Chief AdvisorRashmi Krishnan, IAS

Director, SCERT

GuidanceDr. Pratibha Sharma,

Joint Director, SCERT

Academic Co-ordinator and EditorDr.Rajesh Kumar, Principal, DIET Daryaganj

Sapna Yadav, Sr. Lecturer, SCERT

ContributorsProf. B.K Sharma (Retd.) Professor, NCERTDr. R.P Sharma, Academic Consultant, CBSE

Pundrikaksh Kaundinya, Vice Principal, RPVV KishangajSher Singh, Principal, Navyug School, Lodhi RoadDavendra Kumar, Lecturer, RPVV, Yamuna ViharGirija Shankar, Lecturer, RPVV, Surajmal Vihar

R.Rangarajan , Lecturer, DTEA , Sr.Secondary School , Lodhi RoadNeelam Batra , Lecturer, D.C. Arya, Sr. Sec. School, Lodhi Colony

Chitra Goel, Retd Vice PrincipalDr.Rajesh Kumar, Principal, DIET Daryaganj

Sapna Yadav , Sr. Lecturer , SCERT

Publication OfficerMr. Mukesh Yadav

Publication TeamSh. Navin Kumar, Ms. Radha, Sh. Jai Baghwan

Published by : State Council of Educational Research & Training, New Delhi and printed atEducational Stores, S-5, Bsr. Road Ind. Area, Ghaziabad (U.P.)

Preface

Physics is basic to the understanding of almost all the branches of science and technology. It isinteresting to note that the ideas and concepts of physics are behavioural science too.

The students may or may not continue to study physics beyond the higher secondary stage, butwe feel that they will find the thought process of physics useful in any other branch they maylike to pursuer, be it finance Administration, Social Science, Environment, Engineering, Technology,Biology or Medicine.

Physics discovers new properties of particles and want to create a name for each one. Latest exampleof physics is the innovation of God Practice (Higgs Boson). Higgs boson is a hypothetical elementarypractice, a boson, which is the quantum of the Higgs field. The filed and the particle provide atestable hypothesis for the origin of mass in elementary particles. “God particle”—— that helpsto explain what gives all matter in the universe size and shape.

Physics is not a mugging subject, we can not simply study it by the book. We must attempt questionsfrom workbooks and do a lot of exercises. Only then we will be able to answer various types ofquestions no matter how tough they may seem. Interest plays a vital role in scoring in this subject.We must show keen interest in studying physics. Most people dislike the subject as it is far morecomplicating then the other science subjects. Therefore, they make no effort to master the subject.

This book has some features which will help to service the contents in shorter time. After learningcontent one can answer question for self lest. Some exercises are given with art & pedagogicalremark use of easily understandable language is taken care of suggestions to students from a teacher.

Finally it must be remembered that entire physics is based on observations & experiments withoutwhich a theory does not get acceptance in to the domain of physics.

We are thankful to all those who conveyed these inputs.

We welcome suggestions and comments from our valued users, especially students andteachers at the email id scert.physics@gmail.com & can also post on www.scertdelhi.nic.in.

We wish our readers a happy journey to the exciting realm of physics.

Contents

S.No. Chapter Name Page No.

Preface (iii)

The Physics Student cannot afford to miss it (vii)

Physics Syllabus (xii)

1. Electrostatics 1-23

2. Current Electricity 24-37

3. Magnetic Effect of Current and Magnetism 38-61

4. Electromagnetic Induction and Alternating Current 62-89

5. Electromagnetic Waves 90-96

6. Optics 97-129

7. Dual Nature of Radiation & Matter 130-136

8. Atom Nuclei 137-146

9. Electronic Devices 147-160

10. Communication Systems 161-168

Appendix 169-175

(vi)

Safety Precautions for Students in Physics Laboratory

Designing of all science laboratories according to necessary norms and standards.

Two wide doors for unobstructed exits from the laboratory.

Adequate number of fire extinguishers near science laboratories.

Periodical checking of vulnerable points in the laboratories in relation to possibility of anymishappening.

Periodical checking of electrical fittings/ insulations for replacement and repairs.

Timely and repeated instructions to students for careful handling of equipments in the laboratory.

Display of do's and dont's in the laboratory at prominent places.

Safe and secure storage of all Equipments

Proper labelling and upkeep of Equipments

Careful supervision of students while doing practical work.

Advance precautionary arrangements to meet any emergency situations.

Conduct of any additional experimental work only under supervision and with due advance permission.

Availability of First Aid and basic medical facilities in the school.

Proper location of the laboratories.

How to make the learning of the difficult topics easier?

Do's Und Dont's

1. Do not take the word difficult while teaching and you be positive yourself.

2. Do not place the topic on the board - After completing tell the students that this “whatit is”— For example topics like Potentiometer

3. Try to build the topic from the basics of the basic while teaching.

4. Do not draw the diagram on board before you start the topic – Do build the same asthe discussion continues.

5. Do not forget to place the arrow in circuits and Ray diagrams – A mistake which canbe easily absorbed by the student.

6. Do not postpone the topic for the end of the academic session

7. Give 2-3 revision by asking question's from such identified topics at the beginning ofthe class on the subsequent days.

8. Try to test these identified topics in almost all the tests if possible after prior informationto the students.

9. Do try to create interest on such topics before it is actually discussed.

10. Try to adopt an interactive approach to deal with such topics.

11. Very important a point is to think new and good approaches that may fit your studentswhile dealing with such topics and popularize such method.

(vii)

The Physics Student cannot afford to miss it

1. Identify the chapters in which the weight-age is more.

2. Prepare those identified chapters having more weight-age with an eye to have a sure 5 mark questionand do writing practice also with proper figures. Do the super hit questions/topics like Cyclotron,a.c.generator, Young's experiment, Gauss's theorem, Wheatstone's bridge, potentiometer etc.., manytimes before the examination, so that you do not flop during the examination because of the tiltednature of a question.

3. Instead of leaving the topics like E.M. Waves, Principles of Communication understand to expressall definitions, interpretation of figures, Advantages and disadvantages of various devices and Applicationsetc.

4. Do all the worked examples and the graphs with their Interpretation (which you can easily understand)in a line or two from NCERT and practice them before hand.

5. Go to the examination hall with a positive frame of mind - particularly on the Physics examinationday, at least half an hour before without any books and please do not discuss any question with anyonein this period.

6. Start the answer script with the best known question and complete all the questions that you knowwithout cutting and overwriting.

7. In case you are not having good Interpretation skill, first do the best known five mark questions andtry to create a good impression in the minds of the paper checker.

8. When you approach the numerical question always understand the question, recall the known conceptof the question and never try to list the formula and substitute the values.

9. Present the paper neatly and legibly without cutting and leaving space for anything that you plan todo later, since there will not be any time to do later. If you happen to cut, do it neatly such thatthe cut and the un-cut portions are distinguishable. Thinking and formatting the answer before writingwill improve you on this front.

10. Never leave any question. Write something of what you know of the answer. Remember "What youthink is wrong may be the correct answer" many a times.

DerivationsUnit-1 (Electrostatics)

Electrical / magnetic field at a point on the equatorial or axial line due to an electrical /magneticdipole

Torque experienced by a dipole placed in a uniform electric / magnetic field.

Determine the potential energy of dipole in a uniform electric field

Relation between electric field and electric potential *

Gauss's theorem and its applications

(viii)

Equivalent capacitance when capacitors are connected in parallel / series

Capacitance of parallel plate capacitor

Derive an expression for the capacity of a parallel plate capacitor with (a) dielectric slab of thicknesst < d (b) with conductor between the plates *

Using a labeled diagram, explain the principle and working of Van de Graf f generator

Unit-2 (Current Electricity)

Relation between resistivity and relaxation time

Condition of balance in Wheatstone's bridge

Explain the working and principle of a potentiometer. How can it be used to (a) compare emf ofcell (b) determine internal resistance of a cell *

Unit-3 (Magnetic Effect of Current and Magnetism)

Magnetic field due to a straight conductor / coil carrying current

Force experienced by (a) charge moving in electric field (b) current carrying conductor (c) torqueon coil in magnetic field.

Ampere's circuital law and its application for determining magnetic field in solenoids and toroidal.*

Force between two parallel wires carrying current *

Describe the principle, construction and working of a moving coil galvanometer with a labeled diagram. *

Explain with the help of a labeled diagram, the underlying principle, construction and working ofa cyclotron frequency and total K.E. *

Unit-4 (Electromagnetic Induction and Alternating Current)

Write five differences between dia, Para and Ferro magnetic substances.*

magnetic field at a point on the equatorial or axial line due to an electrical / magnetic dipole

Unit-5 (Electromagnetic Waves)

Determination of (a) coefficient of self induction (b) mutual induction in solenoids *

Energy stored in an (a) inductor (b) capacitor *

Distinguish between resistance, reactance and impedance.

Derive an expression for (a) current in LCR series circuit using phasor diagram and power or LCRcircuit *

Explain with the help of a labeled diagram, the principle, construction and working of a transformer.Why is it used for power transmission? *

Explain with the help of labeled diagram, the principle, construction and working of an AC generator* Explain Hertz's experiment for producing electromagnetic waves

Unit-6 (Optics)

Deduce laws of refraction & reflection on me basis of Huygens's principle. *

Interference by Young's double slit experiment, determination of fringe width and condition for maximaand minima.*

(ix)

Diffraction at a single slit - determination of fringe width of central max. *

Diffraction at a single slit - determination of fringe width of central max. *

Polarisation - Malu's law & Brewster's law *

Mirror formula for concave and convex mirrors

Define critical angle and write condition for total internal reflection. Obtain an expression for refractiveindex in terms of critical angle

Lens formula for convex and concave lenses

Derive an expression for refraction at spherical surfaces.

Deduce lens maker's formula for a biconvex lens *

Obtain an expression for the refractive index of the material of a prism in terms of refracting angleand angle of minimum deviation.

Structure of eye and its defects and rectification

Draw a labeled diagram and determine the magnification and resolving power of (a) simple microscope(b) compound microscope (c) astronomical telescope and (d) reflecting type telescope *

Explain dispersion and rainbow formation

Unit-7 (Dual Nature of Mather)

State the laws of photoelectric effect. Establish Einstein's photoelectric relation *

Explain Davison Germer experiment and show how it proved De Broglie's theory of matter waves. *

Determination of wavelength associated with electron *

Unit-8 (Atom Nuclei)

Short notes on α, β and Y decay

State law of radioactive decay and obtain expression for N *

Bohr's Postulate. Expression for radius, K.E, P.E, Total energy, energy spectrum with energy leveldiagram. *

Unit-9 (Electronic Devices)

Difference between (a) n and p type semiconductors (b) intrinsic and extrinsic semiconductors

Draw the circuit to study the characteristics of p-n junction diode in forward and reverse bias. Sketchthe V × I graph for the same

Explain the use of p-n junction diode as a rectifier. Draw the circuit diagram of (a) full wave rectifierand (b) half wave rectifier. Draw input and output waveforms for them *

Draw the circuit to study the output and input characteristics of a common emitter amplifier. Sketchthe V × I graph for the same. *

With the help of a circuit diagram, explain the working of a pnp / npn transistor as an amplifier incommon emitter mode *

With the help of a circuit diagram, explain the working of a pnp / npn transistor as a switch in commonemitter mode

Discuss the working of a transistor as an oscillator. *

Realization of AND, OR and NOT gates

(x)

Unit-10 (Communication System)

What is a communication system? Describe the constituents of a communication system

Write short notes using block diagram on (a) ground waves (b) sky waves (c) space wave obtainexpression for range for LOG transmission (d) modulation index.*

What do the following terms refer to in communication: transducer, base band, bandwidth, attenuation,modulation, demodulation, noise, modulation index.

What is modulation? Why is modulation required?

What is demodulation? Draw a block diagram to show receiver and demodulation

Draw block diagram for modulation process and determine the bandwidth for amplitude modulation

*very important

Important diagrams

Van de Graft generator – neatly labeled.

Moving coil galvanometer and cyclotron – neatly labeled

Microscope – simple & compound

Telescope – refracting & reflecting

Ray diagram for lens maker's formula & Lens formula

A.C.generator and transformer

Photoelectric effect and bavison Germer experiment

Amplifier (npn & pnp transistor), switch and Oscillator.

Rectifier (full wave & half wave)

Circuit diagram of potentiometer (comparing emf, internal resistance) and Meter Bridge for determiningresistance.

Electrical field due to a point charge, charge on parallel plates Binding energy per nucleon × massno. (graph)

Semi conductor–LED, Photodiode, solar cell, Zener diode, diode and their characteristics

Schematic representation of (a) modulation (b) demodulation (c) wave propagation (d) global satellitecommunication.

(xi)

PHYSICS (Code No. 042)

Senior Secondary stage of school education is a stage of transition from general education to discipline-based focus on curriculum. The present updated syllabus keeps in view the rigour and depth of disciplinaryapproach as well as the comprehension level of learners. Due care has also been taken that the syllabusis comparable to the international standards. Salient features of the syllabus include:

Emphasis on basic conceptual understanding of the content.

Emphasis on use of SI units, symbols, nomenclature of physical quantities and formulations asper international standards.

Providing logical sequencing of units of the subject matter and proper placement of conceptswith their linkage for better learning.

Reducing the curriculum load by eliminating overlapping of concepts/ content within the disciplineand other disciplines.

Promotion of process-skills, problem-solving abilities and applications of Physics concepts.

Besides, the syllabus also attempts to

strengthen the concepts developed at the secondary stage to provide firm foundation for furtherlearning in the subject.

expose the learners to different processes used in Physics-related industrial and technologicalapplications.

develop process-skills and experimental, observational, manipulative, decision making andinvestigatory skills in the learners.

promote problem solving abilities and creative thinking in learners.

develop conceptual competence in the learners and make them realize and appreciate the interfaceof Physics with other disciplines.

(xii)

PHYSICS

COURSE STRUCTURE2011-13

Class XI (Theory)One Paper Three Hours

Max Marks: 70

Class XI Weightage

Unit I Physical World & Measurement 03

Unit II Kinematics 10

Unit III Laws of Motion 10

Unit IV Work, Energy & Power 06

Unit V Motion of System of particles & Rigid Body 06

Unit VI Gravitation 05

Unit VII Properties of Bulk Matter 10

Unit VIII Thermodynamics 05

Unit IX Behaviour of Perfect Gas & Kinetic Theory of gases 05

Unit X Oscillations & Waves 10

Total 70

Unit I: Physical World and Measurement (periods 10)Physics - scope and excitement; nature of physical laws; Physics, technology and society.

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derivedunits. Length, mass and time measurements; accuracy and precision of measuring instruments; errorsin measurement; significant figures.

Dimensions of physical quantities, dimensional analysis and its applications.

Unit II: Kinematics (Periods 30)Frame of reference, Motion in a straight line: Position-time graph, speed and velocity.

Elementary concepts of differentiation and intergration for describing motion.Uniform and non-uniformmotion, average speed and instantaneous velocity. Uniformly accelerated motion, velocity-time, position-time graphs.

Relation for uniformly accelerated motion (graphical treatment).

Scalar and vector quantities; Position and displacement vertors, general vectors and notation; equalityof vectors, multiplication of vectors by a real number; addition and subtraction of vectors. Relativevelocity.

(xiii)

Unit vector; Resolution of a vector in a plane - rectangular components. Scalar and Vector productof vectors. Motion in a plane. Cases of uniform velocity and uniform acceleration-projectile motion.Uniform circular motion.

Unit III: Laws of Motion (Periods 16)Intuitive Concept of force. Inertia, Newton’s first law of motion; momentum and Newton’s secondlaw of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentumand its applications.

Equilibrium of concurrent forces. Static and kinetic friction, laws of friction, rolling friction,lubrication.

Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicleon level circular road, vehicle on banked road).

Unit IV: Work, Energy and Power (Periods 16)Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power.

Notion of potential energy, potential energy of a spring, conservative forces: conservation ofmechanical energy (kinetic and potential energies); non-conservative forces: motion in a verticalcircle; elastic and inelastic collisions in one and two dimensions.

Unit V: Motion of System of Particles and Rigid Body (Periods 18)Centre of mass of a two-particle system, momentum conservation and centre of mass motion.Centre of mass of a rigid body; centre of mass of uniform rod.

Moment of a force, torque, angular momentum, conservation of angular momentum with someexamples.

Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparisonof linear and rotational motions; moment of inertia, radius of gyration.

Values of moments of inertia, for simple geometrical objects (no derivation). Statement of paralleland perpendicular axes theorems and their applications.

Unit VI: Gravitation (Periods 14)Keplar’s laws of planetary motion. The universal law of gravitation.

Acceleration due to gravity and its variation with altitude and depth.

Gravitational potential energy; gravitational potential. Escape velocity. Orbital velocity of a satellite.Geo-stationary satellites.

Unit VII: Properties of Bulk Matter (Periods 28)Elastic behaviour, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear,modulus of rigidity, poisson’s ratio; elastic energy.

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulicbrakes). Effect of gravity on fluid pressure.

Viscosity, Stokes’ law, terminal velocity, Reynold’s number, streamline and turbulent flow. Criticalvelocity. Bernoulli’s theorem and its applications.

(xiv)

Surface energy and surface tension, angle of contact, excess of pressure, application of surfacetension ideas to drops, bubbles and capillary rise.

Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalousexpansion; specific heat capacity; Cp, Cv - calorimetry; change of state - latent heat capacity.

Heat transfer-conduction, convection and radiation, Qualitative ideas of Blackbody radiation greenhouse effect, thermal conductivity, Newton’s law of cooling, Wein’s displacement Law, Stefan’slaw.

Unit VIII: Thermodynamics (Periods 12)Thermal equilibrium and definition of temperature (zeroth law of thermodynamics). Heat, workand internal energy. First law of thermodynamics. Isothermal and adiabatic processes.

Second law of thermodynamics: reversible and irreversible processes. Heat engines and refrigerators.

Unit IX: Behaviour of Perfect Gas and Kinetic Theory (Periods 8)Equation of state of a perfect gas, work done in compressing a gas.

Kinetic theory of gases - assumptions, concept of pressure. Kinetic energy and temperature; rmsspeed of gas molecules; degrees of freedom, law of equipartition of energy (statement only) andapplication to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Unit X: Oscillations and Waves (Periods 28)Periodic motion - period, frequency, displacement as a function of time. Periodic functions. Simpleharmonic motion (S.H.M) and its equation; phase; oscillations of a spring–restoring force andforce constant; energy in S.H.M. Kinetic and potential energies; simple pendulum– derivationof expression for its time period; free and forced and damped oscillations (qualitative ideas only),resonance.

Wave motion. Transverse and longitudinal waves, speed of wave motion. Displacement relationfor a progressive wave. Principle of superposition of waves, reflection of waves, standing wavesin strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect.

Practicals

Note: Every student will perform 15 experiments (8 from Section A and 7 from Section B).Theactivities mentioned are for the purpose of demonstration by the teachers only. These are notto be evaluated during the academic year. For evaluation in examination, students would be requiredto perform two experiments - One from each Section.

(xv)

SECTION AExperiments Total Periods : 60

(Any 8 experiments out of the following to be performed by the Students)

1. To measure diameter of a small spherical/cylindrical body using Vernier Callipers.

2. To measure internal diameter and depth of a given beaker/calorimeter using Vernier Callipersand hence find its volume.

3. To measure diameter of a given wire using screw gauge.

4. To measure thickness of a given sheet using screw gauge.

5. To measure volume of an irregular lamina using screw gauge.

6. To determine radius of curvature of a given spherical surface by a spherometer.

7. To determine the mass of two different objects using a beam balance.

8. To find the weight of a given body using parallelogram law of vectors.

9. Using a simple pendulum, plot L-T and L-T2 graphs. Hence find the effective length of second’spendulum using appropriate graph.

10. To study the relationship betwen force of limiting friction and normal reaction and to findthe co-efficient of friction between a block and a horizontal surface.

11. To find the downward force, along an inclined plane, acting on a roller due to gravitationalpull of the earth and study its relationship with the angle of inclination (O) by plotting graphbetween force and sinθ.

Activities (For the purpose of demonstration only)1. To make a paper scale of given least count, e.g. 0.2cm, 0.5 cm.

2. To determine mass of a given body using a metre scale by principle of moments.

3. To plot a graph for a given set of data, with proper choice of scales and error bars.

4. To measure the force of limiting friction for rolling of a roller on a horizontal plane.

5. To study the variation in range of a jet of water with angle of projection.

6. To study the conservation of energy of a ball rolling down on inclined plane (using a doubleinclined plane).

7. To study dissipation of energy of a simple pendulum by plotting a graph between squareof amplitude and time.

(xvi)

SECTION BExperiments

(Any 7 experiments out of the following to be performed by the students)

1. To determine Young’s modulus of elasticity of the material of a given wire.

2. To find the force constant of a helical spring by plotting a graph between load and extension.

3. To study the variation in volume with pressure for a sample of air at constant temperatureby plotting graphs between P and V, and between P and I/V.

4. To determine the surface tension of water by capillary rise method.

5. To determine the coefficient of viscosity of a given viscous liquid by measuring terminalvelocity of a given spherical body.

6. To study the relationship between the temperature of a hot body and time by plotting a coolingcurve.

7. To determine specific heat capacity of a given (i) solid (ii) liquid, by method of mixtures.

8. (i) To study the relation between frequency and length of a given wire under constant tensionusing sonometer.

(ii) To study the relation between the length of a given wire and tension for constant frequencyusing sonometer.

9. To find the speed of sound in air at room temperature using a resonance tube by two-resonance positions.

Activities (For the purpose of demonstration only)1. To observe change of state and plot a cooling curve for molten wax.

2. To observe and explain the effect of heating on a bi-metallic strip.

3. To note the change in level of liquid in a container on heating and interpret the observations.

4. To study the effect of detergent on surface tension of water by observing capillary rise.

5. To study the factors affecting the rate of loss of heat of a liquid.

6. To study the effect of load on depression of a suitably clamped metre scale loaded at (i)its end (ii) in the middle.

SUGGESTED LIST OF DEMONSTRATION EXPERIMENTS

CLASS XI1. To demonstrate that a centripetal force is necessary for moving a body with a uniform speed

along a circle, and that the magnitude of this force increases with increase in angular speed.

2. To demonstrate inter-conversion of potential and kinetic energy.

3. To demonstrate conservation of linear momentum.

4. To demonstrate conservation of angular momentum.

5. To demonstrate the effect of angle of launch on range of a projectile.

(xvii)

6. To demonstrate that the moment of inertia of a rod changes with the change of position of apair of equal weights attached to the rod.

7. To study variation of volume of a gas with its pressure at constant temperature using a doctors’syringe.

8. To demonstrate Bernoulli’s theorem with simple illustrations

9. To demonstrate that heat capacities of equal masses of different materials are different.

10. To demonstrate free oscillations of different vibrating systems.

11. To demonstrate resonance with a set of coupled pendulums.

12. To demonstrate longitudinal and transverse waves.

13. To demonstrate the phenomenon of beats, due to superposition, of waves produced by two sourcesof sound of slightly different frequencies

14. To demonstrate resonance using an open pipe.

15. To demonstrate the direction of torque.

16. To demonstrate the law of moments.

Recommended Textbooks.

1. Physics Part-I, Textbook for Class XI, Published by NCERT

2. Physics Part-II, Textbook for Class XI, Published by NCERT

Class XII (Theory)Total Periods : 180

One Paper Time: 3 Hours 70 Marks

Unit I Electrostatics 08Unit II Current Electricity 07Unit III Magnetic effect of current & Magnetism 08Unit IV Electromagnetic Induction and Alternating current 08Unit V Electromagnetic Waves 03Unit VI Optics 14Unit VII Dual Nature of Matter 04Unit VIII Atoms and Nuclei 06Unit IX Electronic Devices 07Unit X Communication Systems 05

Total 70

Unit I: Electrostatics (Periods 25)Electric Charges; Conservation of charge, Coulomb’s law-force between two point charges, forcesbetween multiple charges; superposition principle and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric fielddue to a dipole, torque on a dipole in uniform electric fleld.

(xviii)

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitelylong straight wire, uniformly charged infinite plane sheet and uniformly charged thin sphericalshell (field inside and outside).

Electric potential, potential difference, electric potential due to a point charge, a dipole and systemof charges; equipotential surfaces, electrical potential energy of a system of two point chargesand of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics andelectric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel,capacitance of a parallel plate capacitor with and without dielectric medium between the plates,energy stored in a capacitor. Van de Graaff generator.

Unit II: Current Electricity (Periods 22)Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and theirrelation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity. Carbon resistors, colourcode for carbon resistors; series and parallel combinations of resistors; temperature dependenceof resistance.

Internal resistance of a cell, potential difference and emf of a cell,combination of cells in seriesand in parallel.

Kirchhoff’s laws and simple applications. Wheatstone bridge, metre bridge.

Potentiometer - principle and its applications to measure potential difference and for comparingemf of two cells; measurement of internal resistance of a cell.

Unit III: Magnetic Effects of Current and Magnetism (Periods 25)Concept of magnetic field, Oersted’s experiment.

Biot - Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight and toroidal solenoids,Force on a moving charge in uniform magnetic and electric fields. Cyclotron.

Force on a current-carrying conductor in a uniform magnetic field. Force between two parallelcurrent-carrying conductors-definition of ampere. Torque experienced by a current loop in uniformmagnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter andvoltmeter.

Current loop as a magnetic dipole and its magnetic dipole moment. Magnetic dipole momentof a revolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) alongits axis and perpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniformmagnetic field; bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic fieldand magnetic elements. Para-, dia- and ferro - magnetic substances, with examples. Electromagnetsand factors affecting their strengths. Permanent magnets.

(xix)

Unit IV: Electromagnetic Induction and Alternating Currents (Periods 20)Electromagnetic induction; Faraday’s laws, induced emf and current; Lenz’s Law, Eddy currents.

Self and mutual induction.

Alternating currents, peak and rms value of alternating current/voltage; reactance and impedance;

LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits,

wattless current.

AC generator and transformer.

Unit V: Electromagnetic waves (Periods 4)

Need for displacement current, Electromagnetic waves and their characteristics (qualitative ideas

only). Transverse nature of electromagnetic waves.

Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma

rays) including elementary facts about their uses.

Unit VI: Optics (Periods 30)

Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal reflection

and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lens-

maker ’s formula. Magnification, power of a lens, combination of thin lenses in contact combination

of a lens and a mirror. Refraction and dispersion of light through a prism.

Scattering of light - blue colour of sky and reddish apprearance of the sun at sunrise and sunset.

Optical instruments : Human eye, image formation and accommodation correction of eye defects

(myopia, hypermetropia) using lenses. Microscopes and astronomical telescopes (reflecting and

refracting) and their magnifying powers.

Wave optics: Wave front and Huygen’s principle, relection and refraction of plane wave at a

plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle.

Interference Young’s double slit experiment and expression for fringe width, coherent sources

and sustained interference of light. Diffraction due to a single slit, width of central maximum.

Resolving power of microscopes and astronomical telescopes. Polarisation, plane polarised light

Brewster’s law, uses of plane polarised light and Polaroids.

Unit VII: Dual Nature of Matter and Radiation (Periods 8)

Dual nature of radiation. Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric

equation-particle nature of light.

Matter waves-wave nature of particles, de Broglie relation. Davisson-Germer experiment

(experimental details should be omitted; only conclusion should be explained).

(xx)

Unit VIII: Atoms & Nuclei (Periods 18)

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels,

hydrogen spectrum.

Composition and size of nucleus, atomic masses, isotopes, isobars; isotones. Radioactivity-alpha,

beta and gamma particles/rays and their properties; radioactive decay law. Mass-energy relation,

mass defect; binding energy per nucleon and its variation with mass number; nuclear fission,

nuclear fusion.

Unit IX: Electronic Devices (Periods 18)Energy bands in solids (Qualitative ideas only) conductors, insulator and semiconductors;semiconductor diode – I-V characteristics in forward and reverse bias, diode as a rectifier; I-V characteristics of LED, photodiode, solar cell, and Zener diode; Zener diode as a voltageregulator. Junction transistor, transistor action, characteristics of a transistor, transistor as an amplifier(common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR).Transistor as a switch.

Unit X: Communication Systems (Periods 10)Elements of a communication system (block diagram only); bandwidth of signals (speech, TVand digital data); bandwidth of transmission medium. Propagation of electromagnetic waves inthe atmosphere, sky and space wave propagation. Need for modulation. Production and detectionof an amplitude-modulated wave.

PracticalsEvery student will perform atleast 15 experiments (7 from section A and 8 from Section B) Theactivities mentioned here should only be for the purpose of demonstration. One Project of threemarks is to be carried out by the students.

B. Evaluation Scheme for Practical Examination: Total Periods : 60

Two experiments one from each section 8+8 Marks

Practical record (experiments & activities) 6 Marks

Project 3 Marks

Viva on experiments & project 5 Marks

(xxi)

Total 30 Marks

SECTION AExperiments

(Any 7 experiments out of the following to be performed by the students)

1. To find resistance of a given wire using metre bridge and hence determine the specific resistanceof its material

2. To determine resistance per cm of a given wire by plotting a graph of potential differenceversus current.

3. To verify the laws of combination (series/parallel) of resistances using a metre bridge.

4. To compare the emf of two given primary cells using potentiometer.

5. To determine the internal resistance of given primary cell using potentiometer.

6. To determine resistance of a galvanometer by half-deflection method and to find its figureof merit.

7 . To convert the given galvanometer (of known resistance and figure of merit) into an ammeterand voltmeter of desired range and to verify the same.

8. To find the frequency of the a.c. mains with a sonometer.

Activities1. To measure the resistance and impedance of an inductor with or without iron core.

2. To measure resistance, voltage (AC/DC), current (AC) and check continuity of a given circuitusing multimeter.

3. To assemble a household circuit comprising three bulbs, three (on/off) switches, a fuse anda power source.

4. To assemble the components of a given electrical circuit.

5. To study the variation in potential drop with length of a wire for a steady current.

6. To draw the diagram of a given open circuit comprising at least a battery, resistor/rheostat,key, ammeter and voltmeter. Mark the components that are not connected in proper orderand correct the circuit and also the circuit diagram.

(xxii)

SECTION BExperiments

(Any 8 experiments out of the following to be performed by the students)

1. To find the value of v for different values of u in case of a concave mirror and to find thefocal length.

2. To find the focal length of a convex mirror, using a convex lens.

3. To find the focal length of a convex lens by plotting graphs between u and v or between1/u and 1/v.

4. To find the focal length of a concave lens, using a convex lens.

5. To determine angle of minimum deviation for a given prism by plotting a graph betweenangle of incidence and angle of deviation.

6. To determine refractive index of a glass slab using a travelling microscope.

7. To find refractive index of a liquid by using (i) concave mirror, (ii) convex lens and planemirror.

8. To draw the I-V characteristic curve of a p-n junction in forward bias and reverse bias.

9. To draw the characteristic curve of a zener diode and to determine its reverse break downvoltage.

10. To study the characteristic of a common - emitter npn or pnp transistor and to find out thevalues of current and voltage gains.

Activities (For the purpose of demonstration only)1. To identify a diode, an LED, a transistor, and IC, a resistor and a capacitor from mixed collection

of such items.

2. Use of multimeter to (i) identify base of transistor (ii) distinguish between npn and pnp typetransistors (iii) see the unidirectional flow of current in case of a diode and an LED (iv)check whether a given electronic component (e.g. diode, transistor or IC) is in working order.

3. To study effect of intensity of light (by varying distance of the source) on an L.D.R.

4. To observe refraction and lateral deviation of a beam of light incident obliquely on a glassslab.

5. To observe polarization of light using two Polaroids.

6. To observe diffraction of light due to a thin slit.

7. To study the nature and size of the image formed by (i) convex lens (ii) concave mirror,on a screen by using a candle and a screen (for different distances of the candle from thelens/ mirror).

8. To obtain a lens combination with the specified focal length by using two lenses from thegiven set of lenses.

(xxiii)

SUGGESTED INVESTIGATORY PROJECTS

CLASS XII1. To study various factors on which the internal resistance/emf of a cell depends.

2. To study the variations, in current flowing, in a circuit containing a LDR, because of a variation.

(a) in the power of the incandescent lamp, used to ‘illuminate’ the LDR. (Keeping all the lampsat a fixed distance).

(b) in the distance of a incandescent lamp, (of fixed power), used to ‘illuminate’ the LDR.

3. To find the refractive indices of (a) water (b) oil (transparent) using a plane mirror, a equiconvexlens, (made from a glass of known refractive index) and an adjustable object needle.

4. To design an appropriate logic gate combinatin for a given truth table.

5. To investigate the relation between the ratio of (i) output and input voltage and

(ii) number of turns in the secondary coil and primary coil of a self designed transformer.

6. To investigate the dependence, of the angle of deviation, on the angle of incidence, using a hollowprism filled, one by one, with different transparent fluids.

7. To estimate the charge induced on each one of the two identical styro foam (or pith) balls suspendedin a vertical plane by making use of Coulomb’s law.

8. To set up a common base transistor circuit and to study its input and output characteristic andto calculate its current gain.

9. To study the factor, on which the self inductance, of a coil, depends, by observing the effectof this coil, when put in series with a resistor/(bulb) in a circuit fed up by an a.c. source ofadjustable frequency.

10. To construct a switch using a transistor and to draw the graph between the input and outputvoltage and mark the cut-off, saturation and active regions.

11. To study the earth’s magnatic field using a tangent galvanometer.

Recommended Textbooks.

1. Physics, Class XI, Part -I & II, Published by NCERT.

2. Physics, Class XII, Part -I & II, Published by NCERT.

(xxiv)

CHANGES IN PHYSICS SYLLABUS 2012-14

Following are some changes in the Syllabus of Physics at Senior Secondary Level offeredby CBSE for the session 2012-14.

Class XI

(1) Vector is clubbed in Unit II (Kinematics) from Unit IV and Unit V(Multiplicationof Vector)

(2) Add Activity NO 7 in Scetion A “Experiments - Activities (For the purpose ofDemonstration only)

(3) Add the “Suggested List of Demonstration Experiments Class XI”

Class XII

(1) Add the “Suggested List of Demonstration Experiments Class XII”

(NOTE: Motivate the Students to do the Demostration)

(xxv)

Suggestions to Students from a Teacher

Sleep It is important to be well rested. Make sure to get a good night's sleep in the few days before the test.

If you don't sleep well the night before the test, don't worry about it! It is more important to sleepwell two and three nights before. You should still have the energy you need to perform at your best.

Diet Don't change your diet right before the test. Now's not the time to try new foods, even if they are

healthier. You don't want to find out on test morning that yesterday's energy bar didn't go down well.

In the few weeks before the test, try to work a light, healthy breakfast into your daily routine. Ifyou already eat breakfast, good for you - don't change a thing.

Stress Try to be aware of whatever anxiety you're feeling before test day. The first thing to remember is

that this is a natural phenomenon; your body is conditioned to raise the alarm whenever somethingimportant is about to happen. However, because you are aware of what your body and mind are doing,you can compensate for it.

Spend some time each day relaxing. Try to let go of all the pressures that build up during your averageday.

Visualize a successful test day experience. You already know what to expect on test day: when you'llget each test section, how many questions there are, how much time you'll have, etc. You also knowwhere you are strong and where you are weak. Picture yourself confidently answering questions correctly,and smoothly moving past trouble spots - you can come back to those questions later.

Find a family member or trusted friend with whom you can talk about the things that stress you outabout the test. When this person tells you that everything is going to be OK, believe it!

Writing Questions Remember that a few spelling or grammar mistakes are tolerable, but you want to try to eliminate

as many of those as you can.

Try to vary your sentence length and word choice.

Before you begin to write, spend a few minutes brainstorming ideas and outlining the argument youwant to make. Planning will help you to write a well-organized and cohesive essay.

Practice and Review Whatever you do, don't cram for the test! It is a bad strategy because you aren't going to remember

most of what you "learn" while cramming, and the odds are slim that the few things it will help youto remember will happen to be on the test. Save the energy you would have used to cram for testday.

In the few days before the test, do a review of the skills and concepts in which you are strong. Beconfident as you review everything that you know - and remember that confident feeling as you takethe test.

1

1Electric Charge

Charge is the property associated with matter due to which it produces and experiences electrical andmagnetic effects. The charge on a body arises from an excess or deficit of electrons.

Positive charge Negative charge

Glass Rod SilkFur Ebonite

Wool PlasticWool Rubber

Dry hair Comb

Magnitude of electronic charge is e =(1.6 × 10–19C)S.I. unit of charge is coulomb.

Point Charge

A finite size body may behave like a point charge if it produces an inverse square electric field. Forexample an isolated charged sphere behave like a point charge at very large distance.

Methods of Charging

(i) By friction(ii) By conduction (by contact)–In this case transfer of charge takes place by contact.

(iii) By induction (without contact)–In this case charges are Induced by external effect without anyphysical contact.

Properties of Charge

(i) Additivity–Total electric charge of a system = algebraic sum of all the positive and negative chargescontained in that system.

(ii) Conservation–charge can neither be created nor destroyed. It means that total charge of an isolatedsystem always remains constant.

(iii) Quantisation–It is the property due to which all free charges are integral multiple of electroniccharge.Symbolically, q = ± newhere q = total charge on a body

–e = charge on an electron

2

The cause of quantization is that only integral number of electrons can be transferred from onebody to another.

(iv) Charge of body does not depend upon its speed.(v) Charge can not exist without mass though mass can exist without charge.

Coulomb’s Law

According to Coulomb’s law,If two point charges q

1, q

2 are separated by a distance r, the magnitude of the force (F) between

them is given byF = 1 2

2

q qk

rwhere k is constant of proportionality, its value in S.I. System is 9 × 109 Nm2c–2.In vector form

= k

If both charges are of same sign then force will be repulsive other wise attractive.

PermittivityValue of k is also represented as k = 1/4πε

o where ε

o is called the permittivity of free space or vacuum.

Value of εo = 8.854 × 10–12 C2N–1m–2.

∴ Coulomb’s law is written as F =

If the medium between the two charges is other than vacuum, the formula becomes

F =1 2

2m

1 q q.

4 rπεThe force between the charges is reduced. ε

m is called the permittivity of the medium.

Relative permittivity or dielectric constantIt is the ratio of permittivity of the medium to permittivity of free space.

εr =

m

o

εε

εr = 1 for vacuum, 1 for air, 81 for water

Principle of Super Position

Force on any charge due to a number of other charges is the vector sum of all the forces on thatcharge due to the other charges, taking one at a time. The individual forces are unaffected due to thepresence of other charges.

Electric Field

It is the space around a charge (source charge) in which any other charge (test charge) experiencean electric force due to source charge.

Intensity of electric fieldIntensity of the electric field at a given point is defined as the electric force on unit positive chargeplaced at that point.

∴

Direction of electric field is same as the direction of electric force on unit positive charge. Its S.I.Unit is NC–1 or volt per meter.

3

The electric force on a charge q placed in electric field E is given by

F = qE

Electric field intensity due to a point charge.Electric field intensity at any point P due to a point charge q at O, where OP = r is

20

1 qˆE r

4 r=

πε

Superposition principleIt is equal to the vector sum of the electric field intensities due to the individual charges at thesame point.

Electric Field Lines

These are the path of unit positive test charge placed in any given electric field and free to move.

Properties of electric field lines(i) They are hypothetical lines.

(ii) The tangent to an electric field line gives direction of electric field at that point.(iii) The relative closeness of field lines indicates the relative strength of electric field at different

points.(iv) All electric field lines originate from a positive charges and terminate on a negative charges.

They are open curves.(v) The number of electric field lines of any charge is proportional to the magnitude of the charges.

(vi) No two electric field lines ever cross each other because field cannot have two directions atthe same point.

(vii) They are continuous curves.(viii) These are always noamal to the surface of conductor.

(ix) Number of fields lines passing perpendicular to unit area is equal to the magnitude of electricfield.

Electric Field Line In Some Cases(i) Electric field lines of isolated charges.

+q

–q

Isolated (+q) charge Isolated (–q) charge

(ii) Electric field lines of multiples charges

+q –q

+q

+q

4

(iii) Electric field lines of uniform electric field (having same magnitude and direction at everypoint ) are represented by equidistance parallel straight lines with proper direction.

(iv) Electric field lines of non-uniform electric field.

Electric Dipole

An electric dipole is a pair of equal and opposite charges separated by a small distance.It the two charges are (–q) and (+q) and a is the displacement between them, then the vector quantity is known as the ‘electric dipole moment’.

∴ Dipole moment, p 2qa=

Direction of electric dipole moment is from negative charge to positive charge. SI unit of dipolemoments is coulomb-meter (C-m).

Electric Field Of a Dipole At a Point On Its Axis.Consider an electric dipole consisting of two point charges –q and +q separated by small distance 2a.

P is the point where field is to be calculated.

= Field at P due to +q

= Field at P due to –q

∴ Net field at P, using superposition principle

P 1 2E = E + E

, now since

[along A to P, (+ive)]

[along P to B, (-ive)]

5

∴

PE

=

2 20

1 q q–

4 (r – a) (r a)

πε +

=

2 2

2 2 20

q (r a) – (r – a)

4 (r – a )

+ πε

=

2 2 2 2

2 2 20

q r a 2ra – (r a – 2ra)

4 (r – a )

+ + +⋅πε

= 2 2 20

q 4ra

4 (r – a )⋅

πε = 2 2 2

0

p 2r

4 (r – a )⋅

πε

direction is along the axis of the dipole (i.e.from A to P),where p = 2qaIf a << r, a2 becomes negligible in comparison to r2(short elecric dipole), hence a can be neglected.

∴ PE

=

30

1 2p

4 r⋅

πε

i.e. Ep ∝

3

1

r

Electric Field Of a Dipole At a Point On Its Equatorial AxisConsider an electric dipole consisting of two point charges –q and +q separated by small distance 2a.P is the point where field is to be calculated.Electric field at point P due to (+q);

E1 =2 2 2

kq kq

(BP) r a=

+ (along B to P)

It has two components (

1E cosθ

) and (

1E sin θ

) as shown in figure.

E sin 1 θE1

E cos θ1

E cos θ2

E2

E sin 2 θ

θ θ

P

A B

–q a C a +q

r

Similarly, electric field at point P due to (–q);

2 2 2 2

kq kqE

(AP) r a= =

+

(along P to A).

E2 also has two components (

2E cosθ

) and (

2E sinθ

) as shown in figure.

The vertical components (

1E sin θ

) and (

2E sin θ

), are equal and opposite so cancel each other.

∴ Resultant electric field intensity Ep at P

6

=

=

=

=

= 2 2 3/ 20 0

1 (q.2a) 1(where k )

4 (r a ) 4=

πε + πε

=

it is along B to A i.e. opposite to the direction of dipole moment.In vector form

=

If a << r (for short dipole) then a2 becomes negligible in comparison with r2, hence can be dropped.

i.e. p 3

1E

rα

Hence

Electric Dipole Placed in Electric field

(i) If field is uniform-Force on +q, F

1 = qE, in the direction of the field.

Force on –q, F2 = qE, opposite to the direction of the field.

Hence net force on dipole = 0(ii) If field is not uniform then net force is not zero.

(iii) Torque on dipoleIn electric field two forces act on the dipole as shown in figure which form a couple and try

to rotate the dipole.

Therefore, torque acting on the dipole is given by

7

τ = Any one of the two forces (in magnitude)

×

The perpendiculardistance between force vectors.

=

qE 2a sin 2qa E sin× θ = × θ

=

pE sin θ( p 2qa)=∵τ

=

p E×

(vector form)

(iv) Dipole is said to be in stable equilibrium if angle between

p

and

E

is zero and in unstable

equilibrium if angle between

p

and

E

is 180°.

Electric Flux (φφφφφ)

Electric flux is defined as number of electric field lines passing through the area placed normal tothe field direction.

Electric flux dφ through an area element

ds

in an electric field

E

is defined as

dφ =

E.ds

⇒ φ =

E.ds

it is a scalar quantity.Its SI unit is Nm2 C–1.

Gauss’s Law

It states that the electric flux entering or emerging from any closed surface is equal to 1/

ε

0 times the

value of charge enclosed by closed surface.

i.e. φ =enclosed enclosed

0 0

q qor E.ds =

ε ε

Gauss’s law holds good for any closed surface of any shape or size. It does not depend uponthe location of charge inside the close surface.

Application of Gauss's Law

Steps for using gauss law-(1) Assume observation point where electric field is to be determined.(2) Assume a close surface(gaussian surface) which passes from point P.(3) Calculate area of gaussian surface and q

enclosed.

(4) Draw electric filed lines for given distribution.(5) Determined angle between electric field and area vector at every point.(6) Use gauss law.

Electric Field Due to a Line ChargeLet λ = linear charge density

P = Observation pointr = Normal distance of P from line charge.

From symmetry, E will be radially outward. Consider a cylindrical Gaussian surface of radiusr and length l passing through P.

8

++++++++++++++++++++

l

S1

S2

S3

E E EP

Line charge

From Gauss Law E.ds

=

⇒ enclosed1 2 3

0

qE.dS E.dS E.dS+ + =

ε

=

or E =

Direction of this E is radially outward.

Electric Field due to an infinite, non-conducting thin plane sheet of chargeLet P is the observation point.Surface charge density of uniformly charged sheet is σ.

Consider a Gaussian cylindrical surface of length 2l and cross section area S passing throughpoint P as shown in figure.

By Gauss law enclosed

0

qE.ds =

ε

⇒ 1 2 3E.dS E.dS E.dS+ +

=

or =

=

Now, since dS1 and dS

3 are taken at equal distance from the charged sheet so

E1 = E

3 = E(Let)

9

∴ 2ES =

0

Sσε

E =

02

σε

This expression shows that electric field at a point very close to a metal sheet does not dependsupon the distance.

Electric field due to a thin spherical shell of chargeConsider a spherical Gaussian surface of radius r passing through observation point P.

Gauss law

E.ds

=

enclosed

0

q

ε

Chargedshell

S′

S

PR

q

r o

r

P

E

(i) Field outside the cellGaussian surface S is of radius (r >>R)

∴ 2E.4 r .cos 0π =

0

q

ε

or E =

20

1 q

4 rπε

This shows that the shell behaves like a point charge placed at its centre.(ii) On the surface of shell-In this case gaussian surface is sphere itself .Hence

r = R

⇒ E = 20

1 q

4 Rπε(iii) Inside the shell

Charge enclose is zero hence E = 0Variation between E and r:

EEmax

0 r = R

E α 1

r2

r

10

Electric Potential & Capacitance

Electrostatic Potential

The electric potential at a point in an electric field is the work done by an external force in bringing a unitpositive test charge from infinity to that point without any acceleration. potential is a scalar quantity.

potential V =

Unit of electric potential is volt.1V = 1 volt = 1 NmC–1 = 1JC–1

Potential Due to a Point Charge

Let a small positive test charge q0 is brought from to A with a constant velocity.

Let electric field at any point P = E

PQ = an infinitesimally small path element

∴ Electric force due to the field on q0 at P is

= q0

∴ Force to be applied on q0 to move it from P to Q without imparting any acceleration to it

= – q0

∴ corresponding work,

dW = – q0

. Potential at A,

VA = [as ]

VA =

(i.e. electric potential is equal to line integral of electric field)

⇒ VA = 2

kqE

r = ∵

=r

A

1 1 1 kqkq kq V

r r r∞

= − ∞ When q is positive, potential is positive and when q is negative potential is also negative. Potential

due to a point charge is spherically symmetric.

11

(a) Potential DifferencePotential difference between points A & B will be,

VB – V

A =

B

AE.dl−

VB – V

A =

B A

1 1kq

r r

−

(b) Conservation of Electric FieldAs work done by electric field depends only upon the initial and final points of path so electric fieldis conservative in nature.

(c) Variation of V and E With r

V or E

distance r

vE

Potential due to System of Charges

By superposition principle the potential V at a point due to the total charge configuration is thealgebric sum of the potential due to the individual charges.

V = V1 + V

2 + ...V

n

Potential due to a dipole

Let an electric dipole, consist of two equal and opposite charge separated by 2a. P is the observationpoint.

+q

a

a

p

–qφ

α

α

r – r2 1X

r1r

r2

P

Y

r2 = OP

θ

VP = Potential at P due to (+q) + Potential at P due to (–q)

VP =

1 2

1 1kq

r r

−

2 1p

1 2

kq(r r )V

r r

−

If r >> a, then θ ≈ φi.e. r

2 – r

1 =

2a cos 2a cosφ θ

and

r1 ≈ r

2 = r, say i.e. r

1r

2 ≈ r2

12

∴ VP =

VP =

If , then cosθ = 0 i.e. potential at any point on the right bisector of the dipole is zero.(Equatorial axis)If θ = 0, i.e. at a point on the axis of the dipole, (axial axis) V

P becomes maximum and is

= 2

kp

r

Equipotential Surfaces

The surface at which the value of potential at every point is same is called equipotential surface.

Properties of equipotential surface(1) No work is required to be done in moving a charge from one point to another on an equipotential

surface.

(2) No two equipotential surface intersect each other.

(3) For any charge configuration, equipotential surface through a point is normal to the electric fieldat that point.

Some equipotential surfaces:(i) For single point charge equipotential surfaces are concentric spheres.

(ii) Equipotential surfaces in a uniform electric field are planes normal to field.E

(iii) Equipotential surface between two point charges + q and – q is as follows:

+q –q+q –q+q –q

Relation Between Field and Potential

As we know V = E.dr−

hence E =

dV dr is called potential gradient

Negative sign indicates that the direction of the electric field is opposite to the direction in whichpotential is increasing.

13

Electric Potential Energy

Potential energy of a point charge at a point is defined as the amount of work done in bringing thecharge from an infinite distance to thet point . S.I. unit of potential energy is joule (J).Other unitof P.E. is eV(electron volt)

Electric potential energy of a point charge in an external electric fieldPotential energy U of a single charge q at a point P at distance r in an external electric field

U = q(potential at P)=qVp

Work done by or on a charge q in moving it from (V1) to V

2 will be equal to the P.E. lost or gained

by the charge.Thus ∆U = q(V

2 – V

1)

Electric potential energy of two point charges(No external field)Here q

1 and q

2 are to be placed at P

1 and P

2. When q

1 is brought from ∞ to P

1, no work is needed

to be done∴ PE of q

1 = 0

q1 q2P1 P2r12

from ∞When q

2 is brought from ∞ to P

2, the fied of q

1 already exists at P

2, against which work has to be

done.If q

2 is unit charge, this work would be = potential at P

2.

P.E. of q2 = q

2 × V

P2 (potential of q

1)

=

12

12

kqq

r×

= 1 2

12

kq q

r

∴ U =

1 2

12

kq q0

r+

U = 1 2

12

kq q

r

Potential energy of a system of two charge in an external fieldLet point P

1 and P

2 are in external electric field and their potentias are V

1 and V

2 respectively. Potential

energy of a system of two charges q1 and q

2 located at r

1 and r

2 respectively

=

1 21 1 2 2

0 12

q qq V q V

4 r+ +

πε

where r12

is the distance between q1 and q

2.

14

P.E. of an dipole in an external fieldWork done in rotating the dipole through a small angle

dw =

= pE sinθ dθwork done in rotating the dipole from θ 1

to 2 is given by

= =

or w = [ ]2 1pE cos cos− θ − θThis work is stored as the potential energy of the system. Hence

U =

Electrostatics of Conductors

(i) Electrostatic field inside a conductor is zero.(ii) At the surface of a charged conductor, electrostatic field must be normal to the surface at every

point.(iii) Inside the conductor charge is zero. Charge reside only at the surface of cinductor.(iv) Electrostatic potential is constant throughout the volume of the conductor and has the same

value as on its surface.(v) Electric field at the surface of a charged conductor

=

where is the surface charge density and is a unit vector normal to the surface.(iv) The surface charge density (σ) is high where the radius of curvature of the surface of the

conductor is small.(vii) Electrostatic shielding: It is the process in which any object or region is protected from electric

field. It is done by enclosing that object or region by conducting surface.

Dielectric and Polarisation

Dielectrics are non conducting substances which transmit electric effect without any actual conductionof electricity.e.g., vacuum, paper (waxed or oiled), mica, glass, plastic foil, fused ceramic, or air etc.There are two type of dielectric medium.(i) Polar dielectric: In this dielectric center of positive charge and centre of negative charge does

not coincide with each other e.g. water, HCl etc. (liquids).(ii) Non polar dielectric: In this dielectric center of positive charge and centre of negative charge

coincide with each other. e.g. gases.

Dielectric polarisationWhen a dielectric is placed in a uniform electric field then the centres of positive and negativecharges in the molecule are separated, and dielectric is said to be polarised.The polarized dielectric is equivalent to two charged surfaces with induced surface charge densities,say + σ

p and –σ

p. The field produced by these surface charges opposes the external field. The total

field in the dielectric is, thereby, reduced from the case when no dielectric is present.

15

Dielectric strengthThe maximum value of the electric field at which the dielectric withstand without break down, iscalled the dielectric strength of the material.

Electrostatic Capacitance

Capacitance is the capacity to store electric charges by any conductor and such conductor areknown as capacitor.If we give some charge Q to an isolated conductor, its potential increases to V.then Q α Vor, Q = CVwhere C is the constant of proportionality called capacitance.S.I. unit of capacitance is farad, F.(1 farad = 1 coulomb/volt).A conductor is said to have a capacityof 1farad when a charge of 1 coulomb increases its potential by 1 volt.

Capacitance of an isolated conducting sphereIf a conducting sphere of radius r is given a charge Q, then the potential on the surface of the sphere is

V =

0

Q

4 rπε

∴ =

0

Q4 r

V= πε

since Q/V = C

∴ C =

04 rπεEstimation of one farad

If we use C = 1 farad in above relation then r = 9 × 109m hence farad is a very large unit.Farad being a very large unit, other practical units are

1 microfarad = 1µF = 10–6 farad1 milifarad = 1mF = 10–3 farad1 picofarad = 1pF = 10–12 farad

Parallel plate Capacitor

In this capacitor one or more pairs of plates of conductors are seperated by a dielectric medium.

PrincipleThe principle of parallel plate capacitor is based on the fact that potential of a insulated conductoris decreased when an uncharged earthed conductor is kept close to it.In this case charge of insulatedconductor remains same and more charges can be added to it.This is the way to increase thecapacitance of an insulated conductor.

Capacitance of a parallel plate capacitorLet σ = surface charge density = Q/A

Q = Total charge on one face of either plateA = area of one face of either plateE = electric field between the platesε0 = permittivity of vacuum

16

d = seperation between the platesV = potential difference between the two plates

now, V = Ed

V =

V =0 0

Q d Qas

A

σ = ε ε

⇒ Q = 0AV

d

ε

as C = Q/V hence

C =

Thus capacitance of a parallel plate capacitor depends entirely on its geometrical dimensions andthe dielectric used.

Capacitance With Dielectric Between PlatesIf K is the dielectric constant of medium between the plates of capacitor.

C =

Capacitance of a Parallel Plate Capacitor With a Conducting SlabLet t = thickness of metal slabAs V = E (d – t) +0(t)since E = 0 inside the slab.

=

= ( )0

Q Qd t

A A − σ = ε ∵

or,Q

V =

∴ C =

Capacitance of a Parallel Plate capacitor with a dielectric slab

= outside field

= net field inside the dielectric slab

t = thickness of dielectric slabε

r = relative permittivity of the dielectric

Now, V = E0(d – t) + E

mt

+

–

+Q

–Q

d

+

–

+

–

+

–

+

–

+

–

+

–

+

–

t Em

E0

Dielectricslab

17

= 0 00 m

r r

E EE (d t) t E

− + = ε ε

∵

= 0r

1E d t 1

− − ε

V =

0 r

Q 1d t 1

A

− − ε ε

i.e. C =

0

r

A QC

V1d t 1

ε=

− − ε

∵

This analysis shows that on introducing conducting or dielectric slab between the plates ofcapacitors,its capacitance increases.

Combinations of Capacitors

(i) Series combination(ii) Parallel combination

Equivalent capacity in seriesIn series combination,(i) Charge stored on each capacitor is same. (ii)Potential difference across each capacitor is proportional

to its capacitance.as V = V

1 + V

2 + V

3 + ...

=1 2 3

Q Q Q...

C C C+ + +

=

1 2 3

1 1 1Q ...

C C C

+ + +

...(i)

From (i) & (ii)

1

C

=

1 2 3

1 1 1...

C C C+ + +

...(ii)

Equivalent capacity in parallelIn a parallel combination,

(i) Potential difference across each capacitor is same.(ii) Charge stored in each capacitor is proportional to its capacitance.as Q = Q

1 + Q

2 + Q

3+...

Let C = capacity of the combinationQ = CV

or CV = C1V + C

2V + C

3V + ...

or C =

1 2 3C C C ...+ + +Q Q Q

c v1 1 c v2 2 c v3 3

V

+Q1

+Q2

+Q3

–Q1

–Q2

–Q3

V

V

V+ –

18

Energy Stored in a Capacitor

Work done in charging a capacitor is stored in form of potential energy of capacitor.This energyis stored in the electric field between the plates.If small charge dQ is given to a capacitor of potential V. The work required to be done for doingit is given by

dW = V. dQ

∴ U =

=

The energy (U) stored in the capacitor can be written in any one of the following forms:

U =2

21 Q 1 1CV QV

2 C 2 2= =

Total energy stored in series or parallel combination of capacitors is equal to sum of the energiesstored in individual capacitors.

Energy density (u)

u =

21CVTotal energy (U) 2

Volume of capacitor Ad=

= ( )20A1 1Ed

2 d Ad

ε

u =2

0

1E

2ε

Common Potential

When two charged capacitors are connected by conducting wire then charge flow from higher tolower potential. This flow continues till their potentials become equals and this is called commonpotential.

Common Potential =

In this process of charge flows and energy loss takes place in form of heat produced in connectingwire.

VAN-DE Graph Generator

It is a device used to accelerate charge partical. It is used in high energy nuclear physics experiments.

Principle: It is based upon the principle of electrostatic induction and corona discharge. (action ofsharp point).

Corona discharge: When a conductor carries a charge then leakage of charge takes place from itspointed ends.

The process of spraying charge is called corona discharge (Action of sharp point).

19

Construction: Motor drives the pulley P1 and P

1 drives pulley P

2 through an insulating belt. P

2 is

inside an air evacuated spherical metallic shell.

There are two metallic brushes placed near the insulating belt. The lower brush is connected to highvoltage battery.

The upper brush is connected with the inner surface of the spherical shell.

Working: The lower metal brush is kept at a positive potential (104 volt). Due to discharging actionof sharp points, it sprays positive charge on the belt.

++++

++

++++

++++++++++++++

++

++++++ + + +

++

++++++++++++++

––––

Metal brushPulley, P2

Metalbrush

Insulating beltIon source

Insulating columnMotor driven

pulley, P1

As the belt moves, and reaches the sphere, a negative charge is induced on the sharp ends of theupper collecting metal brush and an equal positive charge is induced on the farther end of thatbrush. This positive charge shifts immediately to the outer surface of the shell.

Due to action of sharp points of the upper metal brush, a negatively charges are sprayed on the belt.This neutralizes the positive charge on the belt. This is repeated again and again.

Thus the positive charge on the metallic shell goes on accumulating.

Hence the potential of the spherical shell goes on increasing up to 6-8 million volts.

Answer Yourself

Very Short Questions

Q1. Draw schematically an equipotential surface of a uniform electrostatic field along x axis.Q2. Sketch field lines due to (i) Two equal positive charges near each other (ii) dipole.Q3. Name the physical quantity whose SI unit is volt/meter. Is it a scalar or vector quantity?Q4. Two point charges repel each other with a force F when placed in water of dielectric constant

81. What will the force between them when placed the same distance apart in air?Q5. Net capacitance of three identical capacitors connected in parallel is 12 microfarad. What will

be the net capacitance when two are connected in (i) parallel (ii)series.Q6. A charge q is placed at the centre of an imaginary spherical surface. What will be the electric

flux due to this charge through any half of the sphere?

20

Q7. Sketch the electric field vs distance( from the centre) graph for (i) a long charged rod with linearcharge density ë < 0 (ii) spherical shell of radius R and charge Q > 0.

Q8. Diagrammatically represent the position of a dipole in (i) stable (ii) unstable equilibrium whenplaced in a uniform electric field.

Q9. A charge Q is distributed over a metal sphere of radius R.What is the electric field and electricpotential at the centre?

Q10. The relative permittivity of mica is 6. What is its absolute permittivity?Q11. If q

1q

2 > 0, and if q

1q

2 < 0 what can we say about the nature of force?

Q12. Although ordinary rubber is an insulator, the tyres (rubber) of aircraft are made slightly conducting.Why?

Q13. The force between two charges separated by distance r in air is 10N. When the charges are placedsame distance apart in a medium of dielectric constant K, the force between them is 2N. Whatis the value of K?

Q14. A square ABCD has each side 1 m. Four charges + 0.02 µC, + 0.04 µC, + 0.06 µC and + 0.02 µCare placed at A, B, C and D respectively. Find the potential at the centre of the square.

Short Questions

Q1. Find the number of field lines originating from a point charge of q = 8.854 µC.Q2. What is the work done in rotating a dipole from its unstable equilibrium to stable equilibrium?

Does the energy of the dipole increase or decrease?Q3. Derive an expression for the work done in rotating an electric dipole from its equilibrium position

to an angle è with the uniform electric field.Q4. The figure shows the Q (charge) versus V (potential) graph for a combination of two capacitors.

Identify the graph representing the parallel combination.

Q5. Calculate the work done in taking a charge of 1µC in a uniform electric field of 10 N/C fromBtoC given AB= 5cm along the field and AC= 10cm perpendicular to electric field.

Q6. Draw equipotential surface for a (i) point charge (ii) dipole with same nature of charge.Q7. What is the ratio of electric field intensity at a point on the equatorial line to the field on axial

line when the point is at the same distance from the centre of the dipole?

21

Q8. Show that the electric field intensity can be given as negative of potential gradient.Q9. For an isolated parallel plate capacitor of capacitance C and potential V, what will happen to

(i) charge on the plates (ii) potential difference across the plates (iii) field between the plates(iv) energy stored in the capacitor, when the distance between the plates is increased?

Q10. Obtain an expression for the field due to electric dipole at any point on the equatorial line.Q11. Can two equi potential surfaces intersect each other? Give reasons.

Two charges –q and +q are located at pointsA(0,0,-a) and B (0,0,+a) respectively. How muchwork is done in moving a test charge from point P(7,0,0) to Q (-3,0,0)? (zero)

Q12. Define electrostatic potential and its unit. Obtain expression for electrostatic potential at a pointP in the field due to a point charge.

Q13. What is polarization of charge? With the help of a diagram show why the electric field betweenthe plates of capacitor reduces on introducing a dielectric slab. Define dielectric constant on thebasis of these fields.

Q14. Using Gauss’s theorem in electrostatics, deduce an expression for electric field intensity due toa charged electric shell at a point in (i) inside (ii) on its surface (iii)outside it. Graphically showthe variation of electric field intensity with distance from the centre of shell.

Q15. Three capacitors are connected first in series and then in parallel. Find the equivalent capacitancefor each type of combination.

Q16. Derive an expression for the energy density of a parallel plate capacitor.Q17. What should be the position of charge q=5 µC for it to be in equilibrium on the line joining

two charges q1 = - 4 µC and q

2= 10µC separated by 9cm. Will the position change for any other

value of charge q. (9cm from- 4 µC)Q18. Two point charges 4e and e each, at a separation r in air, exert force of magnitude F. They are

immersed in a medium of dielectric constant 16. What should be the separation between the chargesso that the force between them remains unchanged (1/4 the original separation)

Long Questions

Q1. State the principle of Van De Graff generator. Explain its working with the help of a neat labeleddiagram.

Q2. Derive an expression for the strength of electric field intensity at a point on the axis of a uniformlycharged circular coil of radius R carrying charge Q.

Q3. Derive an expression for potential at any point distant r from the centre O of dipole makingan angle è with the dipole.

Q4. Suppose that three points are set at equal distance r = 90cm from the centre of a dipole, pointAand B are on either side of the dipole on the axis (A closer to +ve charge and B closer toB) point C which is on the perpendicular bisector through the line joining the charges. Whatwould be the electric potential due to the dipole of dipole t 3.6×10-19Cm at points A,B and C.

Q5. Derive an expression for capacitance of parallel plate capacitor with dielectric slab of thicknesst (t < d) between the plates separated by distance d. How would the following (i) energy (ii)charge, (iii) potential be affected if dielectric slab is introduced with battery disconnected, (b)dielectric slab is introduced after the battery is disconnected.

Q6. Derive an expression for torque experienced by dipole placed in uniform electric field. Hencedefine electric dipole moment.

22

Q7. State Gauss’s theorem. Derive an expression for the electric field due to a charged plane sheet.Find the potential difference between the plates of a parallel plate capacitor having surface densityof charge 5x10-8C/m2 with the separation between plates being 4mm.

Q1. A point charge Q is placed at point O as shown in fig. Is the potential difference Va – Vb positive,negative or zero, if Q is (i) positive (ii) negative charge.

O A B

Q2. Electric dipole moment of Cu S04 molecule is 3.2x10-32 Cm. Find the separation between copper

and sulphate ions.Q3. The electric potential V at any point in space is given V=20x3 volt, where x is in meter. Calculate

the electric intensity at point P (1,0,2).Q4. What is electric field between the plates with separation of 2cm, (i) with air (ii) dielectric medium

of dielectric constant K, electric potential of each plate as marked in fig

Q5. Two point charges 6 µC and 2 µC are separated by 3cm in free space. Calculate the work donein separating them to infinity. (3.6joule)

Q6. BC is an equilateral triangle of side10cm. D is the mid point of BC, charge 100 µC, -100 µCand 75 µC are placed at B, C, and D respectively. What is the force experienced by a 1 µCpositive charge placed at A. (9√2×103N)

Q7. In the following fig. calculate the potential difference across capacitor C2 Given potential at Ais 90 V. C1=20 µF., C2=30 µF. and C3= 15 µF.

(20V)

Q8. A point charge develops an electric field of 40 N/C and a potential difference of 10J/C at a point.Calculate the magnitude of the charge and the distance from the point charge. (2.9x10-10C, 25cm)

Q9. For what value of C does the equivalent capacitance between A and B is 1µ. Find the given circuit

(2microfarad)

To be Learnt

23

Q10. What should be the charge on a sphere of radius 4cm, so that when it is brought in contactwith another sphere of radius 2cm carrying charge of 10 µC, there is no net transfer of charge?

Q11. Two capacitors of capacitances C1 and C2 are charged to potentials V1 and V2 respectively.The capacitors are joined through a conducting wire. What is the value of common potential?

1. Direction of electric field is along decreasing potential.2. Formulae of dipole moment3. E = -dV/dr4. Same as 35. Refer NCERT example of potential6. Use coulomb’s law and principle of vector addition.7. Charge remains same in all capacitors in series combination. Q = CV8. E/V = 1/r9. Apply capacitor combination principle10. No charge transfer if potential is same11. Potential = net charge /net capacitor

Pedagogical Remark

24

2 The flow of charge is known as current electricity. The rate of flow of charge is called electric

current.

I =q

t

∵ q = ne where n is an integer

and e = 1.6 × 10–19C.

The SI unit of electric current is ampere (A).

In metals or conductors free electrons move randomely in all possible directions and collide withatoms of the matter. They move in straight line between two successive collisions. When electricfield is applied across a conductor the electrons get accelerated which may not be in the directionof velocity of electron in absence of electric field. Hence the electrons do not follow straight linepath in presence of electric field.

(a) (b)Fig. 2.1

The average velocity of all free electrons in a conductor in presenceof electric field is called drift velocity. The velocity of free electronjust after the collision becomes zero and just before the collisionremains maximum i.e., v = aτ where τ is the time between twosucessive collision and called relaxation time.

Fig. 2.2

Let n be the number of free electrons per unit volume in a conductor of length l and area ofcorss-section A, then the total charge of free electrons,

q = neAl

And on applying potential difference V across it, the electric current,

I =

25

or I = neAvd

where vd = drift velocity

If u1, u2...un be the initial and v1, v2 ... vn that the final velocities of free electrons than

v =

1 2 1 2... ...n nu u u v v v

n

+ + + + +

=

0

ne En

m

n

τ

+

=eE

m

τ

=

.eV

ml

τ

τ decreases with increases in temperature, hence drift velocity decreases with increase in temperature.

∵ I = neAvd

=

neAeV

ml

τ

=

2ne AV

m l

τ or

V

I =2

m l

ne A τ

∵

V

I = R (resistance)

∴ R =

2

m l

ne A τ

Also R =l

Aρ

Where ρ =

2

m

ne τ

called resistivity or specific resistance. It depends

upon the material and temperature.

Ohm's law states that the ratio of potential difference and current flowing though a conductoris constant if all external condition like temperature, etc., are remain unchanged.

V

I

= R

The conductors who obey the Ohm's law are called Ohmic and those do not Obey are called non-ohmic conductor.

26

The I – V graph of ohmic conductor is straight line. Fig. 2.3

Fig. 2.3

When resistances are connected in series the equivalent resistance,

Fig. 2.4

R = R1 + R2 + R3

In series the equivalent resistance is greater than the largest resistance present in the combination.The current in all resistances remains same and potential difference distributes in direct ratio oftheir resistances.

I1 = I2

and =

Fig. 2.5

When resistances are connected across two same points, the combination is called parallel combination.Fig. 2.6.

In paralle combination the equivalent resistance, R in given by

=

in parallel combination the equivalent resistance is less than the smallest resistance present in thecombination.

Fig. 2.6

27

The p.d. aross each resistor is same and current distibutes ininverse ratio of the resistances

V1 = V2

and1

2

I

I =

2

1

R

R

When temperature of Ohmic conductor increases, the mean free path and relaxation time decreases.

Therefore, the resistivity ρ =

2

m

ne τ

increases.

The change in resistivity is directly proportional to the original resistivity and change in temperature(∆θ)

∆ ρ α ρ0∆θor ∆ρ = αρ0∆θwhere α is called the temperature co-efficient of resistivity. α is positive for metals and negativefor semiconducture. The new resistivity,

ρ = ρ0 + ∆ρ= ρ0 + αρ0∆θ= ρ0 (1 + α∆θ)

correspondingly R = R0 (1 + α∆θ)

The carbon resistors are coded by coloured rings. For resistances three coloured rings are used.

The coloures are coded as

Fig. 2.8

Black Brown Red Orange Yellow Green Blue Violet Gray WhiteBl Br R O Y G G W

0 1 2 3 4 5 6 7 8 9

Table 2.1

The code of first second and third coloured bands give thefirst digit, second digit and number of zeroes followed bysecond digit of the resistance value.

Electric cell is the simplest source of electrical energy. Thecell consists of electrolyte and electrodes.

The resistance offered by cell (electrolyte and electrods) iscalled internal resistance of the cell (r). Fig. 2.9

Fig. 2.7

28

The internal resistance depends upon the nature of materialof electrodes and electrolyte.

Concentration of electrolyte. It increases with increase inconcentration.

Surface area of electrodes. It decreases with increase in surface area of electrodes.

Seperation of the electrodes. It increase with increase in seperation of electrodes.

Temperature. It decrease with increase in temperature.

The amount of worce done in circulating a unit positive charge in a closed circuit including the cellis called electro motive force (emf).

Current drawn from the cell

I =

I =

where R = external resistance and r is internal resistance

IR + Ir = ε∵ V = IR

∴ ε = V + Ir

If I = 0, ε = V

∴ emf can also be defined as the terminal voltage of the cell when no current is drawn from thecell.

If n identical cells each of emf ε and internal resistance r are connected in series with externalresistance R, the current I can be given by

I =

If r << R, I will be maximum

and I =

Similarly. If m identical cells are connected in parallel, then

I =

If R << r, I will be maximum and can be given by

I =

The current in the circuit is maximum when internal resistance and external resistance are equal.

29

When the cell is short-circuited, the external resistance becomes zero.

∴ I =

E E

(R r) r=

+

i.e., short-circuit current of a cell is maximum while terminal voltage is zero. Power transfer to the load by the cell will be

P = I2R = 2

2

R

(R r)

ε+

If R = 0 or ∞ the P will be minimum.For maximum value of P

dP

dR = 0 i.e.,

2

2

d R0

dR (R r)

ε = +

⇒ R = r

For R = r, Pmax

=

2E

r

This is called maximum power transfer theorem. The variation P with respect to R is shown n fig. 2.10. Kirchhoff's Laws of Electric Circuits Point rule or junction rule : It states that the

algebric sum of electric point at a point orjunction is zero

ΣI = 0current coming towards the point is taken aspositive and going away from the point is takenas negative.

For a Fig. 2.11,I

1 + I

2 – I

3 – I

4 – I

5 = 0

or I1 + I

2 = I

3 + I

4 + I

5

i.e., current coming = current going.

Loop rule : The law states that the algebric sum of the potential difference in complete traversalof a loop is zero.

ΣV = 0or ΣV = ΣIRFor closed loop ABCDA,

E1 – E

2 = I

1R

1 – I

2R

2...(i)

For closed loop DCEFDE

2 – E

3 = I

2R

2 – I

3R

3...(ii)

Also ABFEAE

1 – E

2 = I

1R

1 – I

2R

2...(iii)

Fig. 2.10

Fig. 2.11

Fig. 2.12

30

Wheat stone Bridge. In 1843 charles wheatstone devised that anarrangment of four resistances which can be used to measure oneof them in terms of the rest. The bridge is said to be balancedwhen deflection in galvanometer connected across the diagonalof the arrangment of four resistances PQRS shown in fig. 2.13.For balanced bridge

P/Q = R/S

Metre bridge is used to measure unknown resistance and specificresistance. The circuit used for this purpose is shown in fig. 2.14.For no deflection in galvanometer,

=

or = R/S

or the unknown resistance,

S =

If r and l′ be the radius and length of the unknown resistance wire, then

s =

or ρ =

or ρ =

Potentiometer : The potentiometer, one of the mostimportant electrical instrument, consists basically of aresistance R through which a steady current flows asshown in fig. 2.15.When emf E

s is connected between A and B and the

tapping is adjusted will no current passes in thegalvanometer G, by 'loop rule'

Es = IR

s...(i)

Now replacing the source Es by unknown potential

Fig. 2.13

Fig. 2.14

Fig. 2.15

31

difference V (to be measured), the balance point is achieved by shitting the tapping for value RX

so thatV = IR

X and hence from eqn (i)

V =

x xs s

s s

R LE E

R L=

[as R α L)

or V = KLX with K = s

s

E

L = Potential gradient. ...(iii)

Though the driving emf E and resistance r and R do not need to be known, they must remain sameduring the whole experiment from the time the first balnace is achieved (otherwise I will notremain same and so theory will not apply)

Usually the first balance (called standardisation) is achieved by fixing the tapping point B at agiven length L

s and adjusting I by means of r so that K= E

s/L

s becomes simple ratio such as

10–2 or 10–3 (V cm–1). A balance point will be achieved only if the source of potential difference V is connected between

A and B with the same polarity and its potential difference V is lesser than that across R. As in this method no current is drawn from the source of potential difference V, the potentiometer

acts like a voltmeter of infinite resistance (i.e. ideal voltmeter) and hence can measure emf of acell of potential difference most accurately.

Answer Yourself

Very Short Questions

Q1. Why is a voltmeter always connected in parallel with a circuit element across which voltage isto be measured?

Q2. If the length of a conductor wire is doubled by stretching it, keeping the p.d. across it constant,by what factor does the drift velocity of eletrons change?

Q3. What will be the the change in the resistance of Eureka wire, when its radius is halved and lengthis reduced to one-fourth of its original length?

Q4. V-I graph for a metallic wire at two different temperatures T1, and T

2,

is shown in the figure. Which of the two temperatures is the higherand why?

Q5. What is the principle of a potentiometer?Q6. On a given resistor, the colour bands are in the sequence : green violet and red what is its

resistance?Q7. How does current density of a metallic conductor is related to the drift speed of electrons through

the conductor,Q8. Draw the circuit diagram of a potentiometer which can be used to determine the internal resistance

(r) of a given cell of emf (E).

Fig. 2.16

32

Fig. 2.18

Fig. 2.17

Q9. Draw the current versus potential difference characteristics of a cell. How can the internal resistanceof the cell be determined from this graph?

Q10. Write the nature of path of free electrons in a conductor in the

(i) presence of electric field

(ii) absence of electric field

Q11. What will be to tolrance of a carbon resistance having only three colour bands of green, orangeand red colours.

Short Answer

Q1. Define the term electrical resistivity of a material. How is it related to its electrical conductivity?Of the factors, length, area of cross-section, nature of material and temperature, which one controlthe resistivity of a conductor?

Q2. A resistance of a wire is 5Ω at the 50°C and 6Ω at 100°C. What will be the resistance of the wireat 0°C?

Q3. State Kirchhoff rules of current distribution in an electrical net work.Using these rules determine the value of the current in R = 20Ω resistancein the electric circuit given in the fig. 2.17.

Q4. A cylinderical metallic wire is stretched to increase its length by 5%. Calculate the percentagechange in its resistance.

Q5. Two wire of equal lengths, one of copper and the other of manganin have the same resistance.Which wire is thicker?

Q6. It is observed that the deflection in a potentiometer set up is in the same sense at both the startingend as well as at the other extreme end of the potentiometer. However, the value of this deflectionis more at the other extreme end than at the starting end. What could be the reason for this? Howcan it be corrected?

Q7. V-l graphs for parallel and series combination of two metallic resistors are asshown in the figure. Which graph represents the parallel combination? Justifyyou answer.

Q8. Calculate the equivalent resistance of the resistance network between the point A and B as shownin the fig. 2.19, when-switch S is closed.

Fig. 2.19

33

Q7. A volmcter V of resistance 400 Ω is used to measured the potentialdifference across a 100 Ω resistor in the circuit shown alongside.(i) What will be reading of the voltmeter?(ii) Calculate the p.d, across 100 Ω resistor before the voltmeter is

connected.

Q8. A potentiometer wire has a length of 10 m and resistance 10 Ω. An accumulator of emf 2V anda resistance box are connected in series with it. Calcuate the resistance to be introduced in the boxso as to get a potential gradient of (i) 0.1V/m, and (ii) 0.001 V/m.

Q9. Why is a potentiometer preferred over a volimeter to measureemf of a ceil? The potentiometer wire AB shown in the figureis 400 cm long. Where should the free end of the galvanometerbe connected on AB, so that the galvanometer shows zerodeflection ?

Q10. The plot of variationof potential difference across a combination ofthree identical cells in series, versus current is as shown. What isthe emf of each cell?

Q11. The figure shows experimental set up of a metre bridge when thetwo unknown resistances X and Y are inserted, the null point Dis obtained 40 cm from the end A, When a resistance of 10Ω isconnected in series with X, the null point D is obtained 10 cm.Find the position of the null point whne the 10 Ω resistance isinserted in series with resistance Y. Determine the value of theresistance X and Y.

Long Questions

Q1. Are the paths of electrons straights lines between successive collisions (with positive ions of themeta!) in the (i) absence of electric field? (ii) presence of electric field? Establish a relationbetween drift velocity 'v

d' of an electron in a conductor of cross-section 'A' carrying current 'l' and

concentration 'n' of free electrons per unit volume of conductor. Hence, obtain the relation betweencurrent density and drift velocity.

Q2. Write the methametical relation for the resistivity of a material in terms of relaxation time, numberdensity, mass and charge of charge carriers in it. Explain, using this relation, why the resistivityof metal increases and that of a semi-conductor decreases with rise in temperature.

Q3. Explain the principle of Wheatstone bridge for determining an unknown resistance. How is itrealised in actual practice in the laboratory?

Fig. 2.20

Fig. 2.21

Fig. 2.22

Fig. 2.23

34

Q4. Obtain an expression for the potential gradient 'and' of potentiometer whose wire of length l hasa resistance r. The driving cells has an emf E is connected in series with an external resistance R.

Q5. Derive an expression for the resistivity of a good conductor, in terms of the relaxation time ofelectrons.

Very Short Questions

Q1. Write the relation between relaxation time and mean free path.

Q2. The I-V graphs of two resistors of value R1 and

R2 and their combinations A and B respectively

are shown in Fig. 2.24. Name the combinationrepresented by graph A and B respectivley.

Q3. If m rows of n identical cells each of emf E and internal resistance r are connected to the externalresistance R, then write an expression for maximum current.

Q4. In a wheatstone bridge the deflection in galvanometer is zero. Write therelation between V

1, V

2, V

3 and V

4 in the given circuit.

Q5. In Fig. 2.25 a wheatstone bridge is shown. If

R > , P, Q, R and S are resistances then what

will be direction of current in galvanometer arm.

Q6. What will happen to the sensitivity of potentiometer if its potential gradient is increased?Q7. Can terminal voltage of a cell be greater than the emf of the cell?Q8. The ratio of the relaxation times and number densities of free electrons of two conductors A and

B are 1 : 2 and 2 : 3 respectively. What will be the ratio fo their spectific resistnace?Q9. The resistance offered by two rehotate A and B are R

A and R

B respectively. Which one is greater?

Fig. 2.26.

(A) (B)Fig. 2.26

Q10. The filament of incandasent buils is ohmic or non-ohmic.

V B

IO

R2

R1

A

Fig. 2.24

Fig. 2.25

To be Learnt

35

Short Questions

Q11. Show that drift speed is independent of area of cross-section of a conductor.Q12. Draw the circuit diagram for V-I characteristics of a conductor of resistance 2Ω. Also explain the

how to determine the resistance from V-I charectristics.Q13. Find the current drawn from the battery in the given circuit.Q14. A melattic wire of mass and density d is being streatched uniformaly. Show the variation of

resistance of the wire with respect to length.

1. Relaxation time =mean free path

r.m.s. relocityof free electronsPedagogical Remark. In absence of electric fielf, the free electrons move rondemly in all presslydirections. Thus, the average velocity of the free electrons is zero. To avoid the directional effectto find effective velocity the root mean square velocity is taken.

2. A represents parallel combination of R1 and R

2. B represents series combination of R

1 and R

2.

Pedagogical Remark. In series combination the equivalent resistance is greater than the individualresistance and in parallel combination the equivalent resistance is less than the individual resistances.And the resistance is the slop of the I-V graph line.

3. Maximum current,

Imax

=2

mnE

Rr

Pedagogical Remark. For maximum current in the mix grouping of cells the internal resistancemust be equal to external resistance.

4.

1

2

V

V

=

3

4

V

V

Pedagogical Remark. In balance wheatstonebridge the ratio of the resistances is given by

AB

BC

R

R

=

AD

DC

R

R

this is also applicable for potential difference and power dissipated.

5. From B to D.Pedagogical Remark. In unbalnaced wheatstone bridge, the current throught galvanometer is notzero. The current flow from higher potential to lower potential.

6. The sensitivity will degreeage.Pedagogical Remark. The potential gradient is fall in potential per unit length and sensitivity is

the smallest potential difference which can be measured.

Vk

l

∆ =

.

G

B

A C

V4V3

V1 V2

V

Fig. 2.27

Pedagogical Remark

36

7. Yes.

Pedagogical Remark. During the charging of the cell, V = E + Ir.

8. ρ =

Pedagogical Remark. When conductors are of different materials the number density of free

electrons and the relaxation time will be different. Therefore, the resistivity depends upon the

nature of the material.

9. RB > R

A

Pedagogical Remark. The sliding rod is metallic whose resistance can be ignored.

10. Ohmic in steady state.

Pedagogical Remark. When switch is on, its temperature increases and becomes non-ohmic after

that it reaches at steady state and becomes ohmic.

11. The drift speed

Vd =

eE.

mτ

=

It shows that it does not depend upon area of cross-section.

Pedagogical Remark. Application of the relation between drift velocity and physical dimensions

of the conductor.

12.

Fig. 2.28

The reciprocal of the stope of V-I graph line is the resistance of the conductor.

Pedagogical Remark. Understanding the difference between I-V and V-I characteristics.

In choice of scale the division values must be multiple of least count of the instrument and choice

of scale in drawing the graph.

37

13. Total resistance of the circutitR = 6 Ω

∴ current drawn from the battery

I =V 16

2.0AR 6

= =

Pedagogical Remark. The current flows due to difference of potential, hence there is no currentin 10Ω resistance and there is no contribution of these resistances in the circuit.

14. ∵ R =A

lρ

and m =

2r dlΠ

∴ R =

2d

m

lρ

Thus, R α l2

Pedagogical Remark. To study single relationship betweenphysical quantities, the variable like area, etc. are to beeliminated.

6Ω

6Ω

6Ω

6Ω10Ω10Ω

12 V

Fig. 2.29

l

R

Fig. 2.30

38

3Oersted's Experiment

Orested's observed that when a magnetic compass needle is placed close to a current carrying wire thenit shows deflection.It shows that current carrying conductor produces magnetic field around it.

Biot-Savart's Law

Biot-Savart law states that the magnitude of the magnetic field dB

at a point P due to a current element(small length element) is given by

dB =

vector form =

where µ0 = permeability of free space

= 4π × 10–7

θ =

SI unit of magnetic field is tesla (T) or Weber/m2.Unit of magnetic field in CGS system is gauss (G).

Note-On the surface of conductor (i.e. at θ = 0) magnetic field intensity is zero.

Direction of magnetic field is determined by right hand thumb rule.

According to this rule if we hold current carrying conductorin our right hand such that the thumb represent the direction ofcurrent, the curvature of the fingers around the conductor then ofrepresents the direction of magnetic field lines.

Magnetic Field due to a Current Carrying Circular Loop at its Centre

Let a be the = radius of loop

I = Current

direction of current

direction of currentmagnetic linesof force

39

Direction of magnetic field at point O is normal to the plane of the paper and inward.

dB = 02

Idlsin 90 (Biot - sa var t law)

4 a

µπ

B =

02

Idl

4 a

µπ

=

02

I(2 a)

4 a

µ ππ

=

0I

2a

µ

(Normal inward)

If there are N turn in the coil then

B =

0NI

2a

µ

Magnetic Field due to Current Carrying Circular Loop at any point on its Axis

Consider a circular coil of radius a with centre o.

dlθ

I

a

O

C

D

I

r

R dBcosα

αα

α

dBcosα

dBsinαdBsinα

dB

dB

Here θ, the angle between dl

& R is 90°.

P is the point where magnetic field is to be calculated. Direction of I is anticlockwise as observedfrom P. Magnetic field at P due to dl is given by-

dB =

( )0

22 2

Idlsin 90.

4 (r a )

µ °π +

= 0

2 2

Idl.

4 (r a )

µπ +

Direction of dB

is as shown in figure.If we consider another current element dl at diametricallyopposite point D then components of dB perpendicular to axis of loop (dB cosθ) will cancel each otherand component of B along the axis (dB sinθ) will add up.Assuming all the current carrying elementalong the loop, total magnetic field at P is given by-

∴ B =

dBsin α

=

02 2

Idlsin.

4 (r a )

µ απ +

=0

2 2 2 2

Idl a. .

4 (r a ) r a

µπ + +

=o

2 2 2 2

I a.2 a

4 (r a ) r a

µ ππ + +

40

∴ B =

If there are N turns, B =2

02 2 3/ 2

NIa

2(r a )

µ+

Direction of B at P is along the axis of the loop from centre to point P.

Variation of r and B

B

r

Above analysis shows that current carrying loop form two poles(North and south)on its two faceshence behave as magnetic dipole.Magnetic dipole moment of such dipole is defined as-

M = (current)(area of loop)

Direction of M is from south pole to north pole.(Using right hand rule)

Ampere's Circuital Law

The line integral of the magnetic field along a closed loop or curve ( B.dl ) is equal to the times the

total current I passing through the closed loop or curve.

i.e. =

Magnetic Field due to an Infinitely long Straight Current Carrying Wire

In the figure P is observation point.Let a circle of radius r be the amperian loop.Using ampere law

(Here angle between B and dl is 0),hence

B dl =

or B (2πr) = µ01

∴ B =

In the given figure, direction of magnetic field at P is normal to plane of paper and upward.

41

A Current Carrying Solenoid and its Magnetic Field Along its Axis

Solenoid is a helical shapped coil in which length of the conductor is much greater than its radius.

Magnetic field of a current carrying solenoid is same as a bar magnet. In a current carrying solenoid,magnetic field-

(i) is along the axis of solenoid inside the solenoid directed from south pole to north pole.

(ii) is zero just outside the solenoid.

Let n = number of turns per unit length of solenoid

I = current in solenoid

Consider a square abcd of side h as amperian loop

Using ampere law

B.dl

=

0 enclosedIµ

Here =

b c d a

a b c dB.dl B.dl B.dl B.dl+ + +

= Bhcos0° + Bhcos90° + 0.hcos0° + Bhcos 90°

= Bh

and current enclosed = I (number of turns passing through amperian loop)

= I(nh)

∴ Bh = µ0Inh

or B = µ0nI

The magnetic field does not depends upon dimension of solenoid.

Toroid

Toroid is an endless solenoid. It can be considered as a solenoidwhich is bent into a circular shape to close on it-self.

Magnetic field is along the axis of solenoid.There is no poleformation in magnetic field configuration of toroid.

P is the observation point and r is the radius of toroid. SupposeI is current as shown and n is the number of turns per unit length.

Consider circle of radius r as amperian loop.Direction of B is along the amperian loop.

Using Ampere's Law,

0 enclosedB.d Il = µ

⇒ 2πrB = µ0

(no. of turns in solenoid) I

∴ 2πrB = µ0

(2π r n) I

∴ B =

0nIµ

Toroid plays an important role in the equipment for plasma confinement in fusion power reactors.

Force on a Moving Charge in a Magnetic Field

If a positive charge 'q' moving with a velocity

v

moving in external magnetic field

B

.Then charge

particle experiences a

F

are F.

hd c

Ia b

B

P

0

r

I

I I

42

=

or F = qv B sin

where θ is the angle between and .

(direction of this force is determined by flemming left hand rule).

Fleming's Left Hand Rule

If we stretch the fore finger, the central finger and the thumb of left hand such that they are mutuallyperpendicular to each other and the fore finger indicates the direction of magnetic field, the centralfinger indicates the direction of electric current (direction of velocity of positive charge) then the thumbrepresents the direction of force experienced by the charged particle.

Motion of a Charged Particle in a Uniform Magnetic Field

(I) When & are perpendicular to each other

(a)

direction of this is perpendicular to both and . Since force is perpendicular to , so chargedparticle executes uniform circular motion at the same place. If r is radius and v is the speed of particle.

∴ qvB =

r =

Angular velocity, ω =

∴ ω =Bq

m

Time period, T = ⇒ T =

This T does not depend on speed of particle.

(II) When V and B are in same directionIn this case, no magnetic force acts on the particle and it moves in straight line path.

43

(III) When v

&

B

are not ⊥⊥⊥⊥⊥ to each other

Let velocity

v

is making an angle 0<φ<90 with

B

B

y

V

Vcosφ

φ

Vsinφ

z

So the particle will move in a circular path because of 1v vsin= φ

and at the same time it will have

a linear velocity because of

2v v cos= φ

. This means it starts moving in a helical path as shown in

figure.

(i) Radius of helix

1mvr

qB=

(ii) Time period to complete one revolution, 2 r

TBq

π=(iii) Path of the helix = V

2 × T

(III) Since F

&

v

are ⊥ to each other, no work is done by the magnetic field.

(IV) Since no work is done on the particle, its kinetic energy is also constant(i.e. speed remainsconstant)

Lorentz Forces

The force experienced by a charge particle due to electric and magnetic field is called Lorentz force.

Electric Lorentz Force

eF

=

qE

Magnetic Lorentz force

mF

=

q(v B)×

Total Lorentz force

F

=

e mF F+

F

=

q(E v B)+ ×

Cyclotron

It is a device used to accelerate charged particle to a very high energies.

PrincipleIt is based on the principle of lorentz force. The electric field accelerates the particle and magnetic fieldmakes it move in a circular orbit.

Construction1. Strong electromagnet (N & S)

2. Two Dees, D1 & D

2. (D-shaped hollow semicircular copper chambers called Dees, placed horizontally

with a small gap in between D1 & D

2 as shown.)

3. High frequency oscillator (f > 106 to 107 Hz)with high voltage (V > 104 to 105 V) .

44

4. Source of charged particles at the centre of the Dees.(Ion source)

The whole assembly is placed in a vacuum chamber maintained at very low pressure of about10–6 mm of Hg.

WorkingSuppose a positive ion P is generated from ion source when D

1 is at a (+) potential & D

2 at

(–) potential. Positive ion will be accelerated due to this polarity and enter in D2 with increased speed.

Inside Dee electric field is zero and due to magnetic field it will move in a circular path.

radius of path r =mv

Bq ...(i)

After moving the semi circular path in time t inside D2, (p) reaches at the edge of D

2 .In this time,

t the polarities of D1 & D

2 are reversed and again particle get accelerated and enter in to D

1.This whole

process repeats again and again and finally the particle obtain a sufficiently high kinetic energy. Theaccelerated particle is removed out of Dees.

Time spent inside a Dee in semicircular path,

t = r/v = m/Bq ...(ii)

If T = Time period of the electric field,then

or T = ...(iii)

∴ The cyclotron's angular frequency ωc is

ωc =

2 Bq

T m

π = ...(iv)

The cyclotron frequency is

f =1 Bq

T 2 m=

πkinetic energy of ions is maximum when it moves along the largest circular path i.e. radius of Dees.

Limitations of cyclotron1. Speed of charged particle can be increased up to a certain limit(Not beyond the speed of light).

2. Very light particle like electron can not be accelerated because it go quickly out of step with electricfield.

3. It cannot accelerate uncharged particles.

45

Force on a Current Carrying Conductor Placed in a Magnetic Field.

l = length of conductor in the magnetic field

v = drift velocity of electrons

θ = angle between v and B

I = current in conductor

Force on the charged particle

F

=

q(v B)×

F

=

q(l / t B)×

F

=

(q / t)( B)l ×

F

=

I( B)l ×

Direction of this force is determined by Fleming's left hand rule.

Force Between two Parallel Current Carrying Straight Wires

R

FF

I1 I2

A C

B D

R

I2

A C

B D

F

I1

F

Consider AB and CD as two infinite conductors seperated by a distance R.

Case-I : When direction of currents are same Force on CD due to AB i.e. F21

Magnetic field B produced by AB at a point on CD is given by

B = 0 1I

2 R

µπ ...(1)

and direction of B is perpendicular to plane of paper and inward.

magnitude of force per unit length on CD is given by

F = I2 × l × B

or F =

o 1 2I I

2 R

µπ

Direction of F is towards conductor AB.

Similarly magnitude of force per unit length on AB due to CD is also same but direction isopposite.We can say that if direction of current in the wires are same then they attract each other

Case-II:When direction of currents are opposite to each other

In this case, magnitude of force is same as above but directions of forces are different. We cansay that if direction of current in the wires are opposite to each other then wire they repell each other.

46

Definition of Ampere

We know that F =

If I1 = I

2 = 1 ampere and R = 1m,

then F = = 2×10–7 Nm–1 in free space

One ampere is defined as the strength of current which when flowing through two parallelinfinitely long wires placed in vacuum seperated by 1 m , produces a force between the two conductorsof 2 × 10–7 newton per metre length.

Torque on a Current Carrying Coil in Magnetic Field

Consider a rectanglar coil PQRS[PQ=b ,PS=l] carrying current I such that the normal of coil makesan angle with the direction of magnetic field B.

Magnetic force on each wire is as below-

Force on PQ, Upward

Force on RS, Downward

Force on PS, Normal inward

Force on QR, Normal outward Force on PQ cancel force on RS. Other twoforces produce torque.

∴ Torque, τ = Force (FPS

or FQR

) × perpendicular distance between these two force vectors= I/B × b sinθ = BIA sin θ

∴ τ = MB sin θ M B (Vector form)τ = ×

where M = IA and A = Area of coil l × b)

If there are n turns in the coil

τ =

NOTE - If given angle is φ which is the angle between magnetic field and the plane of coil thenin above formula put (90 – φ) in place of θ.

47

Moving Coil Galvanometer

It is an instrument used for detection and measurement of small electric current.

N S

T1

T2Phosphorbronze stripMirror

Magnnet

CoilSoftiron cylinder

PrincipleWhen current is passed through coil placed in a magnetic field, it experiences a torque which rotatesthe coil.

ConstructionA coil is suspended about a vertical axis in a radial magnetic field produced by horse shoe magnet. (Inradial magnetic field angle between the area vector A

of the coil and the magnetic field is 90°, i.e.torque on coil is maximum.The coil is wound on soft iron cylinderical core (which increases theintensity of magnetic field at the place of coil). This coil is attached with a phosphor bronze wire. Theother side of the coil is attached with a spring. Spring restores the position of coil by its restoring torque.The deflection is indicated on the scale by a mirror (which behave as pointer) attached to the spring.

If K = torsional constant

θ = angle of rotation

For equilibrium of coil

Magnetic torque = Restoring torque of spring

nIAB = kθ

or I =

KG

nBA θ = θ

Where K

GnBA

= is galvanometer constant.

Hence direction of deflection shows direction of current and magnitude of deflection shows magnitudeof current.

SensitivityCurrent sensitivity(I

S)

It is defined as angle of reflection per unit current

Is =nBA

I K

θ =

The unit of current sensitivity is rad A–1 or division A–1.

48

It can be increased by increasing the area of the coil, the number of turns, the strength of themagnetic field and by decreasing K.

Voltage sensitivity(VS)

It is defined as angle of reflection per unit potential difference applied across the coil

Vs =

g g

nBA

V IR KR

θ θ= = [∵V = IRg where R

g = resistance of galvanometer]

The unit of voltage sensitivity is rad V–1 or division V–1.

Conversion of Galvanometer into Ammeter

A galvanometer can be converted into ammeter by connecting a low resistance parallel to thegalvanometer.(Low resistance connected in parallel is called shunt)The coil of the galvanometer has aresistance G.

Let I = maximum circuit current to be measured by ammeter

Ig = full scale deflection current of galvanometer.

Rg= Resistance of galvanometer.

S = resistance of shunt

As galvanometer and shunt are in parallel so potential differenceacross them will be same

Ig.G = (I – Ig)S i.e.

Conversion of Galvanometer into Voltmeter

To convert a galvanometer into voltmeter a high resistance is connected in series with galvanometer.

If R = resistance connected in series

Rg = resistance of galvanometer coil

V = voltage to be measured by voltmeter.

Ig = Full scale deflection current.

So V = Ig (R + G)

∴ R =g

VG

I−

49

Magnetism and Matter

Atom as a Magnetic Dipole

According to Bohr's model, the electron of charge e has uniform circular motion around a stationary

heavy nucleus. This constitutes a current i,as

i =

e

T

Where T is the time period of revolution. Let r is the orbital radius of the electron and v the orbital

speed.

Then, T =

2 r evi

v 2 r

π π

Thus we can say revolving electron and hence atom behave as a small magnetic dipole.Its magnetic

moment M is given by

Μ =

2 evri( r )

2π =

Μ =neh nh

mvr4 m 2

= π π ∵

where, n is a natural number, n = 1, 2, 3, ... and h = 6.6 × 10–34 Js (Planck's constant).

Taking n = 1,

M =eh

Bohr Magneton4 m

= µ =π B

It is the magnetic moment associated with an atom due to orbital motion of electron.

Magnetic Dipole (Bar Magnet)

It consists of two magnetic poles(Point of maximum attraction) of equal and opposite strengths and

separated by finite distance(magnetic length). The two poles of a magnetic dipole are called north and

south poles. An isolated north or a south pole does not exist.Two poles of a magnet are always of same

pole strength.

N S• •

Magnetic dipole moment

pole strength Magnetic length

M m 2l

= ×

SI unit M is joule/tesla or A m2.It is a vector quantity whose direction is from S-pole to N-pole.

50

Values of pole strengths, when magnet is cut

Magnet Pole Effective MagneticStrength length after moment

cutting (m)

SN m 2l m(2l) = M

S N m/2 2l lm M

(2 )2 2

=

N S m l

N S m/2 l/2

Magnetic Field Lines

These are imaginary curve. These lines defined as path of a unit north pole placed in a given magneticfield and free to move.

Properties(i) It starts from N-pole to S-pole outside the magnet and from S-pole to N-pole inside the magnet.

(ii) Tangent at any point gives the direction of magnetic field at that point.(iii) They are continuous curve.(iv) Number of field lines crossing per unit area is equal to intensity of magnetic field.(v) They do not intersect each other.Note-They are not termed as magnetic lines of force.

Gauss' Theorem in Magnetism

According to Gauss's law in magnetism magnetic flux (φB) through any close surface is always zero

B ds⋅

= 0

It conclude that isolated magnetic pole does not exist.

Analogy between electric and magnetic quantity

Physics quantity Electrostatic Magnetism

1. Basic quantity and nature charge (q) (positive) pole (strength-m) (north pole)

2. Free space constant

3. Field 02

mB

4 r

µ=π

51

Magneticequator

Geographicequator

SM

SM

SG

S

N

4. Force1 2

20

1 q qF

4 r=

πε0 1 2

2

m mF

4 r

µ=π

5. dipole moment

p q(2a)=

M m(2l)=

Axial field

30

1 2p.

4 rπε

03

2M

4 r

µπ

equitorial field

30

1 p.

4 r−

πε

03

M

4 r

µ−

π

6. Torque

p Eτ = ×

M Bτ = ×

7. Force

F qE=

F mB=

8. Potential energy

p.E−

M.B−

1 2pE(cos cos )− θ − θ1 2M (cos cos )− θ − θBMagnetic Field of Earth

Initially Sir Gilbirt suggested that earth itself is a huge

magnet.Earth's magnetic field is approximately uniform

of magnitude 10-5 tesla.The source for earth's magnetism

is perhaps due to the rotation of earth. Due to this

motion electric currents is produced by ions of metallic

fluids in the outer core of the earth. This is known as

the dynamo effect.

NG

= Geographic n-pole, NM

= Magnetic n-pole

SG

= Geographic s-pole, SM

= Magnetic s- pole

An imaginary plane passing through NG and S

G is called Geographic meridian.

An imaginary plane passing through NM

and SM

is called Magnetic meridian.

Elements of Earth's Magnetic Field

Earth's magnetic field (B) at a point on earth's surface can be completely defined by the following three

parameters of earth's magnetic field at that point

52

1. Magnetic declination (θ)

2. Magnetic dip or Inclination (δ)3. Horizontal component of earth's total field (B

H) Declination (θ). It is the angle between earth's

geographic meridian and magnetic meridian at the given place.Dip (δ). It is the angle between freely suspended magnetic needle and the horizontal direction whenneedle is suspended in magnetic meridian at the given place.At poles dip = 90° and at magnetic equator dip = 0. Dip-circle is the instrument used to measureangle of dip at the given place.

Horizontal component of earth's total magnetic field (BH)It is the component of earth's total magnetic field in the horizontal direction at the given place.

BH = VBcos and B Bsinδ = δ

where BV is vertical component and where B =Earth's total magnetic field

Also B = and V

H

Btan

Bδ =

If a plane is making an angle θ with magnetic meridian then angle of dip (δ′) in that plane is calledapparent dip. In that case

tan δ′ =tan

cos

δθ

Some Important DefinitionsMagnetising Intensity ( ) (Magnetisation-force): It is the degree to which a magnetic field canmagnetise a material. It is represented by

= ⇒ B = µH

For solenoid B = µnI so H = nIUnit of H is Am-1

Intensity of magnetisation (I): It is defined as induced magnetic moment per unit volume of thematerial.

= Its unit is also Am–1 so we can write it as B = µI.

Magnetic susceptibility (χχχχχm): It is the ratio of intensity of magnetisation (I) to magnetisation force(H). i.e.

χm = it is a dimensionless quantity.

Relative permeability:– It is the ratio of permeability of medium to permeability of free space.

µr =

Relation between magnetic susceptibility and relative permeability⇒ µ

r = 1 + χ

53

Permanent Magnets And Electromagnets

Retentivity or Residual Magnetization: It is the residual magnetic flux density in a substance whenthe external magnetizing field is zero.

Coercivity (or Coercive force): It is value of the magnetic field in reversed direction that isrequired to reduce the magnetic flux density in a substance from its residual value to zero.

Magnetic Materials (Kind of Magnetic Material and their Properties)

1. Cause Diamagnetism Paramagnetism Ferromagnetism Its net magnetic moment

is zero due to cancellation of moment of different electron.

It possess a permanent magnetic dipole moment either due to unpaired electron or due to electron spin

Its own magnetic moment is not zero Magnetic moment of atom align to produce large magnetic moment

2. If kept near a magnet

Feebly repelled Feebly attracted Strongly attracted

3. Behaviour of a freely Suspended rod in an external magnetic field

rod align itself Normal to field direction

Road align itself along the field direction

Road align itself strongly along the field direction

4. Nature of Magnetic Lines when material is placed in a magnetic field

N S N S

N

N S

N

5. Behaviour of liquids and gaes

Such materials move from stronger to weaker magnetic field

Such materials from weaker to stronger magnetic field s

Such materials move from weaker to stronger magnetic field

6. Magnetisation Get weekly magnetized in the direction opposite to the direction

Get weekly magneised in the direction of magnetizing field.

Get strongly magnetized in the direction of magnetizing field.

7. χ (Magnetic susceptibility)

Negative & small [ ]−1 ≤ χ < 0

Positive & small [ ]ε0 < χ <

Positive & large [ ]1χ >>

8. rµ (RELATIVE

PERMEABILITY)

slightly less than one [ ]0 1rµ≤ <

slightly more than one ( )1 1rµ ε< < +

quite larger than one [ ]1rµ >>

9. µ 0µ µ< 0µ µ> 0µ µ>>

10. Flux density within it

Less than in vacuum Greater than in vacuum Much larger than in vacuum

54

Materials for Making Permanent Magnets(i) The material should have high retentivity and high coercivity

(ii) The material should have a high permeability.(iii) Steel is preferred as it has a slightly smaller retentivity than soft iron but has much larger

coercivity than soft iron.(iv) Other suitable materials for permanent magnets are alnico, cobalt steel and ticonal.For Electromagnet: Material should have high permeability and low retentivity soft iron is suitable

material for making electromagnet.Use of electromagnetic: In electric bell, telephone, loudspeaker, cranes etc.

Answer Yourself

Very Short Questions

Q1. How will the magnetic field intensity at the centre of a circular coil carrying current change, ifthe current through the coil is doubled and the radius of the coil is halved?

Q2. A current is set up in a long copper pipe. Is there a magnetic field (i) inside, (ii) outside the pipe?Q3. In a certain arrangement, a proton does not get deflected while passing through a magnetic field

region. Under what condition is it possible?Q4. When a charged particle enters a magnetic field at right angles to the field, which one of the

following does not change?Velocity of the particle, momentum of the particle, energy of the particle.

Q5. A proton (or an electron) is moving in a uniform magnetic field. What is the path of the proton(or an electron) if it enters (i) parallel to the field, (ii) perpendicular to the field.

Q6. What is the magnetic field at the centre O of the cubical mesh in which current I is entering fromone of the vertex? .

Q7. What is the function of cylinderical soft iron core in a moving coil galvanometer?Q8. State two properties of a material used as a suspension wire in a moving coil galvanometer.Q9. What is meant by ‘figure of merit’ of a galvanometer?Q10. What should be the orientation of a magnetic dipole in a uniform magnetic field so that its

potential energy is maximum?Q11. What should be the orientation of a magnetic dipole in a uniform magnetic field so that its

potential energy is minimum?Q13. Steel is preferred for making permanent magnets whereas soft iron is preferred for making

electromagnets. Explain.

Short Questions

Q1. What is the angle of dip at a place, where the horizontal and vertical components of the earth'smagnetic field are equal?

55

Q2. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the samedirection are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Q3. A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for acurrent of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A?

Q4. In which direction would a compass free to move in the vertical plane point to, if located righton the geomagnetic north or south pole?

Q5. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its northtip pointing down at 22° with the horizontal. The horizontal component of the earth's magneticfield at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic fieldat the place.

Very Short Questions

Q1. An electron is revolving in an orbit with a uniform speed. Name the fields associated with.

Q2. An electron beam projected along +x-axis, experiences a force due to a magnetic field along the+y axis. What is the direction of the magnetic field?

Q3. Consider the circuit shown here where APB and AQB aresemi-circles. What will be the magnetic field at the centre Cof the circular loop?

Q4. An electron and a proton moving with same speed enter the same magnetic field region at rightangles to the direction of the field. For which of the two particles will the radius of the circularpath be smaller?

Q5. A wire of length L is bent in the form of a circle of radius R and carries current I. What is itsmagnetic moment?

Q6. Two wires of equal lengths are bent in the form of two loops. One of the loop is square shapedwhereas the other loop is circular. These are suspended in a uniform magnetic field and the samecurrent is passed through them. Which loop will experience greater torque? Give reasons.

Q7. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) lessthan when the core is empty?

Q8. A certain region of space is to be shielded from magnetic fields. Suggest a method.

Q9. How does the intensity of magnetisation of a paramagnetic material vary with increasing appliedmagnetic field?

Q10. An iron bar is heated to 1000 °C and then cooled in a magnetic field free space. Will it retain itsmagnetism?

Q11. Why should the material used for making permanent magnets have high coercivity?

Q12. Must every magnetic field configuration have a north pole and a south pole? What about the fielddue to a toroid?

To be Learnt

• CA B

P

Q

I

56

Short Question

Q13. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction.Give the magnitude and direction of B

at a point 2.5 m east of the wire.

Q14. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A andmaking an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Q15. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameterof the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B

inside thesolenoid near its centre.

Q16. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil issuspended vertically and the normal to the plane of the coil makes an angle of 30° with thedirection of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude oftorque experienced by the coil?

Q17. A galvanometer coil has a resistance of 12Ω and the meter shows full scale deflection for a currentof 3 mA. How will you convert the galvanometer into a voltmeter of range 0 to 18 V?

Q18. A beam of alpha-particles and of the protons of the same velocity v enter a uniform magnetic fieldat right angles to the field lines. The particles describe circular paths. What is the ratio of the radiiof these two circles.

Q19. An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. Calculate theequivalent magnetic moment if the frequency of revolution of electron is 6.8 × 109 MHz.

Q20. A short bar magnet of magnetic moment m = 0.32J T–1 is placed in a uniform magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a)stable and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Q21. A short bar magnet has a magnetic moment of 0.48 J T–1. Give the direction and magnitude ofthe magnetic field produced by the magnet at a distance of 10cm from the centre of the magneton (a) the axis, (b) the equational lines (normal bisector) of the magnet.

Q22. Answer the following questions:(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east

to west) is set up in a chamber. A charged particle enters the chamber and travels undeflectedalong a straight path with constant speed. What can you say about the initial velocity of theparticle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varyingfrom point to point both in magnitude and direction, and comes out of it following a complicatedtrajectory. Would its final speed equal the initial speed if it suffered no collisions with theenvironment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field innorth to south direction. Specify the direction in which a uniform magnetic field should be setup to prevent the electron from deflecting from its straight line path.

57

Q23. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normalto the plane of the coil. If the current in the coil is 5.0 A, what is the(a) total torque on the coil,(b) total force on the coil,(c) average force on each electron in the coil due to the magnetic field?(The coil is made of copper wire of cross-sectional area 10–5 m2, and the free electron density incopper is given to be about 1029 m–3)

Long Question

Q24. Write the expression for the magnetic moment

( )m

due to a planar square loop of side ‘l’ carryinga steady current I in a vector form.In the given figure this loop is placed in a horizontal plane near a long straight conductor carryinga steady current I

1 at a distance l as shown. Give reasons to explain that the loop will experience

a net force but no torque. Write the expression for this force acting on the loop. (5 marks)

I1

l

l

Q25. A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The currentis uniformly distributed across the cross-section. Apply Amphere’s circuital law to calculate themagnetic field at a poin ‘r’ in the region for (i) r < a and (ii) r > a. (5 marks)

1. Electric field as well as the magnetic field.2. Along +ve z-axis in accordance the Fleming's left hand rule.

3. Zero, because magnetic fields due to APB and AQB are equal in magnitudes but opposite indirections.

4. ∵ Radius of circular path mv

rBq

=

same speed v, the mass of electron is much less than that of proton. Hence re << r

p.

5. Since LR

2=

π, hence magnetic moment M = I.A = I.π R2

2I.L

4=

π

.

6. Torque

nIABsinτ = θ

, as length of wire is same (say L),

for square loop

2 2

s

L LA

4 16 = =

and

for circular loop 2 2

2c

L LA r

2 4 = π = π = π π

. Since, Ac > A

s, the circular loop will experience

greater torque than the square loop.

Pedagogical Remark

58

7. Slightly less, since bismuth is a diamagnetic in nature.

8. Surround the region by soft iron rings. Magnetic field lines will be drawn into the rings, and theenclosed space will be free of magnetic field.

9. It gradually increases.10. No. it will not retain its magnetism because it has been heated once beyond Curie's temperature.11. So that the magnetisation is not erased by external magnetic fields, temperature fluctuations or

minor mechanical damage.12. It is not necessary that every magnetic field configuration have a north pole and a south pole.

Poles are formed only if the source of field has a net non-zero magnetic moment. In case of atoroid and for an inifinitely long conductor carrying current net magnetic moment is zero andhence there is no pole formation.

Short Questions

13. Here I = 50 A and R = 2.5 m

∴ Magnitude of magnetic field

B =

= = 4 × 10–6T

As per right hand rule, the magnetic field at the given point is in vertically upward direction.

14. Here I = 8 A, θ = 30° and B = 0.15 T

∴ Force per unit length of the wire

=

= 0.15 × 8 × sin 30°

= 0.15 × 8 × = 0.6 N m–1.

15. As the solenoid has 5 layers of windings of 400 turns each, it means N = 5 × 400 = 2000. FurtherL = 80 cm = 0.8 m and I = 8.0 A

∴ Magnitude of inside the solenoid near its centre

B =

=74 10 2000 8

0.8

−π × × ×

= 2.5 × 10–2 T.

16. It is given that each side of square coil l = 10 cm = 0.1m, hence area

A = l2 = (0.1)2 = 0.01 m2

N = 20, I = 12 A, = 30° and B = 0.80T

∴ Torque τ = nIAB sinθ

59

= 20 × 0.8 × 0.01 × 12 × sin30°

= 20 × 0.8 × 0.01 × 12 ×

1

2

= 0.96 Nm.

17. Here G = 12Ω, Ig = 3 mA = 3 × 10–3A and voltmeter's range V = 18V

To convert galvanometer into a voltmeter of desired range, we should connect a resistance R inseries with the galvanometer, where

R =

3g

V 18G 12

I 3 10−− = −×

= 6000 – 12 = 5988 Ω.

18. We know that radius of a circular path is given by

r =mv

BqHence, for α-particles and protons of same velocity v entering a uniform magnetic field B normally:

∴

p

r

rα

=

p

p

qm.

m qα

α

But mα = 4mp

and qα = 2qp

∴

p

r

rα

=

14

2×

= 2

rα = 2rp

19. Here radius of orbit R = 0.53Å. = 5.3 × 10–11 m and frequency of revolution ν = 6.8 × 109 MHz= 6.8 × 1015 Hz

∴ Magnetic moment m = IA = (ev).πR2 = 1.6 × 10–19 × 6.8 × 1015 × 3.14 × (5.3 × 10–11)2

= 9.6 × 10–24 A m2.20. Here m = 0.32 J T–1

and B = 0.15 T(a) Stable equilibrium orientation means the magnet is set parallel to the external magnetic field

and in this position, the potential energy of the magnet isU

1 =

m B⋅

= –mBcos0°= –mB = –0.32 × 0.15= –4.8 × 10–2 J.

(b) Unstable equilibrium orientation means the magnet is set anti-parallel to the external magneticfield i.e., θ = π. In unstable equilibrium state, the potential energy of the magnet will be

U2 =

m B⋅

– mBcos π= – 0.32 × 0.15 × (–1)= 4.8 × 10–2 J.

21. Hence m = 0.48 J T–1

and r = 10 cm = 0.1 m

60

(a) Magnetic field at a point on its axis

B =

= 9.6 × 10–5 T along S-N or 0.96 G along S-N direction.(b) Magnetic field at a point on the equitorial line

B =7

03 3

m 10 0.48

4 r (0.1)

−µ ×=π

= 4.8 × 10–5 T along N-S or 0.48 G along N-S direction

22. (a) Initial velocity v

is either parallel or anti-parallel to .

(b) Yes, because magnetic force, can change the direction of velocity v but cannot change itsmagnitude.

(c) magnetic field should be in a vertically downward direction so that in accordance withFleming's left hand rule the deflection of electron due to B is in horizontal plane from northto south and may nullify the deflection due to electrostatic field.

23. (a) Torque on the coil τ = nBIAsin θ = 0 (∵ θ = 0°)

(b) Total force on the coil

F = BIlsin θ = 0 (∵ θ = 0°)

(c) Average force on each electron in the coil due to magnetic field

F = Bevd, where v

d is the drift speed of electrons

= Be. (∵ I = neAvd)

F = B = = 5.0 × 10–25 N.

24. The magnetic moment due to a current loop carrying a steady current I is given by whereA is the area of the loop.

For a planar square loop of side ‘l’, we have

=

∴ magnetic moment =

In the arrangement shown in the figure, the long straight conductor MN carrying a steady currentI

1 produces a magnetic field around it.

Forces acting on paris QR and SP of given loop due to the magnetic field of current carrying longconductor MN are equal, mutually opposite and collinear. Hence they balance one another.

Now force acting on part PQ

61

F1 = 0 1 0 1

1 2 2

I I IB Il I l

l

µ µ= ⋅ ⋅ =π π

The force F1 is perpendicular to wire PQ and is directed towards the conductor MN i.e., the force

is attractive in nature.

Again force acting on part RS

F2 = B

2Il =

0 1 0 1

2 4

I I I Iµ µ=π π

The force F2 is also perpendicular to wire RS but is directed away from the conductor MN i.e.,

the force is repulsive in nature.

∴ Net force acting on loop PQRS

F

=

1 2F F+

or F =

0 1 0 1 0 11 2 2 4 4

I I I I IIF F

µ µ µ− = − =

π π π

(attractive)

However, since

1 2 and F F

are collinear hence these forces will not have a torque. Thus, the loopwill not experience a torque.

25. Consider a long straight wire of a circular cross-section ofradius ‘a’ carrying a steady current I. The current is uniformlydistributed across the cross-section.

(i) To calculate the magnetic field B at a point P distant rfrom the axis of wire (where r < a) consider a circularloop of radius r around the axis of wire. Then applyingAmphere’s circuital law to the closed loop, we find thatB d l⋅

=

.2B dl B dl B r= = π

The current enclosed by the said loop is

Ien

=

2 2

2 2

r rI I

a a

π = ⋅ π

Hence as per Ampere’s law :

B2πr =2

0 0 2( )en

rI I

aµ = µ ⇒ B = 0

22

Ir

a

µπ

(ii) To calculate the magnetic field at a point P' (where r > a) outside the current carrying wire,on applying Ampere’s circuital law, as above, we get

B d l⋅

=

2B dl B dl B r= = ⋅ π

⇒ B × 2π r = µ0

I

⇒ B =

0

2

I

r

µπ

(a)

a

rO P

(b)

dlB

P

P

O

a

I

r

62

4It is the phenomenon of generating e.m.f. in a conductor when there is a change of magnetic flux linkedwith the conductor.

The e.m.f. so developed is called induced e.m.f. If the conductor is a closed circuit, a current alsoflows in the circuit (called induced current).

Magnetic FluxThe magnetic flux (φ) is the measure of total number of magnetic field lines crossing through any

surface A

placed in magnetic field .

It is given by , where θ is the angle between area vector A and B. S.I. unit offlux is Weber.

Faraday's Laws of Electromagnetic Induction

First law: Whenever there is a change in magnetic flux linked with a circuit, an induced e.m.f. andhence induced current is produced in the circuit. It lasts only till the magnetic flux changes.

Second law: The magnitude of the induced e.m.f. is directly proportional to the rate of change ofmagnetic flux linked with the circuit.

e = ⇒ e =

=

Here negative sign shows that induced e.m.f. is always opposite to change in magnetic flux (Lenz's law).

Lenz's Law

According to this law, the direction of the induced e.m.f. is in such a way that it always opposes thereason due to which change in magnetic flux takes place.

Lenz's law is in accordance with the law of conservation of energy.

Fleming's Right Hand Rule

It is used to know the direction of induced currentproduced in a conductor when the magnetic fluxassociated with it changes. According to this rule, ifwe stretch the first finger, central finger and thumb ofour right hand in mutually perpendicular directions suchthat first finger points out in the direction of magneticfied, thumb points out in the direction of motion (orforce) of the conductor, then the central finger givesthe direction of induced current.

Field

InducedCurrent (if any)

Force

63

Methods of Producing Induced emf

(i) By changing strength of magnetic field (B) keeping area (A) and orientation (θ) of coil constant.

(ii) By changing area (A) of coil keeping magnetic field (B) and orientation (θ) of coil constant.(Motional emf)

(iii) By changing orientation (θ) of coil keeping both area (A) and magnetic field (B) constant.

Motional e.m.f.

PQ is movable wire of length l and moving towards right with a velocity v as shown. Magnetic fieldB is outward.

–

+

P B

v

X ∆xQ

l

At any instant ‘t’,

flux linked with the circuit = B.A B xl=

where A = Area in magnetic field

At time (t + ∆t), flux is given by Bl (x +∆x)

∴ Change in flux in time ∆t

= [Bl (x + ∆ x) – Blx] = Bl∆x

∴ Induced emf = Rate of change of flux

e = Bl.

x

t

∆∆ = Blv

Hence, magnitude of induced e.m.f., e = Bvl

Direction of induced emf and corresponding induced current is determined by Flemmings righthand rule which is shown in figure given earlier.

Energy Consideration

We know that motional emf = Blv

If r is the resistance of wire then inducd current i is given by

e Blvi

r r= =

Force experienced by wire (using F = ilB)

F =2 2B v B v

i B Br r

l ll l

= = Direction of this force is opposite to the direction of velocity of wire.

Now the power required to move the wire P is

P =2 2 2 2 2B v B v

F.v .vr r

l l= =

Since wire is pushed by external power P' which is given by

P′ =2 2 2 2

2 B v B vI r r

r r

l l = =

64

since P = P', we can say that mechanical energy is converted into electrical energy. Hence electro-magnetic induction is in accordance with law of conservation of energy.

Also as we know

e =d

and e Irdt

φ =

If q is the amount of charge induced then

=

(a) Induced e.m.f. produced by changing the orientation of the coilConsider a coil of Area A

is placed in magnetic field .

N = no. of turns

ω = angular velocity of rotation of the coil

θ = angle between and at time t (θ = ωt)

Flux passing through the coil is given by

φ =

=

Uaing Faraday law e =

e =d(NBA cos t)

dt

ω−

∴ e = NAB sin tω ωThis equation shows that induced e.m.f e changes sinusoidally with time t.

Maximum value of emf is e =

(b) A.C. GeneratorIt is a device which converts the chemical energy into electrical energy.

Working Principle: It is based upon the principle of electromagnetic induction.

Construction: It has three main components

(i) Permanent magnet NS

(ii) Armature coil PQRS

(iii) Slip rings (S1 & S

2) with carbon brushes (B

1 & B

2)

Working: When the coil is rotated in a magnetic field, the magnetic flux associated with ifchanges. The induced emf produced in it is given by

e =

If R is the resistance of the coil, then the current

I =

I =where I

0 = peak value or amplitude of induced current

65

In the diagram, PQRS is Armature coil: R1, R

2 are the Slip rings;

C1,C

2 are Carbon brushes; R

L is Load resistance

(i) During1st half cycle of rotation of coil, side PQ is moving outward in the magnetic field, sousing Fleming right hand rule direction of induced current is from P to Q.Hence direction ofcurrent in R

L is from M to N.

R

S

N S

P

2R

1R1C

2C

LR

Q

M N

2R

1R1C

2C

LR

Q R

P S

N S

M N

1st half Cycle 2nd half Cycle

(ii) During 2st half cycle of rotation of coil, side PQ is moving inward in the magnetic field so usingFleming right hand rule, direction of induced current is from Q to P.Hence direction of currentin R

L is from N to M.

This analysis shows that direction of current in RL changes with time and hence output is alternating

current i.e. the direction of flow of induced current in the external circuit changes periodically..

Variation of Induced e.m.f. with time is shown as

TimeT3T4

T2

T4

ε0ε

0

–ε0

Eddy Currents

Current induced in bulk piece of a conducting material kept is a clanging magnetic field is called eddycurrent. Its magnitude is usually very large and it flows in closed loops.

Uses of eddy currents(i) Electromagnetic damping in galvanometers.

(ii) Heating and even melting of metals as in the induction furnace.

(iii) Electric or eddy current brakes in electric trains.

(iv) In electric power meters.

We can minimise eddy current, but cannot eliminate them.

66

(a) Self inductance (or coefficient of self induction)

It is the property of a coil due to which it opposes the change in magnetic flux through itself.

If φ = magnetic flux linked with the coil at time t when current flowing through it is I, then φ ∝ I

or φ = LI

where L is constant of proportionality, known as coefficient of self induction, or self inductance orsimply inductance of the coil.

Using Faraday law-

e =d dI

– = –Ldt dt

φ

Dimensions of L = [ML2T–2A–2]

The SI unit of inductance is Henry.

Thus, if rate of change of current of one ampere per second induces an e.m.f. of one volt in a coil,the inductance of that coil is one henry.

(b) Self-inductance of a Long Solenoid

Let the current I be flowing in the solenoid of length l and cross sectional area A.

Let N = total no. of turns

B =

=

Magnetic flux for N turns

φ =0NI

N Al

µ

∴ L =

If core of solenoid is of any other magnetic material, then µ0 is replaced by (µ

0µ

r)

(a) Mutual inductance (M) (or Coefficient of matual induction)

It is the property of a coil due to which it opposes the change in magnetic flux associated throughanother neighbouring coil.

Any change of current in coil-1 results in a change of magnetic flux linked with coil-2 (&vice-versa). This process is called mutual induction.

Let I1 = current in coil-1

φ2 = total flux linked with coil-2 due to I

1,

then φ2 α I

1

∴ φ2 = M

I

1

where M is constant of proportionality called coefficient of mutual induction.

Also if e = induced e.m.f. in coil-2 due to rate of change of I1, then

e = 21d

dt

φ−

l

I

Axis2r

I

67

or e = 1dIM

dt−

Unit of M is Henry.

(b) Mutual Induction between two long solenoidsLet I

1, I

2 = current in solenoids s

1 and s

2

N1,N

2 = No. of turns in s

1 and s

2

φ1. φ

2 = flux linked with s

1 and s

2

l = common length of S1 & S

2

M,M' = coefficient of mutual inductance.

A1 and A

2 = Area of coil s

1 and s

2

case-(i) Mutual induction in coil 2 due to coils 1 In this case current is flowing in coil 1.Magneticfield B along the axis of solenoid

B1 = 1

o 1

NI

lµ

Magnetic flux passing through one turn of coil 2.

= B1A

2

For N2

turns φ2

= 2 1 2N B A ..............(ii)

so we get φ2

=

0 1 12 2

N IN A

l

µ

...(iii)

comparing with φ2

=2 1M I

M2 =

0 1 22

N NA

l

µ

...(iv)

case-(ii) Mutual induction in coil 1 due to coil 2 In this case current is flowing in coil 2.Magneticfield B along the axis of solenoid

B2 =

2o 2

NI

lµ

Magnetic flux passing through one turn of coil 1.

= B2A

2 (Area of smaller coil)

For N1

turns φ1 = N

1B

2A

2..............(v)

we get φ1 =

0 2 21 2

N IN A

l

µ

comparing with

φ1 = M

1I

2

we get M2 =

1210 A

NN

l

µ

......(vi)

From (iv) & (vi), we get

M2 = M

1

l

r1N2

N1 S2

r2

S1

68

Energy (U) Stored in an Inductor

When an inductor carries a current, a magnetic field builds up in it and energy is stored in it in the formof magnetic field energy.

Power consumed by battery

P = eI (e = emf, I = current)

dw

dt =

dw = LIDI

⇒ =

Integrating, Total work done or energy stared is

Energy U =

Coils in Series and Parallel.

When two coils are connected in series, the equivalent self inductance is

Ls = L

1 + L

2

and when they are connected in parallel, the equivalent self inductance is

=

Alternating CurrentIf direction of current in circuit changes continuously with time then current is known as alternatingcurrent.

I = 0 0I sin t or I I cos tω = ωHere, I is instantaneous value of current(value of current at time t) and I

0 is the peak value or

maximum value or amplitude of a.c., ω is called angular frequency of a.c.

Also, ω =

where T is the time period of a.c.For alternating e.m.f. E

E =average value of E or I from 0 to T is represented by

Iavg

=

Mean or Average ValueMean or average value of a.c. over one complete cycle is zero.

Mean value of a.c. for 1st half cycle

=

T / 2

o0T / 2

0

I sin tdt

dt

ω

69

=

T / 2

0

0

2I cos t

T

ω − ω

=

02I 2 T 2cos . cos .0

T T 2 T

π π − + ω

=

[ ]02I1 1

T+

ω

=

02I 2T

T 2 π

=

02I

π

= 0.637I0

Similarly, the mean or average value of a.c. over 2nd half cycle is – 0.637 I0.

Hence the mean or average value of a.c. over one complete cycle is 0.637 I0 – 0.637 I

0 = zero.

Similarly,

avg 0 T/ 2E

−

= o2E

π

avg T/ 2 TE

−

= o2E−π

hence average value of alternating voltage over complete cycle is also zero.So ordinary d.c. instrument(ammeter and voltmeter) will show zero reading.The instruments used

to measure a.c. based on heating effect of current so division on scale on such instrument are not equallyspaced.

R.M.S. Value or Effective Value of Alternating Current and Voltage

The r.m.s. (root mean square) value of a.c. is defined as that value of steady current, which wouldgenerate the same amount of heat in a given resistance in a given time, as is done by a.c. when passedthrough the same resistance for the same time.

The r.m.s. value is also called effective value or virtual value of a.c. It is represented by Irms

or Ieff

or Iν.r.m.s. stands for root, mean and square so,

Irms

=

( )1/ 2T 2

00T

0

I sin t dt

dt

ω

=

12

0I T.

T 2

T2

0

Tsin tdt

2 ω = ∵

⇒ Irms

=0

o

I0.707 I

2=

Similarly,

⇒ Erms

= 0o

E0.707 E

2=

A.C. Circuit Containing Resistance (R) only

Let applied emf is given by

E = 0E sin tω ...(1)

70

Let I = current in the circuit at instant t.

since I =E

R

⇒ I = ...(2)

where I0 = E

0/R, Comparing I

0 = E

0/R with ohm's law we find that behaviour of resistance R in

d.c. and a.c. circuits is the same.Comparing (1) and (2), we find that there is no phase difference in E and I .Therefore, in an a.c.

circuit containing R only, the voltage and current are in the same phase, as shown in phasor diagam.

A.C. Circuit Containing Inductor (L) only

Let applied emf is given by

E = ...(1)

e.m.f. induced across the inductor is given by

To maintain the flow of current in this circuit, induced voltage (e) must be equal and opposite tothe applied voltage (E).

i.e. 0E sin t− ω =

or dI = 0Esin t dt

Lω

Integrating both sides, we get

I = 0Esin tdt

Lω

⇒ I =0E cos t

L

− ω ω

L

Circuit

EI

O

E & I

E0

I0

0 90ºωt

ωt

Let, Io = ...(2)

we get, I = ...(3)

This is the equation of alternating current due to inductor.

71

Comparing (1) with (3), we find that voltage across L leads the current by a phase angle of 90°.This is shown in Fig. using eq. (2)

o

o

E

I

= ωL = Effective resistance on inductor

This is called inductive reactance and is denoted by XL.The units of inductive reactance is ohm.

Thus XL =

L = 2 Lω πν

where ν is the frequency of a.c. supply.Hence inductive reactance is proportional to frequency of A.C.In dc circuits, ν = 0 so X

L = 0

In ac circuits, ν = high so XL = high

It means a pure inductor block ac and allow dc.

A.C. Circuit Containing Capacitor (C) only

Let applied emf is given by E =

0E sin tω

...(1)

For capacitor we know that q = CVhere V = E so

q =

0CE sin tω

⇒

dq

dt

=

0d(CE sin t)

dt

ω

⇒ I =

oE C cos tω ω

⇒ I =

oE C sin( t / 2)ω ω + πLet I

0=

oE Cω...(2)

⇒ I =0I sin( t / 2)ω + π ...(3)

Comparing (1) with (3) we find that in an a.c. circuit containing C only, alternating current I leadsthe alternating e.m.f. by a phase angle of 90°. This is shown in Fig.

C

E

I

O

E & I

E0

0

I090º

ωtt

using eq. (2)

o

0

E 1

I C=

ω

= Effective resistance of capacitor This is called capacitive reactance and is denoted by

XC.The units of capacitive reactance is ohm.

Thus XC =

1 1

C 2 C=

ω πν

Hence capacitive reactance is proportional to frequency of A.C.In dc circuits, ν = 0 so X

L = Infinity

In ac circuits, ν = high so XL = Low

It means a pure inductor block dc and allow ac.

72

A.C. Circuit Containing Resistance, Inductor and Capacitor (RLC Circuit) in Series

As R, L, C are in series, therefore, current through the threeelements has the same amplitude and phase. Let maximumvalue of current is I.

Let potential difference across R,L and C are VR

,VL

and VC then

VR = IR

(Phase difference between VR and I is zero.)

VL = IX

L

(voltage across L leads the current by 90°)V

C = IX

C

(voltage across C lags behind the current by 90°Phasor diagram of all potential drops and current is shown in

figure.Phase difference between VL and V

C is 180°. Let (V

L >V

C)

then net reactive voltage is (VL - V

C) .If V

Z represent the resultant

of VR

,VL , and V

C and making an angle f with current phasor

I.Then

∴ VZ =

= 2 2L C(IR) (IX IX )+ −

VZ = 2 2

L CI R (X X )+ −

The total effective resistance of series LCR circuit is called impedance of the circuit. It is representedby Z, where

Z =2 20

L C0

VR (X X )

I= + −

From Fig. it is clear that in an a.c. circuit containing R, L and C and (VL >V

C) the voltage leads

the current by a phase angle φ, where

tan φ =L C

R

V V

V

−

=

or tan φ =

∴ The alternating e.m.f. in the LCR circuit would be represented by

E =

Here three cases arise:

(i) When XL = X

C, ∴ φ = 0

Hence voltage and current are in the same phase. The a.c. circuit is non-inductive.

I0

C

O

90°

90°

VL

VC

φ

K

A

VR

zV

73

(ii) When XL > X

C, tanφ is positive.

Therefore, φ is positive. Hence voltage leads the current by a phase angle φ.

(iii) When XL < X

C, tanφ is negative.

Therefore, φ is negative. Hence voltage lags behind the current by a phase angle φ.

For R and L circuitFor RL circuit (put X

C=0)

VZ =

2 2R LV V+

and 2 2LZ R X= +

and tan φ = LX

RIn this case voltage leads current.

For R and C circuitFor RC circuit (put X

L=0)

VZ =

2 2R CV V+

and 2 2CZ R X= +

and tanφ = CX

RIn this case voltage lags behind current.The reciprocal of reactance is called susceptance of the a.c. circuit and reciprocal of impedance is called

admittance of a.c. circuit. Both, the susceptance and admittance are measured in mho i.e. ohm–1 or Siemen.

LC Oscillations

Oscillator is a device which convert DC in to AC.The simplest oscillator consist of L and C called tankcircuit.

A capacitor of capacitance C is connected to an inductor of inductanceL through a key K

2. A cell is connected to C through key K

1.

When plug of K1 is put in, charging of capacitor takes place.Energy

stored in capacitor in form of electric field.Now open K

1 and close K

2,the charged capacitor is connected to L.

Step-1 Capacitor starts discharging through L and a discharging currentflow in LC circuit. An induced e.m.f. developes across L.Energy is storedin L.

Step-2 This induced e.m.f. starts recharging the condenser in the opposite direction by inducedcurrent. Energy is stored in C.

+ +C

I

L+ +

C

I

L

Step 1 2Step

Step-3 Again capacitor starts discharging through L and a discharging current flow in LCcircuit.Again an induced e.m.f. developes across L.Energy is stored in L.

XLZ

Rφ

XC

Z

Rφ

L ba

K2

+ C

K1

–

DISCHARGINGCURRENT

CHARGINGCURRENT

s

74

Step-4 This induced e.m.f. starts recharging the condenser in the opposite direction by inducedcurrent. Energy is stored in C.

+ +

C

I

L+ +

C

I

L

3Step 4Step

The entire process is repeated. Thus energy taken once from the cell and given to capacitor keepson oscillating between C and L.Current flow in LC circuit is AC.

In actual practice some energy loss takes place due to heat.Therefore, amplitude of oscillations goes on decreasing. There arecalled damped oscillations as shown in Fig.

Frequency of oscillation is

Resonance in Series LCR Circuit

At a certain frequency current of circuit becomes maximum, In this case circuit is said to be in resonance.This certain frequency is called resonant frequency or natural frequency of circuit.

For maximum current, the impedance (Z) of an RLC series circuit should be minimum

since Z =2

2 1R L

C + ω − ω

For minimum Z, XL

= XC

i.e. rLω = or ωr =

= or =

ωr = Angular resonant frequency

and νr = Resonant frequency.

At resonance value of Z is given by (XL = X

C)

Z = 2R 0+ = R = minimum

In this case circuit is pure resistive and phase differencebetween voltage and current is zero.

The variation of circuit current with the changing frequencyof applied voltage is shown in Fig.

The practical application of series resonance circuit is in radioand T.V. receiver sets.

I

Y R = 10 Ω

R = 100 Ω

O Xω1 ωr ω2ω

I = I0

2

75

Quality Factor (Q Factor) or Sharpness of Resonance

The characteristic of a series resonant circuit is determined by the Q factor or Quality factor of thecircuit. It defines sharpness of current at resonance.

The Q factor of series resonance circuit is defined as the ratio of the voltage across the Inductoror Capacitor at resonance to the applied voltage or voltage applied across R.

i.e. Q = )Racrossvoltage(voltageapplied

CorLacrossvoltage

Q = L L r

R

V IX L

V IR R

ω= =

⇒ Q =L 1

R LC

r

1sin ce

LC

ω =

As R is increased, Q factor of the circuit decreases.

Another Definition of Q FactorGraph shows that for each value of I there are two values of ω. If ω

1 and ω2 are the values for which

the current is

1 2

times its maximum value (and power is half).

If

1 rω = ω − ∆ω

and

2 rω = ω + ∆ω

The difference

2 1 2ω − ω = ∆ω

is called the band width of the circuit.

Q factor of resonance circuit is the ratio of resonance angular frequency to band width of the circuit.

i.e. Q-factor =r

2

ω ∆ω Average Power in RLC Circuit or Inductive Circuit

Let the alternating e.m.f. applied

E =

oE sin tω

If φ is the phase angle between current then alternating current will be

I =

oI sin( t )ω + φ

since power = (e.m.f.) x (current)

P =

o oE sin t I sin( t )ω × ω + φ

=

o oE I sin t[sin t cos cos t sin )ω ω φ + ω φ

P =

2o o o o

sin 2 tE I cos sin t E I sin

2

ωφ ω + φ

Now average power in one cycle of AC

Pavg

=

T T2

0 0o o o oT T

0 0

sin tdt sin 2 tdt

E I cos E I sin

dt 2 dt

ω ωφ + φ

76

or Pavg

=

As

T T2

0 0

Tsin tdt and sin 2 tdt 0

2

ω = ω =

∴ Average power in the inductive circuit over a complete cycle

Pavg

= o o

1E I cos .

2φ

=

Pavg

=

Power Factor of an A.C. Circuit and Wattless Current

In eqn.

P is called true power and ( ) is called apparent power or virtual power.

cosφ is the ratio of true power and apparent power and called power factor of the circuit.

So, Power factor =

∴ power factor = cos φ

= [from phasor diagram]

In a non-inductive circuit, XL = X

C

∴ Power factor = cos φ = 1This is the maximum value of power factor. In a pure inductor or an ideal capacitor,

φ = 90°∴ Power factor = cos φ = cos 90° = 0.Average power consumed in a pure inductor or ideal capacitor, P = E

rms I

rms cos 90° = Zero.

Therefore, current through pure L or pure C, which consumes no power for its maintenance in thecircuit is called Idle current or Wattless current.

At resonance, XL = X

C and φ = 0°.

∴ cos φ = cos 0° = 1Therefore, maximum power is dissipated in a circuit at resonance. Power dissipation is always

through resistance R.

Transformer

A transformer is an electrical device which is used for changing the a.c. voltages and current.A transformer which increases the a.c. voltages is called a step up transformer. A transformer which

decreases the a.c. voltages is called a step down transformer.

PrincipleA transformer is based on the principle of mutual induction.

77

ConstructionA transformer consists of a rectangular soft iron core made of laminated sheets, well insulated from oneanother. Two coils P

1 P

2 and S

1 S

2 are wound on the same core, but are well insulated from each other.

The source of alternating e.m.f. is connected to P1

P2 (the primary coil) and a load resistance R is

connected to S1 S

2 (secondary coil),

Assume that the energy losses is negligible.Theory and workingLet the alternating e.m.f. supplied to primary is

E = 0E sin tωFlux induced in primary coil = φ

B

Because the core is ferromagnetic and extends through the secondary winding, so induced flux alsoextends through the turns of secondary.

The voltage Ep across the primary is equal to the e.m.f. induced in the primary, and the voltage E

sacross the secondary is equal to the e.m.f. induced in the secondary. Since

E =

Bd

dt

φ

⇒

BE n ( nAB)α φ =∵

⇒

p

s

E

E

=

p

s

n

n

Here, np and n

s represent total number of turns in primary and secondary coils respectively

If ns > n

p then E

s > E

p, the transformer is a step up transformer. Similarly, and if n

s < n

p then E

s < E

p,

The transformer is a step down transformer transformer.s

p

n

n

= K = transformation ratio.

Using conservation of energy requires thatI

p E

p = I

s E

s

∴

p

s

I

I

=

s

p

E

E

For a step up transformer, Es > E

p but I

s< I

p

For a step down transformer, Es < E

p but I

s> I

p

i.e. whatever we gain in voltage, we lose in current in the same ratio.Efficiency of a transformer is defined as the ratio of output power to the input power.

i.e. η =

s s

p p

E IOutput power

Input power E I=

Transformer is essentially an a.c. device. It cannot work on d.c. It does not affect the frequency of a.c.

Energy loss in transformer1. Copper loss—-It is the energy loss in the form of heat in the copper coils of a transformer.2. Iron loss—It is the energy loss in the form of heat in the iron core of the transformer due to eddy

current. It can be minimised by taking laminated cores.

P1

P2

InputA.C.

S1

S2

R Output

Laminated Core

78

3. Leakage of magnetic flux—It can be reduced by winding the primary and secondary coils oneover the other.

4. Hysteresis loss—This is the loss of energy due to repeated magnetisation and demagnetisation ofthe iron core when a.c. is fed to it. The loss can be reduced by using a magnetic material whichhas a low hysteresis loss.

5. Magnetostriction—humming noise of a transformer.

Uses of TransformerA transformer is used:

(i) In voltage regulators for T.V., refrigerator, computer, air conditioner etc.(ii) In the induction furnaces.

(iii) A step down transformer is used for welding purposes.(iv) In long distances power transmission of AC.To reduce the power loss, a.c. is transmitted

over long distances at extremely high voltages. This reduces I in the same ratio. Therefore,heat I2 R becomes negligibly low.

Answer Yourself

Very Short Questions

Q1. Determine the direction of the induced current in the loop given below:P

Q

R

× × ×

× × ×

× × ×

(a) (b)

Q2. Predict the polarity of the capacitor in the situation described in the figure below:

Q3. An applied voltage signal consists of a superposition of a d.c. voltage and an a.c. voltage of highfrequency. The circuit consists of an inductor and a capacitor in series. Show that the d.c. signalwill appear across C and the a.c. signal across L.

Q4. What is the phase difference between the voltage across inductor and capacitor in an LCR seriescircuit?

Q5. The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage andcurrent in this circuit?

79

Q6. When a lamp is connected to an alternating voltage supply, it lights with the same brightness aswhen connected to a 12 V d.c. battery. What is the peak value of alternating voltage source?

Q7. Can a transformer be used in d.c. circuits? Why?

Q8. What value of current/voltage do youmeasure with an a.c. ammeter/voltmeter?

Short Questions

Q1. A square loop of side 10 cm and resistance 0.5Ω is placed vertically in the east-west plane.A uniform magnetic field of 0.10T is set up across the plane in the north-east direction. Themagnetic field is decreased to zero in 0.7s at a steady rate. Determine the magnitude ofinduced e.m.f. and current during this time-interval.

Q2. A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0ms–1, at right angles to the horizontal component of earth’s magnetic field, 0.30 × 10–4 Wb m–2.

(a) What is the instantaneous value of the e.m.f. induced in the wire?

(b) What is the direction of the e.m.f.?

(c) Which end of the wire is at the higher electrical potential?

Q3. Draw the graphs showing variation of inductive reactance and capacitive reactance withfrequency of applied AC source. Can the voltage drop across the inductor or the capacitor ina series LCR circuit be greater than the applied voltage of the AC source? Justify youranswer.

XL

( )ω

XC

( )ω

Q4. A 100 ω resistor is connected to a 220V, 50 Hz a.c. supply.

Q5. A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency offree oscillations of the circuit?

Q6. A power transmission line feeds input power at 2300 V to a step-down transformer with itsprimary windings having 4000 turns. What should be the number of turns in the secondary inorder to get output power at 230 V?

Q7. Give two advantages and two disadvantages of a.c. over d.c.

Q8. Capacitors block d.c. but allow a.c. to pass. Why?

Long Questions

Q1. In a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 µ F, R = 40Ω.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

(c) Determine the rms potential drops across the three elements of the circuit. Show that thepotential drop across the LC combination is zero at the resonating frequency.

80

Very Short Questions

Q1. (a) Closed loop is held stationary in the magnetic field between the north and south poles oftwo permanent magnets held fixed. Can we hope to generate current in the loop by usingvery strong magnets?

(b) A closed loop is moved normal to the constant electric field between the plates of a largecapacitor. Is a current induced in the loop (i) when it is wholly inside the region betweenthe capacitor plates (ii) when it is partially outside the plates of the capacitor? The electricfield is normal to the plane of the loop.

Q2. Current in a circuit falls from 5.0 A to 0.0 A in 0.1s. If an average e.m.f. of 200 V induced,give an estimate of the self-inductance of the circuit.

Q3. A solenoid with an air core and the bulb are connected to a dc source. How does the brightnessof the bulb change, when the iron core is inserted into the solenoid?

Q4. Give the direction in which the induced current flows in the coil mounted on aninsulating stand when a bar magnet is quickly moved along the axis of the coil from one sideto the other as shown in the figure.

S NS N

Q5. In any a.c. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneousvoltages across the series elements of the circuit? Is the same true for rms voltage?

Q6. A choke coil in series with a lamp is connected to a d.c. line. The lamp is seen to shine brightly.Insertion of an iron core in the choke causes no change in the lamp's brightness. Predict thecorresponding observations if the connection is to an a.c. line.

Q7. The instataneous current and voltage of an a.c. circuit are given by I = 10 sin 314t A and V = 50

sin 314t2

π + V. What is the power dissipation in the circuit?

Q8. What are the maximum and minimum values of power factor of an a.c. circuit?

Q9. If the frequency of the a.c. source in a LCR series circuit is increased, how does the current inthe circuit change?

Q10. Power factor can often be improved by the use of a capacitor of appropriate capacitance in thecircuit. How?

Q11. An ideal inductor is in turn put across 220 V, 50 Hz and 220 V, 100 Hz supplies. Will the currentflowing through it in the two cases be the same or different?

Q12. The electric mains in the house are marked 220 V, 50 Hz. Write down the equation for instantaneousvoltage.

To be Learnt

81

Short Questions

Q13. A circular coil of radius 10 cm, 500 turns and resistance 2Ω is placed with its plane perpendicularto the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameterthrough 180° in 0.25s. Estimate the magnitudes of the e.m.f. and current induced in the coil.Horizontal component of the earth’s magnetic field at the place is 3.0 × 10–5 T.

Q14. A wheel with 10 metallic spokes each 0.5 m long is rotated with angular speed of 120revolutions per minute in a plane normal to the earth’s magnetic field. If the earth’s magneticfield at the given place is 0.4 gauss, find the E.M.F. induced between the axle and the rimof the wheel.

Q15. A bar magnet M is dropped so that it falls vertically through the coil C. The graph obtainedfor voltage produced across the coil vs. time is shown in figure.

(i) Explain the shape of the graph.

(ii) Why is the negative peak longer than the positive peak?

V RCoil

C

Magnet

p.d. (mV)

Time (ms)

C

DOA

B

Q16. Two circular coils, one of small radius r1 and the other of very large radius r

2 are placed co-

axially with centres coinciding. Obtain the mutual inductance of the arrangement.

Q17. How is the mutual inductance of a pair of coils affected when:

(i) separation between the coils is increased?

(ii) the number of turns of each coil is increased?

(iii) A thin iron sheet is placed between the two coils, other factors remaining the same? Explainyour answer in each case.

Q18. (a) The peak voltage of an a.c. supply is 300V. What is the rms voltage?

(b) The rms value of current in an a.c. circuit is 10A. What is the peak current?

Q19. Obtain the resonant frequency

rω

of a series LCR circuit with L = 2.0 H, C = 32 µF and R = 10Ω. What

is the Q-value of this circuit?

Q20. Suppose the initial charge on the capacitor of C = 30

Fµ

is 6mC. What is the total energy storedin the circuit initially? What is the total energy at later time?

Q21. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µF is connected to a variable-frequency200 V a.c. supply. When the frequency of the supply equals the natural frequency of the circuit,what is the average power transferred to the circuit in one complete cycle?

Q22. Calculate the current drawn by the primary of a transformer, which steps down 200 V to 20 Vto operate a device of resistance 20 W. Assume the efficiency of the transformer to be 80%.

Q23. Can a.c. source be connected to a circuit and yet deliver no power to it? If so under whatcircumstances?

82

Q24. A coil of inductance 0.50 H and resistance 100 is connected to a 240 V, 50 Hz a.c. supply.

(a) What is the maximum current in the coil?(b) What is the time lag between the voltage maximum and the current maximum?

Q25. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away froman electric plant generating power at 440 V. The resistance of the two wire line carrying poweris 0.5Ω per km. The town gets power from the line through a 4000 – 220 V step down transformerat a sub-station in the town.(a) Estimate the line power loss in the form of heat.(b) How much power must the plant supply, assuming there is negligible power loss due toleakage?(c) Characterise the step up transformer at the plant.

Q26. In the following circuit calculate, (i) the capacitance ‘C’ of the capacitor, if the power factor ofthe circuit is unity, and (ii) also calculate the Q-factor of the circuit.

Long Questions

Q27. In figure, the arm PQ is moved from x = 0 to x = 2b and then back to x = 0 with a constant speedv. Consider, there is a magnetic field for 0 < x ≤ b. There is no magnetic field for x > b. If theresistance of the arm PQ is r. Obtain expressions for the flux, the induced E.M.F., the forcenecessary to pull the arm and the power dissipated as Joule loss. Sketch the variation of thesequantities with distance.

× × ×

× × ×

× × ×

S

Rx = 0

x = b x = 2b

P

Q

Q28. An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC.The resistance of the circuit is negligible. Let the instant, the circuit is closed be t = 0.(a) What is the total energy stored initially? Is it conserved during LC oscillation?(b) What is the natural frequency of the circuit?(c) At what time is the energy stored(i) Completely electrical (i.e., stored in the capacitor)?(ii) Completely magnetic (i.e., stored in the inductor)?(d) At what times is the total energy shared equally between the inductor and the capacitor?(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

83

Pedagogical Remark

1. (a) No. However strong the magnet may be, current can be induced only by changing the magneticflux through the loop.

(b) No current is induced in either case. Current cannot be induced by changing the electric flux.

2. E.M.F., e =dt

dtL−

200 =1.0

5L−

Hence, L = 4 H.

3. For steady state DC, there will be no change in the brightness of the bulb, as there is no changein current and hence in the flux. When iron core is inserted, the magnetic flux increases suddenly,the lamp will shine less brightly for a moment and then there will be no change in the brightness ofthe bulb as long as current is not changed.

4. If the observer is looking at the loop from the side, from which the bar enters the loop: Thedirection of current will be clockwise, when bar magnet is moving towards the loop. Thedirection of current will be anticlockwise, when bar magnet is moving away from the loop.

5. Yes, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltagesacross the series elements of the given a.c. circuit. The same is not true for rms voltage, becausevoltages across different elements may not be in phase.

6. For a steady state d.c., inductance of the choke coil has no effect, even if it is increased by insertingan iron-core. For a.c., the lamp will shine dimly because of additional impedance of the choke. Itwill dim further when the iron core is inserted which increases the choke's impedance.

7. As voltage is leading the current by 90° hence power factor cosφ = cos 90° = 0. Consequentlypower dissipation in the circuit is zero.

8. Maximum value of power factor is 1 and its minimum value is zero i.e., 0 cos 1≤ φ ≤ .

9. With increase in frequency, current in a.c. circuit first increases, attains a maximum value atresonant frequency and then again decreases.

10. Because by using a capacitor of appropriate capacitance the effect of inductive reactance can benullified or reduced and consequently power factor increases.

11. Current flowing in two cases will be different. Current will be less across 220 V, 100 Hz becausereactance X

L = L.2π n is more for frequency n = 100 Hz.

12. Equation for instantaneous voltage is

V = 220 2 sin 2 (50) t = 311 sin 314t.π

13. Initial flux through the coil,

φB(initial)

= BA cos θ= (3.0 × 10–5 ) (π10–2)cos 0° = 3π × 10–7 Wb

Final flux (after the rotation),

φB(final)

= (3.0 × 10–5) (π10–2)cos180°

= –3π × 10–7 Wb

84

The induced e.m.f. is, e = Nt

∆φ∆

= 500 × (6 × 10–7)/0.25

= 3.8 × 10–3V

I = E/R = 1.9 × 10–3A

14. The area covered by an angle θ,

A =

The induced E.M.F., |E| =

=2 2r d Br

B2 dt 2

θ ω=

ν = 120rev/min = 2rev/s;

ω = 2πν = 4π rad/s

r = 0.5m;

B = 0.4G = 0.4 × 10–4 T

∴ E =4 22 2 0.4 10 (0.5)

2

−π × × × ×

= 6.28×10–5V

Each spoke will act as a parallel source of E.M.F. Hence, the E.M.F. will be 6.28 × 10–5V

15. As the bar magnet accelerates, its velocity keeps increasing, hence the rate of change of fluxincreases. Hence,

(i) (a) As magnet approaches the coil, E.M.F. increases (A to B)

(b) The magnet enters the coil and rate of change of flux becomes constant, hence, the E.M.F.reduces to zero (B to O)

(c) The magnet emerges out of the coil, hence, the direction of E.M.F. reverses and peak becomessharp as the velocity has raised further (O to C)

(d) As the magnet goes away from the coil, the E.M.F. subsequently decreases to zero (C to D).

(ii) The negative peak is longer than the positive peak, because the velocity of magnet increases with time,giving rise to faster rate of magnetic flux change.

16. The magnetic field at the centre of the outer coil due to current in the outer coil,

B2 = 0 2

2

I

2r

µ

Flux linking to the inner coil,

φ1 = π r

12B

2 = I

2 = M

12I

2

Hence, M21

= M12

= ; where r2 > r

1

85

17. (i) Mutual Inductance (M) decreases because the quantity of flux linking to a coil due to the othercoil will decrease.

(ii) M increases because as the no. of turns increase, the overall flux density also increases andhence the mutual inductance will also increase.

(iii) M increases because iron is ferromagnetic in nature hence, it will increase the flux density.

18. (a) ∵ Vm

= 300 V,

hence, Vrms

=

mV 300

2 2=

= 212.1V

(b) As Irms

= 10 A,

hence Im = rms2 I⋅

=

2

× 10 = 14.14 A

14.1 A.

19. Here L = 2.0 H,

C = 32 µF = 32 × 10–6F

and R = 10Ω.

∴ Resonant frequency

ωr =

6

1 1

LC 2.0 32 10−=

× ×

= 125 s–1.

and Q value =

rL 2.0 125

R 10

ω ×= = 25.

20. Here qm = 6 mC = 6 × 10–3 C.

∴ Total energy stored in the circuit initially = Total energy stored at any later time

= U = 2m1 q

2 C

=

3 2

6

(6 10 )

2 30 10

−

−

×× ×

= 0.6J.

21. When frequency of the supply = natural frequency of circuit

0

1,

2 LCν =

π

XL = X

C and resonance

takes place so that

current rmsrms

V 200I

R 20= = = 10 A

As in resonance condition voltage and current are in same phase condition, hence average

Power = Vrms

Irms

= 200 × 10= 2000 W or 2kW.

86

22. Here Vp = 200 V, V

s = 20 V,

efficiency η = 80% = 80

100 = 0.8

and resistance in secondary ciruit

R = 20

∴ Is = = 1A

As efficiency η =s s

p p

V IOutput

Input V I

⋅=

⋅

⇒ IP =

= 0.125 A.23. It is possible that an a.c. source be connected to a circuit and yet deliver no power to it. This is

possible where there is a phase difference of 90° between the applied voltage and the resultingcurrent and the power factor of the a.c. ciruit is zero.

24. Here L = 0.5H, R = 100Ω, Vrms

= 240Vand ν = 50 Hz

ω = 2πν= 2 × 3.14 × 50 = 314s–1.

(a) Impedance Z = 2 2 2R L+ ω

=

= 186.1W

∴ Im = rmsm 2 VV 2 240

Z Z 186.1

×= =

= 1.82 A(b) If the current lags behind the voltage in phase by φ , then

φ =

= tan–1 (1.5700) = 57.5°

=

∴ Time lag between voltage maximum and current maximum

t =

= 3.2 × 10–3s or 3.2 ms.

87

25. (a) Power required by the townP = 800 kW

= 800 × 103 W = 8 × 105 W

Total resistance of 2 wire line of 15 km at 0.5 Ω per km

R = 2 × 15 × 0.5 = 15 ΩAs supply voltage is through 4000 – 220 V line, the transmission line transmits power at 4000 V

∴ Line current I =

5P 8 10

V 4000

×=

= 200 A

∴ Line power loss in the form of heat

= I2R

= (200)2 × 15

= 6 × 105 W = 600 kW

(b) If there is no power loss to leakage then the total power to be supplied by power plant

= Power needed by town + Power loss during transmission

= 800 kW + 600 kW = 1400 kW

(c) Voltage drop along the transmission line

= IR = 200 × 15 = 3000 V

As the transmission line supplies power to sub-station on 4000 V hence, voltage at generatingstation side of line = 4000 + 3000 = 7000 V

Hence, a step up transformer of 440 V – 7000 V be used at the generator plant.

26. Here it is given that L = 200 mH = 0.2 H, R = 10Ω and ν = 50 s–1

(i) As power factor

cosφ = R/Z = 1,

hence R = Z

or X = XL – X

C = 0

or XL = X

C

⇒ C =

2 2 2

1 1

L 4 L=

ω π ν

=

2 2

1

4 (3.14) (50) 0.2× × ×

= 5.07 × 10–5F = 50.7 µF

(ii) Q-factor of the circuit

= 0L 2 L

R R

ω πν=

=2 3.14 50 0.2

10

× × ×

= 6.28

88

Long Answer

27.

Induced e.m.f.,

e Blv for 0 x b and 0 for b x 2b= − ≤ < ≤ <

The current,

The force on arm PQ, 2 2B v

F for 0 x b and 0 for b x 2br

l= ≤ < ≤ <

The Joule's heating loss, P=I2r so 2 2 2B v

P for 0 x b and 0 for b x 2br

l= ≤ < ≤ <

28. Here L = 20 mH = 20 × 10–3 H,C = 50 µF = 50 × 10–6 F

and qm = q

0 = 10 mC

= 10 × 10–3C = 0.01 C,when t = 0.

(a) ∴ Total energy stored initially

U =2 20

6

q1 (0.01)

2 C 2 50 10−=× ×

= 1 J

Yes, the energy remains conserved during LC oscillations provided that resistance R of the circuitis zero.(b) Natural frequency of the circuit

ν0 =

= 3 6

1

2 3.14 20 10 50 10− −× × × × ×= 159 Hz

and natural angular frequency

ω0 = = 2π × 159 = 1000 rad s–1

(c) Let time period of oscillations be T,

where T =

= 6.3 × 10–3s = 6.3 ms

Then, (i) at times t = 0, T

,2

T, 2T .......... the energy stored is completely electrical and (ii)

at times t = .......... the energy stored is completely magnetic.

89

(d) Let at time t the electrial and magnetic energies are equal i.e.,2

2C

q =

22 01 1

LI2 2 2C

=

q

i.e., at time t, q =

0

2± q

∵ q =

0q cos tω

∵

0

2± q

=

0q cos tω

or cos ωt =

1

2±

or ωt = ,4

π

3,

4

π

....

Substituting ω =

2,

T

π

we have

t =

T 3T 5T 7T, , , ..........

8 8 8 8

(e) If a resistor is inserted in the circuit, due to energy loss amplitude of LC oscillations graduallydecreases with time and finally the oscillations stop altogether.

90

5Weightage-03 marks

1. Conduction current & Displacement current—The current carried by conductors due to flowof charges is called conduction current; the current produced due to changing electric field is.called displacement current.

2. Modified Ampere circuital law—It states that the line integral of the magnetic field B over aclosed path is equal to µ

0 times the sum of conduction current and the displacement current threading

the closed path.

B.dl = ,

where IC – Conduction current

ID – Displacement current.

3. Maxwell’s equations— Gauss’s law of electrostatics

=

Gauss’s law of magnetism

= 0

Faraday’s law of electromagnetic induction

=

Ampere-Maxwell’s law

=

4. Electromagnetic waves—Radiation consisting of self-sustainingoscillating electric and magnetic fieldsat right angles to each other and tothe direction of propagation is calledelectromagnetic wave.

91

5. Important facts about electromagnetic waves

The accelerated electric charges radiate electromagnetic waves.

A linearly polarized electromagnetic wave, propagating in the z-direction with the oscillatingelectric field E along the x-direction and the oscillating magnetic field B along the y-direction.Both electric and magnetic field vary in phase with each other.E

x = E

0 sin (kz – ωt), B

y = B

0 sin (kz – ωt),

k =

2,

πλ

ω is the angular frequency, k is the magnitude of the wave vector (or propagation

vector) and its direction describes the direction of propagation of the wave.The speed of propagation of the wave is (co/k). The velocity of electromagnetic waves depends on electric& magnetic properties of the medium,

and is given by

( )12

1v =

εµ

, where ε and µ are the permittivity and permeability of the dielectric

medium respectively. It does not depend upon the amplitude of the field vectors.

They cannot propagate within a conductor; they are totally reflected when they strike a conductingsurface.

The wave is a transverse wave, since the fields are perpendicular to the direction in whichthe wave travels.

All electromagnetic waves, regardless of their frequency, travel through vacuum at the samespeed, the speed of light.

They travel with a constant velocity of 3 × 108 m s–1 in vacuum. The magnitude of the electric and the magnetic fields in an electromagnetic wave are related as

B0

= (E0/c)

They are not deflected by electric or magnetic field. They can show interference or diffraction. They may be polarized. They need no medium of propagation. The energy from the sun is received by the earth through

electromagnetic waves. A single unit, or quantum, of electromagnetic radiation is called a photon. The wavelength (λ) and the frequency (v) of electromagnetic wave is related as

c = vλ = ω/k, ω = ck, v = ω/2π, λ = 2π/kThe SI unit of frequency is hertz wiht unit symbol HzThe SI unit of wavelength is metre with unit symbol m.

They carry energy with them which is equally divided between electric energy and magneticenergy.

Only electric field vector is responsible for the optical effects of electromagnetic waves soit is named as optical or light vector.

They transport energy and momentum and can exert pressure on the surface on which it falls. Orderly distribution of electromagnetic radiations according to their wavelength or frequency

is called electromagnetic spectrum.

92

93

Answer Yourself

Very Short Questions

Q1. What is displacement current? Give its SI unit.

Q2. What is the main inconsistency in Ampere’s Circuital Law?

Q3. What modification was made to Ampere’s Circuital Law by Maxwell?

Q4. What led Maxwell to predict the existence of electromagnetic waves?

Q5. Distinguish between conduction current and displacement current.

Q6. Write down Maxwell’s equations for steady electric field.

Q7. Write down Maxwell’s equations for steady magnetic field.

Q8. What are electromagnetic waves?

Q9. Name the basic source of em waves.

Q10. State the principle of production of em waves.

Q11. Write two characteristics of electromagnetic waves.

Q12. What is the dimension of E/B?

Q13. Write an expression for speed of electromagnetic waves in free space.

Q14. Write the frequency limit of visible range of electromagnetic spectrum in kHz.

Q15. Name the electromagnetic radiations which have the largest penetrating power.

Q16. Name the electromagnetic radiations used for viewing objects through haze & fog.

Q17. Why does “RADAR” use microwaves for air craft navigation?

Q18. What is wave length range of X-rays?

Q19. Which layer of the earth’s atmosphere is useful in long distance radio transmission?

Q20. What is the cut-off frequency beyond which the ionosphere does not reflect electromagneticradiations?

Short Questions

Q1. Write Maxwell’s equations of electromagnetism and state the law underlying the equations.

Q2. Name the main parts of the electromagnetic spectrum and mention their frequency range andsource of production. Also write their important properties and uses.

Q3. Why is the ozone layer of atmosphere crucial to the existence of life on earth?

Q4. In a microwave oven, the food kept in a closed plastic container gets cooked without meltingor burning the plastic container. Explain how?

Q5. Can an electromagnetic wave be deflected by an electric or magnetic field? Justify your answer.

Q6. Name the constituent radiation of electromagnetic spectrum, which

94

(a) is used in satellite communication.

(b) is used for studying crystal structure.

(c) is emitted during decay of radioactive nuclei.

(d) is absorbed by ozone layer.

Q7. Give atleast three applications of (i) radio wave (ii) infrared radiations (iii) microwaves (iv) ultravioletlight (v) x-rays (vi) gamma rays.

Q8. Arrange the following radiations in the descending order of frequency: gamma rays, infraredradiations, microwaves, yellow light, ultraviolet radiations.

Q9. Find the wavelength of electromagnetic waves of frequency 4 × l09Hz in free space. Give itstwo applications.

Q10. Differentiate between radiowaves and γ-rays.

Very Short Questions

Q1. What oscillates in an Electromagnetic Wave?

Q2. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates?

Q3. Write an expression for the speed of em waves in free space.

Q4. For an electromagnetic wave, write the relationship between amplitudes of electric and magneticfields in free space.

Q5. What is the ratio of speed of infrared and ultraviolet radiations in vacuum?

Short Questions

Q6. What led Maxwell to predict the existence of electromagnetic waves.

Q7. A variable frequency AC source is connected to a capacitor. Will the displacement current increaseor decrease with increase in frequency?

Q8. Why are sound waves not considered em waves?

Q9. What is the phase relationship between the oscillating electric and magnetic fields in anelectromagnetic wave?

Q10. The sunlight reaching the earth has a maximum electric field of 810 V/m. What is the maximummagnetic field associated with it?

Q11. Calculate the relative permittivity of a medium of relative permeability 1.0 if the velocity oflight through the medium is 2 × 108 m/s.

Q12. Draw the diagram of electromagnetic wave. The charging current for a capacitor is 0.25 A. Whatis the displacement current across its plate?

To be Learnt

95

Q13. In a plane electromagnetic wave the electric field oscillates at a frequency 3 × 1010 Hz and amplitude50 V/m. What is (a) the wavelength of the wave and What is (b) the amplitude of the oscillatingmagnetic field? (Ans: 1cm, 1.67 × 10–7 T)

Q14. Why is the electromagnetic wave of transverse nature? A plane electromagnetic wave travels invacuum, along y-direction. Write the (i) ratio of the magnitudes and (ii) the direction of its electricand magnetic field vectors.

Q15. Which of the following , if any, can act as a source of electromagnetic waves? (i) a charge movingwith a constant velocity (ii) a charge moving in a circular orbit (iii) a charge at rest. Give reason.

Q16. Identify the part of electromagnetic spectrum, to which of frequency (i) 1020 Hz (ii) 109 Hz belong.

Q17. Find the wavelength of electromagnetic waves frequency 4 × l09 Hz in free space. Give its twoapplications.

Q18. Suppose that the electric field amplitude of an electromagnetic wave E0 =120 NC–1 and that its

frequency is v=50.0 MHz (a) determine B0, ω, k and λ (b) find expression for E and B.

Q19. Draw a labelled diagram of hertz’s apparatus and explain the principle of the experiment to produceelectromagnetic waves.

Q20. Name the electromagnetic radiation to which the following wavelength belong (a)10–2 m (b) l A0

1. Electric & Magnetic fields oscillate in mutually perpendicular directions.2. Displacement cutrrent = charging current = 0.25 A

3. c =

0 0

1

µ ε

4.

0

0

Ec

B=

5. One, as speed of em waves in vacuum is independent of wavelenght or frequency.6. Electric current produces a magnetic field around it. Maxwell showed that for logical consistency,

a changing electric field must also produce a magnetic field. This effect explains existance ofem waves.

7. ω increases means XC decreases, I

D increases.

8. They are mechanical waves hence require material medium for their propagation, can not beconsidered as em waves.

9. They are in same phase.

10. Sun light is em wave, use

0

0

EC

B=

Pedagogical Remark