a) 10,404.3 volts, (6) i°39'. Note. — The answer as given is absurdly precise, the last three digits would be meaningless under practical conditions. This remark applies also to the answers given for problems 33 and 34. ELEMENTARY THEORY OF ALTERNATING CURRENTS. 219 33. If the transmission line specified in problem 32 were to deliver 100 amperes of current at 10,000 volts to a condenser, what would be the value of the generator voltage? Ans. 9,703.1 volts. 34. The transmission line specified in problem 32 delivers 100 amperes of current at 10,000 volts to a receiving circuit of which the power factor (lagging) is 0.7. What is the value of the generator voltage? Ans. 10,494.6 volts. 35- An alternator delivers current to two inductive receiving circuits in parallel. One receiving circuit takes 20 amperes and its power factor (lagging) is 0.9, and the other receiving circuit takes 25 amperes and its power factor (lagging) is 0.7. What is the power factor of the combination? 36. An alternator delivers lOO amperes at 1,100 volts and 60 cycles per second to an inductive receiving circuit of which the power factor is 0.85. What capacity condenser would be re- quired to compensate for lagging current? What number of leaves of paraffined paper 22 by 27 centimeters would be re- quired for this condenser, thickness of paraffined paper being 0.08 centimeter, allowing i centimeter margin beyond the tinfoil? Take inductivity of paraffined paper equal to 2. Ans. 127.0 microfarads, 114,800 leaves. APPENDIX C. ELECTRICAL MEASUREMENTS. 32. International standards. — The international standard am- pere, as defined in Art. 24 of Chapter II, is based on a very careful determination by Lord Rayleigh of the deposition of metallic silver from a solution of pure silver nitrate by a current measured in terms of its magnetic effect. The international standard ohm, the resistance at 0° C. of a column of pure mercury 106.3 centimeters long of uniform sec- tional area and weighing 14.4521 grams, is based on very careful measurements by the method of Lorentz as outlined in problem 80 on page 85. 33. Working standards. — The international standards as above defined are very inconvenient to use in the laboratory. ZnSOt, aoIaUoa' Fig. 34. Nearly all precise measurements in the laboratory are based on working standards as follows. ELECTRICAL MEASUREMENTS. 221 Resistance standards are made of carefully annealed wire, usually manganin wire, with heavy copper terminals; and when certified by the United States Bureau of Standards they can be depended upon certainly to one part in ten thousand. Electromotive force standards. — The Clark standard cell, which is shown in cross-section in Fig. 34 is extensively used as a Fig. 35. standard of electromotive force,* and its electromotive force when it is at a temperature of t° C. is: E (in volts) = 1.4292 — 0.00123 {t — 18) — 0.000,007 (' — 18)' The Weston standard cell is similar in every respect to the Clark cell except that cadmium amalgam and cadmium sulphate are used instead of zinc amalgam and zinc sulphate. Its electro- Fig. 36. motive force (with saturated solution of cadmium sulphate) when it is at a temperature of t° C. is * Specifications for the Clark cell are given on page 12 of Franklin, Crawford and MacNutt's Practical Physics, Vol. II. 222 LESSONS IN ELECTRICITY AND MAGNETISM. E (in volts) = 1. 0187 — 0.000035 {t — 18) — 0.000,000,65 {t — 18)^ Standard condensers. — A general view of a subdivided condenser is given in Fig. 35, and a diagram of the internal connections is shown in Fig. 36. Such condensers when carefully made, using mica as the dielectric, and certified by the Bureau of Standards may be depended on, certainly, to one tenth of one per cent. Standard inductances. — A fixed standard of inductance is a winding of insulated copper wire wound preferably on a marble Fig. 37. spool. A variable standard of inductance consists of a fixed coil of wire with an inside coil that can be turned so that the coils may stand with any desired angle between their planes. The two coils are connected in series and the inductance of the two coils in series is indicated by a pointer which plays over a divided circle. A variable inductance of this type is shown in Fig- 37- 34. The potentiometer. — The potentiometer is an arrangement ELECTRICAL MEASUREMENTS. 223 for measuring the ratio of two electromotive forces as follows: A battery B produces a constant current * through a stretched wire WW, Fig. 38, and the sliding contacts a, a', A and A' WW ■■ A e -L^A' R ■* H'lKg)-^- w r-^)-^ I'l'i^ £- ■R-~. Fig. 38. E Fig. 39. are moved until neither galvanometer gives a deflection when keys k and K are closed Then e = ri and E = Ri, whence Eje = Rjr = Ljl, where L is the measured length of the portion A A' of the wire WW and I is the measured length of the portion cm'. Therefore, if e is the known electromotive force of a standard cell, the value of E can be calculated. The above described potentiometer is called the simple sUde wire potentiometer, and it is not very accurate because the wire WW is sure to be to some extent non-uniform so that R/r is not equal to L/J. This source of error is very greatly reduced in the Leeds and Northrup potentiometer by using a very care- ,.,.,.B H'— ®- i Wjfi,.^.,., d\j'\f^f\^^ S' S" s luullL M W • A' W E K ■'""'"" I— Jl|f— ^ variable standard of inductance S is connected in the "1 I B B --©-- Fig. 47. Other arm of the bridge as shown in Fig. 47, and two adjustments are made as follows: (o) The resistances are adjusted until the galvanometer G gives no deflection when a steady battery cur- rent floivs though the bridge arrangement, and then (&) The variable standard of inductance is adjusted until the telephone gives minimum sound when the bridge arrangement is supplied with alternating current as indicated by the dotted lines in the figure. Then S/X = a/j3. 38. Measurement of the horizontal component H of the earth's magnetic field. — (a) Comparison of values of H at two places by magnetometer deflections. A very small magnet ns, Fig. 48, a pivoted compass needle, is placed where the horizontal ELECTRICAL MEASUREMENTS. 233 component of the earth's field is H and the needle comes to rest pointing in the direction of H (the plane of the paper is horizon- tal in Fig. 48). A large bar magnet MM is then placed at a M M Fig. 48. measured distance d due magnetic east or west of the compass needle. The large magnet produces at the compass needle a definite westerly (or easterly) field /, and the compass needle is deflected through the angle 4>, where tan ^ H (i) The compass needle is then placed where the horizontal com- ponent of the earth's magnetic field is H' and the deflection <^' produced by the large magnet at the same distance d is observed. Then / tan ' = H' (ii) and dividing equation (ii) by equation (i), member by member, we get IT _ tan H ~ tan 0' (iii) (&) Comparison of values of H at two places by the method of oscillations. — The large bar magnet MM of Fig. 48 is suspended and set oscillating at a place where the horizontal component of the earth's magnetic field is H, and we have ^■jtVK = mlH (iv) 234 LESSONS IN ELECTRICITY AND MAGNETISM. where n is the number of complete vibrations per second, K is the moment of inertia of the suspended magnet referred to the axis of suspension, m is the strength of the poles of the magnet, and / is the length of the magnet (distance between its poles) The large bar magnet is then suspended at another place where the horizontal component of the earth's magnetic field is H', and we have 47rVX = mlH' (v) and from equations (iv) and (v) we get (vi) H' n (c) Gauss's method for measuring m and H. — The value of / in equation (i) is m m l2 / = (-^)- (-^y (vii) according to Art. ii of Chapter I, so that equation (i) becomes tan (^ = H m (-:-)' (-01 (viii) and this equation and equation (iv) contain only m and H as unknown quantities so that the values of m and H can be calculated if all other quantities in (viii) and (iv) have been determined. The distance I between the poles of the large magnet is, of course, somewhat indefinite, and this uncertain quantity can be eliminated if the deflection * of the compass needle is observed with the large magnet at distance D. We thus have a third simultaneous equation [in addition to equations (iv) and (viii)], namely. tan ^ = m H (-0 (-m (ix) weighed quantity of radioactive material, h is a high-voltage bat- ■S ELECTRICAL MEASUREMENTS. 235 Writing M for ml, and using equations (iv), (viii) and (ix), we get the two following moderately convenient simultaneous equations for use in determining M{ = ml) . and H, according to Gauss. M # tan - £)^ tan ^ and MH = 4tVK 39. A simple measurement of radioactivity. — ^An ideal arrange- ment for measuring the radioactivity of a weighed quantity of radioactive material is shown in Fig. 49 in which A A and BB are metal plates, MM is a tery and G is a galvanometer for ^ measuring the current. The only A- '^■^.«^:J,..^„j A difficulty with this arrangement is that the current is usually too ^. , ■' Fig. 49. small, by far, to be measured by a galvanometer, and therefore the arrangement shown in Fig. 50 must be used. The underlying idea may, however, be most easily set forth by thinking of the arrangement in Fig. 49 as follows: When the voltage of the battery & is large enough* (four or five hundred volts) all of the electrons and ions which are formed by the radiations from MM serve as current carriers, and the value of the current may be taken as a measure of the radio- activity of MM. Figure 50 is a gold leaf electroscope arranged for the measure- ment of radioactivity. The metal plate BB, the metal rod R and the gold leaf L are. highly insulated, the plug 5' being of cast sulphur. The plate A A, on which the weighed quantity of radioactive material MM is placed, is connected to ground, the plate BB the rod R and the gold leaf L are charged by * But not sufficiently large to produce ionization by collision in the region be- tween AA and BB. J+ 236 LESSONS IN ELECTRICITY AND MAGNETISM. momentarily connecting the movable wire C, and the time t required for the gold leaf to fall from a given initial position to a given final position (because of the discharge of plate BB) is Fig. SO. observed. The reciprocal of this time t is proportional to the current and it is taken as a measure of the radioactivity of MM. To correct for incomplete insulation of BB, R and L the time /' required for the given movement of the gold leaf before the material MM is in place is observed, and then 0-p)'' the true measure of the radioactivity of MM. If the object is to study the decay of radioactivity of MM, the value of I 7 ~ 77 ) hours or days. is determined at intervals for several APPENDIX D. CORRESPONDING EQUATIONS OF TRANSLATORY MOTION, ROTATORY MOTION AND "ELECTRICAL MOTION." X = vt (i) where * is the distance traveled in t seconds by a body which has a constant velocity v = iaP (4) where x is the distance traveled in t seconds by a body which starts from rest and has a constant acceleration a W = Fx (7) where W is the work done by a force F when the body on which F acts moves distance x in the direction of F. Fv (10) where P is the power developed by a force F when the body on which F acts moves at velocity V in the direction of F. (t> = St (2) where is the angle in radians turned in ( sec- onds by a body which has a constant spin velocity of s radians per second. = hat^ (S) where is the angle turned in t seconds by a body which starts from rest and has a constant spin acceleration a W = T4> (8) « = it (3) where q is the amount of electric charge which flows in t seconds through a circuit in which a constant current i is flowing. q = i X rate of growth X t' (6) If the current in a circuit grows at £t constant rate starting at zero, the a- mount of charge which flows through the circuit in t seconds is equal to 5 X rate of growth of current X <*. W = Eq (9) where W is the work done where W is the work done by torque T when the by an electromotive force body on which T acts E when an electric charge turns through radians q flows through the cir- about the axis of T. cuit on which E acts. P = Ts (II) P = Ei _(12) where P is the power developed by a torque T when the body on which T acts turns about the axis of T' at a speed of ^ radians per second. where P is the power developed by an electro- motive force E when cur- rent » flows through the circuit on which E acts. F = m dv It (13) dt (14) di ■dt (IS) where is the accelera- tion produced by an un- balanced force F which acts on a body of mass m. where — dt is the spin ac- where dt is the rate of celeration produced by a torque T which acts on a wheel of which the spin- inertia is X. growth of current due to electromotive force E acting on a circuit of in- ductance L. ^ 237 238 LESSONS IN ELECTRICITY AND MAGNETISM. W = Jmi^ (i6) where W is the kinetic energy of a body of mass m moving at velocity v. W = iKs' (17) W = iLfl (i8) F = ax (19) A ball is fixed to a spring and F is the force re- quired to pull the ball to a distance x from its equi- librium position. The factor a is called the stiffness coefficient of the spring. ^Trhihu = a (22) A ball is fixed to a spring of which the stiffness co- efficient is o. Then when the ball is pulled to one side and released it per- forms simple harmonic motion of which the num- ber of complete vibrations per second is ». m being the mass of the ball. where W is the kinetic energy of a wheel rotating at a speed of j radians per second, the spin-inertia of the wheel about its axis of spin being K. T = , A body is hung by a wire and T is the torque re- quired to turn the body (and twist the wire) through the angle of radians. The factor b is called the coefficient of tor- sional stiffness of the wire. 47r«n2X = 6 (23) A body is hung by a wire of which the coefficient of torsional stiffness is b. Then when the body is turned so as to twist the wire and released it per- forms harmonic rotatory motion of which the num- ber of complete vibrations per second is n, K being the spin inertia of the body referred to the wire as an axis. where W is the kinetic energy of a current i in a circuit of which the in- ductance is L. (20) or q = CE E = -.a (21) Two metal plates are sepa- rated by a layer of dielec- tric constituting what is called a condenser, and E is the electromotive force required to draw q cou- lombs out of one plate and force it into the other plate. The factor C is called the capacity of the condenser. 47r%2L = ^ (24) A charged condenser of capacity C is connected to a circuit of inductance L and of negligible resis- tance. The discharge from the condenser then surges back and forth through the circuit, we have what is called an oscillatory dis- charge (harmonic), and the number of complete oscillations per second is ». W ■■ 1 ' m ^-t^ (2S) A spring of which the stiff- ness coefficient is i* is bent, the force which is acting on the fully bent spring is F, and the po- tential energy of the bent spring is W. e (26) A wire of which the coeffi- cient of torsional stiffness is 6 is twisted, the torque which is acting on the fully twisted wire is T, and the potential energy of the twisted wire is W. W ■■ iCE* (27) A condenser of which the capacity is C is charged, the electromotive force which is acting on the fully charged condenser is B, and the potential en- ergy of the charged con- denser is W. APPENDIX E. ANSWERS TO PROBLEMS AND LEADING QUESTIONS. On page lo: Problem 8, Ans. 11,550 dyne-centimeters of torque. 1. What is meant by the poles of a magnet? 2. Define the unit pole? How large is the pole or poles sup- posed to be in this definition? What is meant by a concentrated pole? Ask your professor of mathematics to define a con- centrated pole as a finite differential in accordance with his hobby relating thereto. 3. A pole has m units of strength. What does this mean? On page 13 : Problem 9, Ans. 0.0438 centimeter. Additional problems: pb. The poles of a bar. magnet are 30 centimeter apart and their strength is -f- 700 units and — 700 units. Find the torque exerted on the magnet when it is placed in a uniform magnetic field of which the intensity is 0.6 gauss, the axis of the bar magnet being inclined 30° to the direction of the field. Ans. 6,300 dyne- centimeters. 9c. A compass balances on its pivot before it is magnetized. Does it balance after it is magnetized? If not, why not? Problem 10, Ans. 0.0737 V^^ second. 4. What is meant when it is stated that a given region is a magnetic field? 5. The intensity of a magnetic field at a given point is 10 gausses. What does this mean? 6. What is a uniform magnetic field? Give an example. What is a non-uniform magnetic field? Give an example. What is the character of the force action exerted on a complete magnet in a uniform field? In a non-uniform field? 239 240 LESSONS IN ELECTRICITY AND MAGNETISM. On page 15: Problem li, Ans. 2.48 gausses. • 12, Ans. 13.3°. " 13, Ans. H = 0.366 gauss; m = 1372 units pole. On pages 19-20 : Problem 15, Ans. 7,070,000 dynes. 7. Define the unit of magnetic flux. Why is this unit often called the "line of force"? What is meant by an actual line of force in a magnetic field? 8. The distance d in Fig. 24 is assumed to be very small in comparison with the length and breadth of the polar area (the shaded area) in Fig. 23 side view. Why? 9. Work is done in pulling the magnet poles in Fig. 22 or Fig. 24 apart. Where does this work go? Derive an expression for the energy per unit volume of a uniform magnetic field of intensity H. On pages 32-34: Problem 17, Ans. 300 dynes act on S-pole towards the left. 600 dynes act on wire towards the left. " 20, Ans. 0.6 centimeter. " 21, Ans. 2,500 dynes attraction. 10. State the right-handed-screw rule concerning the direction of flow of an electric current in a wire. State the right-handed- screw rule concerning the direction of magnetization of an iron rod by a current flowing around the rod. 11. What is the physical explanation of the side push exerted by a magnetic field on a wire in which current is flowing? 12. How would you hold a magnet so as to blow out an electric arc? 13. Why is the current in one of the armature wires on a two- pole direct-current dynamo equal to one half of the current which enters or leaves the armature winding? 14. The current in a wire is 12 abamperes. What does this mean magnetically? 15. What is the legal definition of the ampere? See page 36. 16. A long straight wire carries a current of 25 amperes. What„ ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 241 is the intensity of the magnetic field due to this wire at points distant 12 centimeters from the wire? 17. If m units of north pole were spread uniformly along a length of / centimeters of a very small steel rod, what would be the trend of the lines of force of the magnetic field due to this pole in the region very near the pole? What would be the intensity of this field at a small distance r from the long slender pole? Explain. 18. If the very long slender pole mentioned in question 17 were placed along the axis of a coil or winding of wire on a long paper tube of radius r, what force would be exerted on the pole by the coil? Express this result in terms of the current in the coil in abamperes, the number of turns of wire on unit length of the paper tube and the strength m of the pole, and explain. 19. A wire is at right angles to a magnetic field. State how you can determine the direction of the side push on the wire when the direction of the current and the direction of the field are given. On pages 42-44 : Problem 24, Ans. 96,540 seconds. " 25, Ans. 1.07 joules per second per ampere (volts). " 26, Ans. 3.17 centigrade degrees. " 27, Ans. 20.3% due to voltaic action. 20. Define the terms electrolysis, electrolyte, electrode, anode and cathode. 21. What kind of an electrolytic cell is a voltaic cell? 22. What is meant by voltaic action and local action in a voltaic cell? 23. What is meant by primary and secondary reactions in an electrolytic cell? Give examples. On pages 48-49 : Problem 28, Ans. 0.025 ampere per square centimeter. " 29, Ans. 13.6 hours. " 30, Ans. 1. 5 1 joules per second per ampere (volts). " 32, Ans. positive plate gains 95.5 grams; negative gains 143.2 grams. 24. What conditions must be realized in a voltaic cell in order 17 242 LESSONS IN ELECTRICITY AND MAGNETISM. that it may be used as a storage cell? What about non-dis- integration of the electrodes? 25. In what terms is the electrochemical equivalent of silver expressed? What is its value? 26. State the two laws of electrolysis. See I and II on page 46. On pages 53-54: Problem 33, Ans. 4.31 ohms; 109.8 joules per second per am- pere (volts). 34, Ans. 0.00558 centigrade degree per second. 35, Ans. 11.25 ohms. 36, Ans. 0.2 ohm. 37, Ans. 0.122 ohm. 38, Ans. 0.00653 ohm. 39, Ans. 0.00206 ohm. 40, Ans. 0.120 ohm. 41, Ans. 8 ohms. 27. State Joule's law. 28. What is an ohm of resistance? What is an abohm? 29. In what terms is resistivity expressed (a) When resistance is expressed in ohms, length in centimeters and sectional area in square centimeters, and (6) When resistance is expressed in ohms length in feet and sectional area in circular mils? (c) What is meant by the conductivity of a substance? On pages 55-S6: Problem 43, Ans. 4.63 ohms at 0° C. ; 6.30 ohms at 90° C. " 44, Ans. 0.0036 per centigrade degree. " 45, Ans. 77.3° C. " 46, Ans. 0.000206 per centigrade degree. 30. In what terms is the temperature coefficient of resistance expressed ? On pages 62-64 : Problem 49, Ans. 85%. " 51, Ans. 76.2 amperes. " 52, Ans. 2.75 cents. 53, Ans. #2.63. " 54, Ans. 1 15.6 volts. ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 243 Problem 55, Ans. 6.2 ohms. 56, Ans. 23.5 cents; 4.08 cents. 57, Ans. 2.08 cents by gas;. 10.7 cents by electric heater. 31. What is meant by the electromotive force £ of a battery? 32. Under what conditions is Ohm's law true? 33. Show that the electromotive force of a battery cell is equal to jz, where j is the number of joules of energy developed by the dissolving of one gram of zinc in the cell and z is the number of grams of zinc consumed per second per ampere. On what assumption is this proof based? 34. What is meant when it is stated that a dry cell is polarized? 33. State Joule's law as applied to a portion of a circuit. State Ohm's law as applied to a portion of a circuit. How would you apply equation (12) on page 57 to a portion of a circuit? 36. The international standard ampere is defined on page 36 and the international standard ohm is defined on page 220. Define the volt in terms of these international standards. 37. What conditions must be met in order that one may use the equation P — Fv to calculate the power P developed by a force F? In the first place the velocity v must be parallel to F, but this condition does not interest us here. The force must act on the body which is moving at velocity v, not on some other body, and the force must act while the body is moving. A horse cannot do work by pulling on a post while a plow is moving, nor can a horse do work by pulling on a plow today if the plow moves only tomorrow! Furthermore, one cannot, in general, multiply any kind of an average value of F during a given time by any kind of an average value of v during the same time to get the average power because the force may have acted chiefly during the first part of the time and the body may have moved chiefly during the last part of the time. Explain why equation (12) on page 57 is not, in general, applicable to a portion of a circuit taking power from alternating-current supply mains. On pages 65-66: Problem 58, Ans. 0.151 ampere; 0.753 volt; 0.317 volt. 244 LESSONS IN ELECTRICITY AND MAGNETISM. Problem 59, Ans. 0.543 ohm. " 60, Ans. 4 ohms. " 61, Ans. 851 mils. " 62, Ans. 470 mils. " 63, Ans. 6,730 volts. " 64, Ans. 100 mils. On pages 74-75: Problem 65, Ans. 24 cells. " 66, Ans. 15,385 ohms. " 67, Ans. 50 volts; 40 volts; 20 volts. " 68, Ans. 0.63 ampere. " 69a, Ans. 0.5 ampere; 0.67 ampere. " 696, Ans. 0.0606 ampere; 20 ohms. " 70, Ans. i/i8oohm. " 71, Ans. 0.004 ohm. " 72, Ans. 393,000 ohms. " 73, Ans. 1.2 X 10* ohms; 48 X 10' ohms. On pages 83-85 : Problem 74, Ans. (a) 8,9iowatts; (6) 420 watts; (c) 104.8 volts; (d) 8,490 watts. " 75, Ans. (o) 220 watts; (b) 132 watts; (d) 66 volts. " 76, Ans. 660,000 lines or maxwells per second. " 77, Ans. 942 volts. " 78, Ans. 1,310,000 lines or maxwells; 851 revolutions per minute. 38. Describe an experiment which shows that an induced electromotive force opposes the flow of current through the armature of an electric motor in operation. 39. What is Lenz's law? Give it in the form first given by Helmholtz. 40. Derive the equation E = IHv, and state the physical meaning of each step. (See page 78). 41. Explain the meaning of equation (16) on page 80. On page 89 : Problem 83, Ans. 11 X 10' lines or maxwells per second; iioo volts. ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 245 Problem 84, Ans. 0.5 ampere; 5 amperes; 1077.5 volts. " 85, Ans. 0.491 amperes; 1058 volts. 42. Give a diagram showing the essentials of the scheme of winding of a 4-pole alternator armature having 8 armature con- ductors. Make the diagram similar to Fig. 69. 43. What is meant by the frequency of an alternating current? What is meant by a cycle? 44. Explain step-up and step-down transformation. 45. Two round steel rods have an air gap between their ends as indicated in Fig. 22. A large sheet of copper is placed in this air gap and then the steel rods are suddenly magnetized. Make a sketch of the sheet of copper and draw lines showing the flow of the induced currents in the sheet. The steel rods being magnetized the large sheet of copper is drawn out of the gap. Make a sketch of the sheet of copper and draw lines showing the approximate flow of the induced currents in the sheet. What force is exerted on these currents by the magnetic field in the gap space and what is the direction of the force? If the velocity of withdrawal of the plate is doubled this force will be doubled. Why? 46. Explain the meaning of Fig. 80. On pages 98-99 : Note. The electromotive force required to make the current in a coil increase is e = Z ^-, where Z is the number of turns at d^ of wire in the coil and -77 is the rate of increase of magnetic at flux through the coil (through the opening of the coil) due to the d^ . increasing current. This is evidently true because Z X -37 is the back electromotive force induced in the coil according to equation (16) on page 80. // is only when the growing flux * is proportional to the increasing current i that the conception of INDUCTANCE is legitimate. Problem 86, Ans. 183 amperes per second; 0.03 second. " 88, Ans. 83.3 amperes per second. 246 LESSONS IN ELECTRICITY AND MAGNETISM. Problem 89, Ans. 5.26 X io~^ henrys; 7.6 X 10' amperes per second. " 90, Ans. 800 amperes per second. " 91, Ans. 2.677 amperes. 47. Differentiate equation (22) on page 96, and, using equation (12) on page 57, prove equation (20) on page 95. Make the physical meaning of each step perfectly clear. 48. A circuit carries a current which rises to a maximum of 5 amperes and the resistance of the circuit is 4 ohms. The value of di ■J- ranges about 2000 amperes per second and the inductance of the circuit is one henry. Would you consider this circuit to be non-inductive? A circuit may be considered as non-inductive under one set of conditions but by no means as non-inductive under another set of conditions. Explain. 49. Does a coil with an iron core have a definite inductance? 50. What is an induction coil? What is an inductance? What is a choke coil? 51. What becomes of the energy §Z,i^ when an inductive circuit is broken? On pages 108-109 : Problem 95, Ans. 0.744 ampere. " 98, Ans. 6.93 X io~' abcoulombs per division; H = 10,032 gausses. 52. What is a condenser? What is meant by charging a condenser? What is meant by discharging a condenser? 53. How does a condenser connected across a break in a circuit eliminate the spark at break? How does the condenser in Fig. 74 cause a quick demagnetization of the iron core of the induction coil? 54. What is a coulomb? What is an ampere-second? What is the difference between an ampere-second and an ampere per second? 55. Describe an experiment showing that two insulated metal plates connected to electric supply mains attract each other. 56. How much electric charge will a one-farad condenser hold? ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 247 Is the word capacity as applied to a condenser used in a sense which is strictly analogous to its use when we speak of the capacity of a water pail? Explain. 57. Define the farad; the microfarad. On pages 111-112: Problem 100, Ans. 654 leaves of mica; 655 leaves of tin foil. " loi, Ans. 0.015 microfarad. 58. What is meant by the inductivity of a dielectric? On pages 114-115: Problem 104, Ans. 8.84 X io~« joules; 8.84 X io~^ joules. " 105, Ans. (a) 20,000 volts; {b) and (c) 8.84 X io~^ joules or 8.84 X 10* ergs; {d) and (e) 4.42 X 10^ dynes. " 106, Ans. (o) 200 volts; {b) and (c) 8.84 X io~^ joules or 88.4 ergs; id) and (e) 44.2 dynes. 59. If you charge a condenser by allowing the full charging voltage E to act on it from the start is all the work done by E represented by the potential energy of the condenser? If not, why not? If not, what becomes of the remainder? How would you charge a condenser so that all the work done by the charging electromotive force is represented by the potential energy of the condenser? 60. Prove equation (35) page 113 as applied to a charged condenser and state the physical meaning of each step. 61. What can you say as to the charges + g and — g on two parallel plates if the plates are perfectly insulated? Suppose that two such charged plates are pulled farther apart, prove that the voltage between the plates must increase in proportion to the distance between the plates. Does the potential energy of the charged plates increase as they are pulled apart? If so where does this increase of energy come from? Derive an expression for the force of attraction of two parallel metal plates in terms of size and distance of plates, inductivity k of the dielectric and voltage E between them. How much force is one joule per centimeter? 248 LESSONS IN ELECTRICITY AND MAGNETISM. On pages 120-121 : Problem 109&. Ans. 167 joules. " logd, Ans. 65,000 per second. 62. Under what conditions is the voltage required to puncture a dielectric proportional to the thickness of the dielectric? What do you mean by the specific strength of a dielectric? In what terms is it expressed? 63. If you could never produce a movement more rapid than one foot per century, how could you set a tuning fork vibrating? Think of this question in connection with Art. 77. On pages 125-126 : Problem no. Ans. 1.78 X io~^ colombs. 64. What is meant by an electric field ? What is meant by the intensity / of an electric field at a point? Answer this question in terms of force exerted by the field on a small charged ball. What is meant by the direction of an electric field at a point? In what terms is the intensity of an electric field expressed? What is meant by a line of force drawn through an electric field? 65. The oppositely charged plates in Fig. 95 attract; would you say that electric field has a certain tension like a magnetic field? See pages 18 and 19. 66. The oppositely charged plates in Fig. 95 represent a store of potential energy [see equations (33), (34) and (35) on page 113]. Where does this energy reside? Does a magnetic field represent a store of energy? Where does the kinetic energy of the current, |L*^ reside in Fig. 81? On page 138 : Problem in, Ans. 1.47 X io~* coulomb. " 112, Ans. 2.6 X io~^ coulomb. " 113, Ans. 11,538 volts per cm. in the oil; 3,846 volts per cm. in the glass. " 114, Ans. (a) 330,000 volts; (&) 139,500 volts. 67. What is meant, mathematically, by electric flux? By electric flux density? See Art. 90. 68. Prove that the electric field intensity between the plates in Fig. 95 is q/aB volts per centimeter where g is the charge in ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 249 coulombs on one of tne plates, a is the area of one face of one plate in square centimeters and B has the value given in Art. 73 or in Art. 90. 69. Knowing the energy of the two charged plates in Fig. 95 [see equations (31) and (33)-(35) on pages no and 113], derive an expression for the energy per cubic centimeter of an electric field (in air, let us say) of / volts per centimeter. Show that this energy per cubic centimeter is equal to the tension of the field in units of force per square centimeter. 70. The lines of force of an electric field meet the surface of an insulator at right angles.* The surface has no charge on it, let us suppose. What is the relation between electric flux density just outside and just inside of the insulator? What is the rela- tion between the field intensity (volts per centimeter) just out- side and just inside of the insulator? 71. Make a sketch showing the approximate trend of the lines of force near an uncharged glass sphere in an electric field, which, but for the presence of the sphere, would be a uniform field. Ditto for an air bubble in oil. 72. Given two short, square-ended cylinders of rubber, {a) The cylinders are set side by side between two parallel jaws and squeezed. Is stress or strain the same in the two cylinders? (&) The cylinders are placed end on end between two jaws and squeezed. Is stress or strain the same in the two cylinders? (c) Are these mechanical cases "directly" or "inversely" anal- ogous to blocks of insulating material side by side between charged plates or end on end between charged plates? On pages 140-141 : Problem 116, Ans. {a) 708 microfarads; (&) 708 coulombs; (c) 0.00157 volts per centimeter. Note. The radius of the earth is about 6375 kilometers. " 118, Ans. 5,000 volts per centimeter. On page 175: Problem i , Ans. 22 gausses. " 2, Ans. 7.5 X I0~* henry. * Not necessarily true, merely assumed for the sake of the following questions. 250 LESSONS IN ELECTRICITY AND MAGNETISM. Problem 3, Ans. 2,655 ampere-turns. " 4, Ans. 37.7 gausses. " 5, Ans. 4.0 amperes. " 6, Ans. 1,920,000 maxwells or lines. 73. If a short thick rod of iron is placed in a magnetic field, which, but for the presence of the rod would be a uniform mag- netic field of intensity H parallel to the rod, the "magnetizing force " is less than H\ but if the iron rod is very long and slender the "magnetizing force" would be equal to H. Explain. 74. Consider a long slender iron rod of sectional area g which is magnetized by being placed in and parallel to a uniform magnetic field of intensity H. Two distinct causes produce magnetic field near one of the poles of the rod, namely, the original cause of the uniform field H and the magnet pole on the end of the rod (of strength m). Show that the total magnetic flux which emanates from the north pole of the rod is ^irm + Hg^. This is, of course, the total flux coming along or through the rod to the m north pole, so that $/g = B = /^ir \- H. The ratio m/g is called the intensity of magnetization of the rod. On page 184 : Problem 8, Ans. {a) 3.44 joules; (6) 3.42 joules. " 9, Ans. 6.63 watts. 75. How is work done in magnetizing a bar of iron by means of a coil of wire? 76. What is meant by magnetic hysteresis? On page 190 : Problem 10, Ans. (a) 22 volts; (&) 49.2 volts. On pages 194-195 : Problem 11, Ans. 0.0694 cycle; 0.00116 second. " 12, Ans. 155.6 volts; 4.4 kilowatts. On pages 197-198 : Problem 13, Ans. 65.1°. 14, Ans. 89.7°. 15, Ans. I744°- ANSWERS TO PROBLEMS AND LEADING QUESTIONS. 25 1 On pages 202-204 ■ Problem 16, Ans. 0.840; 32.9°. 17, Ans. no volts; 100 amperes; 11 kilowatts; 22 kilowatts. " 18, Ans. 220 volts; 150 amperes; 16.5 kilowatts. " 19, Ans. 1,628 volts; A 29.77 kilowatts; B 146.4 kilowatts. " 20, Ans. A 30.89 kilowatts negative; B 122.8 kilo- watts positive. " 22, Ans. 250.0 amperes. On pages 208-209 : Problem 24, Ans. (a) 45.24 ohms; (6) 132.7 ohms; (c) 102.7 cycles per second. " 25, Ans. 0.0270 henry. " 26, Ans. 29.67 microfarads ; i .6 ohms. " 27, Ans. i? = 10.34 ohms; -X" = 3.75 ohms; im- pedance = 11 ohms. " 28, Ans. i? = 5.13 ohms; Z = 5.13 ohms; imped- ance = 7.33 ohms. On pages 213-214 : Problem 29, Ans. (a) 112.6 cycles per second; (6) 6224 volts. " 30, Ans. ^x; = 40 microfarads; 6.634 ohms; 0.6634 ohm. " 31, Ans. a; = 40 microfarads; 1.33 ohms. On pages 218-219: Problem 35, Ans. 0.80. INDEX Abampere, definition, of, 30 Abcoulomb, definition of, 103 Abfarad, definition of, 105 Abhenry, definition of, 96 Abohm, definition of, 51 Abvolt, definition of, 61 Alternating-current transformer, see transformer Alternating currents, elementary theory of, 185-218 Alternator, speed-frequency relation of, 88 the, 85 Ammeter multiplying shunt, 69 the direct-current, 24, 61 Ampere, definition of, 30 the international standard, 36 Anode, definition of, 36 Atomic theory of electricity, 142-169 Attraction, electrostatic, 103 Audion, see pleiotron Ballistic galvanometer, the, 104 uses of, 230 Battery the electric ; see voltaic cell the storage, 44 Blow-out, the magnetic, 25 Bridge, Wheatstone's, 71 Canal rays and cathode rays, 160 Capacity of condenser, definition of, 105 equation of, 11 1 measurement of, 231 Cathode, definition of, 36 rays, 154, 160 Charging by influence, 131 Chemical effect of electric current, 35-49 Circuit, the electric, 3 Clark standard cell, the, 221 Clock diagram, the, 191 Condenser capacity, definition of, 105 equation of, 1 1 1 charged, energy of, 112 spherical, capacity of, 139-140 the, 100-12 1 Conduction, metallic, electron theory of, 166 Conductors and insulators, 3 Connections in parallel, 68 in series, 66 Contact-potential-differences, 166 Coolidge tube, the, 153 Corona discharge, the, 148 Cottrell process, the, 150 Coulomb, definition of, 102 Crookes' tube, 160 Current, electric ; see electric current valve, the vacuum-tube, 155 Cycle, definition of, 87 Dielectric, definition of, 109 inductivity of, 109 strength, 115 Disruptive discharge, iij Doubler, the electric, 132 Drop, voltage, discussion of, 64-65 Dynamo, the alternating-current, see alternator the direct-current, 26 fundamental equation of, 80 Eddy currents and lamination, 92-94 Effective values of electromotive force and current, 189 Electrical measurements, 220-236 Electric charge and the condenser, 100-121 252 INDEX. 253 Electric charge, measurement of, 104 current, chemical effect of, 35-49 direction of, 20 growing and decaying, 97 heating effect of, 50-75 kinetic energy of, 96 magnetic effect of, 1-34 the, 3 doubler, the, 132 field, 122-141 discussion of, 122-141 intensity of, 123 flux, definition of, 136 density, 137 oscillator, frequency equation of, 119 the, 116-117 screening, 127 Electricity, atomic theory of, 142-169 Electrolysis, 35-49 Electromotive force, definition of, 57 induced, 7^-99 equation of, 78, 79, 80 Electron, the, 145 theory, the, 142-169 Electrons, emission of by hot bodies, 152 Electroscope, the gold-leaf, 133 the pith ball, 126 Electrostatic attraction, 103 voltmeter, 103 Faraday's " ice pail " experiment, 129-130 laws of electrolysis, 46 Farad, definition of, 105 Field, electric, 122-141 magnetic, the, 10-14 tension of, 17 Flux density, electric, 137 magnetic, 170 electric, definition of, 136 from magnet pole, 17 magnetic, definition of, 15 measurement of, 107 Flux-turns, definition of, 96 Frequency, definition of, 87 Galvanometer, the ballistic, 104 Gauss's theorem 136 Gold leaf electroscope, 133 Heating effect of electric current, 50- 75 Henry, definition of, 96 Hertz oscillator, the, 117 Hysteresis, magnetic, 181 Impedance, definition of, 208 Induced electromotive force, 76-99 equations of 78, 79, 80 Inductance, calculation of, 98 definition of, 95, 96 measurement of, 231 of a coil, discussion of, 94-99 standards, 222 Induction coil, the, 90 Inductivity of a dielectric, 109 Influence, charging by, 131 electric machine, 134 Insulators and conductors, 3 International standards, 220 Ionization by collision, 146 Ion, the compound, 146 the simple, 146 Iron, magnetism of, 170-184 Joule's law, 50 Kenetron, see current-valve Kirchhoff's rules, 72 Lamination and eddy currents, 92-94 Local action and voltaic action, 38 Magnetic blow-out, the, 25 circuit, the, 172 effect of electric current, 1-34 field tension of, 17 the, 10-14 flux, definition of, 15 density, definition of, 170 measurement of, 107 hysteresis, 181 permeability, 170 254 INDEX. Magnetic reluctance, definition of, 174 testing of iron, 183 Magnetism, 1-34 of iron, 170-184 Magnetomotive force, definition of, 172 of a coil, 172, 176-178 Magnet pole, flux from, 17 poles of, 5-8 the oscillating, 12 Maxwell, the, definition of, 16 Metallic conduction, electron theory of, 166 Microfarad, definition of, 105 Motor, the electric (direct-current), 26 Multiplying coil, voltmeter, 67 shunt, ammeter, 69 Non-inductive circuit, definition of, 96 Oersted's experiment, 20 Ohm, definition of, 51 Ohm's law, 58 Oscillator, electric, frequency equation of, iig the, 116-117 Ozone, the making of, 151 Ozonizer, the, 151 Parallel circuits, combined resistance of, 70 division of current in, 68 connections, 68 Permeability, magnetic, 170 Phase difference, 192 Pleiotron, the, 156 Polarization of a voltaic cell, 59 Pole, unit, definition of, 6 Poles of a magnet, 5-8 Power in alternating current systems, 198-202 measurement of, 62 Potential difference, see electromotive force drop, see voltage drop Potentiometer, the, 222-228 Radioactivity, 161 Radio telegraphy, 117-118 Reactance, definition of, 208 Reluctance, magnetic, definition of, 174 Resistance, measurement of, 71, 228 specific, see resistivity temperature coefficient of, 55 Resistivities, table of, 52 Resistivity, definition of, 51 Resonance, electric, 209-214 Screening, electric, 127 Series connections, 66 Shunt, definition of, 68 Side push of magnetic field on an electric wire, 21-30 Spark at break, loo-ioi discharge, the, 147 Specific resistance, see resistivity Spherical condenser, capacity of, 139- 140 Standard cells, 221 inductances, 222 Storage battery, the, 44 Strength of a dielectric, I IS Telegraphy, wireless, 11 7-1 18 Telephone, the wireless, 158 Tension of magnetic field, 17 Thermions, 152 Thermoelement, the, 168 Thermoelectromotive force, 166 Toepler-Holtz influence machine, 134 Transformation, step-up and step- down, 88, 89 Transformer, the, 88 '"ungar, the, 159 Voltage drop, discussion of, 64-65 Voltaic action and local action, 38 cell, polarization of, 59 the, 36-45 Volt, definition of, 60 Voltmeter multiplying coil, 67 the electrostatic, 103 the direct-current, 61 Weston standard cell, the, 221 Wheatstone's bridge, 71 Wireless telegraphy, 11 7-1 18 Franklin and Charles PUBLISHERS OF Scientific and Technical Books BETHLEHEM. PA. MESSRS. CONSTABLE & CO., Limited 10-12 Orange St., Leicester Square LONDON. W. C. 2 Represent Franklin and 'Charles in Great Britain and Ireland, on tie Continent of Europe and in the British Colonies, Dependencies, and self-governing Dominions, excepting Canada. S*"Any Book in this list will he sent postpaid on approval, to he returned postpaid if not satisfactory. ELECTRICAL ENGINEERING Elements of Electrical Engineering, Franklin and Esty. Volume I — 'Direct Currents Price $5.00 "Volume 11 — Alternating Currents " 4.50 This two-volume treatise was published in 190-8-7 and it has been and still is very extensively used as a text book in our technical schools. Dynamos and Motors, Franklin and Esty Price |4j&0 This is a one-volume arrangement of the simpler parts of Frank- lin and Esty*® EUEMENTS. Elements of Electrical Engineering, W. S. Franklin . . Volume I— D. C. and A. C. Machines and Systems. .Price $5.00 Volume II — Blec. Lighting and Misc. Applications... 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