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PHYSICS
HIGHER SECONDARY
SECOND YEAR
VOLUME - I
Revised based on the recommendation of the
Textbook Development Committee
Untouchability is a sin
Untouchability is a crime
Untouchability is inhuman
TAMILNADU TEXTBOOK
CORPORATION
COLLEGE ROAD, CHENNAI - 600 006
c Government of Tamilnadu
First edition - 2005
Revised edition - 2007
CHAIRPERSON
Dr
...
GUNASEKARAN
Reader
Post Graduate and Research Department of Physics
Pachaiyappa’s College, Chennai - 600 030
Reviewers
P SARVAJANA RAJAN
...
RASARASAN
Selection Grade Lecturer in Physics
Govt
...
G
...
Hr
...
School
Kodambakkam, Chennai - 600 024
S
...
K
...
SANKARI
Selection Grade Lecturer in Physics
Meenakshi College for Women
Kodambakkam, Chennai - 600 024
G
...
Arts College
Villupuram
...
G
...
Girls’ Hr
...
School
Ashok Nagar, Chennai - 600 083
P LOGANATHAN
...
G
...
Girls’ Hr
...
School
Tiruchengode - 637 211
Namakkal District
Dr
...
VIJAYAN
Principal
Zion Matric Hr
...
School
Selaiyur
Chennai - 600 073
Dr
...
PONNUSAMY
Asst
...
R
...
Engineering College
S
...
M
...
rc
s
This book has been prepare b t e D rectorate of School Education on behalf of
d y h i
the Government of Tamilnadu
This book has been printed on 60 G
...
M paper
P i t d b o sta :
r n e y ffe t
Preface
The most important and crucial stage of school education is the
higher secondary level
...
In order to pursue their career in basic sciences and professional
courses, students take up Physics as one of the subjects
...
XII has been reformed,
updated and designed to include basic information on all topics
...
All the topics are presented with clear and concise treatments
...
Understanding the concepts is more important than memorising
...
In order to
make the learning of Physics more interesting, application of concepts in
real life situations are presented in this book
...
Their learning experience would
make them to appreciate the role of Physics towards the improvement
of our society
...
The data has been systematically updated
...
Self-evaluation questions (only samples) are included to sharpen
the reasoning ability of the student
...
They must be prepared to answer the questions and
problems from the text/syllabus
...
S
...
Forces between
multiple electric charges – superposition principle
...
Electric potential – potential difference – electric potential due to
a point charge and due a dipole
...
Electric flux – Gauss’s theorem and its applications to find field
due to (1) infinitely long straight wire (2) uniformly charged infinite
plane sheet (3) two parallel sheets and (4) uniformly charged thin
spherical shell (inside and outside)
Electrostatic induction – capacitor and capacitance – Dielectric
and electric polarisation – parallel plate capacitor with and without
dielectric medium – applications of capacitor – energy stored in a
capacitor
...
UNIT - 2 CURRENT ELECTRICITY (11 periods)
Electric current – flow of charges in a metallic conductor – Drift
velocity and mobility and their relation with electric current
...
V-I characteristics – Electrical
resistivity and conductivity
...
Kirchoff’s law – illustration by simple circuits – Wheatstone’s
Bridge and its application for temperature coefficient of resistance
measurement – Metrebridge – Special case of Wheatstone bridge –
Potentiometer – principle – comparing the emf of two cells
...
IV
UNIT – 3 EFFECTS OF ELECTRIC CURRENT (15 periods)
Heating effect
...
Thermoelectric effects – Seebeck effect – Peltier effect – Thomson
effect
–
Thermocouple,
thermoemf,
neutral
and
inversion
temperature
...
Magnetic effect of electric current – Concept of magnetic field,
Oersted’s experiment – Biot-Savart law – Magnetic field due to an
infinitely long current carrying straight wire and circular coil –
Tangent galvanometer – Construction and working – Bar magnet as an
equivalent solenoid – magnetic field lines
...
Force on a moving charge in uniform magnetic field and electric
field – cyclotron – Force on current carrying conductor in a uniform
magnetic field, forces between two parallel current carrying conductors
– definition of ampere
...
UNIT – 4 ELECTROMAGNETIC INDUCTION AND
ALTERNATING CURRENT (14 periods)
Electromagnetic induction – Faraday’s law – induced emf and
current – Lenz’s law
...
Methods of inducing emf – (1) by changing magnetic induction
(2) by changing area enclosed by the coil and (3) by changing the
orientation of the coil (quantitative treatment) analytical treatment
can also be included
...
(Single phase, three
phase)
...
Alternating current – measurement of AC – AC circuit with
resistance – AC circuit with inductor – AC circuit with capacitor - LCR
series circuit – Resonance and Q – factor: power in AC circuits
...
Emission and Absorption spectrum – Line, Band and continuous
spectra – Flourescence and phosphorescence
...
Scattering of light – Rayleigh’s scattering – Tyndal scattering –
Raman effect – Raman spectrum – Blue colour of the sky and reddish
appearance of the sun at sunrise and sunset
...
Interference – Young’s double slit experiment and expression for
fringe width – coherent source - interference of light
...
Diffraction – differences between interference and diffraction of
light – diffraction grating
...
Bohr’s model – energy quantisation – energy and wave number
expression – Hydrogen spectrum – energy level diagrams – sodium and
mercury spectra - excitation and ionization potentials
...
X-rays – production, properties, detection, absorption, diffraction
of X-rays – Laue’s experiment – Bragg’s law, Bragg’s X-ray
spectrometer – X-ray spectra – continuous and characteristic X–ray
spectrum – Mosley’s law and atomic number
...
Matter waves – wave mechanical concept of the atom – wave
nature of particles – De–Broglie relation – De–Broglie wave length of
an electron – electron microscope
...
Special
theory of relativity – Relativity of length, time and mass with velocity
– (E = mc2)
...
Stability of nuclei-Bain bridge mass spectrometer
...
Artificial radioactivity – radio isotopes – effects and
uses Geiger – Muller counter
...
UNIT – 9 SEMICONDUCTOR DEVICES AND THEIR APPLICATIONS
(26 periods)
Semiconductor theory – energy band in solids – difference
between metals, insulators and semiconductors based on band theory
– semiconductor doping – Intrinsic and Extrinsic semi conductors
...
– P-N Junction diode – Forward and reverse bias characteristics
– diode as a rectifier – zener diode
...
VII
Junction transistors – characteristics – transistor as a switch –
transistor as an amplifier – transistor biasing – RC, LC coupled and
direct coupling in amplifier – feeback amplifier – positive and negative
feed back – advantages of negative feedback amplifier – oscillator –
condition for oscillations – LC circuit – Colpitt oscillator
...
Laws and theorems of Boolean’s algebra – operational amplifier –
parameters – pin-out configuration – Basic applications
...
Non-inverting amplifier – summing and difference
amplifiers
...
Multimeter – construction and uses
...
Amplitude modulation, merits and demerits – applications –
frequency modulation – advantages and applications – phase
modulation
...
Radio transmission
superheterodyne receiver
...
V
...
Vidicon (camera tube) and picture tube – block diagram of a
monochrome TV transmitter and receiver circuits
...
Digital communication – data transmission and reception –
principles of fax, modem, satellite communication – wire, cable and
Fibre - optical communication
...
To determine the refractive index of the material of the prism by
finding angle of prism and angle of minimum deviation using a
spectrometer
...
To determine wavelengths of a composite light using a diffraction
grating and a spectrometer by normal incidence method (By
assuming N)
...
To determine the radius of curvature of the given convex lens
using Newton’s rings experiment
...
To find resistance of a given wire using a metre bridge and hence
determine the specific resistance of the material
...
To compare the emf’s of two primary cells using potentiometer
...
To determine the value of the horizontal component of the magnetic
induction of the earth’s magnetic field, using tangent galvanometer
...
To determine the magnetic field at a point on the axis of a circular
coil
...
To find the frequency of the alternating current (a
...
9
...
(b)
To draw the characteristic curve of a Zener diode and to
determine its reverse breakdown voltage
...
To study the characteristics of a common emitter NPN transistor
and to find out its input, output impedances and current gain
...
Construct a basic amplifier (OP amp) using IC 741 (inverting, non
inverting, summing)
...
Study of basic logic gates using integrated circuits NOT, AND, OR,
NAND, NOR and EX-OR gates
...
1
Electrostatics
1
2
Current Electricity
53
3
Effects of Electric Current
88
4
Electromagnetic Induction
and Alternating Current
5
134
Electromagnetic Waves and
Wave Optics
178
Logarithmic and other tables
228
(Unit 6 to 10 continues in Volume II)
X
1
...
In this chapter, we shall study the
basic phenomena about static electric charges
...
These
charges have forces acting on them and hence possess potential energy
...
1
...
C
...
In the 17th century, William Gilbert
discovered that, glass, ebonite etc, also exhibit this property, when
rubbed with suitable materials
...
These terms are derived from the Greek word
elektron, meaning amber
...
If the charges in a body do not move, then, the
frictional electricity is also known as Static Electricity
...
1
...
(ii)
If an ebonite rod is rubbed with fur, it becomes negatively
charged, while the fur acquires equal amount of positive charge
...
Thus, charging a rod by rubbing does not create electricity, but
simply transfers or redistributes the charges in a material
...
1
...
A charged glass rod is suspended by a silk thread, such that it
swings horizontally
...
It is found that the ends of the two
rods repel each other (Fig 1
...
However, if a charged ebonite rod is
brought near the end of the suspended rod, the two rods attract each
other (Fig 1
...
The above experiment shows that like charges repel and
unlike charges attract each other
...
2 Two charged rods
of opposite sign
Fig
...
1 Two charged rods
of same sign
The property of attraction and repulsion between charged bodies
have many applications such as electrostatic paint spraying, powder
coating, fly−ash collection in chimneys, ink−jet printing and photostat
copying (Xerox) etc
...
1
...
Bodies
which allow the charges to pass through are called conductors
...
g
...
Bodies which do not allow the charges
to pass through are called insulators
...
g
...
2
1
...
4 Basic properties of electric charge
(i)
Quantisation of electric charge
The fundamental unit of electric charge (e) is the charge
carried by the electron and its unit is coulomb
...
6 × 10−19 C
...
It means that the quantity can
take only one of the discrete set of values
...
(ii)
Conservation of electric charge
Electric charges can neither be created nor destroyed
...
But the charges can be
transferred from one part of the system to another, such that the total
charge always remains conserved
...
238
92U
−−−−→
90Th
234
+
4
2He
Total charge before decay = +92e, total charge after decay = 90e + 2e
...
i
...
it remains constant
...
For example, if two charged
bodies of charges +2q, −5q are brought in contact, the total charge of
the system is –3q
...
1
...
Coulomb’s law states that the force of attraction or repulsion
between two point charges is directly proportional to the product of the
charges and inversely proportional to the square of the distance between
3
Let q1 and q2 be two point charges
placed in air or vacuum at a distance r
apart (Fig
...
3a)
...
The direction of forces is along
the line joining the two point charges
...
3a Coulomb forces
q1q 2
r2
where k is a constant of proportionality
...
e
...
854 × 10−12 C2 N−1 m−2
...
If the charges are situated in a medium of permittivity ε, then the
magnitude of the force between them will be,
Fm =
1 q1q2
4πε r 2
Dividing equation (1) by (2)
F
ε
=
= εr
Fm εο
4
…(2)
ε
The ratio ε = εr, is called the relative permittivity or dielectric
ο
constant of the medium
...
ε = εoεr
∴
F
Since Fm = ε , the force between two point charges depends on
r
the nature of the medium in which the two charges are situated
...
1
...
3b Coulomb’s law in
vector form
→
If F 12 is the force exerted on
q1 due to q2,
q1q 2
q2
+
r
F12
where ^12 is the unit vector
r
from q1 to q2
...
r
[Both ^21 and ^12 have the same magnitude, and are oppositely
r
r
directed]
q1q 2
→
F 12 = k r 2
(– ^12)
r
∴
12
or
q1q 2
→
F 12 = − k 2
r12
or
→
→
F 12 = – F 21
^
r 12
So, the forces exerted by charges on each other are equal in
magnitude and opposite in direction
...
1
...
qn
...
The force on q1 due to q2
1 q1q 2
→
^
F 12 = 4πε
r 21
2
ο r21
Similarly, force on q1 due to q3
1 q1q3
→
^
F 13 = 4πε
r 31
2
ο r31
The total force F1 on the charge q1 by all other charges is,
→
→
→
→
F 1 = F 12 + F 13 + F 14
...
2
q1q 3
q1qn
⎤
1 ⎡ q1q 2 ˆ
ˆ
ˆ
⎢ r 2 r21 + r 2 r31 +
...
The presence of an electric field
around a charge cannot be detected unless another charge is brought
towards it
...
Electric Field Intensity (E)
Electric field at a point is measured in terms of electric field
intensity
...
6
F
It is a vector quantity
...
The unit of electric field intensity
qo
−1
...
So, the force exerted by an electric field on a
charge is F = qoE
...
2
...
1
...
A test
charge q o is placed at P at a
distance r from q
...
4 Electric field due to a
point charge
1 q qo
F = 4πε
2
o r
The electric field at a point P is, by definition, the force per unit
test charge
...
→
In vector notation E =
1 q ^
^
r , where r is a unit vector pointing
4πεo r 2
away from q
...
2
...
→
E
→
→
→
→
= E 1 + E 2 + E 3
...
⎥
2
3
⎣1
⎦
7
1
...
3 Electric lines of force
The concept of field lines was introduced by Michael Faraday as
an aid in visualizing electric and magnetic fields
...
The electric field due to simple arrangements of point charges are
shown in Fig 1
...
+q
(a)
Isolated charge
-q
+q
(b)
+q
+q
(c)
Unlike charges
Like charges
Fig1
...
(ii)
Lines of force never intersect
...
(iv)
The number of lines per unit area, through a plane at right angles
to the lines, is proportional to the magnitude of E
...
(v)
1
Each unit positive charge gives rise to ε lines of force in free
o
space
...
o
8
1
...
4 Electric dipole and electric dipole moment
Two equal and opposite charges
separated by a very small distance
constitute an electric dipole
...
6 Electric dipole
chloroform molecules are some examples
of permanent electric dipoles
...
Two point charges +q and –q are kept at a distance 2d apart
(Fig
...
6)
...
∴
Electric dipole moment, p = q2d or 2qd
...
The unit of dipole
moment is C m
...
2
...
AB is an electric dipole of two point charges –q and +q separated
by a small distance 2d (Fig 1
...
P is a point along the axial line of the
dipole at a distance r from the midpoint O of the electric dipole
...
7 Electric field at a point on the axial line
The electric field at the point P due to +q placed at B is,
1
q
E1 = 4πε
2 (along BP)
o (r − d )
9
The electric field at the point P due to –q placed at A is,
1
q
E2 = 4πε
2 (along PA)
o (r + d )
E1 and E2 act in opposite directions
...
The resultant electric
field at P is,
E = E1 + (−E2)
q
1
q
⎡ 1
⎤
−
E = ⎢ 4πε
2
4πεo (r + d )2 ⎥ along BP
...
⎣ (r − d ) ⎦
If the point P is far away from the dipole, then d <
q 4rd
q 4d
=
E = 4πε
4
4πεo r 3
o r
1 2p
E = 4πε
3 along BP
...
10
1
...
6 Electric field due to an electric dipole at a point on the
equatorial line
...
Let 2d be the dipole distance
and p be the dipole moment
...
8a)
...
8
E1sin
P
E2sin
(b) The components of the
electric field
Electric field at a point P due to the charge +q of the dipole,
1
q
E1 = 4πε
2 along BP
...
Resolving E1 and E2 into
their horizontal and vertical components (Fig 1
...
11
The horizontal components E1 cos θ and E2 cos θ will get added
along PR
...
1
...
7 Electric dipole in a uniform electric field
Consider a dipole AB of
dipole moment p placed at an
angle θ in an uniform electric
field E (Fig
...
9)
...
The charge
–q experiences an equal force in
the opposite direction
...
The two equal and unlike
B +q F=qE
2d
θ
E
p
F=-qE
A
-q
C
Fig 1
...
The magnitude of torque is,
τ = One of the forces x perpendicular distance between the forces
= F x 2d sin θ
= qE x 2d sin θ = pE sin θ
(∵ q × 2d = P)
→ → →
In vector notation, τ = p × E
Note : If the dipole is placed in a non−uniform electric field at an
angle θ, in addition to a torque, it also experiences a force
...
2
...
E
2d
B F=qE
+q
p
A
-q
F=-qE
Electric potential energy
of an electric dipole in an
electrostatic field is the work
done in rotating the dipole to
the desired position in the
field
...
10 Electric potential
energy of dipole
τ = pE sin θ
Work done in rotating the dipole through dθ,
dw
= τ
...
dθ
The total work done in rotating the dipole through an angle θ is
W = ∫dw
W = pE ∫sinθ
...
∴
U = – pE cos θ
13
When the dipole is aligned parallel to the field, θ = 0o
∴U
= –pE
This shows that the dipole has a minimum potential energy when
it is aligned with the field
...
Microwave oven
It is used to cook the food in a short time
...
The water molecules in the food which
are the electric dipoles are excited by an oscillating torque
...
This is used to cook food
...
3 Electric potential
+q
Let a charge +q be placed at a
E
B
O
A
point O (Fig 1
...
A and B are two
x
dx
points, in the electric field
...
11 Electric potential
positive charge is moved from A to B
against the electric force, work is done
...
i
...
, dV = WA → B
...
The unit of potential difference is volt
...
The electric potential in an electric field at a point is defined as
the amount of work done in moving a unit positive charge from infinity
to that point against the electric forces
...
Work done in
moving a unit positive charge from A to B is dV = E
...
14
The work has to be done against the force of repulsion in moving
a unit positive charge towards the charge +q
...
dx
E =
−dV
dx
The change of potential with distance is known as potential
gradient, hence the electric field is equal to the negative gradient of
potential
...
The unit of electric intensity can also be
expressed as Vm−1
...
3
...
P is a point at a distance r
from +q
...
12 Electric potential due
to a point charge
A and B at distances x and
x + dx from the point O
(Fig
...
12)
...
4πεo
x2
1
q
dV = − 4 πε
2
...
The electric potential at the point P due to the charge +q is the
total work done in moving a unit positive charge from infinity to that
point
...
dx = 4 π ε r
o
15
1
...
2 Electric potential at a point due to an electric dipole
Two charges –q at A and
+q at B separated by a small
distance 2d constitute an
electric dipole and its dipole
moment is p (Fig 1
...
Let P be the point at a
distance r from the midpoint
of the dipole O and θ be the
angle between PO and the
axis of the dipole OB
...
P
r2
r
180-
A
-q
d
O
r1
p
d
B
+q
Fig 1
...
(1)
Applying cosine law,
r12 = r2 + d2 – 2rd cos θ
⎛
cos θ d 2 ⎞
+
⎟
r12 = r2 ⎜1 − 2d
r
⎝
r2 ⎠
Since d is very much smaller than r,
1
∴
2d
⎞2
r1 = r ⎛1 −
cos θ ⎟
⎜
⎝
⎠
r
16
d2
r2
can be neglected
...
r2
2d
⎛
⎞
cos θ ⎟
= r ⎜1 +
⎝
⎠
r
or
−1/2
1 1⎛
2d
⎞
= ⎜1 +
cos θ ⎟
r2 r ⎝
r
⎠
1/2
∴
(
d2
r2
is negligible)
Using the Binomial theorem and neglecting higher powers,
1 1⎛
d
⎞
= ⎜1 − cos θ ⎟
r2 r ⎝
r
⎠
...
r
2
=
1 p
...
When the point P lies on the axial line of the dipole on the side
of +q, then θ = 0
∴ V =
2
...
p
p
4πεo r 2
When the point P lies on the equatorial line of the dipole, then,
θ = 90o,
∴ V = 0
17
1
...
3 Electric potential energy
The electric potential energy of two
point charges is equal to the work done to
assemble the charges or workdone in
bringing each charge or work done in
bringing a charge from infinite distance
...
14a Electric
potential energy
B
Let us consider a point charge q1,
placed at A (Fig 1
...
The potential at a point B at a distance r from the charge q1 is
q1
V = 4πε r
o
Another point charge q2 is brought from infinity to the point B
...
∴
work done, w = Vq2
q1q 2
Potential energy (U) = 4 π ε r
o
Keeping q2 at B, if the charge q1 is q
3
imagined to be brought from infinity to the point
A, the same amount of work is done
...
14b Potential
energy of system of
previous cases
...
14b), the potential energy (U) is given by
U =
⎡ q1q 2 q1q 3 q 2q 3 ⎤
+
+
⎢
r13
r23 ⎥
4πεo ⎣ r12
⎦
1
1
...
4 Equipotential Surface
If all the points of a surface are at the same electric potential,
then the surface is called an equipotential surface
...
Thus, equipotential surfaces in this
18
A
B
E
+q
(a) Equipotential surface
(spherical)
E
Fig 1
...
15a)
...
If the charge is to be moved between any two points on an
equipotential surface through any path, the work done is zero
...
If VA = VB then WAB = 0
...
as VB – VA =
(ii) In case of uniform field, equipotential surfaces are the parallel
planes with their surfaces perpendicular to the lines of force as shown
in Fig 1
...
1
...
16)
...
The direction of ds is drawn
normal to the surface outward
...
16 Electric flux
electric field over ds is supposed to be a
→ →
constant E
...
The electric flux is defined as the total number of electric lines of
force, crossing through the given area
...
ds = E ds cos θ
The total flux through the closed surface S is obtained by
integrating the above equation over the surface
...
ds
The circle on the integral indicates that, the integration is to be
taken over the closed surface
...
Its unit is N m2 C−1
1
...
1 Gauss’s law
The law relates the flux through any closed surface and the net
charge enclosed within the surface
...
q
φ= ε
o
This closed imaginary surface is called Gaussian surface
...
Charges outside the surface will not contribute to flux
...
4
...
Let P be a point at a distance r
from the wire (Fig
...
17) and E be the
electric field at the point P
...
Consider a very
Fig 1
...
straight charged wire
20
E
By symmetry, the magnitude of the electric field will be the same at
all points on the curved surface of the cylinder and directed radially
outward
...
The electric flux (φ) through curved surface =
φ
=
∫
∫
E ds cos θ
[∵ θ = 0;cos θ = 1]
E ds
= E (2πrl)
(∵ The surface area of the curved part is 2π rl)
Since E and ds are right angles to each other, the electric flux
through the plane caps = 0
∴
Total flux through the Gaussian surface, φ = E
...
1
...
3 Electric field due to an infinite charged plane sheet
Consider
an
infinite plane sheet of
charge with surface
charge density σ
...
1
...
Consider a Gaussian
surface in the form of
cylinder
of
cross−
sectional area A and
length 2r perpendicular
to the sheet of charge
...
18 Infinite plane sheet
21
E
By symmetry, the electric field is at right angles to the end caps
and away from the plane
...
Therefore, the total flux through the closed surface is given by
∫
∫
=
⎡ E
...
ds ⎤
⎢
⎥P ⎢
⎥P 1
⎣
⎦
⎣
⎦
=
φ
EA + EA = 2EA
(∵ θ = 0,cos θ = 1)
If σ is the charge per unit area in the plane sheet, then the net
positive charge q within the Gaussian surface is, q = σA
Using Gauss’s law,
σA
2EA = ε
o
σ
∴ E = 2ε
o
1
...
4 Electric field due to two parallel charged sheets
Consider
two
plane
parallel
infinite sheets with equal and opposite
charge densities +σ and –σ as shown in
Fig 1
...
The magnitude of electric field
on either side of a plane sheet of charge
is E = σ/2εo and acts perpendicular to
the sheet, directed outward (if the
charge is positive) or inward (if the
charge is negative)
...
19 Field due to two
(i) When the point P1 is in between
parallel sheets
the sheets, the field due to two sheets
will be equal in magnitude and in the
same direction
...
The resultant field at P2 is,
E = E1 – E 2 =
σ
σ
– 2ε = 0
...
4
...
E
Consider a charged shell
of radius R (Fig 1
...
Let P be
a point outside the shell, at a
distance r from the centre O
...
The E
electric field E is normal to the
surface
...
ds =
∫
s
P
R
O
r
E
Gaussian
Surface
E
E ds = E (4π r 2 )
Fig1
...
Field at a point
outside the shell
(since angle between E and ds is zero)
By Gauss’s law,
q
E
...
or
Case (ii) At a point on the surface
...
Consider a point P′ inside the
shell at a distance r′ from the centre
of the shell
...
The total flux crossing the
Gaussian sphere normally in an
outward direction is
φ =
∫
s
E
...
20b Field at a point
inside the shell
since there is no charge enclosed by the gaussian surface, according to
Gauss’s Law
q
E × 4πr′ 2 = ε = 0
∴ E = 0
o
(i
...
1
...
6 Electrostatic shielding
It is the process of isolating a certain region of space from
external field
...
During a thunder accompanied by lightning, it is safer to sit
inside a bus than in open ground or under a tree
...
During lightning the electric discharge passes through the body of the
bus
...
5 Electrostatic induction
It is possible to obtain charges without any contact with another
charge
...
It is
used in electrostatic machines like Van de Graaff generator and
capacitors
...
21 shows the steps involved in charging a metal sphere by
induction
...
(b)
When
a
negatively
charged plastic rod is brought close
to the sphere, the free electrons
move away due to repulsion and
start pilling up at the farther end
...
This process of charge distribution
stops when the net force on the free
electron inside the metal is zero
(this process happens very fast)
...
The positive
charge at the near end remains
held due to attractive forces
...
21 Electrostatic Induction
(d) When the sphere is removed from the ground, the positive
charge continues to be held at the near end
...
1
...
1 Capacitance of a conductor
When a charge q is given to an isolated conductor, its potential
will change
...
The potential of a conductor changes by V, due to the
charge q given to the conductor
...
e
...
The capacitance of a conductor is defined as the ratio of the
charge given to the conductor to the potential developed in the
conductor
...
A conductor has a capacitance
of one farad, if a charge of 1 coulomb given to it, rises its potential by
1 volt
...
Principle of a capacitor
Consider an insulated conductor (Plate A) with a positive charge
‘q’ having potential V (Fig 1
...
The capacitance of A is C = q/V
...
An equal amount of
positive charge is induced on the other side of B (Fig 1
...
The
negative charge in B decreases the potential of A
...
But the negative charge on B is nearer
to A than the positive charge on B
...
Thus the capacitance of A is increased
...
22c)
...
Thus the
capacitance of A is considerably increased
...
A capacitor is a device for storing electric
charges
...
22 Principle of capacitor
26
1
...
2 Capacitance of a parallel plate capacitor
The parallel plate capacitor +q
X
consists of two parallel metal plates X
+
+
+
+
+
+
and Y each of area A, separated by a
d
distance d, having a surface charge
density σ (fig
...
23)
...
A charge
Y
+q is given to the plate X
...
23 Parallel plate
a charge –q on the upper surface of
capacitor
earthed plate Y
...
The electric lines of force starting from plate X and ending at the
plate Y are parallel to each other and perpendicular to the plates
...
C =
1
...
3 Dielectrics and polarisation
Dielectrics
A dielectric is an insulating material in which all the electrons are
tightly bound to the nucleus of the atom
...
Ebonite, mica and oil are few examples of dielectrics
...
27
Polarisation
A nonpolar
molecule is one
Electron
in which the
cloud
centre of gravity
-q
-q
+q
+q
of the positive
Electron
cloud
charges
(proE
tons)
coincide
Fig 1
...
Example: O2, N2, H2
...
If a non polar dielectric is placed in an electric field, the centre
of charges get displaced
...
They acquire induced dipole moment p
in the direction of electric field (Fig 1
...
A polar molecule is one in which the centre of gravity of the
positive charges is separated from the centre of gravity of the negative
charges by a finite distance
...
They have
a permanent dipole moment
...
Hence no net dipole moment is observed in the dielectric
...
Hence a net dipole moment is produced
(Fig 1
...
+ -
+ -
(a) No field
(b) In electric field
Fig1
...
The magnitude of the induced dipole moment p is directly
proportional to the external electric field E
...
1
...
4 Polarisation of dielectric material
E0
+ + + -
E
+ -
Ei
+qi
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
-qi
+ -
Consider a parallel plate
capacitor with +q and –q charges
...
If a dielectric slab
is introduced in the space between
them, the dielectric slab gets
polarised
...
26)
...
So, the resultant field,
E < Eo
...
P
Fig1
...
5
...
Consider a parallel plate capacitor having two conducting plates
X and Y each of area A, separated by a distance d apart
...
Let a dielectric slab of thick-ness t and relative permittivity εr be
introduced between the plates (Fig
...
27)
...
27 Dielectric in capacitor
σ
the dielectric slab E′ = ε ε
r o
The total potential difference between the plates, is the work done
in crossing unit positive charge from one plate to another in the field
E over a distance (d−t) and in the field E′ over a distance t, then
V
=
E (d−t) + E′ t
=
σ
σt
(d − t ) +
εo
εo εr
=
σ
εo
t ⎤
⎡
⎢(d − t ) + ε ⎥
⎣
r ⎦
The charge on the plate X, q = σA
Hence the capacitance of the capacitor is,
C
=
εo A
q
σA
=
=
t
t ⎤
V
σ ⎡
⎢(d − t ) + ε ⎥ (d − t ) + ε
εo ⎣
r
r ⎦
Effect of dielectric
In capacitors, the region between the two plates is filled with
dielectric like mica or oil
...
1
...
6 Applications of capacitors
...
(ii)
They are used to reduce voltage fluctuations in power
supplies and to increase the efficiency of power transmission
...
1
...
7 Capacitors in series and parallel
(i)
Capacitors in series
Consider three capacitors of capacitance C1, C2 and C3 connected
in series (Fig 1
...
Let V be the potential difference applied across the
series combination
...
Let V1, V2, V3 be the potential difference across the capacitors C1,
C2, C3 respectively
...
28 Capacitors in series
of the series combination, it should
acquire a charge q when a voltage V is applied across it
...
e
...
(ii)
Capacitors in parallel
Consider three capacitors of capacitances C1, C2 and C3
connected in parallel (Fig
...
29)
...
The potential difference across
each capacitor is the same
...
c1
The total charge in the system of
capacitors is
c2
q = q1 + q2 + q3
q = C1V + C2V + C3V
But q = Cp
...
29 Capacitors
of the capacitances of the individual
in parallel
capacitors
...
5
...
Work has to be done to
store the charges in a capacitor
...
Let q be the charge and V be the potential difference between the
plates of the capacitor
...
U=
1 q2 1
= CV 2
2 C
2
(∵ q = CV)
This energy is recovered if the capacitor is allowed to discharge
...
5
...
30)
...
A charge
given
to
the
system
is
distributed as q1 and q2 on the
surface of the spheres A and B
...
A
B
r1
r2
q2
q1
Fig 1
...
e
...
From the above
equation it is seen that, smaller the radius,
larger is the charge density
...
1
...
The
33
A
+ + + + +
+
+ ++
+
+
+C
++
+
+ + + + + +
Fig 1
...
It is found
experimentally that a charged conductor with sharp points on its
surface, loses its charge rapidly
...
The positive ions are repelled and the
negative ions are attracted by the sharp points and the charge in them
is therefore reduced
...
This principle is made use of in the electrostatic machines for collecting
charges and in lightning arresters (conductors)
...
6
Lightning conductor
This is a simple device used to protect tall buildings from the
lightning
...
The lower end of the rod is connected to a copper plate
buried deeply into the ground
...
When a negatively charged cloud passes over the building,
positive charge will be induced on the pointed conductor
...
This will partly
neutralize the negative charge of the cloud, thereby lowering the
potential of the cloud
...
Thereby preventing the lightning
stroke from the damage of the building
...
Van de Graaff designed an electrostatic
machine which produces large electrostatic potential difference of the
order of 107 V
...
A hollow metallic sphere A is mounted on insulating pillars as
34
+ + + +
A
B
+
E
+ + + +
+ + + +
+
+
+
shown in the Fig
...
32
...
A belt made of
silk moves over the
pulleys
...
Two
comb−shaped conductors
D and E having number
of needles, are mounted
near the pulleys
...
The upper
comb E is connected to
the inner side of the
hollow metal sphere
...
32 Van de Graaff Generator
Because of the high electric field near the comb D, the air gets
ionised due to action of points, the negative charges in air move
towards the needles and positive charges are repelled on towards the
belt
...
As a result of electrostatic induction, the comb E acquires
negative charge and the sphere acquires positive charge
...
The
high electric field at the comb E ionises the air
...
Hence the descending belt
will be left uncharged
...
As a result, the potential of the sphere keeps increasing till
it attains a limiting value (maximum)
...
The leakage of charge from the sphere can be reduced by
enclosing it in a gas filled steel chamber at a very high pressure
...
Solved Problems
1
...
They are brought in contact and then
separated
...
Data
: q1 = −3 × 10−12C, q2 = 8 × 10−12 C, q3 = 4 × 10−12 C
Solution : (i) The charge on each ball
q=
q1 + q 2 + q 3 ⎛ −3 + 8 + 4 ⎞
−12
=⎜
⎟ × 10
3
3
⎝
⎠
= 3 × 10−12 C
(ii) Since the charge is positive, there is a shortage of electrons on
each ball
...
875 × 107
e 1
...
875 × 107
...
2
Two insulated charged spheres of charges 6
...
5m
...
Also calculate the force (i) when the charges are
doubled and the distance of separation is halved
...
5 × 10−7C, r = 0
...
5 × 10 −7 )2
(0
...
52 × 10−2 N
...
= 16 × 1
...
24 N
When placed in water of εr = 80
F
=
1
...
9 × 10−4 N
1
...
Two small equal and unlike charges 2 ×10−8C are placed at A and B
at a distance of 6 cm
...
Data
q1 = +2 ×10−8C,
:
q2 = −2 × 10−8 C
q3 = 1 ×10−8 C at P
XP = 4 cm or 0
...
06 m
F
Solution :
q3= +1 x 10-8C
P
R
F
5cm
5cm
4cm
-8
q1= +2 x 10 C
-8
A
X
3cm
3cm
B
q2= -2 x 10 C
From ∆ APX, AP = 42 + 32 = 5 cm or 5 ×10−2 m
...
2 × 10−4 N along AP
...
To find R, we resolve the force into two components
∴R
1
...
2 × 10−4 ×
=
8
...
(Given : me = 9
...
67 × 10−27 kg ;
G = 6
...
6 × 10−19 C)
Solution :
The gravitational attraction between electron and proton is
Fg = G
me m p
r2
Let r be the average distance between electron and proton in
hydrogen atom
...
1
Fe = 4πε
o
∴
q 1q 2
r2
Fe
q1q 2
1
e2
1
=
=
Fg 4πε o Gme m P
4πε o Gme m P
9 × 109 × (1
...
67 × 10−11 × 9
...
67 × 10−27
2
=
Fe
Fg
= 2
...
27 × 1039 times stronger
than gravitational force
...
5
Two point charges +9e and +1e are kept at a distance of 16 cm from
each other
...
16 m;
Solution :
Let a third charge q be kept at a distance x from +
r
9e and (r – x) from + e
+e
+
q
+9e
+
q 1q 2
F = 4πε r 2
o
1
(r-x)
x
1 9e × q
1
q e
=
= 4πε o
4πε o (r − x )2
x2
∴
x2
=9
(r − x )2
or
x
=3
r −x
x = 3r – 3x
∴
4x = 3r = 3 × 16 = 48 cm
48
=12 cm or 0
...
12 m
from charge 9e
...
6
x=
Two charges 4 × 10−7 C and –8 ×10−7C are placed at the two corners
A and B of an equilateral triangle ABP of side 20 cm
...
Data
:
q1 = 4 × 10−7 C;
q2 = −8 ×10−7 C;
Solution :
r = 20 cm = 0
...
2)2
Electric field E2 along PB
...
04
1
E2 = 4πε
o
∴
E =
E12 + E22 + 2E1E2 cos120o
(
4
2
2
= 9 × 10 2 + 1 + 2 × 2 × 1 − 1 2
)
= 9 3 × 104 = 15
...
7
Calculate (i) the potential at a point due a charge of
4 × 10−7C located at 0
...
Data
: q1 = 4 × 10−7C, q2 = 2 × 10−9 C, r = 0
...
09
(ii) Work done in bringing a charge q2 from infinity to the point is
W = q2 V = 2 × 10−9 × 4 × 104
W = 8 × 10−5 J
1
...
5 × 104 N C−1
...
4 × 10−30 C m
...
Data
: E = 2
...
4 × 10−30 C m
...
4 × 10−30 × 2
...
5 × 10−26 N m
...
9
d
q1
Calculate the electric potential at
+12nc
a point P, located at the centre of
the square of point charges
d
shown in the figure
...
3 m
d
r
+31nc
q3
q2 = −24 n C; q3 = +31n C;
q2
-24nc
d=1
...
3
2
= 0
...
919
V
∴
=
=
352
...
10 Three charges – 2 × 10−9C, +3 × 10−9C, –4 × 10−9C are placed at the
vertices of an equilateral triangle ABC of side 20 cm
...
Data :
q1 = −2 × 10−9C;
/
A
q2 = +3 × 10−9C;
-9
10
-9
x
= 0
...
9 × 109
Ui =
0
...
5 × 10−7 J
9 × 109
Uf =
[−6 × 10−18 – 12 × 10−18 + 8 × 10−18]
0
...
5 × 10−7)
W = − 4
...
11 An infinite line charge produces a field of 9 × 104 N C−1 at a distance
of 2 cm
...
Data
: E = 9 × 104 N C−1, r = 2 cm = 2 × 10–2 m
λ
Solution : E = 2πε r
o
λ = E × 2πεor
= 9 × 104 ×
1
× 2 ×10−2
18 × 109
1
⎛
⎞
⎜∵ 2πε o =
⎟
18 × 109 ⎠
⎝
λ = 10−7 C m−1
1
...
(i) If the radius of the Gaussian surface is doubled,
how much flux will pass through the surface? (ii) What is the value
of charge?
Data
:
φ = −6 × 103 N m2 C−1;
42
r = 10 cm = 10 × 10−2 m
Solution :
(i)
If the radius of the Gaussian surface is doubled, the electric
flux through the new surface will be the same, as it depends
only on the net charge enclosed within and it is independent
of the radius
...
85 × 10−12 × 6 × 103)
o
q = − 5
...
13 A parallel plate capacitor has plates of area 200 cm2 and separation
between the plates 1 mm
...
Data:
d = 1 mm = 1 × 10−3m;
A = 200 cm2 or 200 × 10−4 m2 ;
q = 1 nC = 1 × 10−9 C ;
Solution : The capacitance of the capacitor
C =
εo A
d
=
8
...
177 × 10−9 F = 0
...
65 V
C 0
...
65 × 2
= 11
...
ε = 8
...
14 A parallel plate capacitor with air between the plates has a
capacitance of 8 pF
...
: Co = 8 pF , εr = 6, distance d becomes, d/2 with dielectric
Data
Solution : Co =
Aεo
= 8pF
d
when the distance is reduced to half and dielectric medium fills the
gap, the new capacitance will be
C
=
ε r Aε o
d /2
=
2ε r A ε o
d
= 2εr Co
C
= 2 × 6 × 8 = 96 pF
1
...
Data : C1 = 10µF ;
5µF ; C3 = 4µF
C1
10 F
C3
4 F
C2 =
5 F
C2
Solution : (i) C1 and C2 are
connected in series, the
effective capacitance of the
capacitor of the series combination is
1
1
1
=
+
CS C1 C2
=
∴
CS =
1 1
+
10 5
10 × 5 10
µF
=
10 + 5
3
(ii) This CS is connected to C3 in parallel
...
33 µF
1
...
5 mm
...
How much electrostatic energy is stored by
the capacitor?
Data : A = 90 cm2 = 90 × 10–4 m2 ;
d = 2
...
5 × 10–3 m;
V = 400 V
Solution : Capacitance of a parallel plate capacitor
C=
εo A
d
=
8
...
5 × 10 −3
= 3
...
186 × 10−11 × (400)2
Energy = 2
...
In the same way any question and problem could be framed from the text
matter
...
)
1
...
The
number of electrons it has gained or lost
(a) 5 × 10−7 (gained)
(c) 2 × 10−8 (lost)
1
...
3 N
...
8 N
1
...
5 N
(d) 2 N
Electic field intensity is 400 V m−1 at a distance of 2 m from a point
charge
...
4
(b) 4 cm
(d) 1
...
At what point
on the line joining them the electric field is zero?
(a) 15 cm from the charge q
(b) 7
...
5
A dipole is placed in a uniform electric field with its axis parallel to
the field
...
6
If a point lies at a distance x from the midpoint of the dipole, the
electric potential at this point is proportional to
(a)
1
x2
(b)
1
x3
(c)
1
x4
(d)
1
x 3/2
46
1
...
The electric potential at the centre
O of the square is
(a)
(c)
1
...
9
q 1q 2
(b) 4πε r
o
(d) pE sin θ
The work done in moving 500 µC charge between two points on
equipotential surface is
(a) zero
(b) finite positive
(c) finite negative
(d) infinite
1
...
11 The unit of permittivity is
(a) C2 N−1 m−2
(b) N m2 C−2
(c) H m−1
(d) N C−2 m−2
1
...
129 × 1011
(b) 1
...
25 × 1018
(d) 8
...
13 The electric field outside the plates of two oppositely charged plane
sheets of charge density σ is
(a)
+σ
2ε o
(b)
(c)
σ
εo
(d) zero
47
−σ
2ε o
1
...
The dielectric
constant of the dielectric is
(a) 65
(b) 55
(c) 12
(d) 10
1
...
16 State Coulomb’s law in electrostatics and represent it in vector form
...
17 What is permittivity and relative permittivity? How are they related?
1
...
1
...
Give its unit and obtain an expression
for the electric field at a point due to a point charge
...
20 Write the properties of lines of forces
...
21 What is an electric dipole? Define electric dipole moment?
1
...
1
...
24 Derive an expression for electric field due to an electric dipole (a) at a
point on its axial line (b) at a point along the equatorial line
...
25 Define electric potential at a point
...
1
...
1
...
28 What is electrostatic potential energy of a system of two point charges?
Deduce an expression for it
...
29 Derive an expression for electric potential due to an electric dipole
...
30 Define electric flux
...
48
1
...
Applying this, calculate electric field due to
(i) an infinitely long straight charge with uniform charge density
(ii) an infinite plane sheet of charge of q
...
32 What is a capacitor? Define its capacitance
...
33 Explain the principle of capacitor
...
1
...
1
...
If the dielectric
slab of thickness equal to half the plate separation is inserted between
the plates what happens to (i) capacitance of the capacitor (ii) electric
field between the plates (iii) potential difference between the plates
...
36 Deduce an expression for the equivalent capacitance of capacitors
connected in series and parallel
...
37 Prove that the energy stored in a parallel plate capacitor is
q2
...
38 What is meant by dielectric polarisation?
1
...
1
...
41 The sum of two point charges is 6 µ C
...
9 N, when kept 40 cm apart in vacuum
...
1
...
The charge on one sphere is twice that on the other
...
Calculate the charges and the initial distance between
them
...
43 Four charges +q, +2q, +q and –q are placed at the corners of a square
...
49
1
...
3 m apart
...
1
...
Find the (i) magnitude and direction of the force acting on
each charge
...
46 An electric dipole of charges 2 × 10−6 C, −2 × 10−6 C are separated by
a distance 1 cm
...
(i) axial line 1 m from its centre (ii) equatorial line 1 m from its
centre
...
47 Two charges +q and –3q are separated by a distance of 1 m
...
48 Three charges +1µ C, +3µ C and –5µ C are kept at the vertices of an
equilateral triangle of sides 60 cm
...
1
...
Find the work done in bringing them 4 cm closer, so that, they are
6 cm apart
...
50 Find the electric flux through each face of a hollow cube of side
10 cm, if a charge of 8
...
1
...
12 m has a charge of 1
...
What is the electric field
(i) inside the sphere (ii) on the sphere (iii) at a point 0
...
52 The area of each plate of a parallel plate capacitor is 4 × 10−2 sq m
...
Find the capacitance
of the capacitor
...
53 Two capacitors of unknown capacitances are connected in series and
parallel
...
1
...
5 µF and 0
...
Calculate the charge from the
source and charge on each capacitor
...
55 Three capacitors are connected in parallel to a
100 V battery as shown in figure
...
56 A parallel plate capacitor is maintained at some potential difference
...
To maintain the
plates at the same potential difference, the distance between the plates
is increased by 2
...
Find the dielectric constant of the slab
...
57 A dielectric of dielectric constant 3 fills three fourth of the space
between the plates of a parallel plate capacitor
...
58 Find the charges on the capacitor D
shown in figure and the potential
difference across them
...
59 Three capacitors each of capacitance 9 pF are connected in series (i)
What is the total capacitance of the combination? (ii) What is the
potential difference across each capacitor, if the combination is
connected to 120 V supply?
51
Answers
1
...
2 (c)
1
...
4 (c)
1
...
6 (a)
1
...
8 (b)
1
...
10 (d)
1
...
12 (a)
1
...
14 (c)
1
...
35
(i) increases (ii) remains the same (iii) remains the same
1
...
42
q1 = 33
...
66 ×10−9C, x = 0
...
43
0
...
44
V = 1800 V, E = 4000 Vm−1
1
...
866 × 10−11 Nm
1
...
47
x = 0
...
48
–0
...
49
5
...
50
1
...
51
zero, 105 N C–1, 4
...
52
2
...
53
C1 = 15 µF, C2 = 10µF
1
...
5 µC, q1 = 55 µC, q2 = 82
...
55
0
...
56
εr = 5
1
...
58
q1 = 144 × 10−6C, q2 = 96 × 10−6C, q3 = 48 × 10−6C
V1 = 72 V, V2 = 48 V
1
...
Current Electricity
The branch of Physics which deals with the study of motion of
electric charges is called current electricity
...
The thermodynamic internal energy of the material is
sufficient to liberate the outer electrons from individual atoms,
enabling the electrons to travel through the material
...
Hence, there is zero current
...
The external energy necessary to drive the free
electrons in a definite direction is called electromotive force (emf)
...
The flow of free electrons in a conductor
constitutes electric current
...
1 Electric current
The current is defined as the rate of flow of charges across any
cross sectional area of a conductor
...
The current I is expressed
in ampere
...
The direction of conventional
current is taken as the direction of flow of positive charges or opposite
to the direction of flow of electrons
...
1
...
1)
...
In the absence of an electric field, the
free electrons in the conductor move
randomly in all possible directions
...
1 Current carrying
conductor
They do not produce current
...
The electrons are accelerated and
in the process they collide with each other and with the positive ions
in the conductor
...
Drift velocity is defined as the velocity with which free electrons
get drifted towards the positive terminal, when an electric field is
applied
...
It takes the unit m2V–1s–1
...
It is
very small and is of the order of 0
...
2
...
2 Current density
Current density at a point is defined as the quantity of charge
passing per unit time through unit area, taken perpendicular to the
direction of flow of charge at that point
...
It is expressed in A m–2
* In this text book, the infinitesimally small current and instantaneous
currents are represented by the notation i and all other currents are
represented by the notation I
...
1
...
1)
...
Let n be the
number of free electrons per unit volume
...
The number of conduction electrons in the conductor = nAL
The charge of an electron = e
The total charge passing through the conductor q = (nAL) e
L
The time in which the charges pass through the conductor, t = v
d
The current flowing through the conductor, I =
I = nAevd
(nAL )e
q
= (L /v )
t
d
...
From equation (1),
I
= nevd
A
I
⎡
⎤
⎢∵ J = A ,current density ⎥
⎣
⎦
J = nevd
2
...
4 Ohm’s law
George Simon Ohm established the relationship between potential
difference and current, which is known as Ohm’s law
...
τ
m
But
vd
=
∴
I
= nAe
I
=
nAe 2
τV
mL
eE
τ
m
V⎤
⎡
⎢∵ E = L ⎥
⎣
⎦
where V is the potential difference
...
∴
I α V
55
The law states that, at a constant temperature, the steady
current flowing through a conductor is directly proportional to the
potential difference between the two ends of the conductor
...
e)
I α V
or I =
∴
V = IR or
R =
1
V
R
Y
V
I
Resistance of a conductor is defined as
the ratio of potential difference across the I
conductor to the current flowing through it
...
Its unit is mho (Ω–1)
...
2 V−I graph of an
Since,
potential
difference
V
is
ohmic conductor
...
2) between V and I is a straight line for a conductor
...
2
...
5 Electrical Resistivity and Conductivity
The resistance of a conductor R is directly proportional to the
length of the conductor l and is inversely proportional to its area of
cross section A
...
If l = l m, A = l m2, then ρ = R
The electrical resistivity of a material is defined as the resistance
offered to current flow by a conductor of unit length having unit area
of cross section
...
It is a constant for a
particular material
...
1
...
The materials can be broadly classified into conductors and
insulators
...
They carry
current without appreciable loss of energy
...
The resistivity of metals increase with increase in temperature
...
They offer very high resistance to the flow of current
and are termed non−conductors
...
In between these two classes of materials lie the
semiconductors (Table 2
...
They are partially conducting
...
Example : germanium,
silicon
...
1 Electrical resistivities at room temperature
(NOT FOR EXAMINATION)
Classification
conductors
Semiconductors
Insulators
Material
silver
copper
aluminium
iron
germanium
silicon
glass
wood
quartz
rubber
ρ (Ω m)
Ω
1
...
7 × 10−8
2
...
46
2300
1010 – 1014
108 – 1011
1013
1013 – 1016
2
...
The ability of certain metals, their compounds and
alloys to conduct electricity with zero resistance at very low
temperatures is called superconductivity
...
The phenomenon of superconductivity was first observed by
Kammerlingh Onnes in 1911
...
2 K (Fig 2
...
The first
theoretical explanation of superconductivity
was given by Bardeen, Cooper and Schrieffer
in 1957 and it is called the BCS theory
...
2 K
normal conductor to a superconductor is
T (K)
called the transition temperature or critical
Fig 2
...
At
the
transition
of mercury
temperature the following changes are
observed :
(i)
The electrical resistivity drops to zero
...
Applications of superconductors
(i)
Superconductors form the basis of energy saving power
systems, namely the superconducting generators, which are smaller in
size and weight, in comparison with conventional generators
...
They can be driven at high speed with minimal
expenditure of energy
...
(iv) High efficiency ore–separating machines may be built using
superconducting magnets which can be used to separate tumor cells
from healthy cells by high gradient magnetic separation method
...
(vi) Superconductors can be used as memory or storage
elements in computers
...
3 Carbon resistors
The wire wound resistors are expensive and huge in size
...
Carbon resistor consists of a ceramic core,
on which a thin layer of crystalline
Table 2
...
These
carbon resistors
resistors are cheaper, stable and
Colour
Number
small in size
...
2)
...
The
Orange
3
silver or gold ring at one end
Yellow
4
corresponds to the tolerance
...
The tolerance of silver,
Blue
6
gold, red and brown rings is 10%,
Violet
7
5%, 2% and 1% respectively
...
The
first two rings at the other end of
tolerance ring are significant figures of resistance in ohm
...
Example :
Violet
Orange Silver
Yellow
4 7 000
+ 10%
Fig 2
...
The first yellow ring in Fig 2
...
The next violet ring
corresponds to 7
...
The silver ring
represents 10% tolerance
...
e
...
Fig 2
...
Black
Presently four colour code carbon
resistors are also used
...
59
Brown
Red
1 0 00
Gold
±5 %
Fig 2
...
4 Combination of resistors
In simple circuits with resistors, Ohm’s law can be applied to find
the effective resistance
...
R
2
...
1 Resistors in series
Let us consider the I
resistors of resistances R1,
R2, R3 and R4 connected in
series as shown in Fig 2
...
R2
R1
V1
R3
V2
V3
R4
V4
Fig 2
...
If the potential difference applied
between the ends of the combination of resistors is V, then the
potential difference across each resistor R1, R2, R3 and R4 is V1, V2,
V3 and V4 respectively
...
Hence, IRS = IR1 + IR2 + IR3 + IR4
or
RS = R1 + R2 + R3 + R4
Thus, the equivalent resistance of a number of resistors in series
connection is equal to the sum of the resistance of individual resistors
...
4
...
7
...
When
resistors are in parallel, the potential
difference (V) across each resistor is
the same
...
60
V
I1
R1
R2
I2
A
I
R3
I3
I4
R4
R
Fig 2
...
∴
V
V
V
V
V
=
+
+
+
RP R1 R2 R3 R 4
1
1
1
1
1
=
+
+
+
R P R1 R 2 R3 R 4
Thus, when a number of resistors are connected in parallel, the
sum of the reciprocal of the resistance of the individual resistors is
equal to the reciprocal of the effective resistance of the combination
...
5 Temperature dependence of resistance
The resistivity of substances varies with temperature
...
If Ro
is the resistance of a conductor at 0o C and Rt is the resistance of same
conductor at to C, then
Rt = Ro (1 + αt)
where α is called the temperature
coefficient of resistance
...
Its
unit is per oC
...
8 Variation of
resistance with
temperature
The variation of resistance with temperature is shown in Fig 2
...
Metals have positive temperature coefficient of resistance,
i
...
, their resistance increases with increase in temperature
...
e
...
A material with a negative temperature coefficient is called a
thermistor
...
61
2
...
In order to maintain continuity, the current has to flow
through the electrolyte of the cell, from its negative terminal to positive
terminal
...
This
is termed as the internal resistance of the cell
...
Determination of internal resistance of a cell using voltmeter
+ V
The circuit connections are
made as shown in Fig 2
...
With
key K open, the emf of cell E is
E
found by connecting a high
resistance voltmeter across it
...
9 Internal resistance of a
may be considered as an open
cell using voltmeter
...
Hence the voltmeter
reading gives the emf of the cell
...
The potential
difference across R is equal to the potential difference across cell (V)
...
(1)
Due to internal resistance r of the cell, the voltmeter reads a
value V, less than the emf of cell
...
(2)
Dividing equation (2) by equation (1)
Ir E − V
=
IR
V
or
⎛ E −V
r = ⎜
⎝ V
⎞R
⎟
⎠
Since E, V and R are known, the internal resistance r of the cell
can be determined
...
7 Kirchoff’s law
Ohm’s law is applicable only for simple circuits
...
There
are two generalised laws : (i) Kirchoff’s current law (ii) Kirchoff’s
voltage law
Kirchoff’s first law (current law)
1
Kirchoff’s current law states that the
algebraic sum of the currents meeting at
any junction in a circuit is zero
...
Let 1,2,3,4 and 5 be the
conductors meeting at a junction O in an
Fig 2
...
10)
...
According to Kirchoff’s first law
...
The sum of the currents entering the junction is equal to the sum
of the currents leaving the junction
...
Kirchoff’s second law (voltage law)
Kirchoff’s voltage law states that the algebraic sum of the
products of resistance and current in each part of any closed circuit is
equal to the algebraic sum of the emf’s in that closed circuit
...
In applying Kirchoff’s laws to electrical networks, the direction of
current flow may be assumed either clockwise or anticlockwise
...
If the result is positive, then the assumed direction is the same as
actual direction
...
However,
in the application of Kirchoff’s second law, we follow that the current
in clockwise direction is taken as positive and the current in
anticlockwise direction is taken as negative
...
11a
...
I3
R6
r3
E3
I4 E
R5
I3
D
Fig 2
...
As an illustration of application of Kirchoff’s second law, let us
calculate the current in the following networks
...
2
...
(1)
60
132
132 I3 + 20I1 = 200
...
(3)
100V
I1
E
I2
D
Fig 2
...
(4)
Solving equations (3) and (4), we obtain
Il = 4
...
54 A
64
Illustration 2
Taking the current in the clockwise direction along ABCDA as
positive (FIg 2
...
5 I + 5 I + 0
...
5 I + 5 I + 0
...
5 + 5 + 0
...
5 + 5 + 0
...
25 A
40
∴
10
I
A
50V
5
70V
B
0
...
5
8
The
negative
sign
indicates that the current flows
in the anticlockwise direction
...
7
...
5
0
...
11c Kirchoff’s laws
of
Kirchoff’s
law
is
the
Wheatstone’s bridge (FIg 2
...
Wheatstone’s network consists of
resistances P, Q, R and S connected to form
B
I3
a closed path
...
The current I from
IG
the cell is divided into I1, I2, I3 and I4 across
I1
G
C the four branches
...
The resistance of
galvanometer is G
...
12
Wheatstone’s bridge
Applying
junction D
Kirchoff’s
I2 + Ig – I4 = 0
current
current
...
(3)
Applying Kirchoff’s voltage law to closed path ABCDA
I1P + I3Q – I4S – I2R = 0
65
to
law
to
...
(4)
When the galvanometer shows zero deflection, the points B and
D are at same potential and Ig = 0
...
(5)
I2 = I4
...
(7)
Substituting the values of (5) and (6) in equation (4)
I1P + I1Q – I2S – I2R = 0
I1 (P + Q) = I2 (R+S)
...
If P, Q and R are known,
the resistance S can be calculated
...
7
...
It consists
J
of thick strips of
A
C
l1
l2
copper, of negligible
resistance, fixed to
( )
a wooden board
...
13 Metre bridge
G1 and G2 between
these strips
...
An unknown resistance P is
connected in the gap G1 and a standard resistance Q is connected in
66
the gap G2 (Fig 2
...
A metal jockey J is connected to B through a
galvanometer (G) and a high resistance (HR) and it can make contact
at any point on the wire AC
...
Adjust the position of metal jockey on metre bridge wire so that
the galvanometer shows zero deflection
...
The
portions AJ and JC of the wire now replace the resistances R and S of
Wheatstone’s bridge
...
AJ
= =
Q S r
...
P AJ l1
=
=
Q JC l 2
where AJ = l1 and JC = l2
l1
∴
P = Ql
2
Though the connections between the resistances are made by
thick copper strips of negligible resistance, and the wire AC is also
∴
l1
soldered to such strips a small error will occur in the value of l due
2
to the end resistance
...
2
...
3 Determination of specific resistance
The specific resistance of the material of a wire is determined by
knowing the resistance (P), radius (r) and length (L) of the wire using
the expression ρ =
P πr 2
L
2
...
4 Determination of temperature coefficient of resistance
If R1 and R2 are the resistances of a given coil of wire at the
temperatures t1 and t2, then the temperature coefficient of resistance
of the material of the coil is determined using the relation,
R 2 − R1
α = Rt −R t
1 2
2 1
67
2
...
14)
...
14 Potentiometer
stretched in ten segments,
each of one metre length
...
The ends of the wire are fixed to
copper strips with binding screws
...
Electrical contact with wires is established by
pressing the jockey J
...
8
...
A steady current I
G
HR
flows
through
the
E
potentiometer wire (Fig
Fig 2
...
15)
...
A primary cell is connected in series with the positive
terminal A of the potentiometer, a galvanometer, high resistance and
jockey
...
If the potential difference between A and J is equal to the emf of
the cell, no current flows through the galvanometer
...
AJ is called the balancing length
...
∴ E = Irl,
since I and r are constants, E α l
Hence emf of the cell is directly proportional to its balancing
length
...
68
2
...
2 Comparison of emfs of two given cells using potentiometer
The potentiometer wire
AB is connected in series
( )
Rh
with a battery (Bt), Key (K),
K
Bt
I
rheostat (Rh) as shown in Fig
J
B
2
...
This forms the primary A
E1
circuit
...
The
E2
terminal D is connected to
Fig 2
...
The cell of emf E1 is
connected between terminals C1 and D1 and the cell of emf E2 is
connected between C2 and D2 of the DPDT switch
...
The DPDT switch is pressed towards C1, D1 so that cell E1 is
included in the secondary circuit
...
The balancing length is l1
...
Then, by
the principle of potentiometer,
E1 = Irl l
...
The balancing length l2
for zero deflection in galvanometer is determined
...
(2)
Dividing (1) and (2) we get
E1 l1
=
E2 l2
If emf of one cell (E1) is known, the emf of the other cell (E2) can
be calculated using the relation
...
8
...
The difference of potentials between the two terminals of a
cell in an open circuit is called the electromotive force (emf) of a cell
...
2
...
2
...
If I is the current flowing through a conductor of resistance R in
time t, then the quantity of charge flowing is, q = It
...
q = V It
...
∴
Power =
Work done
VIt
=
= VI
time
t
Electric power is the product of potential difference and current
strength
...
Its unit is
joule
...
1 kWh is known as one unit of electric energy
...
9
...
e energy absorbed in unit time by a circuit
...
A pointer is attached to the movable coil
...
When current flows
through the coils, the deflection of the pointer is directly proportional
to the power
...
10 Chemical effect of current
The passage of an electric current through a liquid causes
chemical changes and this process is called electrolysis
...
17)
...
The plates through
which current enters and leaves
an electrolyte are known as
+
Cathode
electrodes
...
17 Conduction in liquids
towards which negative ions
travel is called anode
...
The negative ions are called anions
...
10
...
First Law : The mass of a substance liberated at an electrode is
directly proportional to the charge passing through the electrolyte
...
According to the law, mass of
substance liberated (m) is
m α q
or
m = zIt
where Z is a constant for the substance being liberated called as
electrochemical equivalent
...
The electrochemical equivalent of a substance is defined as the
mass of substance liberated in electrolysis when one coulomb charge
is passed through the electrolyte
...
If E is the chemical equivalent of a substance, from the second
law
m αE
*Chemical equivalent =
Relative atomic mass
mass of the atom
= 1/12 of the mass C12 atom x valency
Valency
71
2
...
2 Verification of Faraday’s laws of electrolysis
First Law : A battery, a rheostat, a key and an ammeter are
connected in series to an electrolytic cell (Fig 2
...
The cathode is
cleaned, dried, weighed and
then inserted in the cell
...
The current is measured by
the ammeter
...
Hence the mass
Anode
m1 of the substance deposited
is obtained
...
18 Verification of Faraday’s
current I2 is passed for the
first law
same time t
...
It is found that
m1 I1
=
m2 I 2
∴
m αI
...
If the masses of the deposits are m3 and m4
respectively, it is found that
m3 t1
=
m4 t2
∴
m α t
...
Second Law : Two electrolytic cells containing different electrolytes, CuSO4 solution and AgNO3 solution are connected in series with a
battery, a rheostat and an ammeter (Fig 2
...
Copper electrodes are
inserted in CuSO4 and silver electrodes are inserted in AgNO3
...
The current is passed for some time
...
Hence the masses of copper and
silver deposited are found as m1 and m2
...
Bt
A
Thus, the
law is verified
...
11 Electric cells
Rh
Fig 2
...
It was
observed that, whenever the leg of the frog touched the iron railings, it
jumped and this led to the introduction of animal electricity
...
The battery Volta named after him consisted of a
pile of copper and zinc discs placed alternately separated by paper and
introduced in salt solution
...
His experiment established that, a cell could be made by using two
dissimilar metals and a salt solution which reacts with atleast one of the
metals as electrolyte
...
11
...
20)
...
20 Voltaic cell
73
Dilute H2SO4
from copper to zinc outside the cell and from zinc to copper inside it
...
The electrolyte is dilute sulphuric acid
...
At the zinc rod, the zinc atoms get ionized and pass into
solution as Zn++ ions
...
At the same time, two hydrogen ions (2H+) are
discharged at the copper rod, by taking these two electrons
...
As long as excess electrons are available on the
zinc electrode, this process goes on and a current flows continuously in
external circuit
...
Due to opposite charges on the
two plates, a potential difference is set up between copper and zinc,
copper being at a higher potential than zinc
...
08V
...
11
...
The primary cell is capable of
giving an emf, when its constituents, two electrodes and a suitable
electrolyte, are assembled together
...
These cells cannot be
recharged electrically
...
11
...
It
consists of a copper vessel
containing a strong solution of
copper sulphate (Fig 2
...
A zinc
rod is dipped in dilute sulphuric
acid contained in a porous pot
...
Zinc Rod
dilute H2SO4
Porous Pot
CuSO4 Solution
Copper Vessel
Fig 2
...
74
Zn++ ions pass through the pores of the porous pot and reacts
with copper sulphate solution, producing Cu++ ions
...
When Daniel cell is connected in a circuit,
the two electrons on the zinc rod pass through the external circuit and
reach the copper vessel thus neutralizing the copper ions
...
Daniel cell produces
an emf of 1
...
2
...
4 Leclanche cell
A
Leclanche
cell
consists
of
a
carbon
electrode packed in a porous
pot containing manganese
dioxide and charcoal powder
(Fig 2
...
The porous pot is
immersed in a saturated
solution
of
ammonium
chloride
(electrolyte)
contained in an outer glass
vessel
...
Carbon Rod
Mixture of MnO2
and Charcoal
Porous Pot
Zinc Rod
Ammonium
Chloride Solution
Glass Vessel
Fig 2
...
Zn++ ions reacting with ammonium
chloride produces zinc chloride and ammonia gas
...
e
Zn++ + 2 NH4Cl → 2NH3 + ZnCl2 + 2 H+ + 2e–
The ammonia gas escapes
...
In this
process the positive charge of hydrogen ion is transferred to carbon
rod
...
Thus current flows from carbon to zinc
...
The
emf of the cell is about 1
...
25 A
...
11
...
The chemical reactions that take place in secondary cells are reversible
...
The chemical process of obtaining current from a secondary
cell is called discharge
...
The most common secondary cells are lead acid
accumulator and alkali accumulator
...
11
...
The
Glass / Rubber container
container
contains
dilute sulphuric acid
Fig 2
...
Spongy lead (Pb) acts as the negative electrode and lead
oxide (PbO2) acts as the positive electrode (Fig 2
...
The electrodes are
separated by suitable insulating materials and assembled in a way to
give low internal resistance
...
The
electrons flow in the external circuit from negative electrode to positive
electrode where the reduction action takes place
...
This
makes the conventional current to flow from positive electrode to
negative electrode in the external circuit
...
2 Volt and the specific gravity
of the electrolyte is 1
...
The cell has low internal resistance and hence
can deliver high current
...
In the process of charging, the
chemical reactions are reversed
...
11
...
They have very low internal
resistance
...
They can
be recharged a very large number of times without any deterioration in
properties
...
They are used in all
automobiles like cars, two wheelers, trucks etc
...
It should lie between 1
...
12 during charging and discharging
respectively
...
1
If 6
...
(Given : Charge of an electron is
1
...
25 × 1018 ; e = 1
...
2
q ne 6
...
6 × 10−19
=
=
= 1 A
t
t
1
A copper wire of 10−6 m2 area of cross section, carries a current
of 2 A
...
(Given e = 1
...
6 × 10−19 C ; J = ? ; vd =?
Solution : Current density, J =
I
2
=
= 2 × 106A/m2
A 10 −6
J = n e vd
or vd =
2
...
6 × 10−5 m s–1
ne 8 × 1028 × 1
...
5 A
...
5 A ; R = ?
77
Solution : From Ohm’s law
V = IR
2
...
5
The resistance of a copper wire of length 5m is 0
...
If the
diameter of the wire is 0
...
Data : l = 5m ; R = 0
...
05 cm = 5 × 10−4 m
;
r = 2
...
14 × (2
...
9625 × 10−7 m2
ρ=
0
...
9625 × 10−7
5
ρ = 1
...
5
The resistance of a nichrome wire at 0o C is 10 Ω
...
004/oC, find its
resistance at boiling point of water
...
Data : At 0oC, Ro = 10 Ω ; α = 0
...
004 × 100))
Rt
= 14 Ω
As temperature increases the resistance of wire also increases
...
6
Two wires of same material and length have resistances 5 Ω and
10 Ω respectively
...
Data : Resistance of first wire R1 = 5 Ω ;
Radius of first wire = r1
Resistance of second wire R2 = 10 Ω
Radius of second wire = r2
Length of the wires = l
Specific resistance of the material of the wires = ρ
78
Solution : R =
ρl
A
; A = πr 2
ρl
ρl
2 ; R2 =
π r1
π r22
∴ R1 =
R 2 r12
=
or
R1 r22
R2
r1
10
2
=
=
=
r2
R1
5
1
r1 : r2 = 2 :1
2
...
1% longer, what is the
percentage change in resistance?
Data : Initial length of copper wire l1 = l
Final length of copper wire after stretching
l2 = l + 0
...
1
l
100
= l (1 + 0
...
001 l
During stretching, if length increases, area of cross section
decreases
...
001 A2l
Resistance of wire before stretching = R1
...
001 A2l
A1 = 1
...
001l
R2 =
A2
1
...
001) =1
...
002 – 1) = 0
...
002 × 100 = 0
...
8
The resistance of a field coil measures 50 Ω at 20oC and 65 Ω at
70oC
...
Data : At R20 = 50 Ω ; 70oC, R70 = 65 Ω ; α = ?
Solution : Rt = Ro (1 + α t)
R20 = Ro (1 + α 20)
50 = Ro (1 + α 20)
...
(2)
Dividing (2) by (1)
65 1 + 70α
=
50 1 + 20α
65 + 1300 α = 50 + 3500 α
2200 α = 15
α = 0
...
9
An iron box of 400 W power is used daily for 30 minutes
...
Data : Power of an iron box P
= 400 W
rate / unit
= 75 p
consumption time t
= 30 minutes / day
cost / week
= ?
Solution :
Energy consumed in 30 minutes = Power × time in hours
= 400 × ½ = 200 W h
80
Energy consumed in one week = 200 × 7 = 1400 Wh = 1
...
4 × 0
...
1
...
10 Three resistors are connected in series with 10 V supply as shown
in the figure
...
R1 5
R2 3
V1
V2
R3 2
V3
I
10V
Data :
R1 = 5Ω, R2 = 3Ω, R3 = 2Ω ; V = 10 volt
Effective resistance of series combination,
Rs = R1 + R2 + R3 = 10Ω
V
10
Solution : Current in circuit I = R = 10 = 1A
s
Voltage drop across R1, V1 = IR1 = 1 × 5 = 5V
Voltage drop across R2, V2 = IR2 = 1 × 3 = 3V
Voltage drop across R3, V3 = IR3 = 1 × 2 = 2V
2
...
Also find the effective
resistance and total current drawn from the supply
...
9677 Ω
I3
V 15
=
= 5A
Current through R1, I 1 =
R1
3
5
R2
I
2
R3
15V
81
Current through R2, I 2 =
V
15
=
= 3A
R2
5
V
15
Current through R3, I 3 = R = 2 = 7
...
9677 = 15
...
12 In the given network, calculate the effective resistance between
points A and B
(i)
10
5
10
5
5
10
A
B
5
10
5
10
10
5
Solution : The network has three identical units
...
5 Ω
Each unit has a resistance of 7
...
The total network reduces
7
...
5
7
...
5 + 7
...
5 = 22
...
13 A 10 Ω resistance is connected in series with a cell of emf 10V
...
9
...
Find internal resistance of the cell
...
9 V ;
r = ?
10V
⎛ E −V ⎞ R
⎟
⎝ V ⎠
10
Solution : r = ⎜
R
⎛ 10 − 9
...
9 ⎠
= ⎜
V
9
...
101 Ω
Self evaluation
(The questions and problems given in this self evaluation are only samples
...
Students must be prepared to answer any question and problem
from the text matter, not only from the self evaluation
...
1
A charge of 60 C passes through an electric lamp in 2 minutes
...
2
(c) 0
...
3
(b) 1 A
(b) mica
(c) germanium
(d) copper
The current flowing in a conductor is proportional to
(a) drift velocity
(b) 1/ area of cross section
(c) 1/no of electrons
(d) square of area of cross section
...
4
A toaster operating at 240V has a resistance of 120Ω
...
5
(b) 2 W
(c) 480 W
(d) 240 W
If the length of a copper wire has a certain resistance R, then on
doubling the length its specific resistance
(a) will be doubled
(c) will become 4 times
2
...
When two 2Ω resistances are in parallel, the effective resistance is
(a) 2 Ω
2
...
5 Ω
In the case of insulators, as the temperature decreases, resistivity
(a) decreases
(b) increases
83
(c) remains constant
2
...
004 /oC, then its
resistance at 100o C is
(a) 1
...
9
(d) becomes zero
(b) 0 Ω
(c) 4 Ω
(d) 2
...
10 When n resistors of equal resistances (R) are connected in series,
the effective resistance is
(a) n/R
(b) R/n
(c) 1/nR
(d) nR
2
...
12 Explain the flow of charges in a metallic conductor
...
13 Distinguish between drift velocity and mobility
...
2
...
2
...
How are materials classified based
on resistivity?
2
...
List some applications of
superconductors
...
17 The colours of a carbon resistor is orange, orange, orange
...
18 Explain the effective resistance of a series network and parallel
network
...
19 Discuss the variation of resistance with temperature with an
expression and a graph
...
20 Explain the determination of the internal resistance of a cell using
voltmeter
...
21 State and explain Kirchoff’s laws for electrical networks
...
22 Describe an experiment to find unknown resistance
temperature coefficient of resistance using metre bridge?
and
2
...
How will you find this using a
metre bridge?
84
2
...
How can emf of two cells
be compared using potentiometer?
2
...
26 State and Explain Faraday’s laws of electrolysis
...
27 Explain the reactions at the electrodes of (i) Daniel cell (ii) Leclanche
cell
2
...
(i) lead acid accumulator
2
...
30 What is the drift velocity of an electron in a copper conductor
having area 10 × 10−6m2, carrying a current of 2 A
...
2
...
6 × 10−19 C)
2
...
4 mm with a
resistance of 70 Ω
...
2
...
4Ω when two resistors are
connected in series and parallel
...
34 In the given circuit, what is the total resistance and current supplied
by the battery
...
35 Find the effective resistance between A and B in the given circuit
2
2
2
A
B
1
1
85
2
...
37 Calculate the current I1, I2 and I3 in the given electric circuit
...
38 The resistance of a platinum wire at 00 C is 4 Ω
...
0038 /0 C
...
39 A cell has a potential difference of 6 V in an open circuit, but it falls
to 4 V when a current of 2 A is drawn from it
...
2
...
Given P = 1000Ω
Q = 10000 Ω and R = 20 Ω
2
...
Find the electrical energy consumed
...
42 In a house, electric kettle of 1500 W is used everyday for 45
minutes, to boil water
...
3
...
43 A 1
...
Calculate the current and power supplied to it
...
44 In a metre bridge, the balancing length for a 10 Ω resistance in left
gap is 51
...
Find the unknown resistance and specific
resistance of a wire of length 108 cm and radius 0
...
86
2
...
A
5
R2
5
R1
3V
R3
B
5
C
4V
2
...
2
D
C
I1
B
I2
4
5V
(I1 +I2 )
I2
I1
E
10
F
A
Answers
2
...
2
(d)
2
...
4
(c)
2
...
6
(c)
2
...
8
(d)
2
...
10
(d)
2
...
30
1
...
31
80s
2
...
396 µ Ω m
2
...
34
3 Ω and 2A
2
...
33 Ω
2
...
37
0
...
25 A, 0
...
38
5
...
39
1 Ω
2
...
41
1 kWh
2
...
110
2
...
5 mA; 2
...
44
1
...
45
0
...
46
0
...
294A, 3
...
Effects of electric current
The ideas of electric current, electromotive force having been
already discussed in the preceding chapter, we shall discuss in this
chapter the physical consequences of electric current
...
In a source of emf, a
part of the energy may go into useful work like in an electric motor
...
This is the heating effect of current
...
This is thermoelectric effect
...
A steady electric current produces a
magnetic field in surrounding space
...
3
...
When a
voltage V is applied between the ends of the conductor, resulting in the
flow of current I, the free electrons are accelerated
...
The lattice ions or atoms
receive this energy VI from the colliding electrons in random bursts
...
Thus for a steady current I, the amount of heat produced in time t is
H = VIt
...
(2) and
2
V
t
...
By equation (2) Joule’s law implies
H =
88
that the heat produced is (i) directly proportional to the square of the
current for a given R (ii) directly proportional to resistance R for a given
I and (iii) directly proportional to the time of passage of current
...
3
...
1 Verification of Joule’s law
Joule’s law is verified using Joule’s
calorimeter
...
1)
...
A stirrer and a thermometer
T are inserted through two holes in the
lid
...
The
calorimeter is enclosed in a wooden box to
minimise loss of heat
...
1 Joule’s calorimeter
rheostat (Rh) and an ammeter (A) are
connected in series with the calorimeter
...
(i) Law of current
The initial temperature of water is measured as θ1
...
Now a current of I1 is
passed for a time of t (about 20 minutes)
...
The quantity of heat
gained by calorimeter and the contents is calculated as H1 = W (θ2−θ1)
...
The experiment is repeated by passing
currents I2, I3
...
, through the same coil for the same interval of
time t and the corresponding quantities of heat H2, H3 etc
...
It is found that
H1
I12
=
H2
I22
=
H3
I32
89
i
...
e H α I2
i
...
Hence, law of current is verified
...
The corresponding
quantities of heat gained H1, H2, H3 etc
...
It is found
that,
H1
R1
=
H2
H3
=
R3
R2
H
= constant
R
i
...
Hence, law of resistance is verified
...
The
corresponding quantities of heat gained H1, H2, H3 etc
...
It is found that
H1
H2
H3
=
=
t1
t2
t3
H
= constant
t
i
...
Hence, law of time is verified
...
1
...
In
these appliances, Nichrome which is an alloy of nickel and chromium
is used as the heating element for the following reasons
...
It is connected in
series in an electric circuit
...
When large current flows through a circuit due to short
circuiting, the fuse wire melts due to heating and hence the circuit
becomes open
...
(iii) Electric bulb
Since the resistance of the filament in the bulb is high, the
quantity of heat produced is also high
...
Tungsten with a high melting point
(3380oC) is used as the filament
...
Electric arc and electric welding also work on the principle of
heating effect of current
...
These devices are designed in such a way as to
reduce the loss of energy due to heating
...
1
...
Two dissimilar metals connected to form two junctions is called
thermocouple
...
The current through the circuit is called thermoelectric current
...
2 Seebeck effect
91
effect is called thermoelectric effect or Seebeck effect
...
Hence Seebeck effect is reversible
...
2a),
the direction of the current is from copper to iron at the hot
junction (Fig 3
...
The magnitude and sign of thermo emf depends on the materials
of the two conductors and the temperatures of the hot and cold
junctions
...
The direction of the current at the hot junction is from the metal
occurring earlier in the series to the one occurring later in the series
...
The thermo-electric series of metals is :
Bi, Ni, Pd, Pt, Cu, Mn, Hg, Pb, Sn, Au, Ag, Zn, Cd, Fe, Sb
...
The thermoemf of any thermocouple has the temperature
dependence given by the relation,
V = α θ + ½ β θ2,
where θ is the temperature difference between the junctions and
α and β are constants depending on the nature of the materials
...
1
...
3
...
For large difference in
C
n
i
temperature, the graph is a
Temperature of hot junction
parabola
...
3 Graph showing the variation
of thermo emf with temperature
Keeping the temperature of
the cold junction constant, the temperature of the hot junction is
gradually increased
...
Beyond the temperature of inversion,
the thermoemf changes sign and then increases
...
These temperatures are related by the expression
θc + θ i
2
= θn
3
...
5 Peltier effect
In 1834, a French scientist Peltier discovered that when electric
current is passed through a circuit consisting of two dissimilar metals,
heat is evolved at one junction and absorbed at the other junction
...
Peltier effect is the converse of Seebeck effect
...
4 Peltier effect
In a Cu-Fe thermocouple, at the junction 1 (Fig 3
...
When the direction of the current is
reversed (Fig 3
...
Hence Peltier effect is reversible
...
It is denoted by π
...
If H is the quantity of heat absorbed or evolved at one
junction then H = π It
The Peltier coefficient at a junction is the Peltier emf at that
junction
...
93
3
...
6 `Thomson effect
Thomson suggested that when a current flows through unequally
heated conductors, heat energy is absorbed or evolved throughout the
body of the metal
...
3
...
3
...
When no current is flowing, the
point M and N equidistant from C are at the same temperature
...
N shows higher temperature compared
to M
...
It
means from A to C heat is absorbed and from C to B heat is evolved
...
Similar effect is observed in
the case of Sb, Ag, Zn, Cd, etc
...
In the case of Iron (fig
...
5b), when it is heated at the point C and
current is flowing from A to B, M shows higher temperature as
compared to N
...
This is negative Thomson effect
...
If we take a bar of lead and heat it at the middle point C, the
point M and N equidistant from C show the same temperature when
current is flowing from A to B or from B to A
...
Due to this reason, lead is used as one
of the metals to form a thermo couple with other metals for the purpose
of drawing thermo electric diagrams
...
It is denoted by σ
...
3
...
7 Thermopile
Thermopile is a device used to detect thermal radiation
...
Bi
A
5
Sheildi
4
Incident
radiation
3
G
2
1
B
Sb
Fig 3
...
The ends are connected to a
galvanometer G (Fig
...
6)
...
The other set of
junctions (2,4) called cold junction is shielded from the radiation
...
The deflection in the galvanometer is
proportional to the intensity of radiation
...
2
Magnetic effect of current
In 1820, Danish Physicist, Hans Christian Oersted observed that
current through a wire caused a deflection in a nearby magnetic
needle
...
95
3
...
1 Magnetic field around a straight conductor carrying current
A smooth cardboard with iron filings spread
over it, is fixed in a horizontal plane with the help
of a clamp
...
7)
...
When the
cardboard is gently tapped, it is found that the iron
filings arrange themselves along concentric circles
...
Fig 3
...
When a compass needle is
placed, it comes to rest in such a way that its axis is always tangential
to a circular field around the conductor
...
8a) the direction of the magnetic field around the conductor looks
clockwise
...
8
(b) Current Outwards
When the direction of the current is reversed, that it is outwards,
(Fig 3
...
Now,
it is anticlockwise around the conductor
...
This is given by Maxwell’s rule
...
96
3
...
2 Magnetic field due to a circular loop carrying current
A cardboard
is
fixed
in
a
horizontal
plane
...
9
...
Current
is passed through
Fig 3
...
It is observed that the iron filings arrange themselves along the
resultant magnetic field
...
At the centre of
the loop, the line of force is almost straight and perpendicular to the
plane of the circular loop
...
The results of the experiments
are summarized as Biot-Savart law
...
10)
...
P is a point at a
distance r from the mid point O of
AB
...
3
O
r
I
P
X
Fig 3
...
4π
µ I dl sin θ
dB =
4π
r2
µ = µr µo where µr is the relative permeability of the medium and
is the permeability of free space
...
For air
= 1
...
dl sin θ
So, in air medium dB =
...
e plane of the paper) and acts inwards
...
3
...
1 Magnetic induction due to infinitely long straight conductor
carrying current
XY is an infinitely long straight conductor carrying a current I
(Fig 3
...
P is a point at a distance a from the conductor
...
θ is the angle between the current element I dl
and the line joining the element dl and the point P
...
sin θ
r2
...
OPA = φ,
APB = dφ
In ∆ ABC, sin θ =
B
AC
AC
=
dl
AB
C
dl
A
∴ AC = dl sin θ
r
...
(3)
From equations (2) and (3), rdφ=dl sinθ
...
(5)
r
4π
4π
r2
In ∆ OPA, cos φ =
∴
Y
d
2
a
O
P
1
I
a
r
a
r = cos φ
...
11 Straight
conductor
µo I
cos φ dφ
4π a
The total magnetic induction at P due to the conductor XY is
dB =
φ2
B =
∫
−φ
1
φ2
dB =
∫
φ
−1
µo I
cos φ dφ
4π a
µo I
[sin φ1 + sin φ2]
4π a
For infinitely long conductor, φ1 = φ2 = 90o
B =
µo I
2π a
If the conductor is placed in a medium of permeability µ,
∴
B =
µI
2π a
3
...
2 Magnetic induction along the axis of a circular coil carrying
current
B =
Let us consider a circular coil of radius ‘a’ with a current I as
shown in Fig 3
...
P is a point along the axis of the coil at a distance
x from the centre O of the coil
...
C
is the mid point of AB
and CP = r
According to Biot
–
Savart
law,
the
magnetic induction at P
due to the element dl is
A
dl
C
dB Cos
B
R
r
a
I
O
N
P
x
dB Sin
M
A/
B/
dB Cos
Fig
...
12 Circular coil
µo I dl sin θ
, where θ is the angle between Idl and r
4π
r2
Here, θ = 90o
µo I dl
∴
dB =
4π r2
The direction of dB is perpendicular to the current element Idl
and CP
...
dB =
Considering the diametrically opposite element A′B′, the
magnitude of dB at P due to this element is the same as that for AB
but its direction is along PM
...
dB is resolved into two components :- dB sin α along OP and
dB cos α perpendicular to OP
...
So, the total magnetic induction at P due to the entire coil is
µo Idl a
µo Ia
B = ∫ dB sin α = ∫
=
dl
4π r2 r
4π r3 ∫
µ o Ia
=
2πa
4π r 3
=
µ o Ia 2
(∵ r2 = a2 + x2)
3
2(a 2 +x 2 )2
If the coil contains n turns, the magnetic induction is
µo nIa 2
3
B =
2(a 2 +x 2 )2
At the centre of the coil, x = 0
µ o nI
B =
2a
100
3
...
3 Tangent galvanometer
Tangent galvanometer is
a device used for measuring
current
...
A
magnetic needle suspended at
a point where there are two
crossed fields at right angles to
each other will come to rest in
the direction of the resultant of
the two fields
...
The vertical frame is
mounted on a horizontal
circular turn table provided
Fig 3
...
The (This diagram need not be drawn in
vertical frame can be rotated
the examination)
about its vertical diameter
...
The compass box consists of a small pivoted magnet to which a
thin long aluminium pointer is fixed at right angles
...
The scale
consists of four quadrants
...
Since the magnetic field at the centre of the coil is uniform over
a very small area, a small magnetic needle is used so that it remains
in an uniform field even in deflected position
...
Theory
When the plane of the coil is placed parallel to the horizontal
component of Earth’s magnetic induction (Bh) and a current is passed
101
Bh
through the coil, there will be two magnetic
fields acting perpendicular to each other : (1)
the magnetic induction (B) due to the current
in the coil acting normal to the plane of the coil
and (2) the horizontal component of Earth’s
magnetic induction (Bh) (Fig 3
...
Due to these two crossed fields, the
pivoted magnetic needle is deflected through
an angle θ
...
14 Tangent law
...
(2)
2a
Substituting equation (2) in equation (1)
B =
µ o nI
= Bh tan θ
2a
2aBh
∴ I = µ n tan θ
o
I = K tan θ
...
It is a constant at a place
...
of
Since the tangent galvanometer is most sensitive at a deflection
the deflection has to be adjusted to be between 300 and 600
...
4 Ampere’s Circuital Law
Biot – Savart law expressed in an alternative way is called
Ampere’s circuital law
...
If L
B =
102
is the perimeter of the closed curve and Io is the net current enclosed
by the closed curve, then the above equation may be expressed as,
BL = µoIo
...
(2)
∫ B
...
If the current in the wire
is in the opposite direction, the integral would have the opposite sign
...
Although derived
for the case of a number of long straight parallel conductors, the law
is true for conductors and paths of any shape
...
dl for a closed curve is equal to µo times
the net current Io through the area bounded by the curve
...
4
...
Fig 3
...
The magnetic field due to
the solenoid is the vector sum of the
magnetic fields due to current
through individual turns of the
solenoid
...
15 Magnetic field due to a
current carrying solenoid
...
At the interior mid point, the field is
strong and along the axis of the solenoid (i
...
For a point such as P, the field due to the upper part of the solenoid
turns tend to cancel the field due to the lower part of the solenoid turns,
acting in opposite directions
...
The direction of the magnetic field due to circular closed
loops (solenoid) is given by right hand palm-rule
...
The extended thumb, points in the direction
of the magnetic field
...
4
...
Let us consider an infinitely long solenoid having
n turns per unit length carrying a current of I
...
16 Right
compared to its radius), the magnetic field at points
hand palm rule
outside the solenoid is zero
...
17)
...
The lower
Fig 3
...
uniform current sheet going
into the plane of the paper
...
The line integral ∫ B
...
∴
∫
→→
B
...
dl +
c
∫
→→
B
...
dl +
a
∫
→→
B
...
The second and fourth integrals are equal to zero because B is
→
at right angles for every element dl along the path
...
→→
∴
...
dl = Bl
Since the path of integration includes nl turns, the net current
enclosed by the closed loop is
104
Io = Inl
...
(3)
∫ B
...
(4)
The solenoid is commonly used to obtain uniform magnetic field
...
(5)
when a current carrying solenoid is freely suspended, it comes to rest
like a suspended bar magnet pointing along north-south
...
End rule
When looked
from one end, if the
S
N
N
S
current through the
(a)
(b)
Fig 3
...
18a, the nearer end corresponds to south pole
and the other end is north pole
...
18b)
3
...
19 Lorentz force
105
(b)
Y
Let us consider a uniform magnetic field of induction B acting
along the Z-axis
...
19a)
...
H
...
Lorentz formulated the special features of the force F
(Magnetic lorentz force) as under :
(i) the force F on the charge is zero, if the charge is at rest
...
e)
the moving charges alone are affected by the magnetic field
...
(iii) the force is proportional to the magnitude of the charge (q)
(iv) the force is proportional to the magnetic induction (B)
(v) the force is proportional to the speed of the charge (v)
(vi) the direction of the force is oppositely directed for charges of
opposite sign (Fig 3
...
All these results are combined in a single expression as
→
→ →
F = q ( v × B)
The magnitude of the force is
F = Bqv sin θ
Since the force always acts perpendicular to the direction of
motion of the charge, the force does not do any work
...
5
...
Let us consider a uniform magnetic field of induction B acting
along the Z-axis
...
At a point P, the velocity of the particle is v
...
20)
→
→ →
The magnetic lorentz force on the particle is F = q ( v × B)
...
Since the force acts perpendicular to its velocity, the force does not do
any work
...
The force F
acting towards the point O acts as
the centripetal force and makes the
particle to move along a circular
path
...
→
→
Since v and B are at right
angles to each other
Z
B
R
Q
F
B
Y
P
This magnetic lorentz force
provides the necessary centripetal
force
...
20 Motion of a
charged particle
=
...
(2)
m
This equation gives the angular frequency of the particle inside
the magnetic field
...
(3)
From equations (2) and (3), it is evident that the angular
frequency and period of rotation of the particle in the magnetic field do
not depend upon (i) the velocity of the particle and (ii) radius of the
circular path
...
5
...
It was devised by Lawrence
...
Construction
D
...
21)
...
They are placed between the pole
pieces
of
a strong electromagnet
...
The Dees are connected to a
high frequency oscillator
...
21 Cyclotron
Working
When a positive ion of charge q and mass m is emitted from the
source, it is accelerated towards the Dee having a negative potential at
that instant of time
...
By the
time the ion arrives at the gap between the Dees, the polarity of the
Dees gets reversed
...
Thus the particle moves in a spiral path of increasing radius
and when it comes near the edge, it is taken out with the help of a
deflector plate (D
...
The particle with high energy is now allowed to hit
the target T
...
108
Bqv =
∴
mv 2
r
Bq
v
=
= constant
r
m
...
(4)
So, in a uniform magnetic field, the ion traverses all the circles
in exactly the same time
...
Cyclotron
α - particles
...
(ii) At high velocities, relativistic variation of mass of the particle
upsets the resonance condition
...
109
3
...
Let us consider a conductor PQ
of length l and area of cross section A
...
22]
...
Hence, the
electrons are drifted along QP with
drift velocity vd
...
Therefore the current element,
→
→
Il = –nAvdel
Fig 3
...
(1)
The negative sign in the equation indicates that the direction of
current is opposite to the direction of drift velocity of the electrons
...
→
→
→
f = –e (vd × B)
…(2)
The negative sign indicates that the charge of the electron is
negative
...
(3)
The magnetic lorentz force on all the moving free electrons
→ →
F = Nf
Substituting equations (2) and (3) in the above equation
→
→
→
F = nAl { –e (vd × B) }
→
→
→
...
Magnitude of the force
The magnitude of the force is F = BIl sin θ
(i) If the conductor is placed along the direction of the magnetic
field, θ = 0o, Therefore force F = 0
...
Therefore the conductor experiences maximum force
...
The forefinger, the middle finger and the thumb of the left hand
are stretched in mutually perpendicular directions
...
3
...
1 Force between two long parallel current-carrying
conductors
AB and CD are two straight
very long parallel conductors placed
I1
in air at a distance a
...
(Fig 3
...
The magnetic induction
due to current I1 in AB at a distance
B2
outwards
a is
B
D
I2
F
F
a
B1
inwards
µo I1
B1 =
...
3
...
The conductor
long parallel current-carrying
conductors
CD with current I2 is situated in this
magnetic field
...
(2)
2π a
By Fleming’s Left Hand Rule, F acts towards left
...
(3)
2π a
This magnetic field acts perpendicular to the plane of the paper
and outwards
...
Hence force on a segment of length l of AB due to magnetic field
B2 is
F = B2I1l
substituting equation (3)
µ o I1I2l
∴
F =
…(4)
2π a
By Fleming’s left hand rule, this force acts towards right
...
Hence, two
parallel wires carrying currents in the same direction attract each other
and if they carry currents in the opposite direction, repel each other
...
Ampere is defined as that constant current which when flowing
through two parallel infinitely long straight conductors of negligible
cross section and placed in air or vacuum at a distance of one metre
apart, experience a force of 2 × 10-7 newton per unit length of the
conductor
...
7
Torque experienced by a current loop in a uniform magnetic
field
Let us consider a rectangular loop PQRS of length l and
breadth b (Fig 3
...
It carries a current of I along PQRS
...
Let θ be the angle
between the normal to the plane of the loop and the direction of the
magnetic field
...
24 Torque on a current loop
placed in a magnetic field
Fig 3
...
Magnitude of the force F2 = BIb cos θ
The forces F1 and F2 are equal in magnitude, opposite in direction
and have the same line of action
...
→
→
Force on the arm PQ, F3 = I(PQ) × B
→
Since the angle between I(PQ) and B is 90o,
113
Magnitude of the force F3 = BIl sin 90o = BIl
F3 acts perpendicular to the plane of the paper and outwards
...
The forces F3 and F4 are equal in magnitude, opposite in direction
and have different lines of action
...
Hence, Torque
= BIl × PN = BIl × PS × sin θ (Fig 3
...
3
...
1 Moving coil galvanometer
Moving coil galvanometer is a device used for measuring the
current in a circuit
...
Construction
It consists of a rectangular coil of a large number of turns of thin
insulated copper wire wound over a light metallic frame (Fig 3
...
The
coil is suspended between the pole pieces of a horse-shoe magnet by a
fine phosphor – bronze strip from a movable torsion head
...
The other end of the spring is connected to
a binding screw
...
The hemispherical magnetic poles produce a radial magnetic field
in which the plane of the coil is parallel to the magnetic field in all its
positions (Fig 3
...
114
A small plane mirror (m) attached to the suspension wire is used
along with a lamp and scale arrangement to measure the deflection of
the coil
...
26 Moving coil galvanometer
S
Fig 3
...
28)
...
In a radial magnetic field, the plane of the coil is
always parallel to the magnetic field
...
So, they do not experience any force
...
PQ = RS = l, length of the coil and PS = QR = b, breadth of the
coil
Force on PQ, F = BI (PQ) = BIl
...
P
S
F
I
F
B
B
P
b
F
F
R
Q
Torque on the coil
Fig 3
...
29
115
S
Force on RS, F = BI (RS) = BIl
...
These two equal, oppositely directed parallel forces having different
lines of action constitute a couple and deflect the coil
...
29)
= nBIA
When the coil deflects, the suspension wire is twisted
...
This
couple is proportional to the twist
...
nBA
i
...
Since the deflection is directly proportional to the current
flowing through the coil, the scale is linear and is calibrated to give
directly the value of the current
...
7
...
They can
measure current of the order of 10-8 ampere
...
So, in the laboratory, for
experiments like Wheatstone’s bridge, where sensitivity is not required,
pointer type galvanometers are used
...
A lighter aluminium pointer attached
to the coil moves over a scale when current is passed
...
3
...
3 Current sensitivity of a galvanometer
...
A galvanometer is said to be sensitive if it produces large
deflection for a small current
...
This explains why phosphor-bronze wire is used as the suspension wire
which has small couple per unit twist
...
7
...
θ
nBA
θ
=
=
IG
CG
V
where G is the galvanometer resistance
...
(2)
An interesting point to note is that, increasing the current
sensitivity does not necessarily, increase the voltage sensitivity
...
But increasing the number of turns correspondingly
increases the resistance (G)
...
3
...
5 Conversion of galvanometer into an ammeter
A galvanometer is a device used to detect the flow of current in
an electrical circuit
...
Being
a very sensitive instrument, a large current cannot be passed through
the galvanometer, as it may damage the coil
...
As a result, when large current flows in a circuit, only a small
fraction of the current passes through the galvanometer and the
remaining larger portion of the current passes through the low
117
Ig
I
I-Ig
resistance
...
The scale is marked
in ampere
...
30 Conversion of galvanometer
the galvanometer
...
The current Ig will give full scale
deflection in the galvanometer
...
∴
Ig
...
...
When connected in series, the ammeter does not
appreciably change the resistance and current in the circuit
...
118
3
...
6 Conversion of galvanometer into a voltmeter
Voltmeter is an instrument
used to measure potential difference
between the two ends of a current
carrying conductor
...
31 Conversion of
connecting a high resistance in
galvanometer into voltmeter
series with it
...
The value of the resistance
connected in series decides the range of the voltmeter
...
The effective resistance of the voltmeter is
Rv = G + R
Rv is very large, and hence a voltmeter is connected in parallel in
a circuit as it draws the least current from the circuit
...
Otherwise, the voltmeter will draw a large current
from the circuit and hence the current through the remaining part of
the circuit decreases
...
The error is eliminated only when the voltmeter has a high resistance
...
119
3
...
This is evident from the fact that a compass needle when
moved around these two bodies show similar deflections
...
This is
Ampere’s hypothesis
...
For points which are far away from the centre of
the coil, x>>a, a2 is small and it is neglected
...
(1)
The magnetic induction at a point along the axial line of a short
bar magnet is
B =
µo
2M
...
3
2π
x
...
(3)
Hence a current loop is equivalent to a magnetic dipole of
moment M = IA
The magnetic moment of a current loop is defined as the product
of the current and the loop area
...
120
3
...
The revolving electron in a closed path constitutes an
electric current
...
Current, i =
e
where T is the period of revolution of the electron
...
πr2
2π r
evr
µl
=
2
If m is the mass of the electron
µl =
e
(mvr)
2m
mvr is the angular momentum (l) of the electron about the
central nucleus
...
Its value
2m
is 8
...
Bohr hypothesised that the angular momentum
has only discrete set of values given by the equation
...
(2) where n is a natural number
and h is the Planck’s constant = 6
...
substituting equation (2) in equation (1)
121
µl =
e
nh
neh
...
27 × 10–24 Am2
In addition to the magnetic moment due to its orbital motion, the
electron possesses magnetic moment due to its spin
...
Solved problems
3
...
Find the rise in temperature of water, if
the current passes for 20 minutes and the potential difference
across the coil is 9 volt
...
5oC
...
2
Calculate the resistance of the filament of a 100 W,
electric bulb
...
If the voltage drops to
180 V, calculate the power consumed by the heater
...
3
Data : P1 = 1500 W, V1 = 220 V, V2 = 180 V, P2 = ?
V12
R
Solution : (i) P1 =
V12
∴ R = P
1
(220)2
= 32
...
26
R
∴ P2 = 1004 Watt
∴ P2 =
Aliter
V2
V12
, P2 = 2
R
R
P1
V12
= 2
P2
V2
P1 =
∴
2
V2
(180)2
= 1500 ×
(220)2
V12
∴ P2 = 1004 Watt
...
4
P2 = P 1 ×
A long straight wire carrying current produces a magnetic
induction of 4 × 10-6T at a point, 15 cm from the wire
...
Data : B = 4 × 10-6T, a=15 x 10-2m, I=?
Solution : B =
∴I =
µo I
2π a
B × 2π a
µo
=
4 × 10 −6 × 2π × 15 × 10 −2
4π × 10 −7
∴ I = 3A
123
3
...
Calculate the magnetic induction at a point along its axis,
at a distance three times the radius of the coil from its centre
...
9 x 10-5 T
3
...
If the horizontal
component of Earth’s magnetic induction is 4 × 10-5 T, find the
deflection produced in the coil
Data : n = 5; I = 4A; d = 3 × 10–1 m; Bh = 4 × 10–5 T;
a = 1
...
5 × 10−1 × 4 × 10−5
2aB h
tan θ = 2
...
7
In a tangent galvanometer, a current of 1A produces a deflection
of 300
...
Data :
I1 = 1A;
θ1 = 300;
Solution : I1 = k tan θ1 ;
∴
θ2 = 600;
I2 = ?
I2 = k tan θ2
I 2 tan θ 2
=
I1 tan θ1
I2 = I1
tan60o
1× 3
=
= 3 3 = 3A
×
tan30o ⎛ 1 ⎞
⎜
3⎟
⎝
⎠
I2 = 3A
124
3
...
It has 5 layers of
windings of 1000 turns each and carries a current of 5A
...
Data :
l = 2m, N = 5 × 1000 turns, I = 5A, B = ?
Solution : B = µo nI = µo
B =
N
...
57 x 10-2 T
3
...
Find the
force on the particle
...
6 × 10-19 × 5 × 105 ×
F
1
2
= 8 × 10-18N
3
...
A uniform magnetic field of induction
10-3 T acts along the Z-axis
...
(Mass of deuteron is 3
...
6 x 10-19C)
Data : v = 104 ms–1, B = 10–3T, m = 3
...
6 x 10-19C, r = ?
Solution : Bev =
∴ r=
mv2
r
mv
3
...
08 × 10–1
Be
10−3 × 1
...
208m
125
3
...
5 T acts perpendicular
to the plane of the Dees of a cyclotron
...
(mass of proton =
1
...
5 T, mp = 1
...
6 × 10-19C, ν = ?
Bq
Solution: ν = 2π m
p
=
0
...
6 × 10−19
= 0
...
63 × 106 Hz
2 × 3
...
67 × 10−27
∴ ν = 7
...
12 A conductor of length 50 cm carrying a current of 5A is placed
perpendicular to a magnetic field of induction 2 × 10-3 T
...
Data : l = 50 cm = 5×10-1m, I= 5A, B = 2×10-3T; θ = 90o, F = ?
Solution: F
= BIl sinθ
= 2 × 10-3 × 5 × 5 × 10-1 × sin 900
∴ F = 5 × 10-3 N
3
...
They carry equal currents along the same
direction and experience a mutually attractive force of
3
...
Find the current through the conductors
...
6×10-4N, I = ?
Solution: F
=
F
=
∴I2 =
∴ I
µo I1I 2l
2π a
2 × 10−7 I 2l
a
F
...
6 × 10−4 × 10−1
= 36
−7 =
2 × 10 l
2 × 10−7 × 5
= 6A
126
3
...
Find the magnitude and
direction of the resultant force on
the conductor B
...
4A
5A
3A
10 cm
F1 acts towards left
10 cm
−7
F1 =
F2
F1
−7
2 × 10 I1I 2l
2×10 ×3×4×10
=
a
10−1
A
F1 = 24 × 10-5 N
B
C
Between the wires B and C, force of attraction exists
F2 acts towards right
F2 =
2 × 10−7 I1I 2l
2×10−7 ×4×5×10
=
a
10−1
F2 = 40 × 10-5 N
F2 – F1 = 16 × 10-5 N
The wire B is attracted towards C with a net force of 16 × 10-5 N
...
15 A rectangular coil of area 20 cm × 10 cm with 100 turns of wire
is suspended in a radial magnetic field of induction 5 × 10-3 T
...
Data :
n = 100, A = 20 cm × 10 cm = 2 × 10-1 × 10-1 m2
B = 5 × 10-3 T, θ = 150, I = 1mA = 10-3A, C = ?
Solution : θ = 150 =
π
180
× 15 =
π
12
rad
nBIA = Cθ
102 × 5 × 10-3 × 10-3 × 2 × 10-1 × 10-1
nBIA
=
⎛ π ⎞
θ
⎜ ⎟
⎝ 12 ⎠
C = 3
...
16 A moving coil galvanometer of resistance 20 Ω produces full scale
deflection for a current of 50 mA
...
Data :
G = 20 Ω ; Ig = 50 x 10-3 A ; I = 20 A, S = ?
V = 120 V, R = ?
Ig
1
20 × 50 × 10-3
Solution : (i) S = G
...
05
g
20 - 50 × 10
S = 0
...
05 Ω should be connected in parallel
V
(ii) R = Ig – G
=
120
– 20 = 2400-20 = 2380 Ω
50 × 10-3
R = 2380 Ω
A resistance of 2380 Ω should be connected in series with the
galvanometer
...
17 The deflection in a galvanometer falls from 50 divisions to 10
divisions when 12 Ω resistance is connected across the
galvanometer
...
Data :
θ1 = 50 divs, θg = 10 divs, S = 12Ω G = ?
Solution : I α θ1
Ig α θg
In a parallel circuit potential is common
...
Ig = S (I-Ig)
∴ G =
S (I - Ig )
Ig
=
12 (50 - 10)
10
∴ G = 48 Ω
3
...
5 Å
making 1016 revolutions per second
...
128
Data : r = 0
...
5x10-10 m, n = 1016 s-1
Solution :
Orbital magnetic moment µl = i
...
(1)
i =
e
= e
...
(2)
A = πr2
...
n
...
6 × 10-19 × 1016 × 3
...
5 × 10-10)2
= 1
...
256 × 10-23 Am2
Self evaluation
(The questions and problems given in this self evaluation are only samples
...
Students must be prepared to answer any question and problem
from the text matter, not only from the self evaluation
...
1
Joule’s law of heating is
I2
(a) H =
t
R
(c) H = VIt
(b) H = V2 Rt
(d) H = IR2t
3
...
3
Peltier coefficient at a junction of a thermocouple depends on
(a) the current in the thermocouple
(b) the time for which current flows
(c ) the temperature of the junction
(d) the charge that passes through the thermocouple
3
...
The temperature of inversion is
(a) 520oC
(b) 540oC
oC
(c) 500
(d) 510oC
129
3
...
The plane of the coil is rotated through 900
...
6
3
...
8
(b) 600
(d) 00
The period of revolution of a charged particle inside a cyclotron
does not depend on
(a) the magnetic induction
(c) the velocity of the particle
3
...
10 Phosphor – bronze wire is used for suspension in a moving coil
galvanometer, because it has
(a) high conductivity
(b) high resistivity
(c) large couple per unit twist
(d) small couple per unit twist
3
...
12 A galvanometer of resistance G Ω is shunted with S Ω
...
Then, which of the following
statements is true?
(a) G is less than S
(b) S is less than Ra but greater than G
...
13 An ideal voltmeter has
(a) zero resistance
(b) finite resistance less than G but greater than Zero
(c) resistance greater than G but less than infinity
(d) infinite resistance
3
...
15 Explain Joule’s calorimeter experiment to verify Joule’s laws of
heating
...
16 Define Peltier coefficient
3
...
18 State Biot – Savart law
3
...
3
...
3
...
3
...
23 Applying Amperes circuital law, find the magnetic induction due to
a straight solenoid
...
24 Define ampere
3
...
3
...
131
3
...
Problems
3
...
If the temperature of the
cold junction is 20oC, find the temperature of inversion
...
29 Find the magnetic induction at a point, 10 cm from a long straight
wire carrying a current of 10A
3
...
Find the magnetic induction at a point along its axis
at a distance of 20 cm from the centre of the coil
...
31 Three tangent galvanometers have turns ratio of 2:3:5
...
Find the ratio of their radii
...
32 A straight wire of length one metre and of resistance 2 Ω is
connected across a battery of emf 12V
...
Find the force on the
wire
...
33 A circular coil of 50 turns and radius 25 cm carries a current of 6A
...
The
normal to the plane of the coil makes an angle of 600 with the field
...
3
...
5 T is applied normal to the plane of the
Dees of a Cyclotron
...
3 × 10-27 kg and its charge = 1
...
3
...
of 10o
...
36 Two straight infinitely long parallel wires carrying equal currents
and placed at a distance of 20 cm apart in air experience a mutally
attractive force of 4
...
Calculate
the current
...
37 A long solenoid of length 3m has 4000 turns
...
132
3
...
A shunt resistance 1 Ω
is connected across it
...
39 A galvanometer has a resistance of 40 Ω
...
How you will convert the
galvanometer into a voltmeter of range 0 to 20V?
3
...
1 m A/division
...
If a shunt resistance 0
...
Answers
3
...
2
(c)
3
...
4
(a)
3
...
6
(d)
3
...
8
(c)
3
...
10 (d)
3
...
13 (d)
3
...
28
560 C
3
...
55 × 10
3
...
34
2
...
36
-5
3
...
31
6 : 3 √3 : 5
3
...
1 × 10
3
...
166 m A
7 A
3
...
77 A
3
...
39
9960 Ω in series
3
...
Electromagnetic Induction and
Alternating Current
In the year 1820, Hans Christian Oersted demonstrated that a
current carrying conductor is associated with a magnetic field
...
4
...
He explained the possibility of producing emf across the
ends of a conductor when the magnetic flux linked with the conductor
changes
...
The discovery
of this phenomenon brought about a revolution in the field of power
generation
...
1
...
1)
...
A
A
B
Fig 4
...
1
...
Whenever there is a change in the magnetic flux linked with a
closed circuit an emf is produced
...
The phenomenon of producing an induced emf due to the
changes in the magnetic flux associated with a closed circuit is known
as electromagnetic induction
...
Fig
4
...
A strong bar
magnet NS with its north pole
pointing towards the coil is moved
up
and
down
...
G
C
N
S
Fig 4
...
(ii)
The deflection is momentary
...
(iii)
The direction of the flow of current changes if the
magnet is moved towards and withdrawn from it
...
(v)
However, on reversing the magnet (i
...
C
C1
1
C
C2
2
Faraday
demonstrated the electromagnetic induction by
another experiment also
...
3 shows two
coils C1 and C2 placed
close to each other
...
Coil C2 is connected to a
sensitive galvanometer G and kept close to C1
...
3 Electromagnetic Induction
135
sudden momentary deflection
...
This is because when the current in C1 increases from zero
to a certain steady value, the magnetic flux linked with the coil C1
increases
...
This causes the deflection in the galvanometer
...
This indicates that a current is again induced in the coil C2
...
Hence,
the magnetic flux linked with the coil C2 also decreases
...
4
...
3 Faraday’s laws of electromagnetic induction
Based on his studies on the phenomenon of electromagnetic
induction, Faraday proposed the following two laws
...
The induced emf lasts
so long as the change in magnetic flux continues
...
Let φ1 be the magnetic flux linked with the coil initially and φ2 be
the magnetic flux linked with the coil after a time t
...
If dφ is the change in magnetic flux in a time dt,
t
dφ
then the above equation can be written as e α
dt
4
...
4 Lenz’s law
The Russian scientist H
...
Lenz in 1835 discovered a simple
law giving the direction of the induced current produced in a circuit
...
136
If the coil has N number of turns and φ is the magnetic flux
linked with each turn of the coil then, the total magnetic flux linked
with the coil at any time is Nφ
N (φ2 − φ1 )
d
Ndφ
(Nφ) e = –
= –
t
dt
dt
Lenz’s law - a consequence of conservation of energy
∴
e = –
Copper coils are wound on a cylindrical
cardboard and the two ends of the coil are
connected to a sensitive galvanometer
...
4)
...
S
N
Consequently work has to be done to move
the magnet further against the force of repulsion
...
Now the
G
workdone is against the force of attraction
...
The
Fig 4
...
The workdone
in moving the magnet is converted into electrical energy
...
If on the contrary, the direction
of the current were to help the motion of the magnet, it would start
moving faster increasing the change of magnetic flux linking the coil
...
Hence kinetic energy
and electrical energy would be produced without any external work
being done, but this is impossible
...
Thus it is proved
that Lenz’s law is the consequence of conservation of energy
...
1
...
If the forefinger
points along the direction of the magnetic field and the thumb is along
the direction of motion of the conductor, then the middle finger points
in the direction of the induced current
...
137
4
...
Self Induction
The property of a coil which enables
to produce an opposing induced emf in it
when the current in the coil changes is
called self induction
...
4
...
On
Fig 4
...
An induced current flows
through the coil which according to Lenz’s law opposes the further
growth of current in the coil
...
According to Lenz’s law, the induced current will oppose the decay of
current in the coil
...
2
...
φ α I or
φ = LI
where L is a constant of proportionality and is called coefficient
of self induction or self inductance
...
According to laws of
electromagnetic induction
...
The unit of self inductance is henry (H)
...
138
4
...
2 Self inductance of a long solenoid
Let us consider a solenoid of N turns with length l and area of
cross section A
...
If B is the magnetic field at any
point inside the solenoid, then
Magnetic flux per turn = B × area of each turn
But, B =
µoNI
l
Magnetic flux per turn =
µo NIA
l
Hence, the total magnetic flux (φ) linked with the solenoid is given
by the product of flux through each turn and the total number of turns
...
e
φ=
µ o NIA
l
× N
µo N2IA
...
(2)
From equations (1) and (2)
LI =
∴
L =
µo N2IA
l
µο Ν 2 Α
l
If the core is filled with a magnetic material of permeability µ,
then, L =
µΝ 2 Α
l
4
...
3 Energy associated with an inductor
Whenever current flows through a coil, the self−inductance
opposes the growth of the current
...
If e is the induced emf
then,
139
e = – L
dI
dt
The small amount of work dw done in a time interval dt is
dw
= e
...
dt
dt
The total work done when the current increases from 0 to
maximum value (Io) is
Io
w = ∫ dw = ∫ −L I dI
0
This work done is stored as magnetic potential energy in the coil
...
Hence, quantitatively,
= −L ∫ IdI = –
the energy stored in an inductor is
1
L Io2
2
4
...
4 Mutual induction
Whenever there is a change in the
magnetic flux linked with a coil, there is
also a change of flux linked with the
neighbouring coil, producing an induced
emf in the second coil
...
G
S
Cell current
P
+ P and S are two coils placed close to
( )
K
each other (Fig
...
6)
...
6 Mutual induction
battery through a key K
...
On pressing K, current in P starts increasing from
zero to a maximum value
...
Therefore, magnetic flux linked
with S also increases producing an induced emf in S
...
According to Lenz’s law the induced
current in S would oppose the increase in current in P by flowing in
140
a direction opposite to the current in P, thus delaying the growth of
current to the maximum value
...
Therefore magnetic flux linked with S also
decreases and hence, an emf is induced in S
...
4
...
5 Coefficient of mutual induction
IP is the current in coil P and φs is the magnetic flux linked with
coil S due to the current in coil P
...
IfIP = 1A, then, M = φs
Thus, coefficient of mutual induction of two coils is numerically
equal to the magnetic flux linked with one coil when unit current flows
through the neighbouring coil
...
The unit of coefficient
of mutual induction is henry
...
The coefficient of mutual induction between a pair of coils
depends on the following factors
141
(i) Size and shape of the coils, number of turns and permeability
of material on which the coils are wound
...
4
...
When a current is passed through coil P, the magnetic flux
linked with S is small and hence, the coefficient of mutual induction
between the two coils is small
...
4
...
When current is passed through the coil P the magnetic
flux linked with coil S is large and hence, the coefficient of mutual
induction between the two coils is large
...
7 Mutual induction
If the two coils are wound on a soft iron core (Fig 4
...
4
...
6 Mutual induction of two long solenoids
...
The solenoid
S2 is wound closely over the solenoid S1 (Fig 4
...
N1 and N2 are the number of turns in the solenoids S1 and S2
respectively
...
I1 is the current flowing
S2
through the solenoid S1
...
8 Mutual induction
field B1 produced at any point inside the
between two long solenoids
solenoid S1 due to the current I1 is
B1 = µo N I I1
...
142
Total magnetic flux linked with solenoid S2 having N2 turns is
φ2 = B1AN2
Substituting for B1 from equation (1)
φ2
φ2
But
⎛ N
⎞
= ⎜ µ o 1 I 1 ⎟ A N2
l
⎝
⎠
µo N 1N 2 AI 1
=
l
φ2
= MI1
...
(3)
where M is the coefficient of mutual induction between S1 and S2
...
3
From equations (2) and (3)
µo N 1N 2 AI 1
MI1
=
l
µ o N 1N 2 A
M
=
l
If the core is filled with a magnetic material of permeability µ,
µ N1N 2 A
M
=
l
Methods of producing induced emf
We know that the induced emf is given by the expression
dφ
d
=−
(NBA cos θ)
dt
dt
Hence, the induced emf can be produced by changing
e = –
(i) the magnetic induction (B)
(ii) area enclosed by the coil (A) and
(iii) the orientation of the coil (θ) with respect to the magnetic field
...
3
...
The magnetic induction can be changed by moving a magnet
either towards or away from a coil and thus an induced emf is
produced in the coil
...
∴
⎛ dB ⎞
⎟
e = – NA cos θ ⎜
⎝ dt ⎠
143
4
...
2 Emf induced by changing the area enclosed by the coil
PQRS is a conductor bent in the shape as shown in the Fig 4
...
L1M1 is a sliding conductor of length l resting on the arms PQ and RS
...
The closed area of the conductor is L1QRM1
...
Due to the
P
change
in
area
l
L2L1M1M2, there is a
change in the flux
S
linked
with
the R
M1
M2
dx
conductor
...
9 Emf induced by changing the area
produced
...
dA = Bl dx
But
e = –
dφ
dt
∴
e = –
Bldx
= – Bl v
dt
where v is the velocity with which the sliding conductor is
moved
...
3
...
10)
...
Suppose, initially the coil is in vertical position, so that the
angle between normal to the plane of the coil and magnetic field is zero
...
If
φ is the flux linked with the coil at this instant, then
φ = NBA cos θ
144
The induced emf is,
Q
dφ
d
e=–
= −NBA
cos (ωt)
dt
dt
Hence, the induced
emf can be represented as
e = Eo sin ωt
The induced emf e
varies sinusoidally with
time t and the frequency
Q
R
P
N
Q
Q
R
Q
R
∴ e = NBAω sin ωt
...
10 Induced emf by changing the
orientation of the coil
ω ⎞
⎛
being ν cycles per second ⎜ν =
⎟
...
(ii)
When ωt = π/2, the plane of the coil is parallel to B and
hence e = Eo
(iii) When ωt = π, the plane of the coil is at right angle to B and
hence e = 0
...
(v)
When ωt = 2π, the plane of the coil is again perpendicular
to B and hence e = 0
...
4
...
The generator was originally designed by
a Yugoslav scientist Nikola Tesla
...
Essential parts of an AC generator
(i) Armature
Armature is a rectangular coil consisting of a large number of
loops or turns of insulated copper wire wound over a laminated soft
iron core or ring
...
For high power dynamos, field is
provided by electro magnet
...
(iii) Slip rings
The ends of the armature coil are connected to two hollow
metallic rings R1 and R2 called slip rings
...
When the shaft rotates, the
slip rings along with the armature also rotate
...
They
provide contact with the slip rings by keeping themselves pressed
against the ring
...
Working
Whenever, there is a change in
orientation of the coil, the magnetic
flux linked with the coil changes,
producing an induced emf in the coil
...
Suppose the armature ABCD is
initially in the vertical position
...
The side AB of the coil moves
downwards and the side DC moves
146
B
N
To
Power
Line
A
D
B1
B2
C
S
R1
R2
Fig 4
...
4
...
Then according to Flemings right hand rule the
current induced in arm AB flows from B to A and in CD it flows from
D to C
...
In the external
circuit the current flows from B1 to B2
...
Now the current in
2
3
2
4
2
ωt the coil flows along ABCD
...
As the
rotation of the coil continues,
the induced current in the
Fig 4
...
Hence the induced
current is alternating in nature (Fig 4
...
As the armature completes
ν rotations in one second, alternating current of frequency ν cycles per
second is produced
...
4
...
c
...
If a number of armature windings are used in the alternator
it is known as polyphase alternator
...
Thus a polyphase system consists
of a numerous windings which are placed on the same axis but
displaced from one another by equal angle which depends on the
number of phases
...
147
Generation of three phase emf
In a three – phase a
...
generator three
coils are fastened rigidly together and
displaced from each other by 120o
...
Each coil is provided B
with a separate set of slip rings and brushes
...
Three coils
a1 a2, b1 b2 and c1 c2 are mounted on the
same axis but displaced from each other by
120o, and the coils rotate in the
emf
Ea1a2
Eb b
1
2
N
c2
b2
a1
a2
b1
A
c1
S
Fig 4
...
13a)
...
At
Fig 4
...
Thus the emfs induced in all
the three coils are equal in magnitude and
240º
of same frequency
...
13c Angular
displacement between
e
= Eo sin (ωt – 2π/3)
b1 b2
the armature
e
= Eo sin (ωt – 4π/3)
c1 c2
The emfs induced and phase difference in the three coils a1 a2,
b1 b2 and c1 c2 are shown in Fig 4
...
13c
...
5 Eddy currents
Foucault in the year 1895 observed that when a mass of metal
moves in a magnetic field or when the magnetic field through a
stationary mass of metal is altered, induced current is produced in the
metal
...
Hence this current is called
eddy current
...
When a conductor in
the
form
of
a
disc
or a metallic plate as shown
in Fig 4
...
This current acts in a
Fig 4
...
If the metallic plate with holes drilled
in it is made to swing inside the magnetic field, the effect of eddy
current is greatly reduced consequently the plate swings freely inside
the field
...
Applications of Eddy current
(i) Dead beat galvanometer
When current is passed through a galvanometer, the coil
oscillates about its mean position before it comes to rest
...
Now,
when the coil oscillates, eddy currents are set up in the metallic frame,
which opposes further oscillations of the coil
...
Since the
oscillations of the coil die out instantaneously, the galvanometer is
called dead beat galvanometer
...
The material to be melted is placed in a
varying magnetic field of high frequency
...
Due to the heating effect of the current,
the metal melts
...
The eddy current initially
tries to decrease the relative motion between the cylinder and the
rotating magnetic field
...
These motors are used in fans
...
The drum
rotates along with the wheel when the train is in motion
...
Hence, the train comes to rest
...
The magnet rotates inside an aluminium cylinder (drum) which
is held in position with the help of hair springs
...
The drum inturn experiences a
torque and gets deflected through a certain angle depending on the
speed of the vehicle
...
4
...
It transfers
electric power from one
circuit to another
...
Laminated
Steel Core
φ
Primary
Winding
Secondary
Winding
Fig 4
...
15)
...
The a
...
input is applied across
the primary coil
...
Hence, an induced
emf is produced across the secondary
...
Since same flux links with the primary
and secondary, the emf induced per turn of the two coils must be the
same
(i
...
Es I P
(i
...
) E = I
P
s
...
(for step up transformer k > 1 and
for step down transformer k < 1)
In a step up transformer Es > Ep implying that Is < Ip
...
Similarly
a step down transformer decreases the voltage by increasing the
current
...
151
η =
Es I s
output power
= E I
input power
P P
The efficiency η = 1 (ie
...
But practically there are numerous
factors leading to energy loss in a transformer and hence the efficiency
is always less than one
...
This loss can be minimised by using a core with a
material having the least hysterisis loss
...
(2) Copper loss
The current flowing through the primary and secondary windings
lead to Joule heating effect
...
Thick wires with considerably low resistance are used to minimise
this loss
...
This leads to the wastage of energy in the form of heat
...
(4) Flux loss
The flux produced in the primary coil is not completely linked
with the secondary coil due to leakage
...
This loss can be minimised by using a shell type core
...
4
...
1 Long distance power transmission
The electric power generated in a power station situated in a
remote place is transmitted to different regions for domestic and
industrial use
...
There is always some power loss
associated with these lines
...
16 Distance transmission of power
City
Sub-Station
If I is the current through the wire and R the resistance,
a considerable amount of electric power I2R is dissipated as heat
...
However, by transmitting the electrical energy
at a higher voltage, the power loss can be controlled as is evident from
the following two cases
...
Power P = VI
11, 000
P
∴
I =
=
= 50A
220
V
If R is the resistance of line wires,
Power loss = I2R = 502R = 2500(R) watts
Case (ii) 11,000 W power is transmitted at 22,000 V
11,000
P
∴
I = =
= 0
...
5)2 R = 0
...
For transmitting electric power at 11,000 W at 220 V the current
capacity of line wires has to be 50 A and if transmission is done at
22,000 V, it is only 0
...
Thus, for carrying larger current (50A) thick
wires have to be used
...
To
support these thick wires, stronger poles have to be erected which
further adds on to the cost
...
So thicker wires can be replaced by thin wires, thus reducing
the cost of transmission considerably
...
The power is then transmitted through
the transmission lines which forms a part of the grid
...
Outside the city, the power is stepped
down to 110,000 V by a step-down transformer
...
Before distribution to the
user, the power is stepped down to 230 V or 440 V depending upon the
need of the user
...
7 Alternating current
As we have seen earlier a rotating coil in a magnetic field, induces
an alternating emf and hence an alternating current
...
The significance of an alternating emf is that
it can be changed to lower or higher voltages conveniently and efficiently
using a transformer
...
This enables us to utilize the
whole range of electromagnetic spectrum for one purpose or the other
...
For transmission of audio and video signals, the required frequency
range of radio waves is between 100 KHz and 100 MHz
...
4
...
1 Measurement of AC
Since alternating current varies continuously with time, its
average value over one complete cycle is zero
...
c
...
c
...
The rms value is also called effective value of an a
...
and is
denoted by Irms or Ieff
...
17 Variation I, I 2 and Irms with time
T
2
H = ∫ i R dt =
∫ (I
O
t
2
o
sin2 ω t ) R dt
O
T
⎤
I 2R ⎡ T
⎛ 1 − cos 2ω t ) ⎞
dt − ∫ cos 2ω t
...
e) H = I2rms RT
∴
I o 2RT
2
I2rms RT =
Irms =
Io
2
= 0
...
Thus, the rms value of an a
...
707 times the peak value of the
a
...
In other words it is 70
...
155
4
...
2
AC Circuit with resistor
Let an alternating source of emf be connected across a resistor of
resistance R
...
(1)
e
R
e,i
i
O
2
e=E0 sin t
(a)
(c)
i
eR
(b)
Fig 4
...
c
...
Hence, iR = Eo sin ωt
i =
Eo
sin ωt ;
R
i = Io sin ωt
...
c in the circuit
...
From the expressions of voltage and current given by equations (1) and
(2) it is evident that in a resistive circuit, the applied voltage and
current are in phase with each other (Fig 4
...
where Io =
Fig 4
...
4
...
3 AC Circuit with an inductor
Let an alternating source of emf be applied to a pure inductor of
inductance L
...
Due to an alternating emf that is applied to the
inductive coil, a self induced emf is generated which opposes the
applied voltage
...
156
The instantaneous value of applied emf is given by
e = Eo sin ωt
...
In an ideal inductor
circuit induced emf is equal and opposite to the applied voltage
...
Therefore e = −e′
di ⎞
⎛
Eo sin ωt = − ⎜ −L ⎟
dt ⎠
⎝
di
∴
Eo sin ωt =L
dt
Eo
di =
sin ωt dt
L
Integrating both the sides
e=E0 sin t
(a)
e
I
e,i
i
O
2
t
=
=
Eo
L
Eo
L
i =
(b)
Fig 4
...
sin (ωt –
π
2
π
2
)
)
...
Here, ωL is the resistance offered by the coil
...
Its unit is ohm
...
c
...
Conversely the voltage across L leads the current by
the phase angle of π/2
...
19b
...
19c represents the phasor diagram of a
...
circuit containing only L
...
19c si
Phasor diagram
Inductive reactance
XL = ωL = 2π ν L, where ν is the frequency of the a
...
supply
For d
...
ν = 0; ∴ XL = 0
Thus a pure inductor offers zero resistance to d
...
But in an a
...
circuit the reactance of the coil increases with increase in frequency
...
7
...
20a)
...
Y
e
e,i
i
C
i
O
X
90º
e=E0 sin t
ec
Y/
(a)
(b)
(c)
Fig 4
...
(1)
At any instant the potential difference across the capacitor will be
equal to the applied emf
∴ e = q/C, where q is the charge in the capacitor
But
i =
i =
dq
d
=
(Ce)
dt dt
d
(C Eo sin ωt) = ω CEo
...
(2)
158
where
Io =
Eo
(1/ωC )
1
= XC is the resistance offered by the capacitor
...
Its unit is ohm
...
c
...
In
otherwords the emf lags behind the current by a phase angle of π/2
...
20b
...
20c represents the phasor diagram of a
...
circuit containing
only C
...
c
...
In a d
...
circuit
ν = 0
∴
XC = ∞
Thus a capacitor offers infinite resistance to d
...
For an a
...
the
capacitive reactance varies inversely as the frequency of a
...
and also
inversely as the capacitance of the capacitor
...
7
...
21a)
...
21a RLC sereis circuit
VC
90º
φ
VR
V
A
I
4
...
The voltage drop across the resistor is, VR = I R (This is in phase
with I)
159
The voltage across the inductor coil is VL = I XL
(VL leads I by π/2)
The voltage across the capacitor is, VC = IXC
(VC lags behind I by π/2)
The voltages across the different components are represented in
the voltage phasor diagram (Fig
...
21b)
...
The applied voltage ‘V’ equals the vector sum
of VR, VL and VC
...
22 Impedance
diagram
V
= Z = R 2 + (X L − X C )2
I
The expression
B
R 2 + (X L − X C )2
is the net effective opposition
offered by the combination of resistor, inductor and capacitor known as
the impedance of the circuit and is represented by Z
...
The values are represented in the impedance diagram (Fig 4
...
Phase angle φ between the voltage and current is given by
tan φ =
VL −VC I XL − I XC
=
VR
IR
tan φ =
X L − X C net reactance
=
R
resistance
∴
⎛ X L − XC ⎞
⎟
φ = tan–1 ⎜
⎝
R
⎠
∴ Io sin (ωt + φ) is the instantaneous current flowing in the
circuit
...
e
...
e
...
Such a circuit which admits
maximum current is called series resonant circuit or acceptor circuit
...
i
...
By
offering minimum impedance to current at the resonant frequency it is
able to select or accept most readily this particular frequency among
many frequencies
...
This is usually
done by varying the capacitance of a capacitor
...
In other words it refers to the sharpness
of tuning at resonance
...
Q =
voltage across L or C
applied voltage
...
c
...
The applied voltage at resonance is the potential drop across R,
because the potential drop across L is equal to the drop across C and
they are 180o out of phase
...
Applied Voltage = IR
...
Circuit with high Q values
would respond to a very
narrow frequency range and
vice versa
...
Q-factor
can be increased by having a
coil of large inductance but of
small ohmic resistance
...
23 variation of current with
frequency
162
Current frequency curve is quite flat for large values of resistance
and becomes more sharp as the value of resistance decreases
...
23 is also called the frequency response curve
...
7
...
c circuit the current and emf vary continuously with time
...
Thus, we define instantaneous
power of an a
...
circuit as the product of the instantaneous emf and
the instantaneous current flowing through it
...
c circuit
The average power consumed over one complete cycle is
T
∫ ie
dt
0
Pav =
T
=
T
∫ dt
∫ [I
o
sin(ωt + φ )Eo sin ω t ] dt
0
...
cos φ = Erms I rms cos φ
2
2
= apparent power × power factor
Pav =
Pav
Eo I o
cos φ
2
...
Choke coil
A choke coil is an inductance coil of very small resistance used
for controlling current in an a
...
circuit
...
On the other hand there is no dissipation of power
when a current flows through a pure inductor
...
A laminated core is used to minimise eddy
current loss (Fig
...
24)
...
24 Choke coil
Working
The inductive reactance offered by the coil is given by
XL = ωL
In the case of an ideal inductor the current lags behind the emf
π
by a phase angle
...
Hence it may be treated as a series combination of
an inductor and small resistance r
...
(1)
r
is the power factor
...
2
Fig
...
24a A
...
4
...
F
...
c
...
These chokes are known as audio –
frequency (A
...
For radio frequencies, air chokes are used since
a low inductance is sufficient
...
F)
or high frequency (H
...
4
...
4
...
Choke coils can be commonly seen in fluorescent tubes which
work on alternating currents
...
1
Magnetic field through a coil having 200 turns and cross
sectional area 0
...
1 wb m−2 to 0
...
02 s Find the induced emf
...
04 m2, B1 = 0
...
04 wb m−2, t = 0
...
= − NA
...
04 − 0
...
02
e = 24 V
e = −
4
...
48 m flies due north at a
speed of 40 ms−1
...
Data : l = 20
...
48 × 40
e
4
...
0164 volt
A solenoid of length 1 m and 0
...
If
a current of 2A passes through the coil, calculate (i) the
coefficient of self induction of the coil and (ii) the magnetic flux
linked with a the coil
...
05 m;
r = 0
...
14(0
...
616 × 10−3
1
∴ L = 0
...
616 × 10−3 × 2 = 1
...
232 milli weber
165
4
...
5 s in one coil, induces an
emf of 50 mV in the other coil
...
5s;
e = 50 mV = 50 × 10−3V, M = ?
Solution : e = − M
...
5
dI
dt
50 × 10−3
e
e
= − 8−4
= − 6
...
5 ⎠
⎝ dt ⎠
⎝ dt ⎠
M = 6
...
c
...
The coil rotates at an angular speed of 140 rpm in a
uniform magnetic field of 3
...
Find the maximum value
of the emf induced
...
6 × 10−2T
Eo = ?
Solution : Eo = NABω = NAB 2πν
= 104 × 10−2 × 3
...
75 V
4
...
Data :
ν = 25 Hz,
Solution
:
Irms = 30 A
i = Io sin ωt
= Irms
2 sin 2πνt
i
i
4
...
42 sin 157 t
A capacitor of capacitance 2 µF is in an a
...
circuit of frequency
1000 Hz
...
166
Data : C = 2µF, ν = 1000 Hz, Eeff = 10V
1
1
=
C ω C × 2π v
Solution : Xc =
Xc =
1
−6
2 × 10
Irms =
E eff
XC
× 2π × 103
=
= 79
...
6
∴ Irms = 0
...
8
A coil is connected across 250 V, 50 Hz power supply and it
draws a current of 2
...
Find the
self inductance and power factor
...
ν = 50 Hz; Irms = 2
...
5
= 0
...
5 = 100 Ω
rms
From the phasor diagram
sin φ =
∴
XL
Z
XL = Z
...
64)2]
∴
XL = 76
...
8
=
2π v 2π × 50
∴
L =
∴
L = 0
...
9
A bulb connected to 50 V, DC consumes 20 w power
...
c
...
Find the value of the capacitor required so that the
bulb draws the same amount of current
...
4 A
∴ I =
V 50
50
V
∴ Resistance, R =
=
= 125 Ω
0
...
4
⎛ 1 ⎞
R2 + ⎜
⎟
⎝ ωc ⎠
∴ Z =
Z 2 = R2 +
C =
=
Z
φ
2
2
XL
R
⎛ 1 ⎞
R2 + ⎜
⎟
⎝ 2πνC ⎠
=
1
2 2 2
4π ν C
1
2πν Z 2 − R 2
1
2
2
2π × 50 (625) − (125)
=
1
2π × 50 × 612
...
198 µF
4
...
1 H inductor and a 25 µF capacitor
...
Data : R = 24 Ω, L = 0
...
(1)
...
2 V
168
ωt = 314 t
2πν = 314
ν =
314
= 50 Hz
2 × 3
...
1 × 2 π × 50 –
1
1
= L
...
2π v
1
25 × 10−6 × 2π × 50
= 31
...
4 = −96 Ω
XL – XC = −96 Ω
∴
XC – XL = 96 Ω
Z =
R 2 + ( XC − X L )
2
=
242 + 962
=
576 + 9216
= 98
...
4 − 31
...
In the same way any question and problem could be framed from the text
matter
...
)
4
...
2
(b) room heater
(d) choke coil
A coil of area of cross section 0
...
2 Wb/m2
...
3
(b) 10 Wb
(d) zero
Lenz’s law is in accordance with the law of
(a) conservation of charges
(c) conservation of momentum
4
...
5
(b) infinity
(d) very small
The unit henry can also be written as
(a) Vs A−1
(c) Ω s
4
...
The coefficient of self induction of the coil is
(a) 0
...
7
(b) 0
...
8 H
A DC of 5A produces the same heating effect as an AC of
(a) 50 A rms current
(c) 5A rms current
4
...
9
The part of the AC generator that passes the current from the coil
to the external circuit is
(a) field magnet
(b) split rings
(c) slip rings
(d) brushes
4
...
11 Which of the following cannot be stepped up in a transformer?
(a) input current
(b) input voltage
(c) input power
(d) all
4
...
13 Which of the following devices does not allow d
...
to pass through?
(a) resistor
(b) capacitor
(c) inductor
(d) all the above
4
...
(b) the average value of square of current is zero
...
(d) the rms current is
2 time of peak current
...
15 What is electromagnetic induction?
4
...
4
...
Give its unit
4
...
4
...
4
...
4
...
171
4
...
c
...
23 State the methods of producing induced emf
...
24 What is a poly phase AC generator?
4
...
26 Define alternating current and give its expression
...
27 What is capacitive reactance?
4
...
4
...
30 Define power factor
...
31 Why a d
...
c?
4
...
c
...
33 Define quality factor
...
34 A capacitor blocks d
...
c
...
4
...
36 State Lenz’s law and illustrate through an experiment
...
4
...
4
...
4
...
Obtain
an expression for the mutual inductance
...
40 Explain how an emf can be induced by changing the area enclosed
by the coil
...
41 Discuss with theory the method of inducing emf in a coil by
changing its orientation with respect to the direction of the magnetic
field
...
42 What are eddy currents? Give their applications
...
43 Explain how power can be transmitted efficiently to long distance
...
44 Obtain an expression for the current flowing in a circuit containing
resistance only to which alternating emf is applied
...
172
4
...
Find the phase relationship between voltage and
current
...
46 Obtain an expression for the current flowing in the circuit containing
capacitance only to which an alternating emf is applied
...
4
...
4
...
4
...
c
...
4
...
c generator
...
51 Describe the principle, construction and working of three−phase a
...
4
...
Discuss its construction and
working
...
53 A source of altemating emf is connected to a series combination of
a resistor R an inductor L and a capacitor C
...
Problems
4
...
If the magnetic flux linked with the coil changes
from 10–3 Wb to 2 × 10–4 Wb in a time of 0
...
4
...
The train runs at a speed
of 180 Km/hr
...
2 × 10−4 Wb/m2 and the rails are separated by 1m
...
4
...
If a current of 5A flows in the solenoid,
calculate the energy stored in the solenoid
...
57 An iron cylinder 5cm in diameter and 100cm long is wound with
3000 turns in a single layer
...
Calculate the
mutual inductance between the coils (relative permeability of the
core = 500)
...
58 A student connects a long air core coil of manganin wire to a 100V
DC source and records a current of 1
...
When the same coil is
connected across 100V, 50 Hz a
...
source, the current reduces to
1 A
...
4
...
1µF and resistance of 500 Ω in series
...
4
...
If the power output from the secondary at 1100 V
is 12
...
If the resistance of
primary is 0
...
61 A resistance
5 µF are
e = 311 sin
voltage (iii)
impedance
...
5 H and a capacitance of
connected in series with an a
...
supply of
(314t)
...
c
...
62 A radio can tune over the frequency range of a portion of broadcast
band (800 KHz to 1200 KHz)
...
63 A transformer has an efficiency of 80%
...
If the secondary voltage is 240 V
...
4
...
C
...
Calculate the inductance of the
choke required so that the bulb draws the same current of 10 A
...
1
(b)
4
...
3
(d)
4
...
5
4
...
7
(c)
4
...
9
(d)
4
...
11 (c)
4
...
13 (b)
(d)
4
...
54 0
...
55 1 mV
4
...
52575 joule
4
...
37 H
4
...
54 Ω and 0
...
59 0
...
60 220V, (i) 747 W
4
...
9 Ω (v) 482
...
62 87
...
63 40 A, 13
...
64 0
...
, MA
...
D
...
Sc
...
L
...
, F
...
S
...
His father
Mr
...
Chandrasekara Iyer was a teacher
...
He completed his B
...
, degree
with distinction in Presidency College, Chennai in 1904
...
A
...
Raman appeared for the finance examination in February
1907 and again secured the first place
...
Eventhough, Raman worked as an officer in finance department,
he spent the morning and evening hours, out of office hours in
Physics laboratories
...
Raman left
Government Service in July 1917 and joined as a Professor of
Physics in the University of Calcutta
...
The discovery of the Raman effect was not an accident, but
was the result of prolonged and patient research extending over
a period of nearly seven years
...
When, during the voyage made on the occasion
of his first visit to Europe, Raman’s attention was attracted to the
beautiful blue colour exhibited by the water of the deep sea
...
The
experiment of Professor Raman revealed that the scattered light
is different from the incident light
...
For his investigation on the scattering of light and the
discovery of the effect known after him, Raman effect, Nobel
Prize was awarded to Raman on 10th December, 1930
...
C
...
Raman joined the Indian Institute of Science
and Technology, Bangalore as its first Indian director in 1933
...
He continued his research, until death put a full stop to his
activities at the age of 82
...
Electromagnetic Waves and Wave optics
The phenomenon of Faraday’s electromagnetic induction
concludes that a changing magnetic field at a point with time produces
an electric field at that point
...
e) changing electric field with time at a point
produces a magnetic field at that point
...
This
idea led Maxwell to conclude that the variation in electric and magnetic
fields perpendicular to each other, produces electromagnetic
disturbances in space
...
These waves are called electromagnetic waves
...
1
...
In an electromagnetic wave, electric and magnetic field vectors are
at right angles to each
Y
other and both are at
B
B
E
right
angles
to
the
E
direction of propagation
...
These waves are
Fig 5
...
transverse in nature
...
1 shows the variation of electric field E along Y direction and
→
magnetic field B along Z direction and wave propagation in + X
direction
...
1
...
(ii) They do not require any material medium for propagation
...
Hence electromagnetic waves are transverse in nature
...
(v) They travel in vacuum or free space with a velocity
1
...
(vii) The electromagnetic waves being chargeless, are not deflected
by electric and magnetic fields
...
1
...
This experiment is based on the fact
that an oscillating electric charge radiates electromagnetic waves
...
The experimental arrangement is as shown in Fig 5
...
It consists
of two metal plates A and B placed at a distance of 60 cm from each
other
...
Using an
S
Coil
induction coil a high potential
S
difference is applied across the
Detector
small gap between the spheres
...
2 Hertz experiment
difference across S1 and S2, the
air in the small gap between the spheres gets ionized and provides a
path for the discharge of the plates
...
Hertz was able to produce electromagnetic waves of frequency
about 5 × 107 Hz
...
The high frequency oscillation of charges between the
1
plates is given by ν =
2π LC
5
...
4 Electromagnetic Spectrum
After the demonstration of electromagnetic waves by Hertz,
electromagnetic waves in different regions of wavelength were produced
by different ways of excitation
...
FM
Radio waves
1 km
Standard broadcast
Long waves
Fig 5
...
Electromagnetic spectrum covers a wide range of wavelengths (or)
frequencies
...
All electromagnetic waves travel with the velocity
of light
...
The overlapping in certain parts of the spectrum shows that the
particular wave can be produced by different methods
...
1 shows various regions of electromagnetic spectrum with
source, wavelength and frequency ranges of different electromagnetic
waves
...
1
(NOT FOR EXAMINATION)
Sl
...
Name
γ – rays
2
...
Ultra−violet
(UV)
4
...
Infra−red (IR)
6
...
Radio
frequency
waves
Wavelength
Radioactive
nuclei, nuclear
reactions
High energy
electrons suddenly
stopped by a metal
target
Atoms and
molecules in an
electrical discharge
incandescent solids
Fluorescent
lamps
molecules of
hot bodies
Electronic
device
(Vacuum tube)
charges
accelerated through
conducting wires
Frequency
range (m)
1
...
3
3 x 1011 – 1 x 109
10−104
3 x 107 – 3 x 104
181
5
...
5 Uses of electromagnetic spectrum
The following are some of the uses of electromagnetic waves
...
Radio waves : These waves are used in radio and television
communication systems
...
Higher
frequencies upto 54 MHz are used for short waves bands
...
FM band is
from 88 MHz to 108 MHz
...
2
...
Microwave ovens are an interesting
domestic application of these waves
...
Infra red waves :
(i) Infrared lamps are used in physiotherapy
...
(iii) As infrared radiations are not absorbed by air, thick fog, mist
etc, they are used to take photograph of long distance objects
...
4
...
The wavelength range
of visible light is 4000 Å to 8000 Å
...
Ultra− violet radiations
(i) They are used to destroy the bacteria and for sterilizing
surgical instruments
...
(iii) They are used to preserve the food items
...
6
...
(ii) It is used to study the crystal structure in solids
...
γ−
γ−rays : Study of γ rays gives useful information about the
nuclear structure and it is used for treatment of cancer
...
2 Types of spectra
When white light falls on a prism, placed in a spectrometer, the
waves of different wavelengths are deviated to different directions by
the prism
...
Such an image is
called a spectrum
...
These images are the emission lines of sodium having wave lengths
5896Ao and 5890Ao
...
The spectra obtained from different bodies can be classified into
two types (i) emission spectra and (ii) absorption spectra
...
Every source has
its own characteristic emission spectrum
...
1
...
Line spectrum and 3
...
Continuous spectrum
It consists of unbroken luminous bands of all wavelengths
containing all the colours from violet to red
...
Incandescent solids, liquids, Carbon arc, electric filament lamps
etc, give continuous spectra
...
Line spectrum
Line spectra are sharp lines of definite wavelengths
...
It is used to identify the gas
...
e
...
The
Fig 5
...
5
...
183
3
...
Band spectra are obtained from molecules
...
Calcium or Barium salts in a bunsen flame and gases
like carbon−di−oxide, ammonia and nitrogen in molecular state in the
discharge tube give band spectra
...
Using band spectra the molecular
structure of the substance can be studied
...
It is the
characteristic of the absorbing substance
...
continuous absorption spectrum
2
...
band absorption spectrum
1
...
2
...
5 Emission and absorption spectrum of sodium
When light from the carbon arc is made to pass through sodium
vapour and then examined by a spectrometer, a continuous spectrum
of carbon arc with two dark lines in the yellow region is obtained as
shown in Fig
...
5
...
Band absorption spectrum
If white light is allowed to pass through iodine vapour or dilute
solution of blood or chlorophyll or through certain solutions of organic
184
and inorganic compounds, dark bands on continuous bright
background are obtained
...
5
...
1 Fraunhofer lines
If the solar spectrum is closely examined, it is found that it
consists of large number of dark lines
...
Solar spectrum is an example of
line absorption spectrum
...
It emits
continuous spectrum
...
This is at a comparatively lower temperature at about 6000 K
...
When light from the central core of the sun passes through sun’s
atmosphere, certain wavelengths are absorbed by the elements present
in the chromosphere and the spectrum is marked by dark lines
...
5
...
2 Fluorescence
When an atomic or molecular system is excited into higher energy
state by absorption of energy, it returns back to lower energy state in
a time less than 10−5 second and the system is found to glow brightly
by emitting radiation of longer wavelength
...
It may be noted that fluorescence exists as long as the fluorescing
substance remain exposed to incident ultraviolet light and re-emission
of light stops as soon as incident light is cut off
...
2
...
The emission of light continues even
after the exciting radiation is removed
...
185
5
...
Since, light is a form of energy, it is transferred from
one place to another
...
In general, there are two possible modes of propagation of energy
from one place to another (i) by stream of material particles moving
with a finite velocity (ii) by wave motion, wherein the matter through
which the wave propagates does not move along the direction of the
wave
...
5
...
1 Corpuscular theory
According to Newton, a source of light or a luminous body
continuously emits tiny, massless (negligibly small mass) and perfectly
elastic particles called corpuscles
...
The corpuscles are so small that a luminous body does not suffer
any appreciable loss of mass even if it emits light for a long time
...
The sense of
vision is produced, when the corpuscles impinge on the retina of the
eye
...
On account of high speed, they are unaffected by the force
of gravity and their path is a straight line
...
Reflection of the particles is due to repulsion and refraction
is due to attraction
...
But the
experimental results of Foucault and Michelson showed that velocity of
light in a denser medium is lesser than that in a rarer medium
...
5
...
2 Wave theory
According to Huygens, light is propagated in the form of waves,
through a continuous medium
...
The
186
disturbance from the source is propagated in the form of waves through
space and the energy is distributed equally in all directions
...
Initially rectilinear propagation of
light could not be explained
...
The wave theory
could satisfactorily explain all the basic properties, which were earlier
proved by corpuscular theory and in addition, it explains the phenomena
of interference, diffraction and polarisation
...
This is in accordance with the
experimental result of Foucault
...
3
...
He showed that the variation
of electric and magnetic intensities had precisely the same
characteristics as a transverse wave motion
...
5
...
4 Quantum theory
The electromagnetic theory, however failed to account for the
phenomenon of photo electric effect
...
In 1905, Einstein extended this idea
and suggested that light waves consist of small pockets of energy called
Fig 5
...
The energy associated with each photon is E = h ν , where h
is Planck’s constant (h = 6
...
It is now established that photon seems to have a dual character
...
5
...
5
...
The scattering of sunlight by the molecules of the gases in
Earth’s atmosphere is called Rayleigh scattering
...
The
strength of scattering depends on the wavelength of the light and also
the size of the particle which cause scattering
...
This is known as Rayleigh scattering law
...
The blue appearance of sky is due to scattering of
sunlight by the atmosphere
...
This scattered
radiation causes the sky to appear blue
...
Therefore most of the blue
light is scattered away and only the red light which is least scattered
reaches the observer
...
5
...
1 Tyndal scattering
When light passes through a colloidal solution its path is visible
inside the solution
...
The scattering of light by the colloidal particles is called Tyndal
scattering
...
4
...
V
...
The scattered light contains some additional frequencies
188
other than that of incident frequency
...
The lines whose frequencies have been modified in Raman effect
are called Raman lines
...
This series of lines in the scattering of light by the atoms and
molecules is known as Raman Spectrum
...
Let the incident light consist of photons of energy hνo
...
If a photon strikes an atom or a molecule in a liquid, part of
the energy of the incident photon may be used to excite the atom of the
liquid and the rest is scattered
...
2
...
The spectral line will
have higher frequency and it is called Anti−stoke’s line
...
In some cases, when a light photon strikes atoms or molecules,
photons may be scattered elastically
...
are the vibration levels of the ground electronic state
...
7 Raman Spectrum
189
lose energy
...
If νo is the frequency of incident radiation and νs the frequency of
scattered radiation of a given molecular sample, then Raman Shift or
Raman frequency ∆ν is given by the relation ∆ν = νο − νs
...
For Stoke’s lines, ∆ν is positive and for Anti–stoke’s lines
∆ν is negative
...
The different processes giving rise to
Rayleigh, Stoke’s and Anti-stokes lines are shown in Fig 5
...
When a system interacts with a radiation of frequency νo, it may
make an upward transition to a virtual state
...
Most of the molecules of the
system return back to the original state from the virtual state which
corresponds to Rayleigh scattering
...
5
...
3 Applications of Raman Spectrum
(i) It is widely used in almost all branches of science
...
(iii) In industry, Raman Spectroscopy is being applied to study the
properties of materials
...
5
...
Every particle on the surface vibrates
...
5
...
It
is clear that all the particles on such a
Fig 5
...
Such a surface
which envelopes the particles that are in the same state of vibration is
190
known as a wave front
...
A point source of light at a finite distance in an isotropic medium*
emits a spherical wave front (Fig 5
...
A point source of light in an
isotropic medium at infinite distance will give rise to plane wavefront
(Fig
...
9b)
...
9c)
...
9 Wavefront
(c)
5
...
1 Huygen’s principle
Huygen’s principle helps us to locate the new position and shape of
the wavefront at any instant, knowing its position and shape at any
previous instant
...
Huygen’s principle states that, (i) every point on a given wave front
may be considered as a source of secondary wavelets which spread out
with the speed of light in that medium and (ii) the new wavefront is the
forward envelope of the secondary wavelets at that instant
...
5
...
Let AB represent a given wavefront at a time
t = 0
...
To
find the position of the wave front after a time t, circles are drawn with
points P, Q, R
...
These are
the traces of secondary wavelets
...
If the
source of light is at a large distance, we obtain a plane wave front
A1 B1 as shown in Fig 5
...
* Isotropic medium is the medium in which the light travels with same
speed in all directions
...
10 Huygen’s principle
5
...
2 Reflection of a plane wave front at a plane surface
Let XY be a plane reflecting surface and AB be a plane wavefront
incident on the surface at A
...
Hence they represent incident rays
...
The wave front and the surface are
perpendicular to the plane of the paper (Fig
...
11)
...
By the time, the secondary wavelets
from B travel a distance BC, the secondary wavelets from A on the
reflecting surface would travel the same distance BC after reflection
...
From C a tangent
CD is drawn to this arc
...
Therefore CD is the reflected plane wavefront and AD
is the reflected ray
...
(ii)
Angle of incidence i = ∠ PAN = 900 − ∠ NAB = ∠ BAC
Angle of reflection r = ∠ NAD = 900 − ∠ DAC = ∠ DCA
In right angled triangles ABC and ADC
192
0
∠ D = 90
AD and AC is common
∠B =
BC =
∴
The two triangles are congruent
∠ BAC = ∠ DCA
i
...
i = r
Thus the angle of incidence is equal to angle of reflection
...
11 Reflection of a plane wavefront at a plane surface
...
5
...
12)
...
Consider a plane wave
front AB incident on the refracting surface at A
...
Hence they
represent incident rays
...
The
wave front and the surface are perpendicular to the plane of the paper
...
By the time,
the secondary wavelets from B,
reaches C, the secondary wavelets
from the point A would travel a
distance AD = C2t, where t is the
time taken by the wavelets to
travel the distance BC
...
Taking A as centre and
C1
193
Q
N
1
P
C1
B
i
i
X
r
A
r
D
C
Y
2
C2
N1
Fig 5
...
C2
BC
as radius an arc is drawn in the second medium
...
This tangent CD not only envelopes
the wavelets from C and A but also the wavelets from all the points
between C and A
...
Laws of refraction
(i) The incident wave front AB, the refracted wave front CD and
the refracting surface XY all lie in the same plane
...
C1
µ2 is called the refractive index of second medium with respect
1
to first medium
...
sin i BC / AC BC
=
=
=
sin r AD / AC AD
If 1µ2 > 1, the first medium is rarer and the second medium is
C1
denser
...
This means that the velocity of light in rarer
2
medium is greater than that in a denser medium
...
It is clear from above discussions that the refractive index of a
medium µm is given by
µm
=
Ca
velocity of light in vacuum
=
velocity of light in the medium Cm
The frequency of a wave does not change when a wave is reflected
or refracted from a surface, but wavelength changes on refraction
...
e
...
194
5
...
4 Total internal reflection by wave theory
Let XY be a plane surface which separates a rarer medium (air)
and a denser medium
...
A plane wavefront AB passes from denser medium to rarer
medium
...
Let r
be the angle of refraction
...
13 Total internal reflection
cm
Since
< 1 , i is less than r
...
In right angled triangle ADC, there are three possibilities
(i) AD < AC (ii) AD = AC and (iii) AD > AC
(i) AD < AC : For small values of i, BC will be small and so
AD > BC but less than AC (Fig
...
13a)
sin r =
i
...
When AD = AC,
sin r = 1 (or) r = 900
...
e a refracted wavefront is just possible
(Fig
...
13b)
...
The angle of incidence at which the angle of refraction
is 900 is called the critical angle C
...
This is not possible
(Fig 5
...
Therefore no refracted wave front is possible, when the angle
of incidence increases beyond the critical angle
...
This is called total
internal reflection
...
i
...
5
...
The resultant displacement of any particle
is the vector addition of the displacements due to the individual waves
...
14 Superposition principle
→
→
This is known as principle of superposition
...
6
...
Two independent monochromatic sources, emit waves of same
wave length
...
So they are not coherent
...
5
...
2 Phase difference and path difference
A wave of length λ corresponds to a phase of 2π
...
6
...
The waves from
S
these two coherent sources travel
B
in the same medium and
superpose at various points as
shown in Fig
...
15
...
15 Interference phenomenon
thick
continuous
lines
and
troughs are shown by broken lines
...
These points are marked
by crosses (x)
...
At points where the crest of one wave meets the trough of the other
wave, the waves are in opposite phase, the displacement is minimum and
these points appear dark
...
This
type of interference is said to be destructive interference
...
e
...
The redistribution of intensity of light on account of the
superposition of two waves is called interference
...
If a1 and a2 are the amplitude of the two interfering waves, then
I1 ∝ a12 and I2 ∝ a22
∴
I1 a12
=
I 2 a22
For constructive interference, Imax ∝ (a1 + a2)2 and for destructive
interference, Imin ∝ (a1 – a2)2
∴
Imax (a1 + a2 )2
=
Imin (a1 − a2 )2
197
5
...
4 Condition for sustained interference
The interference pattern in which the positions of maximum and
minimum intensity of light remain fixed with time, is called sustained
or permanent interference pattern
...
5
...
5 Young’s double slit experiment
The phenomenon of interference
was first observed and demonstrated
by Thomas Young in 1801
...
16
...
16 Young’s double slit
to each other
...
03 mm and they are about
0
...
Since A and B are equidistant from S, light waves from
S reach A and B in phase
...
According to Huygen’s principle, wavelets from A and B spread
out and overlapping takes place to the right side of AB
...
These
are called interference fringes or bands
...
At P on the screen, waves from A and B travel
equal distances and arrive in phase
...
This is called central bright
fringe
...
This shows clearly that the
bands are due to interference
...
6
...
A screen XY is placed parallel to AB at a distance D from
the coherent sources
...
O is a point on the
screen equidistant from A and B
...
17
...
X
P
Dark fringe
x
A
d C
B
Central
bright
fringe
O
M
Bright fringe
D
Y
Fig 5
...
d
In right angled triangle COP, tan θ =
OP
x
=
CO D
For small values of θ, tan θ = θ
∴
The path difference δ =
xd
D
Bright fringes
By the principle of interference, condition for constructive
interference is the path difference = nλ
199
xd
D
∴
where
n
∴
= nλ
= 0,1,2 … indicate the order of bright fringes
...
Dark fringes
By the principle of interference, condition for destructive
λ
interference is the path difference = (2n−1)
2
where n
=
1,2,3 … indicate the order of the dark fringes
...
Thus, on the screen alternate dark and bright bands are seen
on either side of the central bright band
...
The distance between (n+1)th and nth order consecutive bright
fringes from O is given by
D
D
D
(n + 1)λ − n λ = λ
d
d
d
D
λ
Bandwitdth, β =
d
Similarly, it can be proved that the distance between two
x(n+1) – xn =
Dλ
...
consecutive dark bands is also equal to
Condition for obtaining clear and broad interference bands
(i) The screen should be as far away from the source as possible
...
(iii) The two coherent sources must be as close as possible
...
6
...
These
colours are due to interference between light waves reflected from the
top and the bottom surfaces of thin films
...
Interference in thin films
Consider a transparent thin film of uniform thickness t and its
refractive index µ bounded by two plane surfaces K and K′ (Fig 5
...
A ray of monochromatic light AB incident on the surface K of the
film is partly reflected along BC and partly refracted into the film along
BD
...
The reflected
light then emerges into air along EF which is parallel to BC
...
BC and EF are
reflected rays parallel to
each other and DG and
HJ are transmitted rays
parallel to each other
...
C
i
K
t
Interference due to the
reflected beam
EM is drawn normal
to BC from E
...
18 Interference in thin films
– (BM)in
air
We know, that a distance in air is numerically equal to µ times
the distance in medium
δ = µ (BD + DE) – BM
201
From the figure, it is clear that BD = DE
∴
δ = (2µ
...
µ sin r
In the ∆ BME, sin i =
BM = µ
...
Since the reflection at B is at the surface of a denser medium,
there is an additional path difference
λ
2
...
2µt cos r +
λ
2
= (2n+1)
202
λ
2
∴
2µt cos r = nλ
If light is incident normally i = 0 and hence r = 0
...
Interference due to the transmitted light
The path difference between the transmitted rays DG and HJ is,
in a similar way, δ = 2µt cos r
...
6
...
When a plano convex lens of long focal
length is placed over an optically plane glass plate, a thin air film with
varying thickness is enclosed between them
...
When the air film is illuminated by
monochromatic light normally, alternate bright and dark concentric
circular rings are formed with dark spot at the centre
...
When viewed with white light, the fringes are
coloured (shown in the wrapper of the text book)
...
19 shows an experimental arrangement for producing and
observing Newton’s rings
...
The parallel beam of light emerging
from L1 falls on the glass plate G kept at 45o
...
The reflected beam from the air film is viewed with a microscope
...
203
M
L1
45º
S
G
Air film
L
O
P
Fig 5
...
If t is the thickness of the air film
at a point on the film, the refracted wavelet from the lens has to travel
a distance t into the film and after reflection from the top surface of the
glass plate, has to travel the same distance back to reach the point
again
...
One of the two reflections takes
place at the surface of the denser medium and hence it introduces an
additional phase change of π or an equivalent path difference
λ
2
between two wavelets
...
204
The condition for darkness is,
path difference
∴
δ
= 2t +
λ
2
= (2n+1)
λ
2
2t = nλ
where
n = 0, 1, 2, 3
...
Hence, there is no path difference between the
interfering waves
...
But the wave reflected
from the denser glass plate has suffered a phase change of π while the
wave reflected at the spherical surface of the lens has not suffered any
phase change
...
Around the point of
contact alternate bright and dark rings are formed
...
6
...
20
...
Let t be the thickness of the
air film at S and P
...
Then ST = AO = PQ = t
Let rn be the radius of the nth dark ring which passes through
the points S and P
...
AP = OA
...
20 Radius of
Newton’s rings
nR λ
205
Since R and λ are constants, we find that the radius of the dark
ring is directly proportional to square root of its order
...
e
...
It is clear that the rings get closer as n increases
...
6
...
The radius of nth
dark ring and (n+m)th dark ring are given by
rn2 = nRλ
r2n+m = (n+m) Rλ
and
rn+m2 – rn2 = mRλ
rn +m 2 − rn 2
mR
Knowing rn+m, rn and R, the wavelength can be calculated
...
Let λa and λm represent the wavelength of light in air and
in medium (liquid)
...
7 Diffraction
Sound is propagated in the form of waves
...
Similarly, waves on the surface of water also bend round the edges of
an obstacle and spread into the region behind it
...
Diffraction is a
characteristic property of waves
...
Fresnel showed that the amount of bending produced at an
obstacle depends upon the wavelength of the incident wave
...
As the wavelength of light is very small, compared to that
of sound wave and even tiny obstacles have large size, compared to the
wavelength of light waves, diffraction effects of light are very small
...
A series of coloured
images are observed
...
7
...
In the Fresnel
diffraction, the source and the screen are at finite distances from the
obstacle producing diffraction
...
In the
Fraunhofer diffraction, the source and the screen are at infinite
distances from the obstacle producing diffraction
...
The diffracted rays which
are parallel to one another are brought to focus with the help of a
convex lens
...
5
...
2 Diffraction grating
An arrangement consisting of a large number of equidistant
parallel narrow slits of equal width separated by equal opaque portions
is known as a diffraction grating
...
The modern commercial form of grating contains about 6000
lines per centimetre
...
The
combined width of a ruling and a slit is called grating element (e)
...
Theory
MN represents the section of a plane transmission grating
...
5
...
Let e = a + b
...
According to Huygen’s principle, the
points in the slit AB, CD … etc act as a source of secondary wavelets
which spread in all directions on the other side of the grating
...
21 Diffraction grating
Let us consider the secondary diffracted wavelets, which makes
an angle θ with the normal to the grating
...
It will be seen that
the path difference between waves from any pair of corresponding
points is also (a + b) sin θ
The point P1 will be bright, when
(a + b) sin θ = m λ where m = 0, 1, 2, 3
In the undiffracted position θ = 0 and hence sin θ = 0
...
Hence the wavelets proceeding in the direction of the incident
rays will produce maximum intensity at the centre O of the screen
...
If (a + b) sin θ1 = λ, the diffracted wavelets inclined at an angle
θ1 to the incident direction, reinforce and the first order maximum is
obtained
...
In general, (a + b) sin θ = m λ is the condition for maximum
intensity, where m is an integer, the order of the maximum intensity
...
When white light is used, the diffraction pattern consists of a
white central maximum and on both sides continuous coloured images
are formed
...
Therefore
sin θ = Nmλ is satisfied for m= 0 for all values of λ
...
Hence an undispersed white image is obtained
...
As θ further
increases, (a + b) sin θ passes through λ values of all colours resulting
in the formation of bright images producing a spectrum from violet to
red
...
5
...
3
Experiment
to
determine
the
wavelength of monochromatic light using a
plane transmission grating
...
Initially all the preliminary adjustments of
the spectrometer are made
...
The telescope is
brought in line with collimator to view the direct
image
...
The telescope is
slowly turned to one side until the first order
diffraction image coincides with the vertical
cross wire of the eye piece
...
5
...
209
C
2
2
1
T
1
Direct
ray
T
Fig 5
...
The difference between two positions gives 2θ
...
The wavelength
sin θ
of light is calculated from the equation λ =
...
5
...
4 Determination of wavelengths of spectral lines of white light
Monochromatic light is now replaced by the given source of white
light
...
23
...
23 Diffraction of white light
knowing N, wave length of any line can be calculated from the
relation
sin θ
λ=
Nm
5
...
5 Difference between interference and diffraction
1
...
3
...
Interference
It is due to the superposition of
secondary wavelets from two
different wavefronts produced
by two coherent sources
...
Bright fringes are of same
intensity
Comparing with diffraction, it
has large number of fringes
210
Diffraction
It is due to the superposition
of secondary wavelets emitted
from various points of the
same wave front
...
Intensity falls rapidly
It has less number of fringes
...
8
...
But the
transverse nature of light waves is demonstrated only by the
phenomenon of polarisation
...
8
...
Let a rope AB be passed through two parallel vertical slits S1 and
S2 placed close to each other
...
If the free
end A of the rope is moved up and down perpendicular to its length,
transverse waves are
generated
with
vibrations parallel to
the slit
...
But if S2 is
S1
S2
made horizontal, the
(a)
two
slits
are
perpendicular to each
other
...
S1
S2
i
...
24
...
24 Polarisation of transverse waves
On the otherhand, if longitudinal waves are generated in the rope
by moving the rope along forward and backward, the vibrations will
pass through S1 and S2 irrespective of their positions
...
A similar phenomenon has been observed in light, when light
passes through a tourmaline crystal
...
25 Polarisation of transverse waves
Light from the source is allowed to fall on a tourmaline crystal
which is cut parallel to its optic axis (Fig
...
25a)
...
When the crystal A is rotated, there is no change in the
intensity of the emergent light
...
When both the crystals are rotated together, so
that their axes are parallel, the intensity of light coming out of B does
not change
...
When the axis of B is at
right angles to the axis of A, no light emerges from B (Fig
...
25b)
...
Comparing these observations with the mechanical analogue
discussed earlier, it is concluded that the light waves are transverse in
nature
...
These waves are said to be polarised
...
The phenomenon of restricting the
vibrations into a particular plane is known as polarisation
...
8
...
The plane which is at right angles to
the plane of vibration and which contains the direction of propagation
of the polarised light is known as the plane of polarisation
...
P
In the Fig 5
...
S
H
E
5
...
3 Representation of
light vibrations
G
F
Q
R
Fig 5
...
These are
represented by double arrows and
dots (Fig 5
...
The vibrations in the plane of
the paper are represented by double
Fig 5
...
5
...
4 Polariser and Analyser
A device which produces plane polarised light is called a polariser
...
A polariser can serve as an analyser and vice versa
...
If the
intensity of the emergent light does not vary, when the analyser is
rotated, then the incident light is unpolarised; If the intensity of light
varies between maximum and zero, when the analyser is rotated
213
through 90o, then the incident light is plane polarised; If the intensity
of light varies between maximum and minimum (not zero), when the
analyser is rotated through 90o, then the incident light is partially
plane polarised
...
8
...
Malus, discovered that when a beam of ordinary light is
reflected from the surface of transparent medium like glass or water,
it gets polarised
...
Consider a beam of unpolarised light AB, incident at any angle on
the reflecting glass surface XY
...
The
C
vibrations
which
are
perpendicular to the plane of
Y
the diagram and parallel to
the reflecting surface, shown
by dots (Fig
...
28)
...
Fig 5
...
When the light is allowed to be incident at a particular angle, (for
glass it is 57
...
The
angle of incidence at which the reflected beam is completely plane
polarised is called the polarising angle (ip)
...
8
...
It has been observed
experimentally that the reflected and refracted rays are at right angles to
214
each other, when the light is incident at polarising angle
...
28, ip +900 + r = 1800
r = 900 – ip
sin i p
=µ
From Snell’s law,
sin r
where µ is the refractive index of the medium (glass)
Substituting for r, we get
sin i p
sin i p
=µ ;
=µ
cos i p
sin(90 − i p )
∴
tan ip = µ
The tangent of the polarising angle is numerically equal to the
refractive index of the medium
...
8
...
It consists of a
number of glass plates
placed one over the
other as shown in
Fig 5
...
The plates
Fig
...
29 Pile of plates
are inclined at an angle
of 32
...
A beam of monochromatic light is
allowed to fall on the pile of plates along the axis of the tube
...
5o which is the polarising angle for glass
...
The
larger the number of surfaces, the greater is the intensity of the
reflected plane polarised light
...
5
...
8 Double refraction
Bartholinus discovered that when a ray of unpolarised light is
incident on a calcite crystal, two refracted rays are produced
...
5
...
Hence, two images
of a single object are formed
...
E
O
(a)
E
O
(b)
Fig 5
...
30b)
...
The
stationary image is known as the ordinary image (O), produced by the
refracted rays which obey the laws of refraction
...
The other image is extraordinary image (E), produced by
the refracted rays which do not obey the laws of refraction
...
Inside a double refracting crystal the ordinary ray travels with
same velocity in all directions and the extra ordinary ray travels with
different velocities along different directions
...
Inside the crystal there is a particular direction in which both the
rays travel with same velocity
...
The
refractive index is same for both rays and there is no double refraction
along this direction
...
8
...
Crystals like mica, topaz, selenite and aragonite having two optic
axes are called biaxial crystals
...
8
...
One of the most
common forms of the Nicol prism is made by taking a calcite crystal
whose length is three times its breadth
...
And the two
halves are joined together by a layer of Canada balsam, a transparent
cement as shown in Fig 5
...
For sodium light, the refractive index for
ordinary light is 1
...
486
...
550 for both rays, hence
Canada balsam does not polarise light
...
It splits up into two rays as ordinary ray (O)
and extraordinary ray (E) inside the nicol prism (i
...
The ordinary ray is totally internally reflected at the layer
of Canada balsam and is prevented from emerging from the other face
...
The nicol prism serves as a polariser and also an
analyser
...
31 Nicol prism
B
5
...
11 Polaroids
A Polaroid is a material which polarises light
...
There are different types of polaroids
...
Each crystal is a doubly refracting medium,
which absorbs the ordinary ray and transmits only the extra ordinary
ray
...
Recently, new types of polariod are prepared in which thin film
of polyvinyl alcohol is used
...
217
5
...
12 Uses of Polaroid
1
...
2
...
3
...
4
...
5
...
6
...
In aeroplane one polaroid is fixed outside the
window while the other is fitted inside which can be rotated
...
7
...
8
...
9
...
5
...
13 Optical activity
When a plane polarised light is made to pass through certain
substances, the plane of polarisation of the emergent light is not the
same as that of incident light, but it has been rotated through some
angle
...
The substances
which rotate the plane of polarisation are said to be optically active
...
Optically active substances are of two types, (i) Dextro−rotatory
(right handed) which rotate the plane of polarisation in the clock wise
direction on looking towards the source
...
Light from a monochromatic source S, is made to pass through a
polariser P
...
No light comes out of A
...
5
...
The emerging light is cut off
again, when the analyzer is rotated through a certain angle
...
P
A
S
No light
A
P
S
Light
Optically
active
substance
Fig 5
...
(iii) wavelength of light used
(iv) the temperature of the solutions
...
8
...
Specific rotation for a given wavelength of light at a given
temperature is defined as the rotation produced by one-decimeter
length of the liquid column containing 1 gram of the active material in
1cc of the solution
...
l
...
Sugar is the most common optically active substance and this
optical activity is used for the estimation of its strength in a solution
by measuring the rotation of plane of polarisation
...
1
In Young’s double slit experiment two coherent sources of
intensity ratio of 64 : 1, produce interference fringes
...
Data : I1 : I2 : : 64 : 1
I1 a12 64
=
=
I 2 a 22
1
Solution :
∴
I max
I min = ?
a1 8
= ;
a2 1
a1 = 8a2
2
(8a2 + a2 )2
I max (a 1 + a2 )
=
=
I min (a 1 − a2 )2 (8a2 − a2 )2
=
(9a 2 )2
2
(7a2 )
=
81
49
Imax : Imin : : 81 : 49
5
...
Calculate the fringe width if the
entire apparatus is immersed in a liquid of refractive index 1
...
Data : λ = 6000 Å = 6 × 10−7 m; β = 2mm = 2 × 10−3 m
µ = 1
...
3
λ⎤
⎡
⎢∵ µ = λ ′ ⎥
⎣
⎦
2 × 10−3
= 1
...
5 mm
1
...
33, is illuminated by white light
incident at an angle 30o
...
Calculate the smallest thickness of the film
...
33; i = 30o; λ = 6000 Å = 6 × 10–7 m
n = 1 (Smallest thickness); t = ?
220
sini
sin r =
∴
=
sin i
sin r
=
0
...
3759
1
...
33
µ
Solution :
µ
cos r =
1 − 0
...
9267
2 µt cos r = nλ
λ
=
6 × 10 −7
2 × 1
...
9267
t
2µ cos r
t
=
6 × 10−7
2
...
4
=
= 2
...
The radius
of the 8th dark ring is 3
...
Calculate the wavelength of light
used
...
6 mm = 3
...
5
rn 2
(3
...
What is
the order of the ring?
Data : dn = 2d2 ; n = ?
Solution : dn2 = 4nRλ
d22 = 8Rλ
...
(2)
d 2 n
(1)
⇒ n2 =
d2
(2)
2
221
4d22 n
=
d 22
2
∴
5
...
Two slits 0
...
The screen is placed at 1 m distance from the slits
...
Data : d = 0
...
3 × 10−3 m ; λ = 4500 Å = 4
...
5 × 10−7
0
...
7
A parallel beam of monochromatic light is allowed to incident
normally on a plane transmission grating having 5000 lines per
centimetre
...
Find the wavelength of the light
...
5
=
5
5 × 10 × 2 5 × 105 × 2
λ = 5 × 10−7 m = 5000 Å
...
8
A 300 mm long tube containing 60 cc of sugar solution produces
a rotation of 9o when placed in a polarimeter
...
222
Data : l = 300 mm = 30 cm = 3 decimeter
θ = 9o ; S = 60o ; v = 60 cc
m = ?
θ
θ
Solution : S = l × c = l × (m /v )
m =
=
θ
...
In the same way any question and problem could be framed from the text
matter
...
)
5
...
2
power
fields
power
fields
power
power
is equally transferred along the electric and magnetic
is transmitted in a direction perpendicular to both the
is transmitted along electric field
is transmitted along magnetic field
Electromagnetic waves are
(a) transverse
(b) longitudinal
(c) may be longitudinal or transverse
(d) neither longitudinal nor transverse
5
...
5
...
4
In an electromagnetic wave the phase difference between electric
→
→
field E and magnetic field B is
(a) π/4
(c) π
5
...
5
...
7
(b) expands
(c) remains same
(d) first expands, then contracts
A beam of monochromatic light enters from vacuum into a medium
of refractive index µ
...
8
If the wavelength of the light is reduced to one fourth, then the
amount of scattering is
(a) increased by 16 times
(b) decreased by 16 times
(c) increased by 256 times (d) decreased by 256 times
5
...
What is the value of m?
(a) 2
(b) 4
(c) 8
(d) 10
5
...
The phase difference between
them is
(a) π
(b) 2π
π
(c) 3
(d) π/2
2
5
...
The wave length of the another source is
(a) 4500 Å
(b) 6000 Å
(c) 5000 Å
(d) 4000 Å
224
5
...
005 m wide with 2500 lines
...
13 A diffraction pattern is obtained using a beam of red light
...
14 The refractive index of the medium, for the polarising angle 60o is
(a) 1
...
414
(c) 1
...
468
5
...
16 Mention the characteristics of electromagnetic waves
...
17 Give the source and uses of electromagnetic waves
...
18 Explain emission and absorption spectra
...
19 What is fluoresence and phosphorescence?
5
...
5
...
22 How are Stoke’s and Anti-stoke’s line formed?
5
...
24 Explain the Raman scattering of light
...
25 Explain Huygen’s principle
...
26 On the basis of wave theory, explain total internal reflection
...
27 What is principle of superposition of waves?
5
...
5
...
5
...
225
5
...
32 Distinguish between Fresnel and Fraunhofer diffraction
...
33 Discuss the theory of plane transmission grating
...
34 Describe an experiment to demonstrate transverse nature of light
...
35 Differentiate between polarised and unpolarised light
...
36 State and explain Brewster’s law
...
37 Bring out the difference’s between ordinary and extra ordinary light
...
38 Write a note on : (a) Nicol prism (b) Polaroid
5
...
40 An LC resonant circuit contains a capacitor 400 pF and an inductor
100 µH
...
Calculate
the wavelength of the radiated electromagnetic wave
...
41 In Young’s double slit experiment, the intensity ratio of two coherent
sources are 81 : 1
...
5
...
33
...
5
...
Distance between the centres of adjacent fringes is 0
...
Calculate the distance between the slits, if the screen is 1
...
5
...
6 cm
...
5
...
Calculate the thickness
of the air film
...
46 A soap film of refractive index 4/3 and of thickness 1
...
The reflected
light is examined by a spectroscope in which dark band
226
corresponds to a wavelength of 5000 Å
...
5
...
82 mm and that of the 10th ring 3
...
If the
radius of the plano−convex lens is 1 m
...
5
...
Calculate the
angular separation in second order spectrum of red line 7070 Å
and blue line 5000 Å
...
49 The refractive index of the medium is 3
...
5
...
When observed through polarimeter, the plane of polarisation is
rotated through 10o
...
Specific rotation of sugar is 60o / decimetre / unit concentration
...
1 (b)
5
...
3 (d)
5
...
5 (a)
5
...
7 (a)
5
...
9 (d)
5
...
11 (a)
5
...
13 (c)
5
...
40 377 m
5
...
42 2
...
09 × 1014 Hz, 4429 Å
5
...
44
3 mm
5
...
46
6
5
...
48
15o
5
...
50
0
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