Concepts of Physics PDF

CONCEPTS OF PHYSICS [VOLUME 2] H C VERMA, PhD Retired Professor Department of Physics IIT, Kanpur  Dedicated to Indian Philosophy & Way of Life of which my parents were an integral part ...

CONCEPTS OF PHYSICS [VOLUME 2] H C VERMA, PhD Retired Professor Department of Physics IIT, Kanpur  Dedicated to Indian Philosophy & Way of Life of which my parents were an integral part  FOREWORD A few years ago I had an occasion to go through the book Calculus by L V Terasov. It unravels intricacies of the subject through a dialogue between Teacher and Student. I thoroughly enjoyed reading it. For me this seemed to be one of the few books which teach a difficult subject through inquisition, and using programmed concept for learning. After that book, Dr Harish Chandra Verma’s book on physics, CONCEPTS OF PHYSICS is another such attempt, even though it is not directly in the dialogue form. I have thoroughly appreciated it. It is clear that Dr Verma has spent considerable time in formulating the structure of the book, besides its contents. I think he has been successful in this attempt. Dr Verma’s book has been divided into two parts because of the size of the total manuscript. There have been several books on this subject, each one having its own flavour. However, the present book is a totally different attempt to teach physics, and I am sure it will be extremely useful to the undergraduate students. The exposition of each concept is extremely lucid. In carefully formatted chapters, besides problems and short questions, a number of objective questions have also been included. This book can certainly be extremely useful not only as a textbook, but also for preparation of various competitive examinations. Those who have followed Dr Verma’s scientific work always enjoyed the outstanding contributions he has made in various research areas. He was an outstanding student of Physics Department of IIT Kanpur during his academic career. An extremely methodical, sincere person as a student, he has devoted himself to the task of educating young minds and inculcating scientific temper amongst them. The present venture in the form of these two volumes is another attempt in that direction. I am sure that young minds who would like to learn physics in an appropriate manner will find these volumes extremely useful. I must heartily congratulate Dr Harish Chandra Verma for the magnificent job he has done. Y R Waghmare Professor of Physics IIT Kanpur. (v)  PREFACE Why a new book ? Excellent books exist on physics at an introductory college level so why a new one ? Why so many books exist at the same level, in the first place, and why each of them is highly appreciated ? It is because each of these books has the privilege of having an author or authors who have experienced physics and have their own method of communicating with the students. During my years as a physics teacher, I have developed a somewhat different methodology of presenting physics to the students. Concepts of Physics is a translation of this methodology into a textbook. Prerequisites The book presents a calculus-based physics course which makes free use of algebra, trigonometry and co-ordinate geometry. The level of the latter three topics is quite simple and high school mathematics is sufficient. Calculus is generally done at the introductory college level and I have assumed that the student is enrolled in a concurrent first calculus course. The relevant portions of calculus have been discussed in Chapter 2 so that the student may start using it from the beginning. Almost no knowledge of physics is a prerequisite. I have attempted to start each topic from the zero level. A receptive mind is all that is needed to use this book. Basic philosophy of the book The motto underlying the book is physics is enjoyable. Being a description of the nature around us, physics is our best friend from the day of our existence. I have extensively used this aspect of physics to introduce the physical principles starting with common day occurrences and examples. The subject then appears to be friendly and enjoyable. I have taken care that numerical values of different quantities used in problems correspond to real situations to further strengthen this approach. Teaching and training The basic aim of physics teaching has been to let the student know and understand the principles and equations of physics and their applications in real life. However, to be able to use these principles and equations correctly in a given physical situation, one needs further training. A large number of questions and solved and unsolved problems are given for this purpose. Each question or problem has a specific purpose. It may be there to bring out a subtle point which might have passed unnoticed while doing the text portion. It may be a further elaboration of a concept developed in the text. It may be there to make the student react when several concepts introduced in different chapters combine and show up as a physical situation and so on. Such tools have been used to develop a culture: analyse the situation, make a strategy to invoke correct principles and work it out. Conventions I have tried to use symbols, names, etc., which are popular nowadays. SI units have been consistently used throughout the book. SI prefixes such as micro, milli, mega, etc., are used whenever they make the presentation −6 more readable. Thus, 20 µF is preferred over 20 × 10 F. Co-ordinate sign convention is used in geometrical optics. Special emphasis has been given to dimensions of physical quantities. Numerical values of physical quantities have been mentioned with the units even in equations to maintain dimensional consistency. I have tried my best to keep errors out of this book. I shall be grateful to the readers who point out any errors and/or make other constructive suggestions. H C Verma (vii)  ACKNOWLEDGEMENTS The work on this book started in 1984. Since then, a large number of teachers, students and physics lovers have made valuable suggestions which I have incorporated in this work. It is not possible for me to acknowledge all of them individually. I take this opportunity to express my gratitude to them. However, to Dr S B Mathur, who took great pains in going through the entire manuscript and made valuable comments, I am specially indebted. I am also beholden to my colleagues Dr A Yadav, Dr Deb Mukherjee, Mr M M R Akhtar, Dr Arjun Prasad, Dr S K Sinha and others who gave me valuable advice and were good enough to find time for fruitful discussions. To Dr T K Dutta of B E College, Sibpur I am grateful for having taken time to go through portions of the book and making valuable comments. I thank my student Mr Shailendra Kumar who helped me in checking the answers. I am grateful to Dr B C Rai, Mr Sunil Khijwania & Mr Tejaswi Khijwania for helping me in the preparation of rough sketches for the book. Finally, I thank the members of my family for their support and encouragement. H C Verma (viii)  TO THE STUDENTS Here is a brief discussion on the organisation of the book which will help you in using the book most effectively. The book contains 47 chapters divided in two volumes. Though I strongly believe in the underlying unity of physics, a broad division may be made in the book as follows: Chapters 1–14: Mechanics 15–17: Waves including wave optics 18–22: Optics 23–28: Heat and thermodynamics 29–40: Electric and magnetic phenomena 41–47: Modern physics Each chapter contains a description of the physical principles related to that chapter. It is well supported by mathematical derivations of equations, descriptions of laboratory experiments, historical background, etc. There are "in-text" solved examples. These examples explain the equation just derived or the concept just discussed. These will help you in fixing the ideas firmly in your mind. Your teachers may use these in-text examples in the classroom to encourage students to participate in discussions. After the theory section, there is a section on Worked Out Examples. These numerical examples correspond to various thinking levels and often use several concepts introduced in that chapter or even in previous chapters. You should read the statement of a problem and try to solve it yourself. In case of difficulty, look at the solution given in the book. Even if you solve the problem successfully, you should look into the solution to compare it with your method of solution. You might have thought of a better method, but knowing more than one method is always beneficial. Then comes the part which tests your understanding as well as develops it further. Questions for Short Answer generally touch very minute points of your understanding. It is not necessary that you answer these questions in a single sitting. They have great potential to initiate very fruitful dicussions. So, freely discuss these questions with your friends and see if they agree with your answer. Answers to these questions are not given for the simple reason that the answers could have cut down the span of such discussions and that would have sharply reduced the utility of these questions. There are two sections on multiple-choice questions, namely OBJECTIVE I and OBJECTIVE II. There are four options following each of these questions. Only one option is correct for OBJECTIVE I questions. Any number of options, zero to four, may be correct for OBJECTIVE II questions. Answers to all these questions are provided. Finally, a set of numerical problems are given for your practice. Answers to these problems are also provided. The problems are generally arranged according to the sequence of the concepts developed in the chapter but they are not grouped under section-headings. I don’t want to bias your ideas beforehand by telling you that this problem belongs to that section and hence use that particular equation. You should yourself look into the problem and decide which equations or which methods should be used to solve it. Many of the problems use several concepts developed in different sections of the chapter. Many of them even use the concepts from the previous chapters. Hence, you have to plan out the strategy after understanding the problem. Remember, no problem is difficult. Once you understand the theory, each problem will become easy. So, don’t jump to exercise problems before you have gone through the theory, the worked-out problems and the objectives. Once you feel confident in theory, do the exercise problems. The exercise problems are so arranged that they gradually require more thinking. I hope you will enjoy Concepts of Physics. H C Verma (ix)  Table of Contents Chapters 1–22 Volume–1 Chapter 25 Calorimetry 39 Chapter 23 25.1 Heat as a Form of Energy 39 Heat and Temperature 1 25.2 Units of Heat 39 23.1 Hot and Cold Bodies 1 25.3 Principle of Calorimetry 39 23.2 Zeroth Law of Thermodynamics 1 25.4 Specific Heat Capacity and Molar Heat Capacity 39 25.5 Determination of Specific Heat Capacity 23.3 Defining Scale of Temperature : Mercury in Laboratory 40 and Resistance Thermometers 1 25.6 Specific Latent Heat of Fusion and Vaporization 41 23.4 Constant Volume Gas Thermometer 3 25.7 Measurement of Specific Latent Heat of 23.5 Ideal Gas Temperature Scale 5 Fusion of Ice 41 23.6 Celsius Temperature Scale 5 25.8 Measurement of Specific Latent Heat of 23.7 Ideal Gas Equation 5 Vaporization of Water 42 23.8 Callender’s Compensated Constant 25.9 Mechanical Equivalent of Heat 43 Pressure Thermometer 5 Worked Out Examples 44 23.9 Adiabatic and Diathermic Walls 6 Questions for Short Answer 46 23.10 Thermal Expansion 6 Objective I 46 Worked Out Examples 7 Objective II 46 Questions for Short Answer 11 Exercises 47 Objective I 11 Objective II 12 Chapter 26 Exercises 12 Laws of Thermodynamics 49 26.1 The First Law of Thermodynamics 49 Chapter 24 26.2 Work Done by a Gas 50 26.3 Heat Engines 51 Kinetic Theory of Gases 15 26.4 The Second Law of Thermodynamics 53 24.1 Introduction 15 26.5 Reversible and Irreversible Processes 54 24.2 Assumptions of Kinetic Theory of Gases 15 26.6 Entropy 55 24.3 Calculation of the Pressure of an Ideal Gas 16 26.7 Carnot Engine 55 24.4 rms Speed 16 Worked Out Examples 57 24.5 Kinetic Interpretation of Temperature 17 Questions for Short Answer 60 24.6 Deductions from Kinetic Theory 18 Objective I 61 24.7 Ideal Gas Equation 19 Objective II 61 24.8 Maxwell’s Speed Distribution Law 20 Exercises 62 24.9 Thermodynamic State 20 24.10 Brownian Motion 21 Chapter 27 24.11 Vapour 21 Specific Heat Capacities of Gases 65 24.12 Evaporation 22 27.1 Two Kinds of Specific Heat Capacities of Gases 65 24.13 Saturated and Unsaturated Vapour : 27.2 Relation Between Cp and Cv for an Ideal Gas 66 Vapour Pressure 22 27.3 Determination of Cp of a Gas 67 24.14 Boiling 23 27.4 Determination of Cv of a Gas 68 24.15 Dew Point 23 27.5 Isothermal and Adiabatic Processes 68 24.16 Humidity 24 27.6 Relations between p, V, T in 24.17 Determination of Relative Humidity 24 a Reversible Adiabatic Process 69 24.18 Phase Diagrams : Triple Point 25 27.7 Work Done in an Adiabatic Process 70 24.19 Dew and Fog 25 27.8 Equipartition of Energy 70 Worked Out Examples 26 Worked Out Examples 72 Questions for Short Answer 32 Questions for Short Answer 76 Objective I 33 Objective I 76 Objective II 33 Objective II 77 Exercises 34 Exercises 77 (xi) Chapter 28 Chapter 31 Heat Transfer 81 Capacitors 144 28.1 Thermal Conduction 81 31.1 Capacitor and Capacitance 144 28.2 Series and Parallel Connection of Rods 82 31.2 Calculation of Capacitance 145 28.3 Measurement of Thermal Conductivity of a Solid 83 31.3 Combination of Capacitors 147 28.4 Convection 84 31.4 Force between the Plates of a Capacitor 150 28.5 Radiation 84 31.5 Energy Stored in a Capacitor and 28.6 Prevost Theory of Exchange 84 Energy Density in Electric Field 150 28.7 Blackbody Radiation 85 31.6 Dielectrics 151 28.8 Kirchhoff’s Law 85 31.7 Parallel-plate Capacitor with a Dielectric 152 28.9 Nature of Thermal Radiation 86 31.8 An Alternative Form of Gauss’s Law 153 28.10 Stefan–Boltzmann Law 87 31.9 Electric Field due to a Point Charge q 28.11 Newton’s Law of Cooling 87 Placed in an Infinite Dielectric 154 28.12 Detection and Measurement of Radiation 88 31.10 Energy in the Electric field in a Dielectric 154 Worked Out Examples 89 31.11 Corona Discharge 154 Questions for Short Answer 96 31.12 High-voltage Generator 155 Objective I 97 Worked Out Examples 156 Objective II 97 Questions for Short Answer 164 Exercises 98 Objective I 165 Objective II 165 Chapter 29 Exercises 166 Electric Field and Potential 104 Chapter 32 29.1 What Is Electric Charge ? 104 Electric Current In Conductors 172 29.2 Coulomb’s Law 105 29.3 Electric Field 106 32.1 Electric Current and Current Density 172 29.4 Lines of Electric Force 107 32.2 Drift Speed 173 29.5 Electric Potential Energy 107 32.3 Ohm’s Law 174 29.6 Electric Potential 108 32.4 Temperature Dependence of Resistivity 175 29.7 Electric Potential due to a Point Charge 109 32.5 Battery and emf 176 29.8 Relation between Electric Field and Potential 109 32.6 Energy Transfer in an Electric Circuit 177 29.9 Electric Dipole 110 32.7 Kirchhoff’s Laws 178 29.10 Torque on an Electric Dipole Placed 32.8 Combination of Resistors in Series and Parallel 179 in an Electric Field 112 32.9 Grouping of Batteries 180 29.11 Potential Energy of a Dipole Placed 32.10 Wheatstone Bridge 181 in a Uniform Electric Field 112 32.11 Ammeter and Voltmeter 182 29.12 Conductors, Insulators and Semiconductors 113 32.12 Stretched-wire Potentiometer 183 29.13 The Electric Field inside a Conductor 113 32.13 Charging and Discharging of Capacitors 185 Worked Out Examples 114 32.14 Atmospheric Electricity 186 Questions for Short Answer 119 Worked Out Examples 187 Objective I 119 Questions for Short Answer 196 Objective II 120 Objective I 196 Exercises 121 Objective II 197 Exercises 198 Chapter 30 Chapter 33 Gauss’s Law 127 30.1 Flux of an Electric Field through a Surface 127 Thermal and Chemical Effects of Electric 30.2 Solid Angle 128 Current 206 30.3 Gauss’s Law and Its Derivation 33.1 Joule’s Laws of Heating 206 from Coulomb’s Law 129 33.2 Verification of Joule’s Laws 207 30.4 Applications of Gauss’s Law 131 33.3 Seebeck Effect 207 30.5 Spherical Charge Distributions 134 33.4 Peltier Effect 209 30.6 Earthing a Conductor 135 33.5 Thomson Effect 210 Worked Out Examples 136 33.6 Explanation of Seebeck, Peltier Questions for Short Answer 139 and Thomson Effects 210 Objective I 139 33.7 Electrolysis 211 Objective II 140 33.8 Faraday’s Laws of Electrolysis 211 Exercises 141 33.9 Voltameter or Coulombmeter 213 (xii) 33.10 Primary and Secondary Cells 214 36.14 Determination of M and BH 270 33.11 Primary Cells 214 36.15 Gauss’s Law for Magnetism 270 33.12 Secondary Cell : Lead Accumulator 215 Worked Out Examples 270 Worked Out Examples 215 Questions for Short Answer 275 Questions for Short Answer 217 Objective I 276 Objective I 217 Objective II 277 Objective II 218 Exercises 277 Exercises 218 Chapter 37 Chapter 34 Magnetic Properties of Matter 279 Magnetic Field 221 37.1 Magnetization of Materials : 34.1 Introduction 221 Intensity of Magnetization 279 → 34.2 Definition of Magnetic Field B 221 37.2 Paramagnetism, Ferromagnetism and 34.3 Relation between Electric and Magnetic Fields 222 Diamagnetism 280 34.4 Motion of a Charged Particle 37.3 Magnetic Intensity H 281 in a Uniform Magnetic Field 222 37.4 Magnetic Susceptibility 281 34.5 Magnetic Force on a Current-carrying Wire 223 37.5 Permeability 281 34.6 Torque on a Current Loop 224 37.6 Curie’s Law 282 Worked Out Examples 225 37.7 Properties of Dia-, Para- and Questions for Short Answer 228 Ferromagnetic Substances 282 Objective I 229 37.8 Hysteresis 283 Objective II 229 37.9 Soft Iron and Steel 283 Exercises 230 Worked Out Examples 284 Questions for Short Answer 285 Chapter 35 Objective I 285 Objective II 286 Magnetic Field due to a Current 237 Exercises 286 35.1 Biot–Savart Law 237 35.2 Magnetic Field due to Current in a Straight Wire 238 Chapter 38 35.3 Force between Parallel Currents 239 35.4 Field due to a Circular Current 239 Electromagnetic Induction 288 35.5 Ampere’s Law 241 38.1 Faraday’s Law of Electromagnetic Induction 288 35.6 Magnetic Field at a Point 38.2 Lenz’s Law 289 due to a Long, Straight Current 241 38.3 The Origin of Induced emf 289 35.7 Solenoid 242 38.4 Eddy Current 291 35.8 Toroid 243 38.5 Self-induction 291 Worked Out Examples 243 38.6 Growth and Decay of Current in an LR Circuit 292 Questions for Short Answer 248 38.7 Energy Stored in an Inductor 294 Objective I 248 38.8 Mutual Induction 295 Objective II 249 38.9 Induction Coil 295 Exercises 249 Worked Out Examples 296 Questions for Short Answer 303 Chapter 36 Objective I 304 Objective II 305 Permanent Magnets 255 Exercises 306 36.1 Magnetic Poles and Bar Magnets 255 36.2 Torque on a Bar Magnet Chapter 39 Placed in a Magnetic Field 257 36.3 Magnetic Field due to a Bar Magnet 258 Alternating Current 316 36.4 Magnetic Scalar Potential 258 39.1 Alternating Current 316 36.5 Terrestrial Magnetism 260 39.2 AC Generator or AC Dynamo 316 36.6 Determination of Dip at a Place 261 39.3 Instantaneous and rms Current 317 36.7 Neutral Point 263 39.4 Simple AC Circuits 318 36.8 Tangent Galvanometer 263 39.6 Vector Method to Find the Current 36.9 Moving-coil Galvanometer 264 in an AC Circuit 320 36.10 Shunt 265 39.6 More AC Circuits 320 36.11 Tangent Law of Perpendicular Fields 265 39.7 Power in AC Circuits 322 36.12 Deflection Magnetometer 266 39.8 Choke Coil 323 36.13 Oscillation Magnetometer 268 39.9 Hot-wire Instruments 323 (xiii) 38.10 DC Dynamo 324 43.2 Hydrogen Spectra 369 39.11 DC Motor 325 43.3 Difficulties with Rutherford’s Model 370 39.12 Transformer 325 43.4 Bohr’s Model 370 Worked Out Examples 327 43.5 Limitations of Bohr’s Model 373 Questions for Short Answer 328 43.6 The Wave Function of an Electron 374 Objective I 329 43.7 Quantum Mechanics of the Hydrogen Atom 374 Objective II 329 43.8 Nomenclature in Atomic Physics 375 Exercises 330 43.9 Laser 375 Worked Out Examples 378 Chapter 40 Questions for Short Answer 382 Objective I 383 Electromagnetic Waves 332 Objective II 383 40.1 Introduction 332 Exercises 384 40.2 Maxwell’s Displacement Current 332 40.3 Continuity of Electric Current 333 Chapter 44 40.4 Maxwell’s Equations and Plane Electromagnetic Waves 334 X-rays 388 40.5 Energy Density and Intensity 335 44.1 Production of X-rays 388 40.6 Momentum 336 44.2 Continuous and Characteristic X-rays 388 40.7 Electromagnetic Spectrum and 44.3 Soft and Hard X-rays 390 Radiation in Atmosphere 336 44.4 Moseley’s Law 390 Worked Out Examples 337 44.5 Bragg’s Law 391 Questions for Short Answer 338 44.6 Properties and Uses of X-rays 391 Objective I 338 Worked Out Examples 392 Objective II 339 Questions for Short Answer 393 Exercises 339 Objective I 393 Objective II 394 Chapter 41 Exercises 395 Electric Current through Gases 341 Chapter 45 41.1 Discharge through Gases at Low Pressure 341 41.2 Cathode Rays 343 Semiconductors and Semiconductor Devices 397 41.3 Canal Rays or Positive Rays 343 45.1 Introduction 397 41.4 Discovery and Properties of Electron 344 45.2 Energy Bands in Solids 397 41.5 Thermionic Emission 345 45.3 The Semiconductor 400 41.6 Diode Valve 346 45.4 p-type and n-type Semiconductors 400 41.7 Triode Valve 347 45.5 Density of Charge Carriers and Conductivity 401 41.8 Triode as an Amplifier 349 45.6 p-n Junction 402 Worked Out Examples 349 45.7 p-n Junction Diode 403 Questions for Short Answer 351 45.8 p-n Junction as a Rectifier 405 Objective I 351 45.9 Junction Transistors 406 Objective II 352 45.10 Logic Gates 409 Exercises 352 Worked Out Examples 414 Questions for Short Answer 416 Chapter 42 Objective I 417 Objective II 418 Photoelectric Effect and Wave–Particle Duality 355 Exercises 419 42.1 Photon Theory of Light 355 42.2 Photoelectric Effect 356 Chapter 46 42.3 Matter Waves 359 Worked Out Examples 359 The Nucleus 422 Questions for Short Answer 363 46.1 Properties of a Nucleus 422 Objective I 363 46.2 Nuclear Forces 424 Objective II 364 46.3 Binding Energy 425 Exercises 365 46.4 Radioactive Decay 427 46.5 Law of Radioactive Decay 429 Chapter 43 46.6 Properties and Uses of Nuclear Radiation 431 46.7 Energy from the Nucleus 431 Bohr’s Model and Physics of the Atom 368 46.8 Nuclear Fission 432 43.1 Early Atomic Models 368 46.9 Uranium Fission Reactor 433 (xiv) 46.10 Nuclear Fusion 435 46.11 Fusion in Laboratory 436 Worked Out Examples 437 Questions for Short Answer 440 Objective I 440 Objective II 441 Exercises 442 Chapter 47 The Special Theory of Relativity 446 47.1 The Principle of Relativity 446 47.2 Are Maxwell’s Laws Independent of Frame ? 446 47.3 Kinematical Consequences 447 47.4 Dynamics at Large Velocity 451 47.5 Energy and Momentum 452 47.6 The Ultimate Speed 453 47.7 Twin Paradox 453 Worked Out Examples 455 Questions for Short Answer 456 Objective I 457 Objective II 457 Exercises 458 APPENDIX A 461 APPENDIX B 462 INDEX 463 (xv)   &+$37(5  +($7 $1' 7(03(5$785(  +27 $1' &2/' %2',(6  =(527+ /$: 2) 7+(502'> l) from the centre of the dipole. The angle  S N PON = θ. O P Figure 36.11 The component of magnetic field along OP is r C dV  dV  Br = − =−  PQ  dr  θ = constant S N O ∂V A =− Figure 36.10 ∂r Let SA be the perpendicular from S to OP and NC  µ0 M cosθ  ∂ be the perpendicular from N to OP. As r >> l, =−  2  ∂r  4π r  PS ≈ PA = PO + OA = r + l cosθ. µ0 2M cosθ = ⋅ … (i) Similarly, PN ≈ PC = PO − OC = r − l cosθ. 4π r 3 The magnetic scalar potential at P due to the north The component perpendicular to OP is pole is dV  dV  µ0 m Bθ = − =−  VN = PR  rdθ  4π NP r = constant µ0 m 1 ∂  µ0 M cosθ  = =−   4π r − l cosθ r ∂θ  4π r 2  260 Concepts of Physics 0 M sin earth’s surface. The magnitude and direction of this  3   (ii) field can be obtained approximately by assuming that 4 r the earth has a magnetic dipole of dipole moment 1 about 8.0  10 J T 22 The resultant magnetic field at P is located at its centre (figure B  Br   2 2 B 36.12). The axis of this dipole makes an angle of about 11.5 with the earth’s axis of rotation. The dipole-axis 0 M  4 r 3   2 2 2 cos  sin cuts the earth’s surface at two points, one near the geographical north pole and the other near the 0 M geographical south pole. The first of these points is  4 r 3   1  3 cos. 2  (36.14) called geomagnetic north pole and the other is called geomagnetic south pole. If it makes an angle  with OP, B If we suspend a bar magnet freely at a point near tan    the earth’s surface, it will stay along the magnetic field Br there. The north pole will point towards the direction From (i) and (ii), of the magnetic field. At the geomagnetic poles, the sin tan magnetic field is vertical. If we suspend the bar tan     (36.15) magnet near the geomagnetic north pole, it will become 2 cos 2 vertical with its north pole towards the earth’s surface. Example 36.4 Similarly, if we suspend a bar magnet near the geomagnetic south pole, it will become vertical with its Find the magnetic field due to a dipole of magnetic south pole pointing towards the earth’s surface. moment 1.2 A m at a point 1 m away from it in a 2 Geomagnetic poles may, therefore, be defined as “the direction making an angle of 60 with the dipole-axis. points where a freely suspended bar magnet becomes Solution : The magnitude of the field is vertical”. 0 M B 4 r 3   1  3 cos2 If we treat the assumed magnetic dipole inside the earth as a pair of north and south poles (figure 36.12),   7 T m  1.2 A m 2  10 A  1 m 3   1  3 cos 260 the south pole will be towards the geomagnetic north  pole and the north pole will be towards the  1.6  10 T. 7 geomagnetic south pole. This may be easily remembered by using the fact that the north pole of The direction of the field makes an angle  with the the suspended magnet should be attracted by the south radial line where pole of the assumed dipole. tan 3 tan    Earth’s magnetic field changes both in magnitude 2 2 and direction as time passes. It is fairly constant over a span of a few days, but noticeable changes occur in 36.5 TERRESTRIAL MAGNETISM say, ten years. Studies of magnetic rocks have revealed Earth is a natural source of magnetic field. We that the magnetic field may even reverse its direction. It appears that in the past 7.6  10 years, already 7 have magnetic field present everywhere near the Geographical 171 such reversals have taken place. The latest north pole reversal in earth’s magnetic field is believed to have occurred around 10,000 years ago. Geomagnetic 11.5° north pole The theory of earth’s magnetic field is not yet well- understood. At present, it seems that the field results mainly due to circulating electric currents induced in the molten liquid and other conducting material inside S the earth. N Elements of the Earth’s Magnetic Field The earth’s magnetic field at a point on its surface Geomagnetic is usually characterised by three quantities: south pole (a) declination (b) inclination or dip and (c) horizontal Geographical south pole component of the field. These are known as the Figure 36.12 elements of the earth’s magnetic field.  Permanent Magnets 261 Declination Geographical Vertical North Declination A plane passing through the geographical poles P (( (that is, through the axis of rotation of the earth) and Geographical Bh ' a given point P on the earth’s surface is called the Meridian Bv geographical meridian at the point P. Similarly, the plane passing through the geomagnetic poles (that is, Magnetic through the dipole-axis of the earth) and the point P North is called the magnetic meridian at the point P. In other words, the magnetic meridian is a vertical plane Magnetic through the point P that contains the geomagnetic Meridian poles. The magnetic field due to the earth at P must Figure 36.13 be in this plane (magnetic meridian). The angle made by the magnetic meridian at a Thus, from the knowledge of the three elements, point with the geographical meridian is called the both the magnitude and direction of the earth’s declination at that point. The knowledge of declination magnetic field can be obtained. fixes the vertical plane in which the earth’s magnetic Example 36.5 field lies. The horizontal component of the earth’s magnetic field Navigators often use a magnetic compass needle is 3.6 × 10 T where the dip is 60°. Find the magnitude –5 to locate direction. A compass needle is a short and of the earth’s magnetic field. light magnetic needle, free to rotate about a vertical axis. The needle is enclosed in a small case with a Solution : We have BH = B cosδ glass-top. The needle stays in equilibrium when it is BH 3.6 × 10 T −5 = 7.2 × 10 T. −5 in magnetic meridian. Hence the north direction shown or, B= = cosδ cos 60° by the needle makes an angle equal to the declination with the true north and navigators have to take care 36.6 DETERMINATION OF DIP AT A PLACE of it. Dip Circle Inclination or dip The angle made by the earth’s magnetic field with 90° the horizontal direction in the magnetic meridian, is called the inclination or dip at that point. S In the magnetic nothern hemisphere, the vertical 0° 0° component of the earth’s magnetic field points downwards. The north pole of a freely suspended magnet, therefore, dips (goes down). The knowledge of declination and inclination vernier scale completely specifies the direction of the earth’s magnetic field. E spirit level Horizontal component of the earth’s magnetic field As the name indicates, the horizontal component is component of the earth’s magnetic field in the Figure 36.14 horizontal direction in the magnetic meridian. This The dip at a place can be determined by an direction is towards the magnetic north. apparatus known as dip circle. It consists of a vertical Figure (36.13) shows the three elements. Starting circular scale S and a magnetic needle (a small pointed from the geographical meridian we draw the magnetic permanent magnet) pivoted at the centre of the scale. meridian at an angle θ (declination). In the magnetic The needle can rotate freely in the vertical plane of meridian we draw the horizontal direction specifying the scale. The pointed ends move over the graduations magnetic north. The magnetic field is at an angle δ on the scale which are marked 0°–0° in the horizontal (dip) from this direction. The horizontal component BH and 90°–90° in the vertical direction. The scale S and the total field B are related as together with the needle is enclosed in a glass cover which can be rotated about a vertical axis. The angle BH = B cosδ rotated can be read from a horizontal angular scale E, or, B = BH /cosδ. fixed with the base, and a vernier scale fixed with the 262 Concepts of Physics stand supporting the glass cover. The base can be If the 0–0 line on the scale is not horizontal, the made horizontal by levelling screws fixed with it. A value of dip will have some error. This error may be spirit-level fixed to the apparatus helps in levelling. removed as described below. Determination of Dip Determination of magnetic meridian At the beginning of the experiment, the base of the dip circle is made horizontal with the help of the levelling screws and the magnetic needle is pivoted in its place. The glass cover containing the vertical scale S and the needle is rotated about the vertical axis till Figure 36.16 the needle becomes vertical and reads 90–90 on the vertical scale. In this condition, the plane of the Bring the vertical scale in the magnetic meridian circular scale S is perpendicular to the magnetic and note the readings of the ends of the needle (figure meridian. The horizontal component BH is 36.16a). Now, rotate the circle through 180 about the perpendicular to this plane and hence does not take vertical axis and again note the readings (figure part in rotating the needle. The needle is aligned with 36.16b). The average of these readings will not have the vertical component BV and hence reads 9090. the error due to 0–0 line. The reading of the vernier is noted and the glass cover (c) The magnetic and the geometrical axes of the is rotated exactly through 90 from this position. The needle are different. plane of the circular scale S is now the same as the In the experiment, we read the angles magnetic meridian. corresponding to the ends of the needle. If the magnetic axis is inclined at an angle with the line Measurement of dip joining the ends, the dip obtained is in error. This error When the plane of the vertical scale S is the same can be removed by inverting the needle on its bearing as the magnetic meridian, the earth’s magnetic field and repeating the previous readings. The average of B is in this same plane. In this case, the needle rests these readings is free of this error. in the direction of B. The readings of the ends of the (d) Centre of mass of the needle does not coincide needle on the vertical scale now directly give the value with the pivot. of the dip. If the centre of mass of the needle is not at the pivot, its weight mg will have a torque and will affect Possible errors and their remedies the equilibrium position. To remove this error, one has Errors may occur because of several imperfections to read the dip and then take out the needle. The in the instrument. Some of the possible errors and needle should be demagnetized and then remagnetized their remedies are given below. in opposite direction. Thus, the position of the north (a) The centre of the needle is not at the centre of and south poles are interchanged. The centre of mass the vertical scale. now appears on the other side of the pivot and hence the effect of mg is also reversed. The dip is again determined with this needle and the average is taken. S S N mg mg N Figure 36.17 Figure 36.15 Thus, one should take 16 readings of dip and an If the centre of the needle does not coincide with average of all these gives the true dip. the centre of the scale, the readings do not represent the true dip. The reading of one end of the needle is less than the true value of the dip and the reading of Apparent Dip the other end is greater by the same amount. Thus, both ends are read and the average is taken. If the dip circle is not kept in the magnetic meridian, the needle will not show the correct direction (b) 0–0 line is not horizontal. of earth’s magnetic field. The angle made by the needle  Permanent Magnets 263 with the horizontal is called the apparent dip for this plane. If the dip circle is at an angle α to the meridian, C the effective horizontal component in this plane is B′H = BH cosα. The vertical component is still Bv. If δ′ P A is the apparent dip and δ is the true dip, we have S Bv tanδ′ = B′H T1 B B′H BH cosα T2 or, cotδ′ = = Bv Bv or, cotδ′ = cotδ cosα. … (i) Figure 36.18 Now suppose, the dip circle is rotated through an angle of 90° from this position. It will now make an provided on this base for connecting the galvanometer angle (90° – α) with the meridian. The effective to an external circuit. An aluminium pointer P is horizontal component in this plane is B′′H = BH sinα. If rigidly attached with the compass needle and δ′′ be the apparent dip, we shall have perpendicular to it. The compass needle together with cotδ′′ = cotδ sinα. … (ii) the pointer can rotate freely about the vertical axis. The ends of the pointer move over a graduated, Squaring and adding (i) and (ii), 2 2 2 horizontal circular scale. The graduations are marked cot δ′ + cot δ′′ = cot δ. … (36.16) from 0° to 90° in each quadrant. The scale, the pointer Thus, one can get the true dip δ without locating and the compass needle are enclosed in a closed the magnetic meridian. cylindrical box which is placed with its centre coinciding with the centre of the coil. The box can also Example 36.6 be rotated about the vertical axis. The upper surface of the box is made of glass so that the things inside it At 45° to the magnetic meridian, the apparent dip is 30°. are visible. To avoid the errors due to parallax, a plane Find the true dip. mirror is fixed at the lower surface of the box. While Solution : At 45° to the magnetic meridian, the effective noting the reading of the pointer, the eye should be horizontal component of the earth’s magnetic field is properly positioned so that the image of the pointer is 1 B′H = BH cos 45° = B. The apparent dip δ′ is given by just below the pointer. √2 H When there is no current through the Bv √2 Bv tanδ′ = = = √2 tanδ galvanometer, the compass needle is in magnetic B′H BH north–south direction. To measure a current with the where δ is the true dip. Thus, tangent galvanometer, the base is rotated in such a tan 30° = √2 tanδ way that the plane of the coil is parallel to the compass or, δ = tan −1 1/6. √ needle. The plane then coincides with the magnetic meridian. The box containing the needle is rotated so that the aluminium pointer reads 0°–0° on the scale. 36.7 NEUTRAL POINT The current to be measured is passed through the Suppose at a point, the horizontal component of coil. The current through the coil produces a magnetic the magnetic field due to a magnet is equal and field at the centre and the compass needle deflects under opposite to the earth’s horizontal magnetic field. The its action. The pointer deflects through the same angle net horizontal field is zero at such a point. If a compass and the deflection of both the ends are read from the needle is placed at such a point, it can stay in any horizontal scale. The average of these two is calculated. position. Such a point is called a neutral point. Suppose the current through the coil is i. The 36.8 TANGENT GALVANOMETER radius of the coil is r and the number of turns in it is n. The magnetic field produced at the centre is Tangent galvanometer is an instrument to µ0 in measure an electric current. The essential parts are a B= ⋅ … (i) 2r vertical circular coil C of conducting wire and a small compass needle A pivoted at the centre of the coil This field is perpendicular to the plane of the coil. (figure 36.18). The coil C together with its frame is This direction is horizontal and perpendicular to the fixed to a horizontal base B provided with levelling magnetic meridian and hence to the horizontal screws. Terminals T1 and T2 connected to the coil are component BH of the earth’s magnetic field. The 264 Concepts of Physics resultant horizontal magnetic field is as possible. This is the case when θ = 45°. So, the Br = √  B +B 2 2 tangent galvanometer is most sensitive when the H deflection is around 45°. in a direction making an angle θ with BH (figure 36.19) where Example 36.7 mBr A tangent galvanometer has 66 turns and the diameter N B Br of its coil is 22 cm. It gives a deflection of 45° for 0.10 A current. What is the value of the horizontal component  of the earth’s magnetic field ?  Solution : For a tangent galvanometer BH S i = K tanθ mBr 2r BH Figure 36.19 = tanθ µ0 n tanθ = B/BH. … (ii) µ0 ni or, BH = If m be the pole strength of the needle, the force 2r tanθ on the north pole of the needle is mBr along Br and  T m  × 66 × (0.1 A) −7 4π × 10 on the south pole is mBr, opposite to Br. The needle  A  = will stay in equilibrium when its length is parallel to −2 (22 × 10 m) (tan 45°) Br , because then no torque is produced by the two = 3.8 × 10 T. −5 forces. Thus, the deflection of the needle from its original position is θ as given by (ii). Using (i) and (ii), µ0 in T BH tanθ = 2r 2r BH W or, i= tanθ µ0 n or, i = K tanθ, … (36.17) North South 2r BH where K= is a constant for the given µ0 n galvanometer at a given place. This constant is called S the reduction factor of the galvanometer. The reduction factor may be obtained by passing a known current i Figure 36.20 through the galvanometer, measuring θ and then using 36.9 MOVING-COIL GALVANOMETER (36.17). Equation (i) is strictly valid only at the centre of the coil. The poles of the needle are slightly away from The main parts of a moving-coil galvanometer are shown in figure (36.20). A rectangular coil of several the centre. Thus, the length of the needle should be turns is wound over a soft-iron core. The wire of the small as compared to the radius of the coil. coil is coated with an insulating material so that each turn is insulated from the other and from the iron core. Sensitivity The coil is suspended between the two pole pieces of Good sensitivity means that the change in a strong permanent magnet. A fine strip W of phosphor deflection is large for a given fractional change in bronze is used to suspend the coil. The upper end of current. this strip is attached to a torsion head T. The lower end of the coil is attached to a spring S also made of We have phosphor bronze. A small mirror is fixed on the i = K tanθ suspension strip and is used to measure the deflection or, di = K sec θ dθ 2 of the coil with the help of a lamp–scale arrangement. 2 Terminals are connected to the suspension strip W and di K sec θ dθ 2 dθ the spring S. These terminals are used to pass current or, = = through the galvanometer. i K tanθ sin 2θ 1  di  The current to be measured is passed through the or, dθ = sin 2θ . → 2 i galvanometer. As the coil is in the magnetic field B of → → → Thus, for good sensitivity, sin 2θ should be as large the permanent magnet, a torque Γ = niA × B acts on  Permanent Magnets 265 the coil. Here n = number of turns, i = current in the ig through the galvanometer and remaining is = i − ig → → coil, A = area-vector of the coil and B = magnetic field through the shunt. If the resistance of the at the site of the coil. This torque deflects the coil from galvanometer coil is Rg, we have Rs ig = i. … (36.19) Rs + Rg N S As Rs is much smaller than Rg, only a small fraction goes through the galvanometer. Figure 36.21 Example 36.8 its equilibrium position. The pole pieces are made cylindrical. As a result, A galvanometer having a coil of resistance 20 Ω needs the magnetic field at the arms of the coil remains 20 mA current for full-scale deflection. In order to pass parallel to the plane of the coil everywhere even as a maximum current of 2 A through the galvanometer, the coil rotates. The deflecting torque is then what resistance should be added as a shunt ? Γ = niAB. As the upper end of the suspension strip W Solution : Out of the main current of 2 A, only 20 mA is fixed, the strip gets twisted when the coil rotates. should go through the coil. The current through the coil This produces a restoring torque acting on the coil. If is the deflection of the coil is θ and the torsional constant Rs of the suspension strip is k, the restoring torque is ig = i Rs + Rg kθ. The coil will stay at a deflection θ where Rs niAB = kθ or, 20 mA = 2A Rs + (20 Ω) k or, i= θ. … (36.18) 20 mA Rs nAB or, = 2A Rs + (20 Ω) Hence, the current is proportional to the deflection. −2 k or, (10 ) (Rs + 20 Ω) = Rs The constant is called the galvanometer constant nAB 20 and may be found by passing a known current, or, Rs = Ω ≈ 0.2 Ω. 99 measuring the deflection θ and putting these values in equation (36.18). 36.11 TANGENT LAW OF PERPENDICULAR FIELDS Sensitivity When a compass needle is placed in the earth’s The sensitivity of a moving-coil galvanometer is magnetic field, it stays along the horizontal component defined as θ/i. From equation (36.18), the sensitivity BH of the field. The magnetic forces mBH and –mBH on nAB is ⋅ For large sensitivity, the field B should be the poles do not produce any torque in this case. If an k large. The presence of soft-iron core increases the external horizontal magnetic field B is produced which magnetic field. We shall discuss in a later chapter how is perpendicular to BH, the needle deflects from its soft iron increases magnetic field. position. The situation is the same as that shown in figure (36.19). The resultant of B and BH is Br = √  B + 2 2 36.10 SHUNT BH A galvanometer is usually a delicate and sensitive making an angle θ with BH so that instrument and only a small current is sufficient to B tanθ = ⋅ … (i) deflect the coil to its maximum allowed value. If a BH current larger than this permissible value is passed The forces on the poles are mBr , −mBr which may through the galvanometer, it may get damaged. A small resistance Rs called shunt, is connected in produce a torque and deflect the needle. The needle can stay in a position parallel to the resultant parallel with the galvanometer to save it from such horizontal field Br. Thus, the deflection of the needle accidents. The main current i is divided in two parts, i ig i in equilibrium is θ. Using (i), the external magnetic G field B may be written in terms of BH and θ as is B = BH tanθ. … (36.20) Rs This is known as the tangent law of perpendicular Figure 36.22 fields. 266 Concepts of Physics 36.12 DEFLECTION MAGNETOMETER position of the bar magnet. The magnetic field due to the bar magnet at the site of the needle is, therefore, 90° A P µ0 2Md 0° 0° B= 2 2 2 ⋅ 90° 4π (d − l ) This field is along the length of the magnet, that Figure 36.23 is, towards east or towards west. It is, therefore, A deflection magnetometer (figure 36.23) consists perpendicular to the earth’s field BH. From the tangent of a small compass needle A pivoted at the centre of law, we have, a graduated circular scale. The graduations are B = BH tanθ marked from 0° to 90° in each quadrant. An aluminium pointer P is rigidly fixed with the needle and µ0 2Md or, 2 2 2 = BH tanθ perpendicular to it. The ends of the pointer move on 4π (d − l ) the circular scale. These are enclosed in a cylindrical 2 2 2 box known as the magnetometer box. The upper cover M 4π (d − l ) or, = tanθ. … (36.21) of the box is made of glass so that the things inside BH µ0 2d are visible. A plane mirror is fixed on the lower surface Knowing all the quantities on the right-hand side, so that the pointer may be read without parallax. This one gets M/BH. arrangement is the same as that used in a tangent galvanometer. Possible errors and their remedies The magnetometer box is kept in a wooden frame Errors may occur due to various reasons. having two long arms. Metre scales are fitted on the (a) The pivot of the needle may not be at the centre two arms. The reading of a scale at any point directly of the circular scale (figure 36.25). To remove this error gives the distance of that point from the centre of the both ends of the pointer are read and the mean is compass needle. taken. The basic use of a deflection magnetometer is to determine M/BH for a permanent bar magnet. Here M is the magnetic moment of the magnet and BH is the horizontal component of the earth’s magnetic field. This quantity M/BH can be measured in two standard positions of the magnetometer. One is called Tan-A position of Gauss and the other is called Tan-B position of Gauss. Figure 36.25 Tan-A Position (b) The magnetic centre of the bar magnet may not coincide with its geometrical centre (figure 36.26). N 90° The measured distance d is more or less than the 0° 0° actual distance d1 of the centre of the needle from the W E N S 90° magnetic centre of the magnet. To remove the error S due to this, the magnet is rotated through 180° about d the vertical so that the positions of the north pole and Figure 36.24 the south pole are interchanged. The deflections are In this, the arms of the magnetometer are kept again noted with both ends of the pointer. These along the magnetic east–west direction. The readings correspond to the distance d2. The average of aluminium pointer shows this direction when no extra the sets of deflections corresponding to the distances magnets or magnetic materials are present nearby. d1 and d2 give the correct value approximately. The magnetometer box is rotated in its plane till the d1 pointer reads 0°–0°. The magnet is now kept on one of the arms, parallel to its length (figure 36.24). The S N needle deflects and the deflection θ in its equilibrium d2 position is read from the circular scale. The distance of the centre of the magnet from the centre of the N S compass is calculated from the linear scale on the arm. Figure 36.26 Let d be this distance and 2l be the magnetic length of the magnet. In Tan-A position of the (c) The geometrical axis of the bar magnet may magnetometer, the compass needle is in end-on not coincide with the magnetic axis (figure 36.27). To  Permanent Magnets 267 S N In tan-B position, the compass is in broadside-on position of the magnet. The magnetic field at the N S compass due to the magnet is, therefore, Figure 36.27 µ0 M B= 2 2 3/2 ⋅ avoid error due to this, the magnet is put upside down 4π (d + l ) at the same position and the readings are taken. The field is parallel to the axis of the magnet and (d) The zero of the linear scale may not coincide hence it is towards east or towards west. The earth’s with the centre of the circular scale. To remove the magnetic field is from south to north. Using tangent error due to this, the magnet is kept on the other arm law, of the magnetometer at the same distance from the B = BH tanθ needle and all the readings are repeated. µ0 M or, 2 3/2 = BH tanθ Thus, one gets sixteen values of θ for the same 2 4π (d + l ) distance d. The mean of these sixteen values gives the M 4π 2 2 3/2 correct value of θ. or, = (d + l ) tanθ. … (36.22) BH µ0 Tan-B Position N Applications of a Deflection Magnetometer A variety of quantities may be obtained from the W E basic measurement of M/BH using a deflection S magnetometer. Here are some of the examples. (a) Comparison of the magnetic moments M1 and M2 of two magnets 90° One can find M1/BH and M2/BH separately for the 0° 0° 90° two magnets and then get the ratio M1/M2. There is another simple method known as null method to get M1/M2. The experiment can be done either in Tan-A position or in Tan-B position. The two magnets are placed on the two arms of the magnetometer. The distances of the magnets from the centre of the N S magnetometer are so adjusted that the deflection of the needle is zero. In this case, the magnetic field at the needle due to the first magnet is equal in Figure 36.28 magnitude to the field due to the other magnet. If the In this position, the arms of the magnetometer are magnetometer is used in Tan-A position, kept in the magnetic north to south direction. The box µ0 2 M1d1 µ0 2 M2 d2 is rotated so that the pointer reads 0°–0°. The bar 2 2 2 = 2 2 2 magnet is placed on one of the arms symmetrically 4π (d1 − l1 ) 4π (d2 − l2 ) 2 2 2 and at right angles to it (figure 36.28). The distance d M1 d2(d1 − l1 ) of the centre of the magnet from the centre of the or, = ⋅ M2 d1(d22 − l22) 2 compass needle is calculated from the linear scale. The deflection θ of the needle is noted from the circular If the magnetometer is used in Tan-B position, scale. To remove the errors due to the reasons µ0 M1 µ0 M2 2 2 3/2 = 2 2 3/2 described above, the deflections are read in various 4π (d1 + l1 ) 4π (d2 + l2 ) situations mentioned below. Both ends of the pointer 2 2 3/2 M1 (d1 + l1 ) are read. The magnet is put upside down and readings or, = ⋅ M2 (d22 + l22) 3/2 ar