Chief Advisor
Rashmi Krishnan, IAS
Director, SCERT
Guidance
Dr. Pratibha Sharma,
Joint Director, SCERT
Academic Co-ordinator and Editor
Dr.Rajesh Kumar, Principal, DIET Daryaganj
Sapna Yadav, Sr. Lecturer, SCERT
Contributors
Prof. B.K Sharma (Retd.) Professor, NCERT
Dr. R.P Sharma, Academic Consultant, CBSE
Pundrikaksh Kaundinya, Vice Principal, RPVV Kishangaj
Sher Singh, Principal, Navyug School, Lodhi Road
Davendra Kumar, Lecturer, RPVV, Yamuna Vihar
Girija Shankar, Lecturer, RPVV, Surajmal Vihar
R.Rangarajan , Lecturer, DTEA , Sr.Secondary School , Lodhi Road
Neelam Batra , Lecturer, D.C. Arya, Sr. Sec. School, Lodhi Colony
Chitra Goel, Retd Vice Principal
Dr.Rajesh Kumar, Principal, DIET Daryaganj
Sapna Yadav , Sr. Lecturer , SCERT
Publication Officer
Mr. Mukesh Yadav
Publication Team
Sh. Navin Kumar, Ms. Radha, Sh. Jai Baghwan
Published by : State Council of Educational Research & Training, New Delhi and printed at
Educational 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 is
interesting 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, but
we feel that they will find the thought process of physics useful in any other branch they may
like 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 example
of physics is the innovation of God Practice (Higgs Boson). Higgs boson is a hypothetical elementary
practice, a boson, which is the quantum of the Higgs field. The filed and the particle provide a
testable hypothesis for the origin of mass in elementary particles. “God particle”—— that helps
to 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 questions
from workbooks and do a lot of exercises. Only then we will be able to answer various types of
questions 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 more
complicating 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 learning
content one can answer question for self lest. Some exercises are given with art & pedagogical
remark 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 without
which 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 and
teachers 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
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
97-129
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 any
mishappening.
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.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Do not take the word difficult while teaching and you be positive yourself.
Do not place the topic on the board - After completing tell the students that this “what
it is”— For example topics like Potentiometer
Try to build the topic from the basics of the basic while teaching.
Do not draw the diagram on board before you start the topic – Do build the same as
the discussion continues.
Do not forget to place the arrow in circuits and Ray diagrams – A mistake which can
be easily absorbed by the student.
Do not postpone the topic for the end of the academic session
Give 2-3 revision by asking question's from such identified topics at the beginning of
the class on the subsequent days.
Try to test these identified topics in almost all the tests if possible after prior information
to the students.
Do try to create interest on such topics before it is actually discussed.
Try to adopt an interactive approach to deal with such topics.
Very important a point is to think new and good approaches that may fit your students
while dealing with such topics and popularize such method.
(vi)
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 question
and 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.., many
times before the examination, so that you do not flop during the examination because of the tilted
nature of a question.
3.
Instead of leaving the topics like E.M. Waves, Principles of Communication understand to express
all definitions, interpretation of figures, Advantages and disadvantages of various devices and Applications
etc.
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 examination
day, at least half an hour before without any books and please do not discuss any question with anyone
in this period.
6.
Start the answer script with the best known question and complete all the questions that you know
without cutting and overwriting.
7.
In case you are not having good Interpretation skill, first do the best known five mark questions and
try 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 concept
of 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 to
do later, since there will not be any time to do later. If you happen to cut, do it neatly such that
the cut and the un-cut portions are distinguishable. Thinking and formatting the answer before writing
will improve you on this front.
10. Never leave any question. Write something of what you know of the answer. Remember "What you
think is wrong may be the correct answer" many a times.
Derivations
Unit-1 (Electrostatics)
●
Electrical / magnetic field at a point on the equatorial or axial line due to an electrical /magnetic
dipole
●
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
(vii)
●
●
●
●
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 thickness
t < 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 of
cell (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) torque
on 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 of
a 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 LCR
circuit *
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 maxima
and minima.*
(viii)
●
●
●
●
●
●
●
●
●
●
●
●
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 refractive
index 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 angle
and 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 level
diagram. *
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. Sketch
the 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 rectifier
and (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. Sketch
the V × I graph for the same. *
With the help of a circuit diagram, explain the working of a pnp / npn transistor as an amplifier in
common emitter mode *
With the help of a circuit diagram, explain the working of a pnp / npn transistor as a switch in common
emitter mode
Discuss the working of a transistor as an oscillator. *
Realization of AND, OR and NOT gates
(ix)
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 obtain
expression 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 determining
resistance.
Electrical field due to a point charge, charge on parallel plates Binding energy per nucleon × mass
no. (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 satellite
communication.
(x)
PHYSICS (Code No. 042)
Senior Secondary stage of school education is a stage of transition from general education to disciplinebased focus on curriculum. The present updated syllabus keeps in view the rigour and depth of disciplinary
approach as well as the comprehension level of learners. Due care has also been taken that the syllabus
is 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 as
per international standards.
Providing logical sequencing of units of the subject matter and proper placement of concepts
with their linkage for better learning.
Reducing the curriculum load by eliminating overlapping of concepts/ content within the discipline
and 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 further
learning in the subject.
expose the learners to different processes used in Physics-related industrial and technological
applications.
develop process-skills and experimental, observational, manipulative, decision making and
investigatory 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 interface
of Physics with other disciplines.
(xi)
PHYSICS
COURSE STRUCTURE
2011-13
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 derived
units. Length, mass and time measurements; accuracy and precision of measuring instruments; errors
in 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-uniform
motion, average speed and instantaneous velocity. Uniformly accelerated motion, velocity-time, positiontime graphs.
Relation for uniformly accelerated motion (graphical treatment).
Scalar and vector quantities; Position and displacement vertors, general vectors and notation; equality
of vectors, multiplication of vectors by a real number; addition and subtraction of vectors. Relative
velocity.
(xii)
Unit vector; Resolution of a vector in a plane - rectangular components. Scalar and Vector product
of 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 second
law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum
and 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 (vehicle
on 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 of
mechanical energy (kinetic and potential energies); non-conservative forces: motion in a vertical
circle; 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 some
examples.
Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison
of linear and rotational motions; moment of inertia, radius of gyration.
Values of moments of inertia, for simple geometrical objects (no derivation). Statement of parallel
and 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 hydraulic
brakes). Effect of gravity on fluid pressure.
Viscosity, Stokes’ law, terminal velocity, Reynold’s number, streamline and turbulent flow. Critical
velocity. Bernoulli’s theorem and its applications.
(xiii)
Surface energy and surface tension, angle of contact, excess of pressure, application of surface
tension ideas to drops, bubbles and capillary rise.
Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous
expansion; specific heat capacity; Cp, Cv - calorimetry; change of state - latent heat capacity.
Heat transfer-conduction, convection and radiation, Qualitative ideas of Blackbody radiation green
house effect, thermal conductivity, Newton’s law of cooling, Wein’s displacement Law, Stefan’s
law.
Unit VIII: Thermodynamics
(Periods 12)
Thermal equilibrium and definition of temperature (zeroth law of thermodynamics). Heat, work
and 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; rms
speed of gas molecules; degrees of freedom, law of equipartition of energy (statement only) and
application 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. Simple
harmonic motion (S.H.M) and its equation; phase; oscillations of a spring–restoring force and
force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum– derivation
of 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 relation
for a progressive wave. Principle of superposition of waves, reflection of waves, standing waves
in 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).The
activities mentioned are for the purpose of demonstration by the teachers only. These are not
to be evaluated during the academic year. For evaluation in examination, students would be required
to perform two experiments - One from each Section.
(xiv)
SECTION A
Experiments
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 Callipers
and 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’s
pendulum using appropriate graph.
10. To study the relationship betwen force of limiting friction and normal reaction and to find
the 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 gravitational
pull of the earth and study its relationship with the angle of inclination (O) by plotting graph
between 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 double
inclined plane).
7. To study dissipation of energy of a simple pendulum by plotting a graph between square
of amplitude and time.
(xv)
SECTION B
Experiments
(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 temperature
by 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 terminal
velocity of a given spherical body.
6. To study the relationship between the temperature of a hot body and time by plotting a cooling
curve.
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 tension
using sonometer.
(ii) To study the relation between the length of a given wire and tension for constant frequency
using sonometer.
9. To find the speed of sound in air at room temperature using a resonance tube by tworesonance 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 XI
1.
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.
(xvi)
6.
To demonstrate that the moment of inertia of a rod changes with the change of position of a
pair 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 sources
of 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
Unit
Unit
Unit
Unit
Unit
Unit
Unit
Unit
Unit
Unit
I
II
III
IV
V
VI
VII
VIII
IX
X
Time: 3 Hours
Electrostatics
Current Electricity
Magnetic effect of current & Magnetism
Electromagnetic Induction and Alternating current
Electromagnetic Waves
Optics
Dual Nature of Matter
Atoms and Nuclei
Electronic Devices
Communication Systems
Total
70 Marks
08
07
08
08
03
14
04
06
07
05
70
Unit I: Electrostatics
(Periods 25)
Electric Charges; Conservation of charge, Coulomb’s law-force between two point charges, forces
between multiple charges; superposition principle and continuous charge distribution.
Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field
due to a dipole, torque on a dipole in uniform electric fleld.
(xvii)
Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely
long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical
shell (field inside and outside).
Electric potential, potential difference, electric potential due to a point charge, a dipole and system
of charges; equipotential surfaces, electrical potential energy of a system of two point charges
and of electric dipole in an electrostatic field.
Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and
electric 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 their
relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and nonlinear), electrical energy and power, electrical resistivity and conductivity. Carbon resistors, colour
code for carbon resistors; series and parallel combinations of resistors; temperature dependence
of resistance.
Internal resistance of a cell, potential difference and emf of a cell,combination of cells in series
and in parallel.
Kirchhoff’s laws and simple applications. Wheatstone bridge, metre bridge.
Potentiometer - principle and its applications to measure potential difference and for comparing
emf 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 parallel
current-carrying conductors-definition of ampere. Torque experienced by a current loop in uniform
magnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter and
voltmeter.
Current loop as a magnetic dipole and its magnetic dipole moment. Magnetic dipole moment
of a revolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) along
its axis and perpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniform
magnetic field; bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field
and magnetic elements. Para-, dia- and ferro - magnetic substances, with examples. Electromagnets
and factors affecting their strengths. Permanent magnets.
(xviii)
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, lensmaker ’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).
(xix)
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; IV characteristics of LED, photodiode, solar cell, and Zener diode; Zener diode as a voltage
regulator. 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, TV
and digital data); bandwidth of transmission medium. Propagation of electromagnetic waves in
the atmosphere, sky and space wave propagation. Need for modulation. Production and detection
of an amplitude-modulated wave.
Practicals
Every student will perform atleast 15 experiments (7 from section A and 8 from Section B) The
activities mentioned here should only be for the purpose of demonstration. One Project of three
marks is to be carried out by the students.
B.
Evaluation Scheme for Practical Examination:
Total Periods : 60
8+8 Marks
Two experiments one from each section
Practical record (experiments & activities)
6 Marks
Project
3 Marks
Viva on experiments & project
5 Marks
(xx)
Total 30 Marks
SECTION A
Experiments
(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 resistance
of its material
2. To determine resistance per cm of a given wire by plotting a graph of potential difference
versus 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 figure
of merit.
7 . To convert the given galvanometer (of known resistance and figure of merit) into an ammeter
and voltmeter of desired range and to verify the same.
8. To find the frequency of the a.c. mains with a sonometer.
Activities
1. 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 circuit
using multimeter.
3. To assemble a household circuit comprising three bulbs, three (on/off) switches, a fuse and
a 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 order
and correct the circuit and also the circuit diagram.
(xxi)
SECTION B
Experiments
(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 the
focal 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 between
1/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 between
angle 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 plane
mirror.
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 down
voltage.
10. To study the characteristic of a common - emitter npn or pnp transistor and to find out the
values 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 type
transistors (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 glass
slab.
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 the
lens/ mirror).
8. To obtain a lens combination with the specified focal length by using two lenses from the
given set of lenses.
(xxii)
SUGGESTED INVESTIGATORY PROJECTS
CLASS XII
1.
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 lamps
at 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 equiconvex
lens, (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 hollow
prism 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 suspended
in 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 and
to calculate its current gain.
9.
To study the factor, on which the self inductance, of a coil, depends, by observing the effect
of this coil, when put in series with a resistor/(bulb) in a circuit fed up by an a.c. source of
adjustable frequency.
10.
To construct a switch using a transistor and to draw the graph between the input and output
voltage 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.
(xxiii)
CHANGES IN PHYSICS SYLLABUS 2012-14
Following are some changes in the Syllabus of Physics at Senior Secondary Level offered
by CBSE for the session 2012-14.
Class XI
(1)
(2)
(3)
Vector is clubbed in Unit II (Kinematics) from Unit IV and Unit V(Multiplication
of Vector)
Add Activity NO 7 in Scetion A “Experiments - Activities (For the purpose of
Demonstration only)
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)
(xxiv)
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 sleep
well 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. If
you 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 something
important 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 average
day.
Visualize a successful test day experience. You already know what to expect on test day: when you'll
get each test section, how many questions there are, how much time you'll have, etc. You also know
where 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 out
about 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 you
want 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 you
to remember will happen to be on the test. Save the energy you would have used to cram for test
day.
In the few days before the test, do a review of the skills and concepts in which you are strong. Be
confident as you review everything that you know - and remember that confident feeling as you take
the test.
(xxv)
1
Electric Charge
Charge is the property associated with matter due to which it produces and experiences electrical and
magnetic effects. The charge on a body arises from an excess or deficit of electrons.
Positive charge
Negative charge
Glass Rod
Fur
Wool
Wool
Dry hair
Silk
Ebonite
Plastic
Rubber
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. For
example 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 any
physical contact.
Properties of Charge
(i) Additivity–Total electric charge of a system = algebraic sum of all the positive and negative charges
contained in that system.
(ii) Conservation–charge can neither be created nor destroyed. It means that total charge of an isolated
system always remains constant.
(iii) Quantisation–It is the property due to which all free charges are integral multiple of electronic
charge.
Symbolically, q = ± ne
where
q = total charge on a body
–e = charge on an electron
1
The cause of quantization is that only integral number of electrons can be transferred from one
body 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 q1, q2 are separated by a distance r, the magnitude of the force (F) between
them is given by
q1q 2
F = k 2
r
where 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.
Permittivity
Value of k is also represented as k = 1/4πεo where εo 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
1 q1q 2
F = 4πε . r 2
m
The force between the charges is reduced. εm is called the permittivity of the medium.
Relative permittivity or dielectric constant
It is the ratio of permittivity of the medium to permittivity of free space.
εm
εr = ε
o
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 that
charge due to the other charges, taking one at a time. The individual forces are unaffected due to the
presence of other charges.
Electric Field
It is the space around a charge (source charge) in which any other charge (test charge) experience
an electric force due to source charge.
Intensity of electric field
Intensity of the electric field at a given point is defined as the electric force on unit positive charge
placed 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.
2
The electric force on a charge q placed in electric field E is given by
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
Superposition principle
It is equal to the vector sum of the electric field intensities due to the individual charges at the
same 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 at
F=
= qE1 q rˆ
E
the same point.
4+q
πε 0 r 2
(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 electric
field.
Electric Field Line In Some Cases
(i) Electric field lines of isolated charges.
–q
Isolated (+q) charge
Isolated (–q) charge
(ii) Electric field lines of multiples charges
+q
+q
–q
+q
3
(iii) Electric field lines of uniform electric field (having same magnitude and direction at every
point ) 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 dipole
moments 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
E P = E1 + E 2 , now since
[along A to P, (+ive)]
[along P to B, (-ive)]
4
∴
=
=
=
=
p
2r
q
4ra
⋅ 2
⋅ 2
2 2 =
4πε 0 (r – a 2 ) 2
4πε 0 (r – a )
direction is along the axis of the dipole (i.e.from A to P),
where
p = 2qa
2
If a << r, a becomes negligible in comparison to r2(short elecric dipole), hence a can be neglected.
∴
EP =
i.e.
Ep ∝
Electric Field Of a Dipole At a Point On Its Equatorial 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.
E1 2sin
2
kq+ qaElectric
kq2+–θ2ra
kq––field
E1 kq
sin
cosθθθ
cos
r=
(rq22 +ata 2 point
– 2ra)P due to (+q);
(r
a)
(r
a)
E31P221q=E12⋅ 2p
=
22 2 22 2 –2 2 2 2 2 2 2
3
r4πε (AP)
(BP)
a–r a(r+)(r
a– a+ )a)
r r –(r+a)
0 (r
E1 cos θ
E1 =
(along B to P)
P
E2 cos θ
It has two components (
) as shown in figure.
r
E2
A θ
–q
) and (
E2 sin θ
θ
a
C
a
B
+q
Similarly, electric field at point P due to (–q);
E2 also has two components (
The vertical components (
) and (
) and (
(along P to A).
) as shown in figure.
), are equal and opposite so cancel each other.
∴ Resultant electric field intensity Ep at P
5
=
=
=
=
1
(q.2a)
1
= 4πε (r 2 + a 2 )3/ 2 (where k = 4πε )
0
0
=
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.
E pα
1
r3
Hence
Electric Dipole Placed in Electric field
(i) If field is uniformForce on +q, F1 = 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 dipole
In 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
6
τ = Any one of the two forces (in magnitude)
distance between force vectors.
The perpendicular
=
=
=
(vector form)
(iv) Dipole is said to be in stable equilibrium if angle between
equilibrium if angle between
and
is 180°.
and
is zero and in unstable
Electric Flux (φ
φ)
Electric flux is defined as number of electric field lines passing through the area placed normal to
the field direction.
Electric flux dφ through an area element
in an electric field
is defined as
dφ =
⇒
φ =
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/
value of charge enclosed by closed surface.
×
qE
(εpE
×=2a
θ2qa)
sinθ=
q∵ psin
pEdsE.ds
× E or
enclosed
τE.ds
ε0
2qa
× E sin θ
q enclosed
ε0
E.dsi.e.=
0
times the
φ =
Gauss’s law holds good for any closed surface of any shape or size. It does not depend upon
the 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 qenclosed.
(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 Charge
λ = linear charge density
P = Observation point
r = Normal distance of P from line charge.
From symmetry, E will be radially outward. Consider a cylindrical Gaussian surface of radius
r and length l passing through P.
Let
7
E
E
P S2
E
S1
S3
Line charge
++ + ++ + ++ ++ + ++ ++ + ++ ++
l
From Gauss Law
⇒
=
E.ds
q
E.dS1 + E.dS2 + E.dS3 = enclosed
ε0
=
or
E =
Direction of this E is radially outward.
Electric Field due to an infinite, non-conducting thin plane sheet of charge
Let 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 through
point P as shown in figure.
⇒
q enclosed
E.ds
= ε0
E.dS1 + E.dS2 + E.dS3 =
or
=
By Gauss law
=
Now, since dS1 and dS3 are taken at equal distance from the charged sheet so
E1 = E3 = E(Let)
8
∴
2ES =
E =
This expression shows that electric field at a point very close to a metal sheet does not depends
upon the distance.
Electric field due to a thin spherical shell of charge
Consider a spherical Gaussian surface of radius r passing through observation point P.
Gauss law
=
S′
E
Charged
shell
P
r
S
r o R
P
q
(i) Field outside the cell
Gaussian surface S is of radius (r >>R)
S1E.dsq
σq24εεqσεπε
ε r
enclosed
∴
E.4πr 2 .cos 0 =
or
E =
2
00 0 00
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 =
1 q
4πε 0 R 2
(iii) Inside the shell
Charge enclose is zero hence E = 0
Variation between E and r:
E
Emax
Eα
0
r =R
9
1
2
r
r
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 unit
positive 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
Potential at A,
= – q0
.
VA =
[as
]
VA =
(i.e. electric potential is equal to line integral of electric field)
⇒
∵E =
VA =
1
= kq
r
r
= kq
∞
1 1
−
r ∞
VA
kq
r2
kq
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.
10
(a) Potential Difference
Potential difference between points A & B will be,
B VB – VA = − E.dl
A
VB – VA =
(b) Conservation of Electric Field
As work done by electric field depends only upon the initial and final points of path so electric field
is conservative in nature.
(c)
Variation of V and E With r
Potential due to System of Charges
By superposition principle the potential V at a point due to the total charge configuration is the
algebric sum of the potential due to the individual charges.
V = V1 + V2 + ...Vn
v
Potential
due to a dipole
2a cos
kq(r
− r1θ)
1φ
1E2a2cos
V
kq p − r r
rB rA 1 2 Let an electric dipole, consist of two equal and opposite charge separated by 2a. P is the observation
V or E
point.
distance r
Y
P
r1
r
α
+q
r2
θ
a
α
p
r2 = OP
φ
a
–q
X
r2 – r1
VP = Potential at P due to (+q) + Potential at P due to (–q)
VP = kq
If
i.e.
1 1
−
r1 r2
r >> a,
then θ ≈ φ
r2 – r1 =
and
r1 ≈ r2 = r, say i.e. r1r2 ≈ r2
11
∴
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) VP becomes maximum and is
=
kp
r2
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 field
at 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
Relation Between Field and Potential
As we know
V = − E.dr
hence
E=
is called potential gradient
Negative sign indicates that the direction of the electric field is opposite to the direction in which
potential is increasing.
dV dr
12
Electric Potential Energy
Potential energy of a point charge at a point is defined as the amount of work done in bringing the
charge from an infinite distance to thet point . S.I. unit of potential energy is joule (J).Other unit
of P.E. is eV(electron volt)
Electric potential energy of a point charge in an external electric field
Potential 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 V2 will be equal to the P.E. lost or gained
by the charge.
Thus
∆U = q(V2 – V1)
Electric potential energy of two point charges(No external field)
Here q1 and q2 are to be placed at P1 and P2. When q1 is brought from ∞ to P1, no work is needed
to be done
∴
PE of q1 = 0
When q2 is brought from ∞ to P2, the fied of q1 already exists at P2, against which work has to be
done.
kq
kqq
qfrom
q ∞
q012V
+×1 + q1122V2 + If1 2q is unit charge, this work would be = potential at P .
2
rr1212
4πε 0 r122
P.E. of q2 = q2 × VP2 (potential of q1)
q1
P1
q2
P2
r12
=
=
∴
kq1q 2
r12
U=
U=
kq1q 2
r12
Potential energy of a system of two charge in an external field
Let point P1 and P2 are in external electric field and their potentias are V1 and V2 respectively. Potential
energy of a system of two charges q1 and q2 located at r1 and r2 respectively
=
where r12 is the distance between q1 and q2.
13
P.E. of an dipole in an external field
Work done in rotating the dipole through a small angle
dw =
= pE sinθ dθ
work done in rotating the dipole from θ 1 to 2 is given by
=
=
or
w = − pE [cos θ 2 − cos θ1 ]
This 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 conduction
of 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 polarisation
When a dielectric is placed in a uniform electric field then the centres of positive and negative
charges 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 + σp and –σp. 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.
14
Dielectric strength
The maximum value of the electric field at which the dielectric withstand without break down, is
called the dielectric strength of the material.
Electrostatic Capacitance
Capacitance is the capacity to store electric charges by any conductor and such conductor are
known as capacitor.
If we give some charge Q to an isolated conductor, its potential increases to V.
then
Qα V
or,
Q = CV
where 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 capacity
of 1farad when a charge of 1 coulomb increases its potential by 1 volt.
Capacitance of an isolated conducting sphere
If a conducting sphere of radius r is given a charge Q, then the potential on the surface of the sphere is
V=
Q0 r
4Qπε
= 4πε 0 r
4Vπε 0 r
∴
=
∴
C=
since
Q/V = C
Estimation 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 farad
1 milifarad = 1mF = 10–3 farad
1 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.
Principle
The principle of parallel plate capacitor is based on the fact that potential of a insulated conductor
is decreased when an uncharged earthed conductor is kept close to it.In this case charge of insulated
conductor remains same and more charges can be added to it.This is the way to increase the
capacitance of an insulated conductor.
Capacitance of a parallel plate capacitor
Let
σ
Q
A
E
ε0
=
=
=
=
=
surface charge density = Q/A
Total charge on one face of either plate
area of one face of either plate
electric field between the plates
permittivity of vacuum
15
now,
d = seperation between the plates
V = potential difference between the two plates
V = Ed
V=
V=
⇒
as
Q
A
d
ε0
as σ =
Q
ε0
ε 0 AV
d
C = Q/V hence
Q=
C=
Thus capacitance of a parallel plate capacitor depends entirely on its geometrical dimensions and
the dielectric used.
Capacitance With Dielectric Between Plates
If K is the dielectric constant of medium between the plates of capacitor.
C=
Capacitance of a Parallel Plate Capacitor With a Conducting Slab
Let
As
since
t = thickness of metal slab
V = E (d – t) +0(t)
E = 0 inside the slab.
=
Q
= Aε (d − t )
0
or,
Q
=
V
∴
C=
∵σ =
Q
A
Capacitance of a Parallel Plate capacitor with a dielectric slab
= outside field
+
= net field inside the dielectric slab
t = thickness of dielectric slab
εr = relative permittivity of the dielectric
+
+
+Q +
+
+
+
+
Dielectric
slab
Em
dt
E0
Now,
V = E0(d – t) + Emt
–
16
–
–
–
–Q
–
–
–
–
= E 0 (d − t) +
E0
t
εr
= E0 d − t 1 −
∵ Em =
E0
εr
1
εr
V=
i.e.
C=
This analysis shows that on introducing conducting or dielectric slab between the plates of
capacitors,its capacitance increases.
Combinations of Capacitors
(i) Series combination
(ii) Parallel combination
Equivalent capacity in series
In series combination,
(i) Charge stored on each capacitor is same. (ii)Potential difference across each capacitor is proportional
to its capacitance.
Q
–Q1Q
11Q
ε 1A+QCQ113V+ ...
C
1
1 +1C02 +1
1
as+ ...∵ C = Q
V = V1 + V2 + V3 + ...
+ d+−+t 1+−+ ...
Q
C
C
C
V
1
1
2
3
c
v
c
v
c
v
C
C
C
ε3r 3 3
Q Q Q
dε 0−At11 11 −+Q222 2 –Q
εr V 2
= C + C + C + ...
1
+Q3
V
–Q3
–
+
2
3
=
...(i)
=
...(ii)
From (i) & (ii)
V
Equivalent capacity in parallel
In a parallel combination,
(i) Potential difference across each capacitor is same.
(ii) Charge stored in each capacitor is proportional to its capacitance.
as
Q = Q1 + Q2 + Q3+...
Let
C = capacity of the combination
Q = CV
or
CV = C1V + C2V + C3V + ...
or
C=
17
Energy Stored in a Capacitor
Work done in charging a capacitor is stored in form of potential energy of capacitor.This energy
is 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 doing
it is given by
dW = V. dQ
∴
U=
=
The energy (U) stored in the capacitor can be written in any one of the following forms:
1 Q2 1
1
= CV 2 = QV
2 C 2
2
Total energy stored in series or parallel combination of capacitors is equal to sum of the energies
stored in individual capacitors.
U=
Energy density (u)
1
CV 2
Total energy (U)
=2
u=
Volume of capacitor
Ad
=
u=
1 ε0A
2 d
( Ed )2
1
Ad
1
ε0E2
2
Common Potential
When two charged capacitors are connected by conducting wire then charge flow from higher to
lower potential. This flow continues till their potentials become equals and this is called common
potential.
Common Potential =
In this process of charge flows and energy loss takes place in form of heat produced in connecting
wire.
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 of
sharp point).
Corona discharge: When a conductor carries a charge then leakage of charge takes place from its
pointed ends.
The process of spraying charge is called corona discharge (Action of sharp point).
18
Construction: Motor drives the pulley P1 and P1 drives pulley P2 through an insulating belt. P2 is
inside an air evacuated spherical metallic shell.
There are two metallic brushes placed near the insulating belt. The lower brush is connected to high
voltage 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 action
of sharp points, it sprays positive charge on the belt.
+++++++
++
++++++++
–– +
–– +
+
+
+
+
+
+
+
+
+
+
++
++
+
++
+++
Pulley, P2
++
+++ +++++
Metal brush
Insulating belt
Ion source
Insulating column
Motor driven
pulley, P1
Metal
brush
As the belt moves, and reaches the sphere, a negative charge is induced on the sharp ends of the
upper collecting metal brush and an equal positive charge is induced on the farther end of that
brush. 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.
Q2.
Q3.
Q4.
Draw schematically an equipotential surface of a uniform electrostatic field along x axis.
Sketch field lines due to (i) Two equal positive charges near each other (ii) dipole.
Name the physical quantity whose SI unit is volt/meter. Is it a scalar or vector quantity?
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?
19
Q7. Sketch the electric field vs distance( from the centre) graph for (i) a long charged rod with linear
charge 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 when
placed in a uniform electric field.
Q9. A charge Q is distributed over a metal sphere of radius R.What is the electric field and electric
potential at the centre?
Q10. The relative permittivity of mica is 6. What is its absolute permittivity?
Q11. If q1q2 > 0, and if q1q2 < 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 placed
same distance apart in a medium of dielectric constant K, the force between them is 2N. What
is 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 µC
are 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 from
BtoC 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?
20
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 much
work 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 point
P 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 between
the plates of capacitor reduces on introducing a dielectric slab. Define dielectric constant on the
basis of these fields.
Q14. Using Gauss’s theorem in electrostatics, deduce an expression for electric field intensity due to
a charged electric shell at a point in (i) inside (ii) on its surface (iii)outside it. Graphically show
the 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 capacitance
for 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 q2= 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 charges
so 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 labeled
diagram.
Q2. Derive an expression for the strength of electric field intensity at a point on the axis of a uniformly
charged 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 making
an angle è with the dipole.
Q4. Suppose that three points are set at equal distance r = 90cm from the centre of a dipole, point
Aand B are on either side of the dipole on the axis (A closer to +ve charge and B closer to
B) point C which is on the perpendicular bisector through the line joining the charges. What
would 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 thickness
t (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. Hence
define electric dipole moment.
21
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 density
of charge 5x10-8C/m2 with the separation between plates being 4mm.
To be Learnt
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 done
in 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 µC
and 75 µC are placed at B, C, and D respectively. What is the force experienced by a 1 µC
positive charge placed at A.
(9√2×103N)
Q7. In the following fig. calculate the potential difference across capacitor C2 Given potential at A
is 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)
22
Q10. What should be the charge on a sphere of radius 4cm, so that when it is brought in contact
with 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?
Pedagogical Remark
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Direction of electric field is along decreasing potential.
Formulae of dipole moment
E = -dV/dr
Same as 3
Refer NCERT example of potential
Use coulomb’s law and principle of vector addition.
Charge remains same in all capacitors in series combination. Q = CV
E/V = 1/r
Apply capacitor combination principle
No charge transfer if potential is same
Potential = net charge /net capacitor
23
2
●
The flow of charge is known as current electricity. The rate of flow of charge is called electric
current.
q
I =
t
∵
q = ne where n is an integer
e = 1.6 × 10–19C.
and
The SI unit of electric current is ampere (A).
●
In metals or conductors free electrons move randomely in all possible directions and collide with
atoms of the matter. They move in straight line between two successive collisions. When electric
field is applied across a conductor the electrons get accelerated which may not be in the direction
of velocity of electron in absence of electric field. Hence the electrons do not follow straight line
path in presence of electric field.
(a)
(b)
Fig. 2.1
●
The average velocity of all free electrons in a conductor in presence
of electric field is called drift velocity. The velocity of free electron
just after the collision becomes zero and just before the collision
remains maximum i.e., v = aτ where τ is the time between two
sucessive 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 of
corss-section A, then the total charge of free electrons,
q = neAl
And on applying potential difference V across it, the electric current,
I =
24
or
where
I = neAvd
vd = drift velocity
If u1, u2...un be the initial and v1, v2 ... vn that the final velocities of free electrons than
v =
=
=
eE τ
m
=
●
τ decreases with increases in temperature, hence drift velocity decreases with increase in temperature.
∵
I = neAvd
=
=
V
ueV
neAeV
+2ττu2 eA
τn ...u + v1 + v2 + ...vn
l+
ne
1m
. EV nτ or
2n
ne
Iml
τ A
mml
lm n
0+
n
∵
=
V
= R (resistance)
I
∴
R =
Also
R = ρ
Where
ρ =
l
A
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 conductor
is constant if all external condition like temperature, etc., are remain unchanged.
= R
The conductors who obey the Ohm's law are called Ohmic and those do not Obey are called nonohmic conductor.
25
●
The I – V graph of ohmic conductor is straight line. Fig. 2.3
●
When resistances are connected in series the equivalent resistance,
Fig. 2.3
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 of
their 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 the
combination.
Fig. 2.6
26
The p.d. aross each resistor is same and current distibutes in
inverse ratio of the resistances
V1 = V2
I1
I2 =
and
Fig. 2.7
●
When temperature of Ohmic conductor increases, the mean free path and relaxation time decreases.
Therefore, the resistivity ρ =
increases.
The change in resistivity is directly proportional to the original resistivity and change in temperature
(∆θ)
∆ ρ α ρ0∆θ
∆ρ = αρ0∆θ
or
where α is called the temperature co-efficient of resistivity. α is positive for metals and negative
for semiconducture. The new resistivity,
ρ = ρ0 + ∆ρ
= ρ0 + αρ0∆θ
= ρ0 (1 + α∆θ)
Rm2
R 12 τ
ne
R = R0 (1 + α∆θ)
correspondingly
●
The carbon resistors are coded by coloured rings. For resistances three coloured rings are used.
The coloures are coded as
Fig. 2.8
Black
Bl
Brown
Br
Red
R
Orange
O
Yellow
Y
Green
G
Blue
Violet
Gray
G
White
W
0
1
2
3
4
5
6
7
8
9
Table 2.1
●
The code of first second and third coloured bands give the
first digit, second digit and number of zeroes followed by
second digit of the resistance value.
●
Electric cell is the simplest source of electrical energy. The
cell consists of electrolyte and electrodes.
The resistance offered by cell (electrolyte and electrods) is
called internal resistance of the cell (r).
27
Fig. 2.9
●
The internal resistance depends upon the nature of material
of electrodes and electrolyte.
●
Concentration of electrolyte. It increases with increase in
concentration.
●
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 cell
is called electro motive force (emf).
●
Current drawn from the cell
I =
I =
R = external resistance and r is internal resistance
where
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 the
cell.
●
If n identical cells each of emf ε and internal resistance r are connected in series with external
resistance 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.
28
●
When the cell is short-circuited, the external resistance becomes zero.
∴
●
I=
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
ε2R
P= IR =
(R + r) 2
If R = 0 or ∞ the P will be minimum.
For maximum value of P
2
⇒
For
●
●
Ed2 E ε=2 RE
(R
r) + r)
r2
r + (R
dR
●
dP
= 0
dR
R= r
i.e.,
R = r, Pmax =
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 or
junction is zero
ΣI = 0
current coming towards the point is taken as
= 0positive and going away from the point is taken
as negative.
Fig. 2.10
For a Fig. 2.11,
I1 + I2 – I3 – I4 – I5 = 0
or
I1 + I2 = I3 + I4 + I5
i.e., current coming = current going.
Fig. 2.11
●
Loop rule : The law states
of a loop is zero.
ΣV
or
ΣV
For closed loop ABCDA,
E1 – E2
For closed loop DCEFD
E2 – E3
Also ABFEA
E1 – E2
that the algebric sum of the potential difference in complete traversal
= 0
= ΣIR
= I1R1 – I2R2
...(i)
= I2R2 – I3R3
...(ii)
= I1R1 – I2R2
...(iii)
Fig. 2.12
29
●
●
Wheat stone Bridge. In 1843 charles wheatstone devised that an
arrangment of four resistances which can be used to measure one
of them in terms of the rest. The bridge is said to be balanced
when deflection in galvanometer connected across the diagonal
of 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 specific
resistance. The circuit used for this purpose is shown in fig. 2.14.
For no deflection in galvanometer,
Fig. 2.13
=
or
= R/S
or the unknown resistance,
S=
Fig. 2.14
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 most
important electrical instrument, consists basically of a
resistance R through which a steady current flows as
shown in fig. 2.15.
When emf Es is connected between A and B and the
tapping is adjusted will no current passes in the
galvanometer G, by 'loop rule'
Es = IRs
...(i)
Now replacing the source Es by unknown potential
30
Fig. 2.15
difference V (to be measured), the balance point is achieved by shitting the tapping for value RX
so that
V = IRX and hence from eqn (i)
[as R α L)
V=
Es
= Potential gradient. ...(iii)
Ls
Though the driving emf E and resistance r and R do not need to be known, they must remain same
during the whole experiment from the time the first balnace is achieved (otherwise I will not
remain same and so theory will not apply)
Usually the first balance (called standardisation) is achieved by fixing the tapping point B at a
given length Ls and adjusting I by means of r so that K= Es/Ls 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 a
cell of potential difference most accurately.
or
●
●
●
●
V = KLX with K =
Answer Yourself
Rx
L
E s = xVery
E s Short Questions
Rs
Ls
Q1. Why is a voltmeter always connected in parallel with a circuit element across which voltage is
to 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 length
is reduced to one-fourth of its original length?
Q4. V-I graph for a metallic wire at two different temperatures T1, and T2,
is shown in the figure. Which of the two temperatures is the higher
and why?
Fig. 2.16
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).
31
Q9. Draw the current versus potential difference characteristics of a cell. How can the internal resistance
of 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, orange
and 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 control
the 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 wire
at 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 resistance
in the electric circuit given in the fig. 2.17.
Fig. 2.17
Q4. A cylinderical metallic wire is stretched to increase its length by 5%. Calculate the percentage
change 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 starting
end as well as at the other extreme end of the potentiometer. However, the value of this deflection
is more at the other extreme end than at the starting end. What could be the reason for this? How
can it be corrected?
Q7. V-l graphs for parallel and series combination of two metallic resistors are as
shown in the figure. Which graph represents the parallel combination? Justify
you answer.
Fig. 2.18
Q8. Calculate the equivalent resistance of the resistance network between the point A and B as shown
in the fig. 2.19, when-switch S is closed.
Fig. 2.19
32
Q7. A volmcter V of resistance 400 Ω is used to measured the potential
difference 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.
Fig. 2.20
Q8. A potentiometer wire has a length of 10 m and resistance 10 Ω. An accumulator of emf 2V and
a resistance box are connected in series with it. Calcuate the resistance to be introduced in the box
so 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 measure
emf of a ceil? The potentiometer wire AB shown in the figure
is 400 cm long. Where should the free end of the galvanometer
be connected on AB, so that the galvanometer shows zero
deflection ?
Fig. 2.21
Q10. The plot of variationof potential difference across a combination of
three identical cells in series, versus current is as shown. What is
the emf of each cell?
Fig. 2.22
Q11. The figure shows experimental set up of a metre bridge when the
two unknown resistances X and Y are inserted, the null point D
is obtained 40 cm from the end A, When a resistance of 10Ω is
connected in series with X, the null point D is obtained 10 cm.
Find the position of the null point whne the 10 Ω resistance is
inserted in series with resistance Y. Determine the value of the
resistance X and Y.
Fig. 2.23
Long Questions
Q1. Are the paths of electrons straights lines between successive collisions (with positive ions of the
meta!) in the (i) absence of electric field? (ii) presence of electric field? Establish a relation
between drift velocity 'vd' 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 between
current density and drift velocity.
Q2. Write the methametical relation for the resistivity of a material in terms of relaxation time, number
density, mass and charge of charge carriers in it. Explain, using this relation, why the resistivity
of 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 it
realised in actual practice in the laboratory?
33
Q4. Obtain an expression for the potential gradient 'and' of potentiometer whose wire of length l has
a 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 of
electrons.
To be Learnt
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 combination
represented by graph A and B respectivley.
B
V
R2
R1
A
O
I
Fig. 2.24
Q3. If m rows of n identical cells each of emf E and internal resistance r are connected to the external
resistance R, then write an expression for maximum current.
Q4. In a wheatstone bridge the deflection in galvanometer is zero. Write the
relation between V1, V2, V3 and V4 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.
Fig. 2.25
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 RA and RB respectively. Which one is greater?
Fig. 2.26.
(A)
(B)
Fig. 2.26
Q10. The filament of incandasent buils is ohmic or non-ohmic.
34
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.
Pedagogical Remark
mean free path
r.m.s. relocity of free electrons
Pedagogical Remark. In absence of electric fielf, the free electrons move rondemly in all pressly
directions. Thus, the average velocity of the free electrons is zero. To avoid the directional effect
to find effective velocity the root mean square velocity is taken.
A represents parallel combination of R1 and R2. B represents series combination of R1 and R2.
Pedagogical Remark. In series combination the equivalent resistance is greater than the individual
resistance 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.
Maximum current,
1.
Relaxation time =
2.
B
R
V13mnE
AD ∆V
kAB
=
V
R224BC
DC
Rr l
3.
V1
V2
G
A
Imax =
Pedagogical Remark. For maximum current in the mix grouping of cells the internal resistance
must be equal to external resistance.
C
4.
V3
V4
=
Pedagogical Remark. In balance wheatstone
bridge the ratio of the resistances is given by
V
=
this is also applicable for potential difference and power dissipated.
5.
6.
Fig. 2.27
From B to D.
Pedagogical Remark. In unbalnaced wheatstone bridge, the current throught galvanometer is not
zero. The current flow from higher potential to lower potential.
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.
35
.
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 > RA
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
.τ
m
=
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.
36
13.
Total resistance of the circutit
R= 6Ω
∴ current drawn from the battery
I=
V 16
=
= 2.0A
R 6
Fig. 2.29
Pedagogical Remark. The current flows due to difference of potential, hence there is no current
in 10Ω resistance and there is no contribution of these resistances in the circuit.
14.
∵
R= ρ
and
m=
∴
R=
l
A
R
Thus,
R α l2
Pedagogical Remark. To study single relationship between
physical quantities, the variable like area, etc. are to be
eliminated.
Π
ρlr22dld
m
6Ω
6Ω
10Ω
10Ω
6Ω
12 V
6Ω
37
l
Fig. 2.30
3
Oersted's Experiment
Orested's observed that when a magnetic compass needle is placed close to a current carrying wire then
it 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
=
µ0 = permeability of free space
= 4π × 10–7
where
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.
direction of current
According to this rule if we hold current carrying conductor
in our right hand such that the thumb represent the direction of
current, the curvature of the fingers around the conductor then of
represents the direction of magnetic field lines.
direction of current
magnetic lines
of force
Magnetic Field due to a Current Carrying Circular Loop at its Centre
Let a be the = radius of loop
I = Current
38
Direction of magnetic field at point O is normal to the plane of the paper and inward.
dB =
µ 0 Idl sin 90
(Biot - sa var t law)
4π
a2
B=
=
=
(Normal inward)
If there are N turn in the coil then
B=
Magnetic Field due to Current Carrying Circular Loop at any point on its Axis
Consider a circular coil of radius a with centre o.
Here
θ,
the
angle
between
&
is 90°.
dl
θ
µdBsin
NI
I Idl sin 90
µ
α°
dl
R
0I
dlπa) 2 P2R is the point
.0 .2 (22α
dB
dBcosα
where magnetic
field is to be calculated. Direction of I is anticlockwise as observed
π
42a
2a
4πa (r
a 2+
(r C+afrom
a )2 ) P. Magnetic field at αP due to dl is given byI
(
)
O
r
dBsinα
dBsinα
α
α
I
dB =
D
dBcosα
=
dB
µ0
Idl
. 2
4π (r + a 2 )
Direction of dB is as shown in figure.If we consider another current element dl at diametrically
opposite point D then components of dB perpendicular to axis of loop (dB cosθ) will cancel each other
and component of B along the axis (dB sinθ) will add up.Assuming all the current carrying element
along the loop, total magnetic field at P is given by∴
B=
=
=
µ0
Idl
a
. 2
.
2
2
4π (r + a ) r + a 2
µo
I
a
.2πa
= 4π (r 2 + a 2 ) 2
r + a2
39
∴
B=
µ 0 NIa 2
If there are N turns,
B=
2(r 2 + a 2 )3/ 2
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 faces
hence 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.
40
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
=
Here
=
= Bhcos0° + Bhcos90° + 0.hcos0° + Bhcos 90°
= Bh
and current enclosed = I (number of turns passing through amperian loop)
= I(nh)
∴
Bh = µ 0Inh
or
B = µ 0nI
b does not depends upon dimension of solenoid.
h
c d
d
a field
c
I The magnetic
µ
nI
IB.dl
B
vF0B.d
enclosed
B.dl
l =+ µ 0 IB.dl
+P B.dl + B.dl
enclosed
a
I
b
c
Toroid
a
r
d
b
B
0 Toroid is an endless solenoid. It can be considered as a solenoid
which is bent into a circular shape to close on it-self.
Magnetic field is along the axis of solenoid.There is no pole
formation in magnetic field configuration of toroid.
II
P is the observation point and r is the radius of toroid. Suppose
I 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,
⇒
∴
2πrB = µ0 (no. of turns in solenoid) I
2πrB = µ0 (2π r n) I
∴
B=
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
particle experiences a
are F.
moving in external magnetic field
41
.Then charge
=
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 mutually
perpendicular to each other and the fore finger indicates the direction of magnetic field, the central
finger indicates the direction of electric current (direction of velocity of positive charge) then the thumb
represents 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 charged
particle 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=
Time period,
T=
Bq
m
⇒ T=
This T does not depend on speed of particle.
(II) When V and B are in same direction
In this case, no magnetic force acts on the particle and it moves in straight line path.
42
(III) When v &
Let velocity
are not ⊥ to each other
is making an angle 0<φ<90 with
So the particle will move in a circular path because of v1 = v sin φ and at the same time it will have
a linear velocity because of
. This means it starts moving in a helical path as shown in
figure.
(i) Radius of helix
2πr
(ii) Time period to complete one revolution, T =
Bq
(iii) Path of the helix = V2 × T
(III) Since F &
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 remains
constant)
y Lorentz Forces
mvmvcos
q(v
qE
q(E
F
φ
B
B
v
1 × B)
r m2e=+=×+FvB)
qB V The force experienced by a charge particle due to electric and magnetic field is called Lorentz force.
Vsinφ
φ
Electric Lorentz Force
=
Magnetic Lorentz force
=
Total Lorentz force
=
B
z
Vcosφ
=
Cyclotron
It is a device used to accelerate charged particle to a very high energies.
Principle
It is based on the principle of lorentz force. The electric field accelerates the particle and magnetic field
makes it move in a circular orbit.
Construction
1.
2.
3.
Strong electromagnet (N & S)
Two Dees, D1 & D2. (D-shaped hollow semicircular copper chambers called Dees, placed horizontally
with a small gap in between D1 & D2 as shown.)
High frequency oscillator (f > 10 6 to 10 7 Hz)with high voltage (V > 10 4 to 10 5 V) .
43
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 about
10 mm of Hg.
–6
Working
Suppose a positive ion P is generated from ion source when D1 is at a (+) potential & D2 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.
mv
radius of path
r=
...(i)
Bq
After moving the semi circular path in time t inside D2, (p) reaches at the edge of D2 .In this time,
t the polarities of D1 & D2 are reversed and again particle get accelerated and enter in to D1.This whole
process repeats again and again and finally the particle obtain a sufficiently high kinetic energy. The
accelerated 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
2π Bq
=
ωc =
...(iv)
T
m
The cyclotron frequency is
1
Bq
=
f=
T 2πm
kinetic energy of ions is maximum when it moves along the largest circular path i.e. radius of Dees.
Limitations of cyclotron
1.
2.
3.
Speed of charged particle can be increased up to a certain limit(Not beyond the speed of light).
Very light particle like electron can not be accelerated because it go quickly out of step with electric
field.
It cannot accelerate uncharged particles.
44
Force on a Current Carrying Conductor Placed in a Magnetic Field.
l=
v=
θ=
I=
Force on the charged particle
length of conductor in the magnetic field
drift velocity of electrons
angle between v and B
current in conductor
=
=
=
=
Direction of this force is determined by Fleming's left hand rule.
Force Between two Parallel Current Carrying Straight Wires
B µ lo /I×/1t)(
q(v
F
×ItB)
B)
q(l
(q
I(
l×
B)B)
2×
I12πR
F
B
D
I1
Consider
AB and CD I2as 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
F
F
Magnetic field B produced by AB at a point on CD is given by
I2
F
R
R
A
D
A
C
C
B=
µ 0 I1
2πR
...(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=
Direction of F is towards conductor AB.
Similarly magnitude of force per unit length on AB due to CD is also same but direction is
opposite.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 can
say that if direction of current in the wires are opposite to each other then wire they repell each other.
45
Definition of Ampere
We know that
F=
If
I1 = I2 = 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 parallel
infinitely long wires placed in vacuum seperated by 1 m , produces a force between the two conductors
of 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 makes
an angle with the direction of magnetic field B.
Magnetic force on each wire is as belowForce on PQ,
Upward
Force on RS,
Downward
Force on PS,
Normal inward
Force on QR,
forces produce torque.
Normal outward Force on PQ cancel force on RS. Other two
∴ Torque, τ = Force (FPS or FQR) × perpendicular distance between these two force vectors
= I/B × b sinθ = BIA sin θ
∴
τ = MB sin θ τ = M × B (Vector form)
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 then
in above formula put (90 – φ) in place of θ.
46
Moving Coil Galvanometer
It is an instrument used for detection and measurement of small electric current.
Principle
When current is passed through coil placed in a magnetic field, it experiences a torque which rotates
the coil.
Construction
A coil is suspended about a vertical axis in a radial magnetic field produced by horse shoe magnet. (In
of the coil and the magnetic field is 90°, i.e.
radial magnetic field angle between the area vector A
torque on coil is maximum.The coil is wound on soft iron cylinderical core (which increases the
intensity of magnetic field at the place of coil). This coil is attached with a phosphor bronze wire. The
other 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.
T1
K
θ = Gθ If Phosphor
K = torsional constant
T2
bronze strip
nBA
Mirror
θ = angle of rotation
For equilibrium of coil
Magnetic Storque = Restoring torque of spring
N
Magnnet
nIAB = kθ
Soft
or
iron cylinder
Where G =
Coil
I=
K
is galvanometer constant.
nBA
Hence direction of deflection shows direction of current and magnitude of deflection shows magnitude
of current.
Sensitivity
Current sensitivity(IS)
It is defined as angle of reflection per unit current
Is =
θ nBA
=
I
K
The unit of current sensitivity is rad A–1 or division A–1.
47
It can be increased by increasing the area of the coil, the number of turns, the strength of the
magnetic field and by decreasing K.
Voltage sensitivity(VS)
It is defined as angle of reflection per unit potential difference applied across the coil
Vs =
θ
θ
nBA
=
=
[∵ V = IRg where Rg = resistance of galvanometer]
V IR g KR g
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 the
galvanometer.(Low resistance connected in parallel is called shunt)The coil of the galvanometer has a
resistance 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 difference
across 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
So
∴
R
Rg
V
Ig
V
=
=
=
=
=
R=
resistance connected in series
resistance of galvanometer coil
voltage to be measured by voltmeter.
Full scale deflection current.
Ig (R + G)
V
−G
Ig
48
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=
Where T is the time period of revolution. Let r is the orbital radius of the electron and v the orbital
speed.
Then,
T=
Thus we can say revolving electron and hence atom behave as a small magnetic dipole.Its magnetic
moment M is given by
Μ=
neh
nh
∵ mvr =
4πm
2π
where, n is a natural number, n = 1, 2, 3, ... and h = 6.6 × 10–34 Js (Planck's constant).
Μ=
n = 1,
Magnetic
dipole
moment
2eπr
evr
ev Taking
i(πr 2 ) = i
2 2πr× Magnetic length
=Tvpole strength
M m 2l
M=
eh
= µ B = Bohr Magneton
4 πm
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•
SI unit M is joule/tesla or A m2.It is a vector quantity whose direction is from S-pole to N-pole.
49
Values of pole strengths, when magnet is cut
Magnet
Pole
Strength
Effective
length after
cutting
Magnetic
moment
(m)
N
S
m
2l
m(2l) = M
N
S
m/2
2l
m
M
(2 l ) =
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 magnetic
field 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
1. Basic quantity and nature
Electrostatic
Magnetism
charge (q) (positive)
pole (strength-m) (north pole)
2. Free space constant
B=
3. Field
50
µ0 m
4π r 2
4. Force
F=
1 q1q 2
4πε 0 r 2
F=
µ 0 m1m 2
4π r 2
5. dipole moment
Axial field
equitorial field
6. Torque
7. Force
8. Potential energy
Magnetic Field of Earth
pE(cos
B
(cos
Scos
τ−
p
F
M
p.E
=01µ=02M
pM
mB
m(2
×
Ea)
qE
µM
1q(2
M
2p
pθl)θ1 1−−cos
M.B
×
B
2)
2)
G θθ
.3 3.3 S3 Initially Sir Gilbirt suggested that earth itself is a huge
−
M
π4π
4πε
πε
0r r0 r r
magnet.Earth's magnetic field is approximately uniform
SM
Magnetic
of magnitude
10-5 tesla.The source for earth's magnetism
equator
S
is perhaps due to the rotation of earth. Due to this
motion electric currents is produced by ions of metallic
Geographic
N
equator 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 SG 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
51
1. Magnetic declination (θ)
2. Magnetic dip or Inclination (δ)
3. Horizontal component of earth's total field (BH) 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 when
needle 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 measure
angle 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 = B cos δ and BV = Bsin δ
where BV is vertical component and where B =Earth's total magnetic field
BV
BH
If a plane is making an angle θ with magnetic meridian then angle of dip (δ′) in that plane is called
apparent dip. In that case
Also
and tan δ =
B=
tan δ′ =
tan δ
cos θ
Some Important Definitions
Magnetising Intensity ( ) (Magnetisation-force): It is the degree to which a magnetic field can
magnetise a material. It is represented by
=
⇒ B = µH
For solenoid
B = µnI so H = nI
Unit of H is Am-1
Intensity of magnetisation (I): It is defined as induced magnetic moment per unit volume of the
material.
=
Its unit is also Am–1 so we can write it as B = µI.
χ m): It is the ratio of intensity of magnetisation (I) to magnetisation force
Magnetic susceptibility (χ
(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
⇒
µr = 1 + χ
52
Magnetic Materials (Kind of Magnetic Material and their Properties)
1. Cause
2.
3.
4.
5.
6.
7.
8.
9.
Diamagnetism
Paramagnetism
Ferromagnetism
Its net magnetic moment
is zero due to cancellation
of moment of different
electron.
Its
own
magnetic
moment is not zero
Magnetic moment of
atom align to produce
large magnetic moment
Strongly attracted
Road align itself strongly
along the field direction
If kept near a
magnet
Behaviour of a
freely Suspended
rod in an external
magnetic field
Nature
of
Magnetic
Lines
when material is
placed
in
a
magnetic field
Behaviour
of
liquids and gaes
Feebly repelled
It possess a permanent
magnetic dipole moment
either due to unpaired
electron or due to
electron spin
Feebly attracted
rod align itself Normal to
field direction
Road align itself along
the field direction
µr (RELATIVE
PERMEABILITY)
slightly less than one
[0 ≤ µr < 1]
slightly more
53 than one
1 < µr < (1 + ε ) 
quite larger than one
[ µr >> 1]
µ
µ < µ0
µ > µ0
µ >> µ0
Less than in vacuum
Greater than in vacuum
Much larger than in
vacuum
N
S
N
N
N
S
N
S
materials
from Such materials move
Such materials move Such
to
stronger from weaker to stronger
from stronger to weaker weaker
Permanent magnetic
Magnets
And Electromagnets
magnetic field
magnetic field s
field
Magnetisation
Getflux
strongly
magnetized
Get
weekly
magneised
in
Get
weekly
magnetized
Retentivity or Residual Magnetization: It is the residual magnetic
density
in a substance when
direction
of in the direction of
in the direction opposite the
the external magnetizing field is zero.
magnetizing field.
magnetizing field.
to the direction
Coercivity
(or Coercive
is value
of the magneticPositive
field in
reversed direction that is
χ (Magnetic
Negative
& small force): ItPositive
& small
& large
required to reduce
flux density
residual
value to zero.
susceptibility)
χ < 0magnetic
1]
[ −1 ≤ the
]
[0 < χin< εa] substance from its
[χ >>
10. Flux density
within it
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, if
the 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 from
one 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's
magnetic field are equal?
54
Q2. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same
direction 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 a
current 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 right
on the geomagnetic north or south pole?
Q5. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north
tip pointing down at 22° with the horizontal. The horizontal component of the earth's magnetic
field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field
at the place.
To be Learnt
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 are
semi-circles. What will be the magnetic field at the centre C
of the circular loop?
P
I A
Q4. An electron and a proton moving with same speed enter the same magnetic field region at right
•
C
B
angles
to the direction of the field. For which of the two particles will the radius of the circular
path be smaller?
Q
Q5. A wire of length L is bent in the form of a circle of radius R and carries current I. What is its
magnetic moment?
Q6. Two wires of equal lengths are bent in the form of two loops. One of the loop is square shaped
whereas the other loop is circular. These are suspended in a uniform magnetic field and the same
current 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) less
than 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 applied
magnetic field?
Q10. An iron bar is heated to 1000 °C and then cooled in a magnetic field free space. Will it retain its
magnetism?
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 field
due to a toroid?
55
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 and
making 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
diameter
of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the
solenoid near its centre.
Q16. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is
suspended vertically and the normal to the plane of the coil makes an angle of 30° with the
direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of
torque experienced by the coil?
Q17. A galvanometer coil has a resistance of 12Ω and the meter shows full scale deflection for a current
of 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 field
at right angles to the field lines. The particles describe circular paths. What is the ratio of the radii
of these two circles.
Q19. An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. Calculate the
equivalent 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 0.15
T. 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 of
the magnetic field produced by the magnet at a distance of 10cm from the centre of the magnet
on (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 undeflected
along a straight path with constant speed. What can you say about the initial velocity of the
particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying
from point to point both in magnitude and direction, and comes out of it following a complicated
trajectory. Would its final speed equal the initial speed if it suffered no collisions with the
environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in
north to south direction. Specify the direction in which a uniform magnetic field should be set
up to prevent the electron from deflecting from its straight line path.
56
Q23. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal
to 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 in
copper is given to be about 1029 m–3)
Long Question
Q24. Write the expression for the magnetic moment
due to a planar square loop of side ‘l’ carrying
a steady current I in a vector form.
In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying
a steady current I1 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)
Q25. A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current
is uniformly distributed across the cross-section. Apply Amphere’s circuital law to calculate the
magnetic field at a poin ‘r’ in the region for (i) r < a and (ii) r > a.
(5 marks)
Pedagogical Remark
2
2
(τm=I.L
) nIABsin
θL2 I1
L
=
A s =l
4π 4
l
=
1.
2.
3.
4.
16
Electric field as well as the magnetic field.
Along +ve z-axis in accordance the Fleming's left hand rule.
Zero, because magnetic fields due to APB and AQB are equal in magnitudes but opposite in
directions.
∵ Radius of circular path r =
mv
Bq
same speed v, the mass of electron is much less than that of proton. Hence re << rp.
5.
Since R =
6.
Torque
L
, hence magnetic moment M = I.A = I.π R2
2π
.
, as length of wire is same (say L),
for square loop
and
2
L
L2
. Since, Ac > As, the circular loop will experience
=
2π
4π
greater torque than the square loop.
for circular loop A c = πr 2 = π
57
7.
8.
9.
10.
11.
12.
Slightly less, since bismuth is a diamagnetic in nature.
Surround the region by soft iron rings. Magnetic field lines will be drawn into the rings, and the
enclosed space will be free of magnetic field.
It gradually increases.
No. it will not retain its magnetism because it has been heated once beyond Curie's temperature.
So that the magnetisation is not erased by external magnetic fields, temperature fluctuations or
minor mechanical damage.
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 a
toroid and for an inifinitely long conductor carrying current net magnetic moment is zero and
hence 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. Further
L = 80 cm = 0.8 m and I = 8.0 A
∴ Magnitude of
inside the solenoid near its centre
B=
4π × 10−7 × 2000 × 8
=
0.8
= 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
58
= 20 × 0.8 × 0.01 × 12 × sin30°
= 20 × 0.8 × 0.01 × 12 ×
= 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 in
series with the galvanometer, where
R=
= 6000 – 12 = 5988 Ω.
18. We know that radius of a circular path is given by
mv
r=
Bq
Hence, for α-particles and protons of same velocity v entering a uniform magnetic field B normally:
∴
=
But
and
mα = 4mp
qα = 2qp
∴
=
=2
1q
1rV
18
rα = 2rp
4mα×α⋅−B. Gp=
− 12
−3
2
2Im
rpg p q α 3 ×19.
10 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 is
= –mBcos0°
U1 =
= –mB = –0.32 × 0.15
= –4.8 × 10–2 J.
(b) Unstable equilibrium orientation means the magnet is set anti-parallel to the external magnetic
field i.e., θ = π. In unstable equilibrium state, the potential energy of the magnet will be
U2 =
– 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
59
(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
µ 0 m 10−7 × 0.48
=
B=
4π r 3
(0.1)3
= 4.8 × 10–5 T along N-S or 0.48 G along N-S direction
22.
23.
(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 its
magnitude.
(c) magnetic field
should be in a vertically downward direction so that in accordance with
Fleming's left hand rule the deflection of electron due to B is in horizontal plane from north
to south and may nullify the deflection due to electrostatic field.
(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 vd is the drift speed of electrons
= Be.
F= B
24.
(∵ I = neAvd)
= 5.0 × 10–25 N.
=
The magnetic moment
due to a current loop carrying a steady current I is given by
A is the area of the loop.
For a planar square loop of side ‘l’, we have
where
=
∴ magnetic moment
=
In the arrangement shown in the figure, the long straight conductor MN carrying a steady current
I1 produces a magnetic field around it.
Forces acting on paris QR and SP of given loop due to the magnetic field of current carrying long
conductor MN are equal, mutually opposite and collinear. Hence they balance one another.
Now force acting on part PQ
60
µ 0 I1
µ II
⋅ I ⋅l = 0 1
2 πl
2π
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
F1 = B1 Il =
F2 = B2Il =
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=
or
(attractive)
However, since
are collinear hence these forces will not have a torque. Thus, the loop
will not experience a torque.
25. Consider a long straight wire of a circular cross-section of
radius ‘a’ carrying a steady current I. The current is uniformly
distributed across the cross-section.
(i) To calculate the magnetic field B at a point P distant r
from the axis of wire (where r < a) consider a circular
loop of radius r around the axis of wire. Then applying
2 I
2I I
circuital law to the closed loop, we find that
µ 0⋅ I2π
IrAmphere’s
Id2 l=Fµ
F
r1⋅F
1 = B
1π
dl
Bµ0Ir01Idl
r= µ 0 II1
F2=I=2=
− .2
Fµ21112ππ0B+−Iπand
⋅
I
4π
4π
=
πar 2 a 42π
aπ2
r
O
The current enclosed by the said loop is
B
dl
P
Ien =
a
P
r
O
P
Hence as per Ampere’s law :
(a)
µ Ir
r2
I ⇒ B= 0 2
2
2πa
a
(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
B2πr = µ 0 ( I en ) = µ 0
(b)
=
⇒
⇒
B × 2π r = µ0 I
B=
61
4
It is the phenomenon of generating e.m.f. in a conductor when there is a change of magnetic flux linked
with the conductor.
The e.m.f. so developed is called induced e.m.f. If the conductor is a closed circuit, a current also
flows in the circuit (called induced current).
Magnetic Flux
The magnetic flux (φ) is the measure of total number of magnetic field lines crossing through any
surface A placed in magnetic field .
, where θ is the angle between area vector A and B. S.I. unit of
It is given by
flux 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. and
hence 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 of
magnetic 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 the
reason 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 current
produced in a conductor when the magnetic flux
associated with it changes. According to this rule, if
we stretch the first finger, central finger and thumb of
our right hand in mutually perpendicular directions such
that first finger points out in the direction of magnetic
fied, thumb points out in the direction of motion (or
force) of the conductor, then the central finger gives
the direction of induced current.
62
Force
Field
Induced
Current (if any)
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 field
B is outward.
P
–
B
v
l
Q
X
∆x e Blv
i= =
∆t r
r
+
∆x
At any instant ‘t’,
flux linked with the circuit = B.A = Blx
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.
= Blv
Hence, magnitude of induced e.m.f., e = Bvl
Direction of induced emf and corresponding induced current is determined by Flemmings right
hand 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
Force experienced by wire (using F = ilB)
Blv
B2 l 2 v
i
l
B
=
l
B
=
F=
r
r
Direction of this force is opposite to the direction of velocity of wire.
Now the power required to move the wire P is
B2l 2 v
B2l 2 v 2
.v =
P = F.v =
r
r
Since wire is pushed by external power P' which is given by
2
P′ = I r =
Blv
r
2
r=
63
B2l 2 v 2
r
since P = P', we can say that mechanical energy is converted into electrical energy. Hence electromagnetic induction is in accordance with law of conservation of energy.
Also as we know
dφ
and e = Ir
e=
dt
If q is the amount of charge induced then
=
(a) Induced e.m.f. produced
by changing the orientation of the coil
Consider 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=
d(NBA cos ωt)
dt
∴
e = NABω sin ωt
This equation shows that induced e.m.f e changes sinusoidally with time t.
Maximum value of emf is
e=
e= −
(b) A.C. Generator
It 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 & S2) with carbon brushes (B1 & B2)
Working: When the coil is rotated in a magnetic field, the magnetic flux associated with if
changes. The induced emf produced in it is given by
e=
If R is the resistance of the coil, then the current
I=
I=
where I0 = peak value or amplitude of induced current
64
In the diagram, PQRS is Armature coil: R1, R2 are the Slip rings;
C1,C2 are Carbon brushes; RL is Load resistance
(i) During1st half cycle of rotation of coil, side PQ is moving outward in the magnetic field, so
using Fleming right hand rule direction of induced current is from P to Q.Hence direction of
current in RL is from M to N.
S
S
P
S
N
Q
M
C1
R1
R2
C2
RL
Q
R
R1
R2
S
N
R
C1
P
N
M
1st half Cycle
C2
RL
N
2nd half Cycle
(ii) During 2st half cycle of rotation of coil, side PQ is moving inward in the magnetic field so using
Fleming right hand rule, direction of induced current is from Q to P.Hence direction of current
in RL 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
ε
ε0
0
–ε0
T
4
T
2
3T
4
T Time
Eddy Currents
Current induced in bulk piece of a conducting material kept is a clanging magnetic field is called eddy
current. 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.
65
(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 or
simply inductance of the coil.
Using Faraday lawdφ
dI
= –L
e= –
dt
dt
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=
2r
Axis
l
=
I
Magnetic flux for N turns
φ= N
∴
I
µ 0 NI
A
l
L=
If core of solenoid is of any other magnetic material, then µ0 is replaced by (µ0µr)
(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 through
another neighbouring coil.
Any change of current in coil-1 results in a change of magnetic flux linked with coil-2 (&viceversa). This process is called mutual induction.
Let
I1 = current in coil-1
φ2 = total flux linked with coil-2 due to I1,
then
φ2 α I1
∴
φ2 = M I1
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
dφ 21
e= −
dt
66
e = −M
or
dI1
dt
Unit of M is Henry.
(b) Mutual Induction between two long solenoids
Let
I1, I2 = current in solenoids s1 and s2
N1,N2 = No. of turns in s1 and s2
φ1. φ2 = flux linked with s1 and s2
l = common length of S1 & S2
M,M' = coefficient of mutual inductance.
A1 and A2 = Area of coil s1 and s2
case-(i) Mutual induction in coil 2 due to coils 1 In this case current is flowing in coil 1.Magnetic
field B along the axis of solenoid
N1
I1
B1 = µ o
l
Magnetic flux passing through one turn of coil 2.
= B1A2
N
µµN
M
INµNNN1I1 l
µ
No0220 11210I22 2A
1A 2A 2
lll l
r1
N1
For N2 turns
φ2 = N 2 B1A 2
so we get
φ2 =
comparing with
φ2 =
..............(ii)
...(iii)
r2
N2
M2 =
...(iv)
S1
S2
case-(ii) Mutual induction in coil 1 due to coil 2 In this case current is flowing in coil 2.Magnetic
field B along the axis of solenoid
B2 =
Magnetic flux passing through one turn of coil 1.
= B2A2 (Area of smaller coil)
For N1 turns
φ1 = N1B2A2
we get
φ1 = N1
..............(v)
µ0 N 2I2
A2
l
comparing with
φ1 = M1I2
we get
M2 =
......(vi)
From (iv) & (vi), we get
M2 = M1
67
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 form
of 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 = L1 + L2
and when they are connected in parallel, the equivalent self inductance is
=
Alternating Current
If direction of current in circuit changes continuously with time then current is known as alternating
current.
I = I0 sin ωt or I = I0 cos ωt
Here, I is instantaneous value of current(value of current at time t) and I0 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 Value
Mean or average value of a.c. over one complete cycle is zero.
Mean value of a.c. for 1st half cycle
T/2
0
=
Io sin ωtdt
T/2
0
dt
68
=
=
=
=
=
= 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 I0 = zero.
Similarly,
2E o
=
π
2E o
= −
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 equally
spaced.
R.M.S. Value or Effective Value of Alternating Current and Voltage
2 1/ 2
T
2I
T 2π mean square) value of a.c. is defined as that value of steady current, which would
π 2TT/r.m.s.
2The
0T 2T
2ω
2IE∵
cos
0avg
1I−−0+T0sin
ωtttdt
cos
. dt=+ cos (root
.0
[
2T]−/ T
2ω
−/1
sin
(
)
0
0
TT
πω 2π ω generate
T
T 2 2 theTsame amount of heat in a given resistance in a given time, as is done by a.c. when passed
T
0
dtthrough the same resistance for the same time.
0
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 =
=
⇒
I0 T
.
T 2
1
2
Irms =
I0
= 0.707 Io
2
Erms =
E0
= 0.707 E o
2
Similarly,
⇒
A.C. Circuit Containing Resistance (R) only
Let applied emf is given by
E = E 0 sin ωt
...(1)
69
Let I = current in the circuit at instant t.
since
I=
⇒
E
R
I=
...(2)
where I0 = E0/R, Comparing I0 = E0/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 to
the applied voltage (E).
i.e.
− E 0 sin ωt =
or
dI =
E0
sin ωt dt
L
Integrating both sides, we get
E0
sin ωtdt
L
E 0 − cos ωt
I=
L
ω
I=
⇒
E&I
E
E0
L
0
Circuit
ωt
90º
I
O
ωt
I0
Let,
Io =
...(2)
we get,
I=
...(3)
This is the equation of alternating current due to inductor.
70
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)
= ωL = Effective resistance on inductor
This is called inductive reactance and is denoted by XL.The units of inductive reactance is ohm.
Thus
XL =
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 XL = 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 =
...(1)
For capacitor we know that q = CV
here
V = E so
⇒
q =
⇒
I =
⇒
I =
=
Let
I0 =
...(2)
C
ω
L
dq
d(CE
E
1
1
sin
1
ω
t)
=
2
πν
L
ω
CE
I0o0osin(
C
sin
ω
ω
sin(
cos
tω
t +tωπt /+2)π⇒
/ 2)
0 sin
I =
...(3)
==0
I 0oC ω
ω
dt
dt
2C
πνC
Comparing (1) with (3) we find that in an a.c. circuit containing C only, alternating current I leads
the alternating e.m.f. by a phase angle of 90°. This is shown in Fig.
E&I
E
E0
I0
I
90º
ωt
0
O
t
using eq. (2)
= Effective resistance of capacitor This is called capacitive reactance and is denoted by
XC.The units of capacitive reactance is ohm.
Thus
XC =
Hence capacitive reactance is proportional to frequency of A.C.
In dc circuits, ν = 0
so XL = Infinity
In ac circuits, ν = high so XL = Low
It means a pure inductor block dc and allow ac.
71
A.C. Circuit Containing Resistance, Inductor and Capacitor (RLC Circuit) in Series
As R, L, C are in series, therefore, current through the three
elements has the same amplitude and phase. Let maximum
value 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 = IXL
(voltage across L leads the current by 90°)
VC = IXC
(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 VC is 180°. Let (VL >VC)
then net reactive voltage is (VL - VC) .If VZ represent the resultant
of VR ,VL , and VC and making an angle f with current phasor
I.Then
∴
VL
Vz
90°
O
φ
A
90°
VZ =
K
I0
VC
VR
=
(IR) + (IX L − IX C )
2
2
C
VZ = I R 2 + (X L − X C ) 2
The total effective resistance of series LCR circuit is called impedance of the circuit. It is represented
by Z, where
V0
2
2
Z = I = R + (X L − X C )
0
From Fig. it is clear that in an a.c. circuit containing R, L and C and (VL >VC) the voltage leads
the current by a phase angle φ, where
tan φ =
VL − VC
VR
=
tan φ =
or
∴ The alternating e.m.f. in the LCR circuit would be represented by
E=
Here three cases arise:
(i) When XL = XC,
∴φ=0
Hence voltage and current are in the same phase. The a.c. circuit is non-inductive.
72
(ii) When XL > XC, tanφ is positive.
Therefore, φ is positive. Hence voltage leads the current by a phase angle φ.
(iii) When XL < XC, tanφ is negative.
Therefore, φ is negative. Hence voltage lags behind the current by a phase angle φ.
For R and L circuit
For RL circuit (put XC=0)
VZ =
and Z = R 2 + X 2L
XL
R
In this case voltage leads current.
and
tan φ =
For R and C circuit
For RC circuit (put XL=0)
R
φ
VZ =
and Z = R 2 + X C2
XC
Z
XC
R
In 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.
and
tanφ =
LC Oscillations
XL
Oscillator is a device which
convert DC in to AC.The simplest oscillator consist of L and C called tank
I
2
2 I
Z
VR + VCL circuit.
L φ
L
b
a
A capacitor
of capacitance C
C
C is connected to an inductor of inductance
+ +
L through a keyL K2. A cell is connected to C through key K1.
DISCHARGING
+ +
R
CURRENT
When plug of K1 is put in, charging of capacitor takes place.Energy
stored in capacitor in form of electric field.
C
+ –
Step 1
Step 2
K
Now open K1 and close K2,the charged capacitor is connected to L.
s
Step-1 Capacitor starts discharging through L and a discharging current
CHARGING
CURRENT
flow in LC circuit. An induced e.m.f. developes across L.Energy is stored
K
in L.
Step-2 This induced e.m.f. starts recharging the condenser in the opposite direction by induced
current. Energy is stored in C.
2
1
Step-3 Again capacitor starts discharging through L and a discharging current flow in LC
circuit.Again an induced e.m.f. developes across L.Energy is stored in L.
73
Step-4 This induced e.m.f. starts recharging the condenser in the opposite direction by induced
current. Energy is stored in C.
I
I
C
+ +
C
L
L
+ +
Step 3
Step 4
The entire process is repeated. Thus energy taken once from the cell and given to capacitor keeps
on 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 are
called 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
1
R + ωL −
ωC
since
Z=
For minimum Z,
XL = XC
i.e.
ωr L =
=
2
2
or ωr =
or
=
ωr = Angular resonant frequency
and νr = Resonant frequency.
At resonance value of Z is given by (XL = XC)
Z=
R 2 + 0 = R = minimum
In this case circuit is pure resistive and phase difference
between voltage and current is zero.
The variation of circuit current with the changing frequency
of applied voltage is shown in Fig.
The practical application of series resonance circuit is in radio
and T.V. receiver sets.
74
R = 10 Ω
Y
I
I0
I= 2
R = 100 Ω
O
ω1 ωr ω2
ω
X
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 the
circuit. It defines sharpness of current at resonance.
The Q factor of series resonance circuit is defined as the ratio of the voltage across the Inductor
or Capacitor at resonance to the applied voltage or voltage applied across R.
voltage across L or C
i.e.
Q = applied voltage ( voltage across R )
VL IX L ω r L
=
=
VR
IR
R
L 1
⇒
Q=
R LC
As R is increased, Q factor of the circuit decreases.
Q=
Another Definition of Q Factor
Graph shows that for each value of I there are two values of ω. If ω1 and ω2 are the values for which
the current is
times its maximum value (and power is half).
If
and
The difference
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 =
2ωφt)
Ioo12sin(
E
Isin
ω
φI∆ω
tωcos
t +φφ+) cos ωsin
t sin
ω
=−=
ω
ωω
−
=++×t[sin
∆ω
2∆ω
2 ω1
1ω
o2rsin
o) sin(
1rr tω
φ
ω
+
φ
Esin
I
cos
sin
t
E
I
sin
o o ce ω r =
o o
2
2∆ω
LC Power in
Average
RLC Circuit or Inductive Circuit
Let the alternating e.m.f. applied
E=
If φ is the phase angle between current then alternating current will be
I=
since power = (e.m.f.) x (current)
P=
=
P=
Now average power in one cycle of AC
T
T
sin ωtdt
sin 2ωtdt
2
Pavg = E o Io cos φ
+ E o Io sin φ
0
T
dt
0
75
0
T
2 dt
0
or
Pavg =
T
As
T
T
sin ωtdt = and sin 2ωtdt = 0
2
0
0
2
∴ Average power in the inductive circuit over a complete cycle
1
Pavg = E o Io cos φ.
2
=
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 = XC
∴ Power factor
= cos φ = 1
This 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 = Erms Irms cos 90° = Zero.
Therefore, current through pure L or pure C, which consumes no power for its maintenance in the
circuit is called Idle current or Wattless current.
At resonance, XL = XC and φ = 0°.
∴
cos φ = cos 0° = 1
Therefore, 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.
Principle
A transformer is based on the principle of mutual induction.
76
Construction
A transformer consists of a rectangular soft iron core made of laminated sheets, well insulated from one
another. Two coils P1 P2 and S1 S2 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
S1
P1
connected to S1 S2 (secondary coil),
Input
Output
R
Assume that the energy losses is negligible.
A.C.
Theory and working
P2
S2
Let the alternating e.m.f. supplied to primary is
E = E 0 sin ωt
Laminated Core
Flux induced in primary coil = φB
Because the core is ferromagnetic and extends through the secondary winding, so induced flux also
extends 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 Es
across the secondary is equal to the e.m.f. induced in the secondary. Since
E=
⇒
⇒
=
Here, np and ns represent total number of turns in primary and secondary coils respectively
dOutput
n I> n then Es > Ep, the transformer is a step up transformer. Similarly, and if ns < np then Es < Ep,
nEIEφpspαsBn power
(∵ φB =IfE
nAB)
= ss s p
The transformer
is a step down transformer transformer.
dt
power
E p Ip
nEE
IInput
pssp
= K = transformation ratio.
Using conservation of energy requires that
Ip Ep = Is Es
∴
=
For a step up transformer, Es > Ep but Is< Ip
For a step down transformer, Es < Ep but Is> Ip
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=
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 transformer
1.
2.
Copper loss—-It is the energy loss in the form of heat in the copper coils of a transformer.
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.
77
3.
4.
5.
Leakage of magnetic flux—It can be reduced by winding the primary and secondary coils one
over the other.
Hysteresis loss—This is the loss of energy due to repeated magnetisation and demagnetisation of
the iron core when a.c. is fed to it. The loss can be reduced by using a magnetic material which
has a low hysteresis loss.
Magnetostriction—humming noise of a transformer.
Uses of Transformer
A 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 I 2 R becomes negligibly low.
Answer Yourself
Very Short Questions
Q1. Determine the direction of the induced current in the loop given below:
P
×××
×××
R
×××
Q
(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 high
frequency. The circuit consists of an inductor and a capacitor in series. Show that the d.c. signal
will 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 series
circuit?
Q5. The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage and
current in this circuit?
78
Q6. When a lamp is connected to an alternating voltage supply, it lights with the same brightness as
when 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. The
magnetic field is decreased to zero in 0.7s at a steady rate. Determine the magnitude of
induced 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 5.0
ms–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 with
frequency of applied AC source. Can the voltage drop across the inductor or the capacitor in
a series LCR circuit be greater than the applied voltage of the AC source? Justify your
answer.
XC
XL
(ω)
(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 of
free oscillations of the circuit?
Q6. A power transmission line feeds input power at 2300 V to a step-down transformer with its
primary windings having 4000 turns. What should be the number of turns in the secondary in
order 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 the
potential drop across the LC combination is zero at the resonating frequency.
79
To be Learnt
Very Short Questions
Q1. (a) Closed loop is held stationary in the magnetic field between the north and south poles of
two permanent magnets held fixed. Can we hope to generate current in the loop by using
very strong magnets?
(b) A closed loop is moved normal to the constant electric field between the plates of a large
capacitor. Is a current induced in the loop (i) when it is wholly inside the region between
the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric
field 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 brightness
of 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 an
insulating stand when a bar magnet is quickly moved along the axis of the coil from one side
to the other as shown in the figure.
S N
S N
Q5. In any a.c. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous
voltages 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 the
corresponding 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
π
sin 314t +
V. What is the power dissipation in the circuit?
2
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 in
the circuit change?
Q10. Power factor can often be improved by the use of a capacitor of appropriate capacitance in the
circuit. How?
Q11. An ideal inductor is in turn put across 220 V, 50 Hz and 220 V, 100 Hz supplies. Will the current
flowing 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 instantaneous
voltage.
80
Short Questions
Q13. A circular coil of radius 10 cm, 500 turns and resistance 2Ω is placed with its plane perpendicular
to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter
through 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 120
revolutions per minute in a plane normal to the earth’s magnetic field. If the earth’s magnetic
field at the given place is 0.4 gauss, find the E.M.F. induced between the axle and the rim
of the wheel.
Q15. A bar magnet M is dropped so that it falls vertically through the coil C. The graph obtained
for 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?
B
p.d. (mV)
O
A
Time (ms)
D
C
µ
ωFr
Magnet
V
R
Q16. Two circular coils, one of small radius r1 and the other of very large radius r2 are placed coaxially with centres coinciding. Obtain the mutual inductance of the arrangement.
Q17. How is the mutual inductance of a pair of coils affected when:
Coil (i) separation between the coils is increased?
C
(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? Explain
your 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 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
is 6mC. What is the total energy stored
in 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-frequency
200 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 V
to 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 what
circumstances?
81
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 from
an electric plant generating power at 440 V. The resistance of the two wire line carrying power
is 0.5Ω per km. The town gets power from the line through a 4000 – 220 V step down transformer
at 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 to
leakage?
(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 of
the 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 speed
v. Consider, there is a magnetic field for 0 < x ≤ b. There is no magnetic field for x > b. If the
resistance of the arm PQ is r. Obtain expressions for the flux, the induced E.M.F., the force
necessary to pull the arm and the power dissipated as Joule loss. Sketch the variation of these
quantities with distance.
S
× × ×
P
× × ×
× × ×
R
x=0
Q
x = 2b
x=b
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?
82
Pedagogical Remark
1.
(a) No. However strong the magnet may be, current can be induced only by changing the magnetic
flux through the loop.
(b) No current is induced in either case. Current cannot be induced by changing the electric flux.
dt
2. E.M.F.,
e = −L
dt
5
200 = − L
0.1
Hence,
L = 4 H.
3. For steady state DC, there will be no change in the brightness of the bulb, as there is no change
in 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 of
the 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: The
direction of current will be clockwise, when bar magnet is moving towards the loop. The
direction 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 voltages
across the series elements of the given a.c. circuit. The same is not true for rms voltage, because
voltages 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 inserting
iron-core.
V = 220 2 sin 2an
π (50)
t = 311For
sina.c.,
314t.the lamp will shine dimly because of additional impedance of the choke. It
will 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. Consequently
power dissipation in the circuit is zero.
8.
Maximum value of power factor is 1 and its minimum value is zero i.e., 0 ≤ cos φ ≤ 1 .
9.
With increase in frequency, current in a.c. circuit first increases, attains a maximum value at
resonant frequency and then again decreases.
10. Because by using a capacitor of appropriate capacitance the effect of inductive reactance can be
nullified 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 because
reactance XL = L.2π n is more for frequency n = 100 Hz.
12. Equation for instantaneous voltage is
13. Initial flux through the coil,
φB(initial) =
=
Final flux (after the rotation),
φB(final) =
=
BA cos θ
(3.0 × 10–5 ) (π10–2)cos 0° = 3π × 10–7 Wb
(3.0 × 10–5) (π10–2)cos180°
–3π × 10–7 Wb
83
∆φ
∆t
= 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,
The induced e.m.f. is, e = N
A=
The induced E.M.F.,
|E| =
= B
ν
ω
r
B
15.
(i)
(ii)
16.
=
=
=
=
r 2 dθ Br 2ω
=
2 dt
2
120rev/min = 2rev/s;
2πν = 4π rad/s
0.5m;
0.4G = 0.4 × 10–4 T
2π × 2 × 0.4 × 10−4 × (0.5) 2
∴
E=
2
= 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
As the bar magnet accelerates, its velocity keeps increasing, hence the rate of change of flux
increases. Hence,
(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 becomes
sharp 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).
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.
The magnetic field at the centre of the outer coil due to current in the outer coil,
B2 =
µ 0 I2
2r2
Flux linking to the inner coil,
φ1 = π r12B2 =
Hence,
M21 = M12 =
I2 = M12I2
; where r2 > r1
84
17. (i) Mutual Inductance (M) decreases because the quantity of flux linking to a coil due to the other
coil will decrease.
(ii) M increases because as the no. of turns increase, the overall flux density also increases and
hence 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 =
(b) As
= 212.1V
Irms = 10 A,
hence
Im =
=
19. Here
and
∴ Resonant frequency
2 ⋅ Irms
× 10 = 14.14 A
14.1 A.
L = 2.0 H,
C = 32 µF = 32 × 10–6F
R = 10Ω.
ωr =
= 125 s–1.
L
−1
3 2× 1251
ω
1 r× 10
2.0
V2
300
) ,
m
=
=
ν(6
=
=
0
−6
−6
R
10
LC
2 302π
2 LC
2×
× 10
2.0
× 32 × 10
and
Q value =
= 25.
20. Here
qm = 6 mC = 6 × 10–3 C.
∴ Total energy stored in the circuit initially = Total energy stored at any later time
1 q 2m
= U=
2 C
=
= 0.6J.
21. When frequency of the supply = natural frequency of circuit
XL = XC and resonance
takes place so that
current I rms =
Vrms 200
= 10 A
=
R
20
As in resonance condition voltage and current are in same phase condition, hence average
Power = Vrms Irms = 200 × 10
= 2000 W or 2kW.
85
22. Here
efficiency
Vp = 200 V, Vs = 20 V,
η = 80% =
80
= 0.8
100
and resistance in secondary ciruit
R = 20
∴
Is =
As efficiency
η=
⇒
IP =
= 1A
Output Vs ⋅ Is
=
Input
Vp ⋅ I p
= 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 resulting
current and the power factor of the a.c. ciruit is zero.
24. Here
L = 0.5H, R = 100Ω, Vrms = 240V
and
ν = 50 Hz
ω = 2πν
= 2 × 3.14 × 50 = 314s–1.
(a) Impedance
Z=
R 2 + L2ω 2
=
= 186.1W
Vm
2 Vrms
2 × 240
=
=
Z
Z
186.1
= 1.82 A
(b) If the current lags behind the voltage in phase by φ , then
∴
Im =
φ=
= tan–1 (1.5700) = 57.5°
=
∴ Time lag between voltage maximum and current maximum
t=
= 3.2 × 10–3s or 3.2 ms.
86
25. (a) Power required by the town
P = 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 Ω
As supply voltage is through 4000 – 220 V line, the transmission line transmits power at 4000 V
∴ Line current
= 200 A
I=
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 generating
station side of line = 4000 + 3000 = 7000 V
Hence, a step up transformer of 440 V – 7000 V be used at the generator plant.
5
1 Here it is given that L = 200 mH = 0.2 H, R = 10Ω and ν = 50 s–1
P1 8=× 10126.
2=
22 2
ω
4π ν× (50)
L (i)2 ×As
4 ×L(3.14)
0.2power factor
V
4000
cosφ = R/Z = 1,
hence
R= Z
or
X = XL – XC = 0
or
XL = XC
∴
⇒
C=
=
= 5.07 × 10–5F = 50.7 µF
(ii) Q-factor of the circuit
=
ω 0 L 2πνL
=
R
R
=
2 × 3.14 × 50 × 0.2
10
= 6.28
87
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, F =
B2 l 2 v
for 0 ≤ x < b and 0 for b ≤ x < 2b
r
The Joule's heating loss, P=I2r so P =
B2 l 2 v 2
for 0 ≤ x < b and 0 for b ≤ x < 2b
r
28. Here
L = 20 mH = 20 × 10–3 H,
C = 50 µF = 50 × 10–6 F
and
qm = q0 = 10 mC
= 10 × 10–3C = 0.01 C,
when
t = 0.
(a) ∴ Total energy stored initially
1 q 02
(0.01)2
=
=1J
2 C 2 × 50 × 10−6
Yes, the energy remains conserved during LC oscillations provided that resistance R of the circuit
is zero.
(b) Natural frequency of the circuit
U=
ν0 =
1
=
2 × 3.14 × 20 × 10 −3 × 50 × 10−6
= 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,
at times t =
T
, T,
2
2T .......... the energy stored is completely electrical and (ii)
.......... the energy stored is completely magnetic.
88
(d) Let at time t the electrial and magnetic energies are equal i.e.,
q2
=
2C
i.e., at time t,
q=
∵
q=
∵
=
or
cos ωt =
or ωt =
ω=
we have
Substituting
π
,
4
....
t=
(e) If a resistor is inserted in the circuit, due to energy loss amplitude of LC oscillations gradually
decreases with time and finally the oscillations stop altogether.
2
2πqcos
13Tωt5T 7T
q3T
, ,,02 =, 1 ,q0 ..........
±10LI
4 28 28 2C8
28T
89
5
Weightage-03 marks
1.
Conduction current & Displacement current—The current carried by conductors due to flow
of 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 a
closed path is equal to µ0 times the sum of conduction current and the displacement current threading
the closed path.
B.dl
where
3.
Maxwell’s equations—
Gauss’s law of electrostatics
=
,
IC – Conduction current
ID – Displacement current.
●
=
●
Gauss’s law of magnetism
= 0
●
Faraday’s law of electromagnetic induction
=
●
●
4.
Ampere-Maxwell’s law
=
Electromagnetic waves—
Radiation consisting of self-sustaining
oscillating electric and magnetic fields
at right angles to each other and to
the direction of propagation is called
electromagnetic wave.
90
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 oscillating
electric 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.
Ex = E0 sin (kz – ωt), By = B0 sin (kz – ωt),
k=
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
●
●
●
2π 1
v =,
λ εµ 12
( )
●
●
●
●
●
●
●
●
●
●
●
●
, 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 conducting
surface.
The wave is a transverse wave, since the fields are perpendicular to the direction in which
the wave travels.
All electromagnetic waves, regardless of their frequency, travel through vacuum at the same
speed, 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/k
The SI unit of frequency is hertz wiht unit symbol Hz
The SI unit of wavelength is metre with unit symbol m.
They carry energy with them which is equally divided between electric energy and magnetic
energy.
Only electric field vector is responsible for the optical effects of electromagnetic waves so
it 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.
91
92
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 electromagnetic
radiations?
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 and
source 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 melting
or 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
93
(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) ultraviolet
light (v) x-rays (vi) gamma rays.
Q8. Arrange the following radiations in the descending order of frequency: gamma rays, infrared
radiations, microwaves, yellow light, ultraviolet radiations.
Q9. Find the wavelength of electromagnetic waves of frequency 4 × l09Hz in free space. Give its
two applications.
Q10. Differentiate between radiowaves and γ-rays.
To be Learnt
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 magnetic
fields 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 increase
or 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 an
electromagnetic wave?
Q10. The sunlight reaching the earth has a maximum electric field of 810 V/m. What is the maximum
magnetic field associated with it?
Q11. Calculate the relative permittivity of a medium of relative permeability 1.0 if the velocity of
light 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. What
is the displacement current across its plate?
94
Q13. In a plane electromagnetic wave the electric field oscillates at a frequency 3 × 1010 Hz and amplitude
50 V/m. What is (a) the wavelength of the wave and What is (b) the amplitude of the oscillating
magnetic field? (Ans: 1cm, 1.67 × 10–7 T)
Q14. Why is the electromagnetic wave of transverse nature? A plane electromagnetic wave travels in
vacuum, along y-direction. Write the (i) ratio of the magnitudes and (ii) the direction of its electric
and magnetic field vectors.
Q15. Which of the following , if any, can act as a source of electromagnetic waves? (i) a charge moving
with 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 two
applications.
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 produce
electromagnetic waves.
Q20. Name the electromagnetic radiation to which the following wavelength belong (a)10–2 m (b) l A0
Pedagogical Remark
E0 1
= cC
Bµ
0 0ε0
1.
2.
Electric & Magnetic fields oscillate in mutually perpendicular directions.
Displacement cutrrent = charging current = 0.25 A
3.
c=
4.
5.
6.
9.
One, as speed of em waves in vacuum is independent of wavelenght or frequency.
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 of
em waves.
ω increases means XC decreases, ID increases.
They are mechanical waves hence require material medium for their propagation, can not be
considered as em waves.
They are in same phase.
10.
Sun light is em wave, use
7.
8.
95