where ε is called the absolute permittivity of force between charges by a factor of εr, its
the medium. The force between the same two relative permittivity.
charges placed in free space or vacuum at
distance r is given by, While using Eq. (10.5) we assume that
the medium is homogeneous, isotropic and
Fvac = 1 § q1q 2 · --- (10.3) infinitely large.
4SH 0 ¨© r2 ¸¹ 10.4.3 Definition of Unit Charge from the
Coulomb’s Law:
Dividing Eq. (10.3) by (10.2)
The force between two point charges q1
Fvac = 1 § q1q 2 · =H and q2, separated by a distance r in free space, is
Fmed 4SH 0 ¨© r2 ¸¹ H0 written by using Eq. (10.2),
1 § q1q 2 · 1 q1q 2 q1q
4SH ©¨ r2 ¸¹ 4SH 0 r2 r2
F u 9 u 109 u 2
ε ε0 = 8.85 Í10-12 C2 N-1 m-2
diTehleectrraictiocoεn0stainst the relative permittivity If q1= q2= 1C and r = 1.0 m
or of the medium and is
denoted by εr or K. Then F = 9.0Í109 N
K or Hr H Fvac --- (10.4) From this, we define, coulomb (C) the unit
H0 Fmed of charge in SI units.
Thus, One coulomb is the amount of charge
which, when placed at a distance of one
(i) εr is the ratio of absolute permittivity of a metre from another charge of the same
medium to the permittivity of free space. magnitude in vacuum, experiences a force of
9.0 × 109 N. This force is a tremendously large
(ii) εr is the ratio of the force between two point force realisable in practical situations. It is,
charges placed a certain distance apart in therefore, necessary to express the charge in
free space or vacuum to the force between smaller units for practical purpose. Subunits
the same two point charges when placed at of coulomb are used in electrostatics. For
the same distance in the given medium. example, micro-coulomb (10-6 C, µC), nano-
coulomb (10-9 C, nC) or pico-coulomb. (10-12 C,
εr is a dimensionless quantity. pC) are normally used units.
(iii) εr is also called specific inductive capacity.
The force between two point charges q1 Do you know ?
and q2 placed at a distance r in a medium of
relative permittivity εr, is given by Force between two charges of 1.0 C each,
separated by a distance of 1.0 m is 9.0×109 N
F 1 q1q 2 --- (10.5) or, about 10 million metric tonne. A normal
For w4SaHt0eHrr, r2 (10.4) truck-load is about 10 metric tonne. So, this
force is equivalent to about one million truck-
εr = 80 then from Eq. loads. A tremendously large force indeed !
Fvac Hr 80
Fwater
Fwater Fvac Example 10.2: Charge on an electron is
80 1.6×10-19 C. How many electrons are required
This means that when two point charges to accumulate a charge of one coulomb?
are placed some distance apart in water, the Solution: 1.6Í10-19 C = 1 electron
§ 1 · th 1
©¨ 80 ¸¹ 1.6 u10 19
force between them is reduced to of the ?1C electrons
force between the same two charges placed at 0.625u1019 6.25u1018 electrons
the same distance in vacuum.
6.25×1018 electrons are required to
Thus, a material medium reduces the accumulate a charge of one coulomb.
192
It is now possible to measure a very Here, q1 = q2 = +1.6×10-19 C, r = 10-15m
small amount of current in otto-amperes
which measures flow of single electron. ? Fe 9 u109 (1.6 u10 19 ) (1.6 u10 19 )
(10 15 )2
10.4.4 Coulomb’s Law in Vector Form: 9 u1.6 u1.6 u101 N
As shown in Fig 10.4, q1 and q2 are two Fe 2.3u102 N force --- (10.8.a)
similar point charges situated at points A and B. between the
The gravitational
r12 is the distance of separation between them. protons is given by
F21 denotes the force exerted on q2 by q1 Fg = G m1m 2
r2
1 q1q 2
F21 4SH 0 u u r21 --- (10.6) 6.674 u10 11 u1.67 u10 27 u1.67 u10 27
2
(10 15 )2
r 21 Fg 1.86 u10 34 N
--- (10.8.b)
Comparing 10.8. (a) and 10.8.(b)
Fe 2.30 u10 2 N = 1.23u1036
Fg 1.86 u10 34 N
1
Fig. 10.4: Coulomb’s law in vector form Thus, the electrostatic force is about
36 orders of magnitude stronger than the
gravitational force.
r21 is the unit vector along AB , away Comparison of gravitational and
from B. Similarly, the force F12 exerted on q1 electrostatic forces:
by q2 is given by
Similarities
F12 1 u q1q 2 u r12 --- (10.7) 1. Both 1forces obey inverse square law :
4SH 0 r12 2
F ∝ r2
2. Both are central forces : act along the
r12 is the unit vector along BA , away from line joining the two objects.
A. F12 acts on q1at A and is directed along BA, Differences
away from A. The unit vectors r12 and r21 are 1. Gravitational force between two objects
oppositely directed i.e., r12 r21 hence, is always attractive while electrostatic
F 21 = - F12 force between two charges can be
either attractive or repulsive depending
Thus, the two charges experience force on the nature of charges.
2. Gravitational force is about 36
of equal magnitude and opposite in direction. orders of magnitude weaker than the
electrostatic force.
These two forces form an action- reaction pair.
As F21 and F12 act along the line joining 10.5 Principle of Superposition:
the two charges, the electrostatic force is a
central force. The principle of superposition states that
Example 10.3: Calculate and compare the when a number of charges are interacting,
electrostatic and gravitational forces between
two protons which are 10-15 m apart. Value of the resultant force on a particular charge is
G = 6.674×10-11 m3 kg-1 s-2 and mass of the
proton is 1.67×10-27 kg given by the vector sum of the forces exerted by
individual charges.
Consider a number of point charges q1, q2,
q3 ------- kept at points A1, A2, A3--- as shown
Solution: The electrostatic force between the in bFyigq. 21i0s.5F.12T.hTehfeorvcaelueexeorfteFd12onis the charge
q1 calculated
protons is given by Fe 1 q1q2
4SH0 r2
193
by ignoring the presence of other charges. F AC 2µC 4 cm 3µC B
Similarly, we find F13 , F14 etc, one at a time, F AB A 3 cm
using the coulomb’s law. θ
F14
F15 F13
Fig. a: Position of charges. 4µC
q2 q1 C
A2 A1 Solution : Given,
F12
AB = 4.0 cm, BC = 3.0 cm
q3 q5 ? AC = 42 32 5.0 cm
A3 q4 A5 Magnitude of force F AB on A due to B is,
A4 § 1 · 2 u10 6 u 3u10 6
Fig. 10.5: Principle of superposition. F AB ¨ 4SH 0 ¸ (4 u10 2 )2
Total force F1 on charge q1 is the vector © ¹
sum of all such forces.
9 u109 u 6 u 10 12
16 u10 4
F1 F12 F13 F14 ...
= 3.37u10
1 ª q1q q1q 3 º = 33.7 N
4SH 0 « r13 2 ...»» ,
2 r12 r13 This force acts at point A and is directed
¼» along BA (Fig. (b)).
« r12 2
«¬
where r12 , r13 etc., are unit vectors directed to F
q1 from q2, q3 etc., and r12, r13, r14,etc., are the F AC
distances from q1 to q2, q3 etc respectively.
F AB A
Let there be N point charges q1, q2,q3 Fig. b: Forces acting at point A.
etc., qN. The force F exerted by these charges Magnitude of force F AC on A due to C is,
on a test charge q0 can be written using the
summation notation Σ as follows, F AC § 1 · 2 u10 6 u 4 u10 6
Ftest F1 F2 F3 FN --- (10.9)
¦N = 1 ¦N q0qn r 0 n --- (10.10) ¨ 4SH 0 ¸ (5u10 2 )2
4SH 0 r02n © ¹
Fn n =1
9 u109 u 8.0 u10 12
n =1
Where r0n is a unit vector directed from the 25 u 10 4
nth charge to the test charge qo and r0n is the 72
25
separation between them, r0n = r0n r0n u10 28.8 N
Can you tell? This force acts at point A and is directed
along CA . (Fig. 10.6.(b))
Three charges, q each, are placed at the
vertices of an equilateral triangle. What will F = F AB + F AC
be the resultant force on charge q placed at Magnitude of resultant force is,
the centroid of the triangle?
F = ¬ªFA2C FA2B 2FAC FAB cosT º¼ 1/2
Example 10.4: Three charges of 2µC, 3µC
and 4µC are placed at points A, B and C 59.3 N
respectively, as shown in Fig. a. Determine the
force on A due to other charges.
194
Direction of the resultant force is 16.9° electric field of a charge is also a vector and
is directed along the direction of the coulomb
north of west. (Fig. c) force, experienced by a test charge.
N
The magnitude of electric field at a distance
F r from a point charge Q is same at all points on
F AC the surface of a sphere of radius r as shown in
Fig. 10.6. Its direction is along the radius of
W 16.9° A the sphere, pointing away from its centre if the
F AB charge is positive.
Fig. c: Direction of the resultant force. E
10.6 Electric Field:
Space around a charge Q gets modified so E Q
that when a test charge is brought in this region, 4SH0r 2
it experiences a coulomb force. This region +
around a charged object in which coulomb
force is experienced by another charge is called
electric field. Fig. 10.6: Electric field due to a point charge
(+Q).
Mathematically, electric field is defined as
the force experienced per unit charge. Let Q and SI unit of electric intensity is newton per
q be two charges separated by a distance r. coulomb (NC-1). Practically, electric field
is expressed in volt per metre(Vm-1). This is
The coulomb force between them is given discussed in article 10.6.2.
by F 1 Qq r , where, r is the unit vector
4SH 0 r2
Dimensional formula of E is,
along the line joining Q to q. E = F [LMT 2 ]
q0 [IT]
Therefore, electric field due to charge Q is
given by, E = [LMT 3 I 1]
E F Q r --- (10.11) 10.6.1 Electric Field Intensity due to a Point
q 4SH0r 2
Charge in a Material Medium:
The coulomb force acts across an empty
space (vacuum) and does not need any Consider a point charge q placed at point O
intervening medium for its transmission. in a medium of dielectric constant K as shown
in Fig. 10. 7.
The electric field exists around a charge
irrespective of the presence of other charges.
Since the coulomb force is a vector, the q0
A precise definition of electric field is: k
Electric field is the force experienced by a q
test charge in presence of the given charge at
the given distance from it. Fig. 10.7: Field in a material medium.
Consider the point P in the electric field of
E lim F point charge q at distance r from it. A test charge
qo0 q q0 placed at the point P will experience a force
which is given by the Coulomb’s law,
Test charge is a positive charge so
small in magnitude that it dose not affect the
surroundings of the given charge.
195
F 1 q q0 r
4SH 0 K r2
where r is the unit vector in the direction of Fig. 10.9 (a):
force i.e., along OP. uniform electric
field.
By the definition of electric field intensity
E F 1 q r
q0 4SH 0 K r2 Fig. 10.9 (b):
non uniform
The direction of E will be along OP when electric field.
q is positive and along PO when q is negative.
10.6.2 Practical Way of Calculating Electric
The magnitude of electric field intensity in Field
a medium is given by
A pair of charged parallel plates is
E 1q --- (10.12 ) arranged as shown in Fig. 10.10. The electric
4SH0K r2 field between them is uniform. A potential
difference V is applied between two parallel
For air or vacuum K = 1 then plates separated by a distance ‘d’. The electric
field between them is directed from plate A to
E = 1 q plate B as shown.
4SH 0 r2
The coulomb force between two charges
and electric field E of a charge both follow the
inverse square law, (F∝1/r2, E∝1/r2) Fig. 10.8.
F∝1/r2, E∝1/r2 A B
E Fig 10.10: Electric field
or between two parallel
plates.
F
r A charge +q placed between the plates
experiences a force F due to the electric field. If
Fig. 10.8: Variation of Coulomb force/ we have to move the charge against the direction
Electric field due to a point charge. of field, i.e., towards the positive plate, we have
to do some work on it. If we move the charge
1. Uniform electric field: A uniform electric +q from the negative plate B to the positive
field is a field whose magnitude and plate A, the work done against the field is
direction is same at all points. For example, W = Fd; where ‘d’ is the separation between the
field between two parallel plates. Fig 10.9.a plates. The potential difference V between the
two plates is given by
2. Non uniform electric field: A field whose
magnitude and direction is not the same at W = Vq, but W = Fd
all points. For example, field due to a point
charge. In this case, the magnitude of field ∴Vq = Fd ∴ F/q = V/d = E
is same at distance r from the point charge
in any direction but the direction of the ∴ Electric field can be defined as
field is not same. Fig 10.9.b
E = V/d --- (10.13)
This is the commonly used definition of
electric field.
196
Example 10.5: Gap between two electrodes of magnitudes and are opposite in direction,
the spark-plug used in an automobile engine
is 1.25 mm. If the potential of 20 V is applied EA = - EC . EA + EC = 0 . Thus, the field at P is
across the gap, what will be the magnitude of
electric field between the electrodes? only to the charge at B and can be written as
Ep EB 2 u10 6
4SH0 (BP)2
Solution: Ep 2 u10 6 u 9 u109
EV (5 / 2 )2
d
2 u 9 u103 u 2
20V V 25
1.25u10 3 m m
E 16 u10 3 1.6 u104 36 u 103
25
This electric field is sufficient to ionize 1.44 u1015 NC-1 along BP
the gaseous mixture of fuel compressed in the
To calculate BP
cylinder and ignite it. 5
BP (BA) cos (45)q 2
Can you tell? Example 10.7: A simplified model of hydrogen
atom consists of an electron revolving about a
Why a small voltage can produce a proton at a distance of 5.3×10-11m. The charge
reasonably large electric field? on a proton is +1.6×10-19 C. Calculate the
intensity of the electric field due to proton at
Example 10.6: Three point charges are placed this distance.
at the vertices of a right isosceles triangle as
shown in the Fig. a. What is the magnitude and Solution:
direction of the resultant electric field at point P E q
4SHor 2
which is the mid point of its hypotenuse?
q 1.6 u10 19 C
C
+10 µC
EA EB
r 5.3u10 11 m
5cm P 1 9.0 u109 Nm2C 1
4SH o
EC
1.6 u10 19
+2 µC +10 µC E (5.3u10 11 )2 u 9.0 u109 5.1u1011 NC 1
B
5cm A The force between electron and proton in
Fig (a): Position of charges. hydrogen atom can be calculated by using the
EB electric field. We have, E F F qE
q
EA ?
F = -1.6×10-19 C× 5.1×1011 NC-1
= -8.16×108 N.
P This force is attractive.
EC Using the Coulomb's law,
Fig (b): Electric field at point P. F 1 q1q2
Solution: Electric field is the force an a unit 4SH0 r 2
positive charge, the fields at P due to the charges
at A, B and C are shown in the Fig. b. EA is the 9.0 u109 Nm2C 1 u ( 1.6 u 10 19 C ) u ( 1.6 u 10 19 C )
field at P due to charge at A and EC is the field (5.3u10 11 m)2
at P due to charge at C. Since P is the midpoint
of AC and the fields at A and C are equal in 8.6 u108 N
Knowing electric field at a point is useful
to estimate the force experienced by a charge
at that point.
197
10.6.3 Electric Lines of Force: (6) Electric lines of force are crowded in a
Michael Faraday (1791-1867) introduced region where electric intensity is large.
the concept of lines of force for visualising (7) Electric lines of force are widely separated
electric and magnetic fields. An electric line of from each other in a region where electric
force is an imaginary curve drawn in such a intensity is small
way that the tangent at any given point on this
curve gives the direction of the electric field (8) The lines of force of an uniform electric
at that point. See Fig.10.11. If a test charge is field are parallel to each other and are
placed in an electric field it would be acted upon equally spaced.
by a force at every point in the field and will The lines of force are purely a geometric
move along a path. The path along which the
unit positive charge moves is called a line of construction which help us visualise the nature
force. of electric field in a region. The lines of force
have no physical existence.
Fig. 10.11: Electric line of force.
A line of force is defined as a curve such Fig. 10.13 (a): Lines
that the tangent at any point to this curve gives of force due to positive
the direction of the electric field at that point. charge.
The density of field lines indicates the
strength of electric fields at the given point in Fig. 10.13 (b): Lines of
space. Figure 10.12. force due to negative
charge.
(High field)
Fig. 10.13 (c): Lines
of force due to
opposite charge.
(Low field) Fig. 10.12: density Fig. 10.13; (d):
of field lines and Lines of force
strength of electric due to similar
field. charge.
Characteristics of electric lines of force Fig. 10.13 (e): Lines
of force terminate
(1) The lines of force originate from a on a conductor.
positively charged object and terminate on
a negatively charged object. Fig. 10.13 (f): Intensity
of a electric field is more
(2) The lines of force neither intersect nor meet at point A and less at B.
each other, as it will mean that electric field More lines cross the area
has two directions at a single point. at A and less at the same
area at B.
(3) The lines of force leave or terminate on a Fig. 10.13: The lines of force due to various
conductor normally. geometrical arrangement of electrical charges.
(4) The lines of force do not pass through Can you tell?
conductor i.e. electric field inside a
conductor is always zero, but they pass Lines of force are imaginary, can they
through insulators. have any practical use?
(5) Magnitude of the electric field intensity
is proportional to the number of lines
of force per unit area of the surface held
perpendicular to the field.
198
10.7 Electric Flux: imagine a small charge +q present at a point O
inside closed surface. Imagine an infinitesimal
As discussed previously, the number of area dA of the given irregular closed surface.
lines of force per unit area is the intensity of the
electric field E .
=AreNa uenmcbloesrinogf lines of force
... E the lines of force - (10.14)
Fig. 10.15: Gauss' law.
The magnitude of electric field intensity at
point P on dS due to charge +q at point O is,
Fig. 10.14: Flux through area S. E = 1 § q ·
Number of lines of force = (E).(Area) 4SH 0 ©¨ r2 ¸¹
When the area is inclined at an angle θ with the
direction of electric field, Fig. 10.14, the electric The direction of E is away from point O.
flux can be calculated as follows.
Let θ be the angle subtended by normal drawn
and Let tvheectoarngdleS between electric field E to area dS and the direction of E. Electric flux,
area be θ, then the electric flux dφ, passing through area dS, = Ecosθ dS
passing through area dS is given by q cosT dS
dφ = (component of dS along E ).(area S 4SH0r 2
of )
d § q ·§ dS cosT ·
dφ = E (dS cos θ) ¨ 4SH 0 ¸ ¨© r2 ¹¸
© ¹
dφ = EdS cos θ
dφ = E .d S --- (10.15) dI § q · dZ --- (10.17)
¨ 4SH 0 ¸
Total flux through the entire surface © ¹
) ³ dI ³ E.dS E.S --- (10.16) where, dZ = dS cosT is the solid angle
s r2
The SI unit of electric flux can be calculated subtended by area dS at point O.
using, Total electric flux, φ , crossing the given
closed surface can be obtEained by integrating
) E S =(V/m) m2 =Vm Eq. (10.17) over its area. Thus,
10.8 Gauss' Law:
Karl Friedrich Gauss (1777-1855) one )E ³ dI ³ E.ds ³ q dZ q ³ dZ
of the greatest mathematician of all times, s 4SH 0 4SH 0
formulated a law expressing the relationship
between the electric charge and its electric field But ³ dZ 4S solid angle subtended by
which is called the Gauss’ law. Gauss' law is
analongous to Coulomb’s law in the sense that it entire closed surface at point O
too expresses the relationship between electric
field and electric charge. Gauss' law provides Total flux q ( 4S )
equivalent method for finding electric intensity. 4SH 0
It relates values of field at a closed surface and
the total charges enclosed by that surface. ³)E E ds q / H0 --- (10.18)
s
Consider a closed surface of any shape
which encloses number of positive electric This is true for every electric charge
charges (Fig. 10.15). To prove Gauss’ theorem, enclosed by a given closed surface.
Total flux due to charge q1, over the given
closed surface = + q1/ε0
199
Total flux due to charge q2, over the given Example 10.8: A charge of 5.0 C is kept at the
closed surface = + q2/ε0 centre of a sphere of radius 1 m. What is the flux
passing through the sphere? How will this value
Total flux due to charge qn, over the given change if the radius of the sphere is doubled?
closed surface = + qn/ε0
Solution: Flux per unit area is given by Eq.
Positive sign in Eq. (10.18) indicates that 10.16.
the flux is directed outwards, away form the
charge. If the charge is negative, the flux will According to Gauss law, the total flux
be is directed inwards as shown in Fig 10.16
(b). If a charge is outside the closed surface the through the sphere ) ³ E.ds , where the
net flux through it will be zero Fig 10.16 (c).
integration is over the surface of the sphere. As
the electric field is same all over the sphere i.e.
| E | = constant and the direction of E as well as
that of ds is along the radius, we get
flux = ) = | E | 4S R2
E q 9 u109 u 5.0 C
4SH 0 r 2 (1.0 m)2
Fig. 10.16 (a): Flux due to positive charge. E 9u109 u 5 4.5u1010 NC-1
I ES
?flux = 4.5u1010 u 4S (1)2
5.65u1011Vm
Thus the total flux is independent of radius.
Fig. 10.16 (b): Flux due to negative charge. E ∝ 1/r2, and area ∝ r2. This can also be seen
from Gauss' law, where the net flux crossing a
closed surface is equal to q/ε0 where q is the net
charge inside the closed surface. As the charge
Fig. 10.16 (c): Flux due to charge outside a inside the sphere is unchanged, the flux passing
closed surface is zero. through a sphere of any radius is the same.
Thus, if the radius of the sphere is increased
According to the superposition principle, by a factor of 2, the net flux passing through
the total flux φ due to all charges enclosed its surface remains unchanged. As shown in
within the given closed surface is Fig. 10.17, same number of lines of force cross
both the surfaces. The total flux is independent
of shape of the closed surface because Eq. 10.18
does not involve any radius.
= q1 q2 q3 qn i=n qi Q
H0 H0 H0 H0 H0
=
¦) E
+ + + ---- + Hi=1 0
Statement of Gauss' law
The flux of the net electric field through a
closed surface equals the net charge enclosed by
the surface divided by ε0
where Q is the total charge within the surface. Fig. 10.17: Flux is independent of the shape
and size of closed surface.
Gauss' law is applicable to any closed
surface of regular or irregular shape.
200
Do you know ? dipole moment p .
Gaussian surface Dipole mpo=mqe(n2tlis) de fined as
All the lines of force originating from
a point charge penetrate an imaginary --- (10.19)
three dimensional surface. The total flux
ΦE = q/ε0. The same number of lines of force A dipole moment is a vector whose
will cross the surface of any shape. The
total flux through both the surfaces is the magnitude is q (2l) and the direction is from
same. Calculating flux involves calculating
the negative to the positive charge. The unit of
³ E ds , hence it is convenient to consider a
dipole moment is Columb-meter (Cm) or Debye
regular surface surrounding the given charge
distribution. A surface enclosing the given (D). 1D = 3.33×10-30 Cm. If two charges +e and
charge distribution and symmetric about it is
a Gaussian surface. -e are separated by 1.0A0, the dipole moment is
For example. if we have a point charge 1.6×10-29 Cm or 4.8 D. For example, a water
the Gaussian surface will be a sphere. If the
charge distribution is linear, the Gaussian molecule has a permanent dipole moment of
surface would be a cylinder with the charges
distributed along its axis. Gaussian surface Natural dipole:
offers convenience of calculating the The water molecule is non-linear, i.e.,
integral ³ E ds . the two hydrogen atoms and one oxygen atom
are not in a straight line. The two hydrogen-
Remember that a Gaussian surface oxygen bonds in water molecule are at an
is purely imaginary and does not exist angle of 105°. The positive charge of a water
physically. molecule is effectively concentrated on the
hydrogen side and the negative charge on
10.9 Electric Dipole: the oxygen side of the molecule. Thus, the
positive and negative charges of the water
A pair of equal and opposite charges molecule are inherently separated by a small
separated by a finite distance is called an electric distance. This separation of positive and
dipole. It is shown in Fig. (10.18). negative charges gives rise to the permanent
dipole moment of a water molecule.
Fig. 10.18: Electric dipole. x-y axial line,
Molecules of water,
P-Q equatorial line. ammonia, sulphur di-
oxide, sodium chloride
Line joining the two charges is called the etc. have an inherent
dipole axis. A line passing through the dipole separation of centers of positive and
negative charges. Such molecules are called
axis is called axial line. A line passing through polar molecules.
the centre of the dipole and perpendicular to the Polar molecules are the molecules in
which the center of positive charge and
axial line is called the equatorial line as shown the negative charge is naturally separated.
in Fig. 10.18. Molecules such as H2, Cl2, CO2 CH4 and
many others have their positive and negative
Strength of a dipole is measured in terms charges effectively centered at the same
point and are called non-polar molecules.
of a quantity called the dipole moment. Let
q be the magnitude of each charge and 2l be Non-polar molecules are the
the distance from negative c(h2al r)geis to positive molecules in which the center of positive
charge. Then the product q called the charge and the negative charge is one and
the same. They do not have a permanent
electric dipole. When an external electric
field is applied to such molecules the centers
of positive and negative charge are displaced
and a dipole is induced.
201
6.172×10-30 Cm or 1.85 D. Its direction is from Example 10.9: An electric dipole of length 2.0
oxygen to hydrogen. See box on Natural dipole. cm is placed with its axis making an angle of
30° with a uniform electric field of 105 N/C. as
10.9.1 Couple Acting on an Electric Dipole in shown in figure. If it experiences a torque of
a Uniform Electric Field:
Consider an electric dipole placed in a 10 3 Nm, calculate the magnitude of charge on
the dipole.
uniform electric field E. The axis of electric
dipole makes an angle θ with the direction of
electric field as shown in Fig. 10.19 a.
P
Solution: Given
W 10 3, Nm, E 105 N / C,
Fig. 10.19 (a): Dipole in uniform electric field. 2l 2.0 u10 2 m, T 300
W qE2l sinT
10 3Nm q105 N / C 2.0 u10 2 m § 1 ·
¨© 2 ¹¸
? q = 10 3 3 u10 2 C
103
= 1.73u10 2 C
Fig. 10.19 (b): Couple acting on a dipole. 10.9.2 Electric Intensity at a Point due to an
Figure 10.19. b shows the couple acting on
Electric Dipole:
an electric dipole in uniform electric field.
Case 1 : At a point on the axis of a dipole.
The force acting on charge - q at A is
Consider an electric dipole consisting of
F A = - qE in the direction opposite to that two charges -q and +q separated by a distance
2l as shown in Fig. 10.20. Let P be a point at a
of E and the force acting on charge +q at B distance r from the centre C of the dipole. The
is F B = + qE in the direction of E . Since F A electric intensity Ea at P due to the dipole is the
vector sum of the field due to the charge - q at
= - F B , the two equal and opposite forces A and + q at B.
separated by a distance form a couple. Moment
of the couple is called torque and is defined by
W d u F where, d is the perpendicular distance
b..e.MtwaegennittuhdeetwofoTeoqruqaulea=nd opposite forces.
Magnitude of force × Perpendicular distance
... Torque on the dipole = W BP u qE Fig. 10.20: Electric field of a dipole along
... τ = qE2lsinθ its axis.
--- (10.20)
Electric field intensity at P due to the
b..u. tτ p q u 2l --- (10.21) charge -q at A
= pEsinθ
1 ( +ql))2PPQPDD
In vector form W p u E --- (10.22) = EA 4SH 0 ( r
If θ =90° sin θ =1, then τ = pE
When the axis of electric dipole is where PD is unit vector directed along PD
perpendicular to uniform electric field, torque Electric intensity at P due to charge +q at B
of the couple acting on the electric dipole is 1 q
4SH 0 -l
maximum, i.e., τ = pE. It θ = 0 then τ = 0, this is = EB ( r ) 2PPPQQQ
the minimum torque on the dipole. Torque tends
to align the dipole axis along the direction of where PQ is a unit vector directed along PQ .
electric field.
202
The magnitude of EB is greater than that of where PQ is the unit vector directed along PQ
E A . (Because BP < AB) or BP
Electric field at PE Aisatnhde,EsuB m of E A and E B
Resultant field Ea at P on the axis, due to
the dipole is ∴ Eeq = E A + E B
Ea = EB + EA Consider ∆ ACP
The magnitude of Ea is given by (AP)2 = (PC)2 + (AC)2 = r2 + l2 = (BP)2
| Ea | 1 ªq q º ? EA 1q --- (10.25)
4SH 0 «¬ (r l)2 (r l)2 »¼ 4SH0 (r 2 l 2 ) --- (10.26)
| Ea |= q ª r2 + l2 2lr - r2 2lr l2 º EB 1 q
4SH 0 « (r2 l 2 )2 » 4SH0 (r2 l 2 )
¬ ¼
| Ea |= 2(2lq)r EA = EB
4SH0 (r 2 l 2 )2
The resultant of fields E A and EB acting
But 2lq = p, the dipole moment at point P can be calculated by resolving these
| Ea |= 1 2 pr vectors E A and EB along the equatorial line
4SH 0 (r2 l 2 )2 and along a direction perpendicular to it.
--- (10.23)
Ea , is directed along PQ, which is the
direction of the dipole moment p i.e. from the Fig. 10.21 (b):
Components of
negative to the positive charge, parallel to the the field at point
P.
axis. If r>> l, l2 can be neglected compared to r2,
Ea = 1 2p --- (10.24)
4SH 0 r3
The field will be along the direction of the Consider Fig. 10.21 (b). Let the y-axis
coincide with the equator of the dipole x-axis
dipole moment p . will be parallel to dipole axis, as shown. The
Case 2: At a point on the equatorial line. As origin is at point P.
shown in Fig. 10.21 (a)
The y-components of EA and EB are EAsinθ
Fig. 10.21 (a): and EBsinθ respectively. They are equal in
magnitude but opposite in direction and cancel
r Electric field of a each other. There is no contribution from them
dipole at a point on towards the resultant.
the equatorial line. The x-components of EA and EB are EAcosθ
and EBcosθ respectively. They are of equal
Electric field at point P due to charge -q at magnitude and are in the same direction
A is: E A 1 ( q) PA ∴| Eeq | EA cosT EB cosT --- (10.27)
4SH 0 (A P)2 By using Eq. 10.25 and 10.26
where PA is the unit vector direction along PA . Eeq = 2EAcosT
Similarly, Electric field at P due to charge + q at 2 § 4SH q l 2 ) · l
¨ (r 2 ¸ r2 l2
1 ( q) © 0 ¹
4SH 0 (BP)2
B is: E B PQ 2ql
4SH0 (r2 l 2 )3/2
203
If r>>l then 2 is very small compared to r2 (b) Surface charge density (σ)
l
Eeq 1p 1p --- (10.28) Suppose a charge q is uniformly distributed
4SH0 (r2 )3/2 4SH0 r3
over a surface of area A . As shown in Fig. 10.23,
The direction of this field is along - p (anti- then the surface charge density σ is defined as
parallel to p ) as shown in Fig. 10.21 (c).
V q --- (10.30)
A
Fig. 10.21 (c): SI unit of σ is (C/m2)
Electric field at
point P is anti- For example, charge distributed uniformly
on a thin disc or a synthetic cloth. If the charge
parallel to p . is not distributed uniformally over the surface
of a conductor, then charge dq on small area
Comparing Eq. 10.28 and 10.24 we find element dA can be written as dq = σ dA.
that the electric intensity at an axial point is
twice that at a point on the equatorial position,
lying at the same distance from the centre of
the dipole.
10.10 Continuous Charge Distribution:
A system of charges can be considered as Fig. 10.23: Surface charge.
a continuous charge distribution, if the charges (c) Volume charge density (ρ)
are located very close together, compared
to their distances from the point where the Suppose a charge q is uniformly distributed
intensity of electric field is to be found out.
throughout a volume V, then the volume charge
The charge distribution is continuous density ρ is defined as the charge per unit
in the sense that, a system of closely spaced
charges is equivalent to a total charge which volume.
is continuously distributed along a line or a
surface or a volume. To find the electric field q --- (10.31)
due to continuous charge distribution, we define U
following terms for different types of charge
distribution. V
S.I. unit of ρ is (C/m3)
For example, charge on a solid plastic
sphere or a solid plastic cube.
(a) Linear charge density (λ). If the charge is not distributed uniformaly
over the volume of a material, then charge dq
As shown in Fig. 10.22 charge q is over small volume element dV can be written
uniformly distributed along a liner conductor of as dq = ρ dV.
length l. The linear charge density λ is defined
as,
O q --- (10.29) Fig. 10.24:Volume charge.
l Electric field due to a continuous charge
SI unit of λ is (C/m). distribution can be calculated by adding electric
fields due to all these small charges.
For example, charge distributed uniformly
Can you tell?
on a straight thin rod or a thin nylon thread. If
The surface charge density of Earth is
the charge is not distributed uniformly over the σ = -1.33 nC/m2. That is about 8.3×109
electrons per square meter. If that is the
length of thin conductor then charge dq on small case why don't we feel it?
element of length dl can be written as dq = λdl
Fig. 10.22: Linear charge.
204
Do you know ? iii. One can get static shock if charge
transferred is large.
Static charge can be useful
Static charges can be created whenever iv. Dust or dirt particles gathered on
computer or TV screens can catch static
there is a friction between an insulator and charges and can be troublesome.
other object. For example, when an insulator
like rubber or ebonite is rubbed against Precautions against static charge
a cloth, the friction between them causes i. Home appliances should be grounded.
electrons to be transferred from one to the ii. Avoid using rubber soled footwear.
other. This property of insulators is used iii. Keep your surroundings humid. (dry air
in many applications such as Photocopier,
Inkjet printer, Panting metal panels, can retain static charges).
Electrostatic precipitation/separators etc.
Static charge can be harmful Internet my friend
i. When charge transferred from one body
1. h t t p s : / / w w w. p h y s i c s c l a s s r o o m .
to other is very large sparking can take com>class
place. For example lightning in sky.
ii. Sparking can be dangerous while 2. https://courses.lumenlearing.com>elect
refuelling your vehicle. 3. https://www.khanacademy.org>science.
4. https://www.topper.com>guides>physics
ExercisesExercises
1. Choose the correct option. (A) 1:4 (B) 1:2
i. A positively charged glass rod is brought (C) 1:1 (D) 1:16
close to a metallic rod isolated from iv. Two charges of 1.0 C each are placed
ground. The charge on the side of the
one meter apart in free space. The force
metallic rod away from the glass rod will
between them will be
be (A) 1.0 N (B) 9×109 N
(A) same as that on the glass rod and equal (C) 9×10-9 N (D) 10 N
in quantity
v. Two point charges of +5 µC are so
(B) opposite to that on the glass of and
placed that they experience a force of
equal in quantity
80×10-3 N. They are then moved apart,
(C) same as that on the glass rod but lesser
so that the force is now 2.0×10-3 N. The
in quantity
distance between them is now
(D) same as that on the glass rod but more (A) 1/4 the previous distance
in quantity
(B) double the previous distance
ii. An electron is placed between two (C) four times the previous distance
parallel plates connected to a battery. If (D) half the previous distance
the battery is switched on, the electron vi. A metallic sphere A isolated from ground
will is charged to +50 µC. This sphere is
(A) be attracted to the +ve plate
brought in contact with other isolated
(B) be attracted to the -ve plate
metallic sphere B of half the radius of
(C) remain stationary
sphere A. The charge on the two sphere
(D) will move parallel to the plates
will be now in the ratio
iii. A charge of + 7 µC is placed at the centre (A) 1:2
of two concentric spheres with radius 2.0 (C) 4:1 (B) 2:1
(D) 1:1
cm and 4.0 cm respectively. The ratio of vii. Which of the following produces uniform
the flux through them will be
electric field?
205
(A) point charge
(B) linear charge
(C) two parallel plates iii. Four charges of + 6×10-8 C each are
(D) charge distributed an circular any
placed at the corners of a square whose
viii. Two point charges of A = +5.0 µC and
B = -5.0 µC are separated by 5.0 cm. sides a are 3 cm each. Calculate the
A point charge C = 1.0 µC is placed
at 3.0 cm away from the centre on the resultant force on each charge and show
perpendicular bisector of the line joining
the two point charges. The charge at C its direction on a digram drawn to scale.
will experience a force directed towards
[Ans: 6.89×10-2 N]
(A) point A
(B) point B iv. The electric field in a region is given by
(C) a direction parallel to line AB
(D) a direction along the perpendicular E = 5.0 kN/C. Calculate the electric flux
bisector Through a square of side 10.0 cm in the
2. Answer the following questions. following cases
i. What is the magnitude of charge on an
(a) the square is along the XY plane
electron?
ii. State the law of conservation of charge. [Ans: = 5.0×10-2 Vm]
iii. Define a unit charge.
iv. Two parallel plates have a potential (b) The square is along XZ plane
difference of 10V between them. If the [Ans: Zero]
plates are 0.5 mm apart, what will be the
strength of electric charge. (c) The normal to the square makes an
v. What is uniform electric field?
vi. If two lines of force intersect of one angle of 450 with the Z axis.
point. What does it mean?
vii. State the units of linear charge density. [Ans: 3.5×10-2 Vm]
viii. What is the unit of dipole moment?
ix. What is relative permittivity? v. Three equal charges of 10×10-8 C
respectively, each located at the corners
of a right triangle whose sides are 15 cm,
20 cm and 25 cm respectively. Find the
force exerted on the charge located at the
90° angle.
[Ans: 4.59.×10-3 N]
vi. A potential difference of 5000 volt is
applied between two parallel plates 5 cm
a part a small oil drop having a charge of
9.6 ×10-19 C falls between the plates. Find
(a) electric field intensity between the
plates and (b) the force on the oil drop.
[Ans: (a) 1.0.×105 N/C
3. Solve numerical examples. (b) 9.6.×10-14 N]
i. Two small spheres 18 cm apart have vii. Calculate the electric field due to a charge
equal negative charges and repel each of -8.0×10-8 C at a distance of 5.0 cm
other with the force of 6×10-3 N. Find the from it.
total charge on both spheres. [Ans: -2.88×10-2 N/C]
[Ans: q = 2.938×10-7 C] ***
ii. A charge +q exerts a force of magnitude
-0.2 N on another charge -2q. If they
are separated by 25.0 cm, determine the
value of q.
[Ans: q = 0.8333 µC]
206
11. Electric Current Through Conductors
Can you recall?
1. Do you recall that the flow of charged particles in a conductor constitutes a current?
2. An electric current in a metallic conductor such as a wire is due to flow of electrons, the
negatively charged particles in the wire.
3. What is the role of the valence electrons which are the outermost electrons of an atom?
11.1 Introduction: the plane P from time t to t + ∆t, i.e. during the
time interval ∆t. Then the current is given by
The valence electrons become de-localized
when a large number of atoms come together I (l) = lim 'q --- (11.2)
in a metal. These are the conduction electrons Here, the 'ctouror'ent t is expressed as the limit
or free electrons constituting an electric current
when a potential difference is applied across the of the ratio ∆q/ ∆t as ∆t tends to zero.
conductor.
The current during lightening could be
11.2 Electric current : as high as 10,000 A, while the current in the
house hold circuit could be of the order of a few
Consider an imaginary gas of both amperes. Currents of the order of a milliampere
negatively and positively charged particles. (mA), a microampere (µA) or a nanoampere
Fig. 11.1 shows the negatively and positively (nA) are common in semiconductor devices.
charged particles flowing randomly in various
directions across a plane P. In a time interval 11.3 Flow of current through a conductor :
t, let the amount of positive charge flowing in
the forward direction be q+ and the amount of A current can be generated by positively
negative charge flowing in the forward direction or negatively charged particles. In an
be q-. electrolyte, both positively and negatively
charged particles take part in the conduction.
In a metal, the free electrons are responsible
for conduction. These electrons flow and
generate a net current under the action of an
applied electric field. As long as a steady
field exists, the electrons continue to flow
in the form of a steady current. Such steady
electric fields are generated by cells and
batteries.
Fig. 11.1: Flow of charged particles. Do you know ?
Thus the net charge flowing in the forward Sign convention : The direction of the
direction is q = q+- q-. For a steady flow, this current in a circuit is drawn in the direction
quantity is proportional to the time t. The ratio in which positively charged particles would
S qtI uisnidteo fifntehdeacsIurt=hreeqntct uirsreanmt pIe. re (A), t-h-a- t(1o1f.t1h)e move, even if the current is constituted by
charge and time is coulomb (C) and second (s) the negatively charged particles, electrons,
respectivly. which move in the direction opposite to that
Let I be the current varying with time. Let the electric field. We use this as a convention.
∆q be the amount of net charge flowing across
11.4 Drift speed :
Imagine a copper rod with no current
flowing through it. Fig 11.2 shows the
schematic of a conductor with the free electrons
207
in random motion. There is no net motion of t = VLd --- (11.4)
these electrons in any direction. If electric field
is applied along the length of the copper rod, From the Eq. (11.1), and Eq. (11.3), the current
and a current is set up in the rod, these electrons
still move randomly, but tend to 'drift' in a I= q = n A Le = n AVd e --- (11.5)
particular direction. Their direction is opposite Hence t L / Vd
to that of the applied electric field.
I J
Direction of electric field : Direction of an Vd = nAe = ne --- (11.6)
electric field at a point is the direction of the where J = I/A is current density. J is uniform
force on the test charge placed at that point. over the cross sectional area A of the wire. Its
The electrons under the action of the unit is A/m2
applied electric field drift with a drift speed tVhde. Here, J = I --- (11.7)
The drift speed in a copper conductor is of A
order of 10-4 m/s-10-5 m/s, whereas the electron From Eq. (11.6),
random speed is of the order of 106 m/s. J =(ne)V d --- (11.8)
Fig. 11.2: Free electrons in random motion For electrons, ne is negative and J and V d
inside the conductor. have opposite directions, V d is the drift velocity.
How is the current through a conductor Example 11.1: A metallic wire of diameter
related to the drift speed of electrons? Figure 0.02m contains 1028 free electrons per cubic
11.3 shows a part of conducting wire with its meter. Find the drift velocity for free electrons,
free electrons having the drift speed Vd in the having an electric current of 100 amperes
direction opposite to the electric field E . flowing through the wire.
(Given : charge on electron = 1.6 × 10-19C)
Solution: Given
e = 1.6 × 10-19 C
n = 1028 electrons/m3
D = 0.02m r = D/2 = 0.01m
I = 100 A
Vd = J = n I
ne Ae
where A is the cross sectional area of the wire.
Fig. 11.3: Conducting wire with the applied
A = πr2 = 3.142 × (0.01)2
electric field.
= 3.142 × 10-4m2
It is assumed that all the electron move with
100
the same drift speed Vthdeacnrdosthsaste,ctthioencu(Arr)eonft I is Vd = 3.142 u10-4 u10281.6 u 10-19
the same throughout the
102 4 9
wire. Consider the length L of the wire. Let n be
5.027
the number of free electrons per unit volume of
the wire. Then the total number of electrons in Vd = 10-3 × 0.1989 = 1.9 × 10-4 m/s
the length L of the conducting wire is nAL. The Example 11.2: A copper wire of radius 0.6 mm
carries a current of 1A. Assuming the current to
total charge in the length L is, be uniformly distributed over a cross sectional
area, find the magnitude of current density.
q = n A L e --- (11.3)
where e is the electron charge.
This is total charge that moves through
any cross section of the wire in a certain time
interval t,
208
Solution: Given Reciprocal of resistance is called
conductance.
r = 0.6 mm = 0.6 ×10-3 m
I = 1A C = 1 --- (11.10)
J=? R
Area of copper wire = πr2
The unit of conductance is siemens or (Ω)-1
= 3.142 × (0.6)2 × 10-6 Example 11.3: A Flashlight uses two 1.5V
batteries to provide a steady current of 0.5A in
= 3.142 × 0.36 × 10-6 the filament. Determine the resistance of the
glowing filament.
= 1.1311 × 10-6 m2
J = I 1 Solution:
A 1.1311u10 6
V 3
J = 0.884 × 106 A/m2 R = I 0.5 6.0:
11.5 Ohm’s law :
The relationship between the current ∴ Resistance of the glowing filament is 6.0 Ω.
through a conductor and applied potential
difference was first discovered by German Physical origin of Ohm’s law :
scientist George Simon Ohm in 1828 AD. This
relationship is known as Ohm’s law. We know that electrical conduction in a
conductor is due to mobile charge carriers, the
It states that “The current I through electrons. It is assumed that these conduction
a conductor is directly proportional to the electrons are free to move inside the volume
potential difference V applied across its two of the conductor. During their random motion,
ends provided the physical state of the conductor electrons collide with the ion cores within the
is unchanged”. conductor. It is assumed that electrons do not
collide with each other. These random motions
The graph of current versus potential average to zero. On the application of an
difference across the conductor is a straight line electric field E , the motion of the electrons is a
as shown in Fig. 11.4 combination of the random motion of electrons
due to collisions and that due to the electric field
Ideal Ohm's law
E . The electrons drift under the action of the
Fig. 11.4: I-V curve for a conductor.
field E and move in a direction opposite to the
direction of the field E .
Consider an electron of mass m subjected
to an electric field E . The force experienced by
the electron will be F = e E . The acceleration
experienced by the electron will then be
In general, I ∝ V a = eE --- (11.11)
m
or V= I R or R =V , --- (11.9)
I
where R is a proportionality constant and is The type of collision the conduction
electrons undergo is such that the drift velocity
called the resistance of the conductor. The unit attained before the collision has nothing to do
with the drift velocity after the collision. After
of resistance is ohm (Ω), the collision, the electron will move in random
direction, but will still drift in the direction
1Ω = 1volt
1ampere
If potential difference of 1volt across opposite to E .
a conductor produces a current of 1ampere Let τ be the average time between two
successive collisions. Thus on an average, the
through it, then the resistance of the conductor
is 1Ω.
209
electrons will acquire a drift speed Vd = aτ , vacuum tubes, junction diodes, thermistors etc.
where a is the acceleration given by Eq (11.11). Resistance R for such non-linear devices at a
particular value of the potential difference V is
Also, at any given instant of time, the average given by,
drift speed of the electron will also be Vd = aτ . R = lim 'V = dV
From Eq. (11.11), 'Io0 'I dI
--- (11.16)
eEW
Vd =aW m --- (11.12) where ∆V is the potential difference between
the two values of potential
From the Eq. (11.6) and Eq. (11.12),
Vd =J = eEτ --- (11.13) V 'V to V+ 'V ,
ne m 2 2
which gives and ∆I is the corresponding change in the
§ m · current.
¨© e 2 nW ¸¹
E = J --- (11.14) 11.7 Electrical Energy and Power:
or, E = ρJ, where ρ is the resistivity of the Consider a resistor AB connected to a
material and
cell in a circuit shown in Fig. 11.6 with current
m flowing from A to B. The cell maintains a
ne2W
U --- (11.15) potential difference V between the two terminals
of the resistor, higher potential at A and lower
For a given material, m, n, e2 and τ will at B. Let Q be the charge flowing in time ∆t
be constant and ρ will also be constant, ρ
through the resistor from A to B. The potential
is independent of E , the externally applied difference V between the two points A and B, is
electric field.
equal to the amount of work W, done to carry a
11.6 Limitations of the Ohm’s law: unit positive charge from A to B. It is given by
Ohm’s law is obeyed by various materials V = W , W = VQ --- (11.17)
and devices. The devices for which potential Q
difference (V) versus current (I) curve is a
straight line passing through origin, inclined
to V-axis, are called linear devices or ohmic
devices (Fig. 11.4). Resistance of these devices
is constant. Several conductors obey the Ohms
law. They follow the linear I-V characteristic.
Fig. 11.6: A simple circuit with a cell and a
resistor.
The cell provides this energy through the
charge Q, to the resistor AB where the work is
Fig. 11.5: I-V curve for non-Ohmic performed. When the charge Q flows from the
devices. higher potential point A to the lower potential
point B, i.e. through a decrease in potential of
The devices for which the I-V curve is not value V, its potential energy decreases by an
a straight line as shown in Fig. 11.5 are called amount
non-ohmic devices. They do not obey the
Ohm’s law and the resistance of these devices ∆U = QV = I ∆tV --- (11.18)
is a function of V or I; e.g. liquid electrolytes,
where I is current due to the charge Q flowing
in time ∆t. Where will this energy go? By
the principle of conservation of energy, it is
210
converted into some other form of energy. 11.8 Resistors:
In the limit as ∆t 0, Resistors are used to limit the current
following through a particular path of a circuit.
dU --- (11.19) Commercially available resistors are mainly of
dt = I.V two types :
Carbon resistors and Wire wound
Here, dU the time rate of transfer of resistors. High value resistors are mostly carbon
energy datndisispgoiwveenr, by, resistors. They are small and inexpensive. The
values of these resistors are colour coded to
dU mark their values in ohms. The colour coding
P = dt = I.V --- (11.20) is standardized by Electronic Industries
Association (EIA). One such resistor is shown
We can also say that this power is in Fig. 11.7.
transferred by the cell to the resistor or any
other device in place of the resistor, such as a
motor, a rechargeable battery etc.
Because of the presence of an electric Fig. 11.7: Carbon composition resistor.
field, the free electrons move across a resistor Colour code:
and there would be an increase in their kinetic
energy as they move. When the electrons collide Colours 1st 2nd Multiplier Tolerance
with the ion cores the energy gained by them
is shared among the ion cores. Consequently, digit digit
vibrations of the ions increase, resulting in
heating up of the resistor. Thus, some amount Black 00 ×100
of energy is dissipated in the form of heat in a
resistor. The energy dissipated in time interval Brown 1 1 ×101 ±1%
∆t is given by Eq. (11.18). The energy dissipated
per unit time is actually the power dissipated Red 22 ×102 ±2%
and is given by Eq. (11.20).
Using Eq. (11.20), and using Ohm’s law, V=IR, Orange 3 3 ×103
∴P =V 2 =I2R --- (11.21) Yellow 4 4 ×104
R
Green 5 5 ×105
It is the power dissipation across a resistor Blue 66 ×106
which is responsible for heating it up. For Violet 7 7 ×107
example, the filament of an electric bulb heats Gray 88 ×108
up to incandescence, radiating out heat and White 9 9 ×109
light. For Gold ×10-1 ±5%
Example 11.4 : An electric heater takes 6A For Silver × 10-2 ±10%
current from a 230V supply line, calculate
the power of the heater and electric energy No colour - ±20%
consumed by it in 5 hours.
Easy Bytes:
Solution : Given
I = 6A, V = 230V Finding it difficult to memorize the colour
We know that, code sequence? No need to worry, we have a
P = I × V = (6A) (230V) = 1380 W one liner which will help you out “B. B. Roy in
P = 1.38 kW Great Britain has Very Good Wife”
Energy consumed = Power × time
= (1.38 kW) × (5 h) B B R O Y G B V GW
= 6.90kWh (1.0 Kwh = 1 unit of power)
= 6.9 units of electrical energy. This funny one liner makes it easy to recall
the sequence of digits and multipliers.
211
In the four band resistor colour code Because of series combination, the supply
illustrated in the above table, the first three voltage between two resistors R1 and R2 is V1
bands (closest together) indicate the value in adfRlni2ov.dwiid.seVe.d2t,hinarronseudesgrptihheecestthiccvueoermlrryeebnsiiantsnarteodtimrotnaRh,ie1nssausnptahdpmelyetshavecmourlerterasiegninsettaoilsrIl
ohms. The first two bands indicate two numbers the resistors.
and third band often called decimal multiplier.
The fourth band separated by a space from the
three value bands, (so that you know which end
to start reading from), indicates tolerance of the
resistor.
Example
i. Colour code of resistor is
Yellow Violet Orange Gold
Value : 4 7 103 ±5%
i.e. 47×103 = 47000Ω = 47kΩ ±5%
The value of the resistor is 47kΩ ±5% Fig. 11.9: Series combination of two
ii. From given values of resistor; find the colour resistors R1 and R2.
bands of this resistor According to Ohm’s law,
330Ω = 33×10
3 3 101 R 1 = V1 , R2 = V2 --- (11.22)
I I
Orange Orange Brown tolerance band
11.8.1 Rheostat: Total voltage V=V1+V2 --- (11.23)
A rheostat shown in Fig. 11.8 is an From equation .... (11.22) and (11.23)
adjustable resistor used in applications that
require adjustment of current or resistance in we write
an electric circuit. The rheostat can be used to
adjust potential difference between two points V = II(RR1 +R2 ) --- (11.24)
in a circuit, change the intensity of lights and ∴V = --- (11.25)
control the speed of motors, etc. Its resistive s
element can be a metal wire or a ribbon, carbon
films or a conducting liquid, depending upon Thus the equivalent resistance of the series
the application. In hi-fi equipment, rheostats are
used for volume control. circuit Rs = R1+R2
When a number of resistors are connected
in series, the equivalent resistance is equal to
the sum of individual resistances.
For n number of resistors,
i =n
∑Rs = R1+ R2 + R2+...........+Rn= Ri -- (11.26)
i =1
II. Parallel Combination of Resistors:
In the parallel combination, the resistors
are connected in such a way that the same
Fig . 11.8: Rheostat. voltage is applied across each resistor.
11.8.2 Combination of Resistors: A number of resistors are said to be
connected in parallel if all of them are connected
I. Series combination of Resistors: between the same two electrical points each
having individual path as shown in Fig. 11.10.
In series combination of resistors, these are
connected in single electrical path as shown in In parallel combination the total current I
Fig 11.9. Hence the same electric current flows is divided into I1 and I2 as shown in the circuit
through each resistor in a series combination.
212
diagram Fig.11.10, whereas voltage V across Example 11.5: Calculate i) total resistance and
them remains the same, ii) total current in the following circuit.
R1 = 3Ω, R2 = 6Ω, R3 = 5Ω, V = 14V
Fig. 11.10 : Two resistors in parallel Circuit diagram
combination.
Solution:
I = I1+ I2 -- (11.27) i) Total resistance = R = R +R
T P3
wcuhrererentIf1lioswciunrgretnhtrofulogwhiRng2. through R1 and I2 is R1 R2 3u6
RP = R1 + R2 9 2:
When Ohm’s law is applied to R RToT=ta2l + 5 = 7Ω = 7Ω
1 Resistance
V= I1R1 i.e. I1 = V --- (11.28a) ii) Total current :
law applied to R2 R1 I = V 14V
Ohm’s
RT 7:
V =I2R2 i.e. I2 =V --- (11.28b)
R2 I = 2A
11.9 Specific Resistance (Resistivity):
From Eq. (11.27) and Eq. (11.28),
∴ I =V +V , At a particular temperature, the resistance
R1 R2 of a given conductor is observed to depend on
the nature of material of conductor, the area of
If, I = V , its cross-section, and its length.
Rp
It is found that resistance R of a conductor
V =V +V , of uniform cross section is
Rp R1 R2 i. directly proportional to its length l,
i.e. R∝ l
∴ 1 = 1 + 1 , --- (11.29) ii. inversely proportional to its area of
Rp R1 R2 cross section A,
wcohmerbeinRaptiiosnth. e equivalent resistance in parallel i.e. R∝ l
A
If n irnespisatroarlslelR, t1h, eRe2q,uiRv3a..l.e.n..t...r,esRisntanacree
connected From i and ii
of the combination is given by R =ρ l --- (11.31)
A
1 +1 +............ 1 n 1
R1 R2 Rn i =1 R
∑1= +1 = --(11.30) where ρ is a constant of proportionality and
R3
Rp it is called specific resistance or resistivity
Thus when a number of resistors are of the material of the conductor at a given
connected in parallel, the reciprocal of the temperature.
equivalent resistance is equal to the sum of the From Eq. (11.31), we write
reciprocals of individual resistances. U RA --- (11.32)
l
213
SI unit of resistivity is ohm-meter. Conductivity : Reciprocal of resistivity is
Resistivity of a conductor is numerically the called conductivity of a material, σ = (ρ)-1.
resistance per unit length, and per unit area of
cross-section of material of the conductor. SI unit of σ is: (Ωm)-1 i.e. siemens/meter (Sm-1)
i.e. when, R = 1Ω, A =1m2 and l = 1m,
then, ρ =1Ωm
Table 11.1 : Resistivity of various materials
Material Resistivity ρ Material Resistivity ρ
(Ω.m) (Ω.m)
Conductors Semiconductors
Silver 1.59 × 10-8 Carbon 3.5 × 10-5
Copper 1.72 × 10-8 Germanium 0.5
Gold 2.44 × 10-8 Silicon 3 × 104
Aluminium 2.82 × 10-8
Tungsten 5.6 × 10-8 Insulators 1011-1013
Iron 9.7 × 10-8 Glass 1011-1015
Mercury 95.8 × 10-8 Mica 1013-1016
Nichrome (alloy) 100 × 10-8 Rubber (hard) 1016
Teflon 3 × 108
Wood (maple)
Example 11.6: Calculate the resistance per Again, the SI unit of ρ is
metre, at room temperature, of a constantan unit(E ) = V/m = V m=Ω.m
(alloy) wire of diameter 1.25mm. The resistivity unit(J ) A/m 2 A
of constantan at room temperature is 5.0 × 10-7
Ωm. In terms of conductivity σ of a material, from
Solution: ρ = 5.0 × 10-7 Ωm
d = 1.25 × 10-3 m (11.33),
r = .625 × 10-3 m
J 1 E VE --- (11.34)
U
Cross-sReRceetsisoiissnttaiavlniActyerepUaer=mπRerltA2er = Rl tFhoer
a particular resistor, we had (Eq. 11.9)
given by
resistance R
R = V
I
i.e. RU 5 u 10 7 Compare this with the above Eq (11.33).
l A (0.625u10 3 )2 u 3.142 11.10 Va riation of Resistance with
R
l 0.41 :m 1 Temperature:
Resistivity of a material varies with
temperature. It is a property of material. Fig.
11.11 shows the variation of resistivity of
∴ Resistance per metre= 0.41 Ωm-1 copper as a function of temperature (K). It can
be seen that the variation is linear over a certain
Resistivity ρ is a property of a material, range of temperatures. Such a linear relation
can be expressed as,
while the resistance R refers to a particular
object. Similarly, the electric field E at a point
is specified in a material with the potential
difference across the resistance, and the current ρ = ρ0[1+ α(T - T0)], --- (11.35)
density J in a material instead of the current I where T0 is the chosen reference temperature and
in the resistor. Then for an isotropic material, ρfo0rienxtahme prlees,isTtoivciatyn at the chosen temperature,
be 0 oC.
U E or E U J ---- (11.33)
J
214
Fig. 11.11: Resistivity as a function of R = 2.5 × 1.32
temperature (K). T
In the above Eq. (11.35), RT = 3.3Ω
D U U0 R - R0 --- (11.36) Superconductivity :
U0 (T -T0 ) R0 (T -T0 ) We know that the resistivity of a
metal decreases as the temperature decreases.
Here, α is called the temperature coefficient In case of some metals and metal alloys,
the resistivity suddenly drops to zero at a
of resistivity. Table (11.1) shows the resistivity of particular temperature (Tc). This temperature
is called critical temperature, for example,
some of the metals. The temperature coefficient mercury loses its resistance completely to
zero at 4.2K.
of resistance is defined as the increase in Superconductivity can be harnessed so as to
be useful for mankind. It is already in use in
resistance per unit original resistance at the obtaining very high magnetic field (a few
Tesla) in superconducting magnet. These
chosen reference temperature, per degree rise in magnets are used in research quality NMR
spectrometers. For its operation, the current
temperature. The unit of α is oC-1 or oK-1 (per carrying coils are required to be kept at a
temperature less than the critical temperature
degree celcius or per degree kelvin). of the coil material.
From Eq. (11.36) 11.11 Electromotive Force (emf):
When charges flow through a conductor,
R = R [1+ α (T - T )] --- (11.37)
00 a potential difference has to be established
between the two ends of the conductor. For a
For small difference in temperatures, steady flow of charges, this potential difference
is required to be maintained across the two ends
1 dR of the conductor, the terminals. There is a device
D R0 dT --- (11.38) that does so by doing work on the charges,
thereby maintaining the potential difference.
Do you know ? Such a device is called an emf device and it
provides the emf ε. The charges move in the
Here, the temperature difference is more conductor owing to the energy provided by the
important than the temperature alone. emf device. The device supplies this energy
Therefore, as the sizes of degrees on the through the work it does.
Celsius scale and the Absolute scale are
identical, any scale can be used. You must have used some of these emf
Example 11.7: A piece of platinum wire has devices. Power cells, batteries,Solar cells, fuel
resistance of 2.5 Ω at 0o C. If its temperature cells, and even generators, are some examples
coefficient of resistance is 4 ×10-3/oC. Find the of emf devices familiar to you.
resistance of the wire at 80o C.
Solution: Fig. 11.12: Circuit with emf device.
R0= 2.5 Ω
α = 0.004/oC
T - 0 = T = 80o C
RT = R0(1+ αT )
RT = 2.5 (1+ 0.004 × 80) = 2.5 (1 + 0.32)
215
Fig. 11.12 shows a circuit with an emf from the potential difference across its two
device and a resistor R. Here, the emf device terminals (V).
keeps the positive terminal (+) at a higher
electric potential than the negative terminal (-). V= ε - (I) (r) --- (11.40)
The negative sign is due to the fact that the
The emf is represented by an arrow from current I flows through the emf device from the
the negative terminal to the positive terminal of
a device such as a Voltaic cell. When the circuit negative terminal to the positive terminal.
is open, there is no net flow of charge carriers
within the device. When connected in a circuit, By the application of Ohm’s law Eq. (11.9),
there is a flow of carriers from one terminal
to the other terminal inside the emf device. V = IR
The positive charge carriers move towards the
positive terminal which acts as cathode inside Hence IR = ε - Ir --- (11.41)
the emf device. Thus the positive charge carriers
move from the region of lower potential energy, Or
to the region of higher potential energy which is
cathode inside the emf device. Here, the energy I = ε --- (11.42)
source is chemical in nature. In a Solar cell, it is R +r
the photon energy in the Solar radiation.
Thus, the maximum current that can be
drawn from the emf device is when R = 0, i.e.
Imax = ε --- (11.43)
r
This is the maximum allowed current
from an emf device (or a cell). This decides the
Now suppose that a charge dq flows maximum current rating of a cell or a battery.
through the cross section of the circuit (Fig.
11.12), in time dt. 11.12 Cells in Series:
It is clear that the same amount of charge In a series combination, cells are connected
in single electrical path, such that the positive
dq flows throughout the circuit, including the terminal of one cell is connected to the negative
terminal of the next cell, and so on. The terminal
emf device. It enters the negative terminal (low voltage of battery/cell is equal to the sum of
voltages of individual cells in series, as shown
potential terminal) and leaves the positive in Fig 11.13 a.
terminal (higher potential terminal). Hence, Figure shows two 1.5V cells in series. This
combination provides total voltage of 3.0V
the device must do work dw on the charge dq, (1.5×2).
so that it moves in the above manner. Thus we
define the emf of the emf device.
ε = dw --- (11.39)
dq
The SI unit of emf is joule/coulomb
(J/C).
In an ideal device, there is no internal Fig. 11.13 (a): Cells in parallel.
resistance to the motion of charge carriers. The
emf of the device is then equal to the potential
difference across the two terminals of the
device. In a real emf device, there is an internal
resistance to the motion of charge carriers. Fig. 11.13 (b): Cells in parallel.
If such a device is not connected in a circuit, The equivalent emf of n number of cells
there is no current through it. In that case the in series combination is the algebraic sum of
emf is equal to the potential difference across their individual emf. The equivalent internal
the two terminals of the emf device connected resistance of n cells in a series combination is
in a circuit, there is no current through it. If a the sum of their individual internal resistance.
current (I) flows through an emf device, there is V = I . ri --- (11.44)
an internal resistance (r) and the emf (ε) differs ¦ H i ¦
i i
216
• Advantages of cells in series. Considering Eq. (11.49) and Eq. (11.50) we can
(i) The cells connected in series produce a write,
larger resultant voltage. H = H1r2 H 2 r1
r1 +r2
(ii) Cells which are damaged can be easily eq
identified, hence can be easily replaced. r1r2
r1 +r2
11.13 Cells in parallel: req
i.e. 1 11
Consider two cells which are connected
req
in parallel. Here, positive terminals of all the r1 r2
cells are connected together and the negative
terminals of all the cells are connected together. Heq H1 H 2
In parallel connection, the current is divided --- (11.51)
aFmigo. n1g1.1th3eb.bCraonncshiedseri.peo. iIn1tsanBd1 Ia2nadsBs2hohwavninign req r1 r2
potenFtoiarlsthVeB1fairnsdt cVeBl2l, respectively. For n number of cells connected in parallel with
the potential
difference emf ε1, ε2, ε3, ......, εn and internal resistance r1,
r2, r3, ......, rn
across its terminals is,
1 1 1 1 1
V = VB1 - VB2 = ε1 - I1r1 --- (11.45) req r1 r2 r3 ........... rn --- (11.52)
? I1 = H1 -V --- (11.46) and H eq H1 H2 ............. Hn
Point B1 and B2r1are connected exactly similarly req r1 r2 rn
to the second cell.
Hence, considering the second cell we write, --- (11.53)
VWCo=emkVbnBio1nw-inVgtBh2tah=te Iεl2a=-stII1t2+hrr2eI ;2eIe2quaHt2iro 2nVs, --- (11.47) Substitution of emfs should be done
algebraically by considering proper ± signs
? I = H1 V H2 V --- (11.48) according to polarity.
r1 r1 r2 r2
• Advantages of cells in parallel : For cells
§ H1 H2 · V § 1 1 · connected in parallel in a circuit, the circuit
¨ r1 r2 ¸ ¨ r1 r2 ¸ will not break open even if a cell gets
© ¹ © ¹ damaged or open.
Thus, V § 1 1 · § H1 H2 · I • Disadvantages of cells in parallel : The
¨ r1 r2 ¸ ¨ r1 r2 ¸ voltage developed by the cells in parallel
© ¹ © ¹ connection cannot be increased by
increasing number of cells present in circuit.
§ r1 +r2 · H1r2 H 2r1 11.14 Types of Cells:
¨ r1r2 ¸ r1r2
? V © ¹ I Electrical cells can be divided into several
categories like primary cell, secondary cell,
fuel cell, etc.
V= H1r2 H 2r1 I r1r2 ---(11.49) A primary cell cannot be charged again. It
r1 + r2 r1 + r2 can be used only once. Dry cells, alkaline cells
are different examples of primary cells. Primary
cells are low cost and can be used easily.
If we replace the cells by a single cell But these are not suitable for heavy loads.
Secondary cells are used for such applications.
connected between points B1 and B2 with the The secondary cell are rechargeable and can be
emf εeq and the internal resistance req as in Fig. reused. The chemical reaction in a secondary
(11.13b), cells is reversible. Lead acid cell, and fuel cell
then,
V = εeq - Ireq --- (11.50)
217
are some examples of secondary cells. Lead ii. Current in each resistor :
acid battery is used widely in vehicles and other
applications which require high load currents. Total current I in the circuit is,
Solar cells are secondary cells that convert
Solar energy into electrical energy. I = ε = 15 = 2.1A
RT + r 6 +1
Fuel cells vehicles (FCVs) are electric
vehicles that use fuel cells instead of lead acid Consider resistors between A and B.
batteries to power the vehicles. Hydrogen is
used as a fuel in fuel cells. The by- product 4reΩsisrtLeoseritstIo1rsbeantdheI2cbuerrethnet through one of the
after its burning is water. This is important current in the other
in terms of reducing emission of greenhouse
gases produced by traditional gasoline fueled that iIs1,× I 4==II2f×ro4m symmetry of the two arms.
vehicles. The hydrogen fuel cell vehicles are B1ut 2
thus more environment friendly. + I2 = I = 2.1A
Example 11.8: A network of resistors is I1
connected to a 15 V battery with internal ∴ I1 = I2 =1.05A
resistance 1 Ω as shown in the circuit
diagram. that is, the current in each 4Ω resistor is 1.05A,
Calculate the current in 1Ω resistor between B and C
(i) The equivalent resistance,
(ii) Current in each resistor, would be 2.1A.
(iii) Voltage drops VAB, VBC and VDC.
Now, consider the resistances between C
1Ω D and D
rΩesirsetLsoiers.ttoIr3sbaentdheI4cubreretnhte through one of the 6
current in the other
I3 × 6 = I4 × 6
∴ curIr3e=ntI4in= 1.05A
That is, each 6
Ω resistor is 1.05A
r =1 Ω iii. Voltage drop across BC is VBC
15 V V = I × 1 = 2.1 × 2.1 = 2 V
Solution : BC
i) Equivalent Resistance (ReqR) C=DR=AB66++×R66BC+3R:DC Voltage drop across CD is VCD
RAB = VCD = I × RCD = 2.1 × 3 =6.3V
4×4 [Note : Total voltage drop across AD is
4+4 (4.2 V+2.1V+6.3 V) =12.6 V, while its emf is
15 V. The loss of the voltage is 2.4 V].
2:, Internet my friend
RBC = 1Ω https://www.britannica.com/science/
RT= Req = 2 + 1 + 3 = 6 Ω superconductivityphysics
∴ Equivalent Resistance is 6 Ω
218
ExercisesExercises
1. Choose correct alternative number of cells. The number of rows
should be
i) You are given four bulbs of 25 W, 40 W,
60 W and 100 W of power, all operating (A) 2 (B) 4
at 230 V. Which of them has the lowest
resistance? (C) 5 (D) 100
(A) 25 W (C) 40 W viii) Five dry cells each of voltage 1.5 V are
connected as shown in diagram
(C) 60 W (D) 100 W
ii) Which of the following is an ohmic
conductor?
(A) transistor (B)vacuum tube What is the overall voltage with this
arrangement?
(C) electrolyte (D) nichrome wire
iii) A rheostat is used (A) 0V (B) 4.5V
(A) to bring on a known change of (C) 6.0V (D) 7.5V
resistance in the circuit to alter the current
2. Give reasons / short answers
(B) to continuously change the resistance
in any arbitrary manner and there by alter i) In given circuit diagram two resistors are
the current connected to a 5V supply.
(C) to make and break the circuit at any
instant
(D) neither to alter the resistance nor the
current
iv) The wire of length L and resistance R is
stretched so that its radius of cross-section a] Calculate potential difference across
is halved. What is its new resistance?
the 8Ω resistor.
(A) 5R (B) 8R b] A third resistor is now connected
(C)4R (D) 16R in parallel with 6Ω resistor. Will the
potential difference across the 8Ω resistor
v) Masses of three pieces of wires made of the larger, smaller or the same as before?
the same metal are in the ratio 1:3:5 and Explain the reason for your answer.
their lengths are in the ratio 5:3:1. The
ratios of their resistances are ii) Prove that the current density of a metallic
conductor is directly proportional to the
(A) 1:3:5 (B) 5:3:1 drift speed of electrons.
(C) 1:15:125 (D) 125:15:1 3. Answer the following questions.
vi) The internal resistance of a cell of emf i) Distinguish between Ohmic and non-
2V is 0.1Ω it is connected to a resistance ohmic substances; explain with the help
of 0.9Ω. The voltage across the cell will of example.
be
ii) DC current flows in a metal piece of non-
(A) 0.5 V (B) 1.8 V uniform cross-section. Which of these
quantities remains constant along the
(C) 1.95 V (D) 3V conductor: current, current density or
drift speed?
vii) 100 cells each of emf 5V and internal
resistance 1Ω are to be arranged so as
to produce maximum current in a 25Ω
resistance. Each row contains equal
219
4. Solve the following problems. vii) A silver wire has a resistance of 4.2 Ω
i) What is the resistance of one of the rails of at 27° C and resistance 5.4 Ω at 100° C.
a railway track 20 km long at 20° C? The Determine the temperature coefficient of
cross section area of rail is 25 cm2 and the resistance.
rail is made of steel having resistivity at [Ans: 3.91×10-3/°C]
20° C as 6×10-8 Ω m. viii) A 6m long wire has diameter 0.5 mm. Its
[Ans: 0.48 Ω]
resistance is 50 Ω. Find the resistivity and
ii) A battery after a long use has an emf conductivity.
24 V and an internal resistance 380 Ω. [Ans: 1.636×10-6Ω/m, 6.112×105m/Ω]
Calculate the maximum current drawn
from the battery? Can this battery drive ix) Find the value of resistances for the
starting motor of car? following colour code.
[Ans: 0.063 A] 1. Blue Green Red Gold
iii) A battery of emf 12 V and internal [Ans: 6.5 kΩ ± 5%]
resistance 3 Ω is connected to a resistor.
If the current in the circuit is 0.5 A, 2. Brown Black Red Silver
[Ans: 1.0 kΩ ± 10%]
a] Calculate resistance of resistor. 3. Red Red Orange Gold
b] Calculate terminal voltage of the [Ans: 2.2 kΩ ± 5%]
battery when the circuit is closed.
4. Orange White Red Gold
[Ans: a) 21 Ω, b) 10.5 V] [Ans: 3.9 kΩ ± 5%]
iv) The magnitude of current density in a 5. Yellow Violet Brown Silver
copper wire is 500 A/cm2. If the number
of free electrons per cm3 of copper is [Ans: 4.70 kΩ ± 10%]
8.47×1022 calculate the drift velocity of
the electrons through the copper wire x) Find the colour code for the following
(charge on an e = 1.6×10-19 C) value of resistor having tolerance ± 10%
a) 330Ω b) 100Ω c) 47kΩ
[Ans: 3.69×10-4 m/s] d) 160Ω e) 1kΩ
v) Three resistors 10 Ω, 20 Ω and 30 Ω are xi) A current 4A flows through an automobile
connected in series combination. headlight. How many electrons flow
through the headlight in a time 2hrs.
i] Find equivalent resistance of series
combination. [Ans : 1.8 ×1023]
ii] When this series combination is xii) The heating element connected to 230V
connected to 12V supply, by neglecting
the value of internal resistance, obtain draws a current of 5A. Determine the
potential difference across each resistor.
amount of heat dissipated in 1 hour
(J = 4.2 J/cal.).
[Ans: i) 60 Ω, ii) 2 V, 4 V, 6 V] [Ans : 985.7 kcal]
vi) Two resistors 1k Ω and 2k Ω are connected
in parallel combination.
***
i] Find equivalent resistance of parallel
combination
ii] When this parallel combination is
connected to 9 V supply, by neglecting
internal resistance calculate current
through each resistor.
[Ans: i) 0.66 kΩ, ii) 9 mA, 4.5 mA]
220
12. Magnetism
Can you recall? force?
4. If you freely hang a bar magnetic horizontally,
1. What is a bar magnet?
2. What are the magnetic lines of force? in which direction will it become stable?
3. What are the rules concerning the lines of
quadrupole. In this Chapter the main focus
12.1 Introduction:
The history of magnetism dates back will be on elementary aspects of magnetism and
to earlier than 600 B.C., but it is only in the terrestrial magnetism.
twentieth century that scientists began to
understand it and developed technologies based 12.2 Magnetic Lines of Force and Magnetic
on this understanding. William Gilbert (1544-
1603) was the first to systematically investigate Field:
the phenomenon of magnetism using scientific
method. He also discovered that Earth is a weak You have studied properties of electric lines
magnet. Danish physicist Hans Oersted (1777-
1851) suggested a link between electricity and of force earlier in the Chapter on electrostatics.
magnetism. James Clerk Maxwell (1831-1879)
proved that electricity and magnetism represent In a similar manner, magnetic lines of force
different aspects of the same fundamental force
field. originate from the north pole and end at the
In electrostatics you have learnt about the south pole of a bar magnet. The magnetic
relationship between the electric field and force
due to electric charges and electric dipoles. lines of force of a magnet have the following
Analogous concepts exist in magnetism except
that magnetic poles do not exist in isolation, properties:
and we always have a magnetic dipole or a
i) The magnetic lines of force of a magnet
Do you know ?
or a solenoid form closed loops. This is in
Some commonly known facts about
magnetism. contrast to the case of an electric dipole,
(i) Every magnet regardless of its size and
where the electric lines of force originate
shape has two poles called north pole
and south pole. from the positive charge and end on
(ii) If a magnet is broken into two or more
pieces then each piece behaves like an the negative charge, without forming a
independent magnet with somewhat
weaker magnetic field. complete loop (see Fig. 12.4).
Thus isolated magnetic monopoles
do not exist. The search for magnetic ii) The direction of the net magnetic field B
monopoles is still going on.
(iii) Like magnetic poles repel each other, at a point is given by the tangent to the
whereas unlike poles attract each other.
(iv) When a bar magnet/ magnetic needle is magnetic line of force at that point in the
suspended freely or is pivoted, it aligns
itself in geographically North-South direction of line of force.
direction.
iii) The number of lines of force crossing per
unit area decides the magnitude of the
magnetic field B .
iv) The magnetic lines of force do not intersect.
This is because had they intersected, the
direction of magnetic field would not be
unique at that point.
Try this
You can take a bar magnet and a small
compass needle. Place the bar magnet at a
fixed position on a paper and place the needle
at various positions. Noting the orientation
of the needle, the magnetic field direction at
various locations can be traced.
Density of lines of force i.e., the number of
lines of force per unit area normal to the surface
221
around a particular point determines the strength Equator:- A line passing through the centre
of the magnetic field at that point. The number of a magnet and perpendicular to its axis is
of lines of force is called magnetic flux (φ ). SI called magnetic equator. The plane containing
unit of magnetic flux (φ ) is weber (Wb). For a all equators is called the equatorial plane. The
specific case of uniform magnetic field which locus of points, on the equatorial plane, which
is normal to the finite area A, the magnitude of are equidistant from the centre of the magnet
magnetic field strength B at a point in the area is called the equatorial circle. The popularly
A is given by known ‘equator’ in Geography is actually
Magnetic Field = magnetic flux an ‘equatorial circle’. Such a circle with any
diameter is an equator.
area
i. e. B = φA --- (12.1) Magnetic length (2l):- It is the distance
SI unit of magnetic field (B) is expressed between the two poles of a magnet.
as weber/m2 or Tesla.
1 Tesla = 104 Gauss. 5
However, magnetic lines are only a crude Magnetic length (2l) = 6 × Geometric length
way of representing magnetic field. It is a --- (12.2)
pictorial representation of the strength of the
magnetic field (B). It is better defined in terms 12.3.1 Magnetic field due to a bar magnet
of Lorentz force law which you will learn in std at a point along its axis and at a point
XII. along its equator:
12.3 The Bar magnet:
A bar magnet is said to have magnetic pole Fig. 12.2 (a): Magnetic field at a point
strength +qm and - qm at the north and south along the axis of the magnet.
poles, respectively. The separation of magnetic
poles inside the magnet is 2l. As the bar magnet Beq Beq P0 m
has two poles, with equal and opposite pole 4S r3
strength, it is called a magnetic dipole. This is
analogous to an electric dipole. The magnetic Fig. 12.2 (b): Magnetic
dipole moment, therefore, becomes m = qm.2l field along the equatorial
( 2l is a vector from south pole to north pole) point.
in analogy with the electric dipole moment.
SI unit of pole strength (q ) is A m. Consider a bar magnet of dipole length 2l
m and magnetic dipole moment m as shown in
Fig. 12.2 (a). We will now find magnetic field at
SI unit of magnetic dipole moment m is A m2. a point P along the axis of the bar magnet.
Axis:- It is the line passing through both the
poles of a bar magnet. Obviously, there is only Let r be the distance of point P from the
one axis for a given bar magnet. centre O of the magnetic dipole.
Fig. 12.1: Bar magnet OS = ON = l
... NP SP
r2 l2
We now use the electrostatic analogy to
obtain the magnetic field due to a bar magnet
at a large distance r >> l. Consider the electric
field due to an electric dipole with a dipole
moment p.
222
The Electrostatic Analogue: compared with charge q in electrostatics.
As suggested by Maxwell, electricity Accordingly, we can write the equivalent
and magnetism could be studied analogously. physical quantities in electrostatics and
The pole strength (qm) in magnetism can be magnetism as shown in table 12.1.
Table 12.1: The Electrostatic Analogue
Quantity Electrostatics Magnetism
Basic physical quantity Electrostatic charge Magnetic pole
Field Electric Field E
Magnetic Field B
Constant 1 P0
Dipole moment 4SH 0 4S
Force p = q ( ) amlo=ngqSm ( 2l ) (bar magnet)
Energy (In external field) 2l N pole
of a dipole
Coulomb’s law along (-ve) (+ve) charge
Axial field for a short F =qE F = qm B
dipole
U = - p.E U = -m.B
F § 1 · q1q 2 No analogous law as
¨ 4SH 0 ¸ r2 magnetic monopoles
© ¹ do not exist
2p along p P0 2m
4SH 0 r 3 4S r3
Equatorial field for a short p opposite to p P0 m
dipole 4SH 0 r 3 4S r3
You have studied the electric field due to an Similarly, the equatorial magnetic field
electric dipole of length 2l (p = 2ql) at a distance Beq P0 m
4S r3
r along the dipolar axis (Eq. 10.24) which is --- (12.4)
given by, Negative sign shows that the direction of
Ea 1 2 p , r !! l Beq is opposite to m .
4SH 0 r For the same distance from centre O of a
3
The electric field on the equator (Eq. 10.28) bar magnet,
is antiparallel to p and is given by Baxis = 2Beq --- (12.5)
E eq 1 p , r !! l 12.3.2 Magnetic field due to a bar magnet at
4SH 0 r3
an arbitrary point:
Using the analogy given in Table 12.1, minoimtsFeimngt.a1gm2n.e3wtiScithhfoiewcledsn.atrMbeaaargtmneOati.gcnPemtiosomfamennyatgpnmoeitniicst
we can thus write the axial magnetic field of a
bar magnet at a distance r, r >> l, 2l being the creosmolpvoenden(atsboaulot ntgherceanntrde paoelforpnthegenrdmi,catuhglenaerpto)toiinntrtPo.
length of bar magnet, For the component mcosθ
Ba P0 2m --- (12.3) is an axial point.
4S r3
223
Beq = P0 m
4S r3
10 7 u 0.5 5 u 10 8 0.625u10 5 Wb / m2
8 u 10 3
P (0.2)3
Fig. 12.3: Magnetic field at an arbitrary point. 12.4 Gauss' Law of Magnetism:
ppeoripnetAnaldtsiotch,uelasrafomtorerd,itsththeaenpcoecionrmt.PpUoissninaegnntethqeumaretsosiurnilaθtsl The Gauss' law for electric field is known
to you. It states that the net electric flux through
a closed Gaussian surface is proportional to the
net electric charge enclosed by the surface (Eq.
(10.18)). The Gauss' law for magnetic fields
states that the net magnetic flux Φ through a
closed Gaussian surface is zero, i.e.,B
of axial and equatorial fields, we get (Gauss' law for magnetic
Ba Po 2m cosT --- (12.6) fields)
4S r3
The magnetic force lines of (a) bar
magnet, (b) current carrying finite solenoid,
directed along m cosθ and and (c) electric dipole are shown in Fig.12.4(a),
12.4(b) and 12.4(c), respectively. The curves
Beq Po msinT --- (12.7) labelled (i) and (ii) are cross sections of three
4S r3 dimensional closed Gaussian surfaces.
directed opposite to msinθ
Thus, the magnitude of the resultant magnetic
field B, at point P is given by
B Ba 2 Beq2
∴ B 4PSo rm3 >2 cosT @2 >sinT @2
?B Po m 3cos2 T 1 --- (12.8)
4S r3
Lre.t α be the angle made by the direction of B with
by using eq (12.6) and eq (12.7),
Then,
--- (12.9)
of and m Fig. 12.4 (a): Bar magnet.
The angle between directions B Fig. 12.4 (b): Current (I) carrying solenoid.
is then T D .
Example 12.1: A short magnetic dipole has
magnetic moment 0.5 A m2. Calculate its
magnetic field at a distance of 20 cm from the
centre of magnetic dipole on (i) the axis (ii) the
equatorial line (Given µ0 = 4 π Í10-7 SI units)
Solution :
m = 0.5 Am2 , r = 20 cm = 0.2 m
Ba = P0 2m 10 7 u 2 u 0.5 1u10 7
4S r3 (0.2)3 8 u 10 3
1 u 10 4 1.25u10 5 Wb / m2
8
224
This fact clearly indicates that there is
some magnetic field present everywhere on the
Earth . This is called Terrestrial Magnetism. It
is extremely useful during navigation.
Magnetic parameters of the Earth are
described below. The magnetic lines of force
enter the Earth's surface at the north pole and
emerge from the south pole.
Unless and otherwise stated, the directions
mentioned (South, North, etc.) are always,
Geographic.
Fig. 12.4 (c): Electric dipole. Fig. 12.5: Earth's magnetism.
Magnetic Axis :- The Earth is considered to
If we compare the number of lines of force be a huge magnet. Magnetic north pole (N)
entering in and leaving out of the surface (i), it of the Earth is located below Antarctica while
is clearly seen that they are equal. The Gaussian the south pole (S) is below north Canada. The
surface does not include poles. It means that the straight line NS joining these two poles is called
flux associated with any closed surface is equal the magnetic axis, MM'.
to zero. When we consider surface (ii), in Fig. Magnetic equator :- A great circle in the plane
12.4 (b), we are enclosing the North pole. As perpendicular to magnetic axis is magnetic
even a thin slice of a bar magnet will have North equatorial circle,AA'. It happens to pass through
and South poles associated with it, the closed India near Thiruvananthapuram.
Gaussian surface will also include a South pole. Geographic Meridian:- A plane perpendicular
However in Fig. 12.4(c), for an electric dipole, to the surface of the Earth (vertical plane)
the field lines begin from positive charge and perpendicular to geographic axis is geographic
end on negative charge. For a closed surface meridian. (Fig.12.6)
(ii), there is a net outward flux since it does Magnetic Meridian:- A plane perpendicular to
include a net (positive) charge. According to the surface of the Earth (Vertical plane) and passing
Gauss' law of electrostatics as studied earlier, through the magnetic axis is magnetic meridian.
Direction of resultant magnetic field of the
, where q is the positive charge Earth is always along or parallel to magnetic
meridian. (Fig.12.6)
enclosed. Thus, situation is entirely different Magnetic declination:- Angle between the
geographic and the magnetic meridian at a
from magnetic lines of force, which are shown place is called ‘magnetic declination’ (α). The
declination is small in India. It is 0° 58′ west at
in Fig. 12.4(a) and Fig. 12.4(b). Thus, Gauss' law Mumbai and 0041′ east at Delhi. Thus, at both
these places, magnetic needle shows true North
of magnetism can be written as .
From the above we conclude that for
electrostatics, an isolated electric charge exists
but an isolated magnetic pole does not exist. In
short, only dipoles exist in case of magnetism.
12.5 Earth’s Magnetism:
It is common experience that a bar magnet
or a magnetic needle suspended freely in air
always aligns itself along geographic N-S
direction. If it has a freedom to rotate about
horizontal axis, it inclines with some angle with
the horizontal in the vertical N-S plane.
225
accurately (Fig.12.6). (magnetic equator) B= BH along South to
North, BV= 0 and φ =0
φ
Magnetic maps of the Earth:-
Fig. 12.6: Magnetic declination.
Magnetic inclination or angle of dip (φ):- φ) Magnetic elements of the Earth (wBiHt,hαtiamned.
Angle made by the direction of resultant vary from place to place and also
magnetic field with the horizontal at a place
is inclination or angle of dip at the place (Fig. The maps providing these values at different
12.7).
locations are called magnetic maps. These
are extremely useful for navigation. Magnetic
maps drawn by joining places with the same
value of a particular element are called Iso-
magnetic charts.
Lines joining the places of equal horizontal
components (BH) are known as ‘Isodynamic
lines’
Lines joining the places of equal declination
(α) are called Isogonic lines.
Lines joining the places of equal inclination
or dip (φ) are called Aclinic lines.
Example 12.2: Earth's magnetic field at the
equator is approximately 4×10-5 T. Calculate
Earth's dipole moment. (Radius of Earth =
6.4×106 m, µ0 = 4π×10-7 SI units)
Solution: Given
Fig. 12.7: Magnetic inclination. Beq = 4 ×10-5 T
Earth’s magnetic field:- Magnetic force r = 6.4 ×106 m
experienced per unit pole strength is magnetic
field B at that place. It can be resolved in Assume that Earth is a bar magnet with N and
components along the horizontal, BH and along S poles being the geographical South and North
poles, respectively. The equatorial magnetic
vertical, BV . The vertical component can be field due to Earth's dipole can be written as
conveniently determined. The two components
can be related with the angle of dip (φ) as, Beq P0m
4S r3
BH = B cosφ, BV = B sinφ m 4S Beq u r3 / P0
BV 4 u10 5 u (6.4 u106 )3 u107
BH
tan I --- (12.10) 1.05u1020 A m2
Example 12.3: At a given place on tmheisEakretpht,
a bar magnet of magnetic moment
B2 BV 2 BH 2
? B BV 2 BH 2 --- (12.11) horizontal in the East-West direction. P and
Special cases fQieladreofthtehitswmo angenuettraalnpdoiBnHts due to magnetic
is the horizontal
1) At the magnetic North pole, B = BV , component of the Earth's magnetic field.
directed upward, BH = 0 and φ =900. (A) Calculate the angles dibreetcwtieoennofpmos.ition
2) At the magnetic south pole, B = BV ,- vectors of P and Q with the
directed downward, BH = 0 and φ = 2700. (B) Points P and Q are 1 m from the centre of
3) Anywhere on the magnetic great circle the bar magnet and BH 3.5u10 5 T . Calculate
226
magnetic dipole moment of the bar magnet. (B)
tan2T 2
Neutral point is that point where the
resultant magnetic field is zero. ?sec2T 1 tan2T 1 2 3
Solution: oo(fpApm)oasAgitnseettsoieceBnfHieflardtotmhBe the figure, the ?cos2T 1
direction due to the bar 3
magnet is points P and Q.
r 1m and B BH 3.5u10 5 T (Given)
Also, T D = 900 at P and it is2700 at Q. we have,
B P0 m 3cos2T 1
4S r3
? m BH u r3
§ P0 ·
©¨ 4S ¸¹ 3cos2T 1
3.5u10 5 u13
10 7 u 3 1 1
3
?m 350 247.5 A m2
2
Always remember:
In this Chapter we have used B as a symbol
for magnetic field. Calling it magnetic
induction is unreasonable. We have used
the words magnetic field which are used in
spoken language.
ExercisesExercises
1. Choose the correct option. iii) The horizontal and vertical component
of magnetic field of Earth are same at
i) Let r be the distance of a point on the some place on the surface of Earth. The
axis of a bar magnet from its center. The magnetic dip angle at this place will be
magnetic field at r is always proportional
to (A) 30o (B) 45o
(A) 1/r2 (B) 1/r3 (C) 1/r (C) 0o (D) 90o
(D) not necessarily 1/r3 at all points iv) Inside a bar magnet, the magnetic field
ii) Magnetic meridian is the plane lines
(A) perpendicular to the magnetic axis of (A) are not present
Earth (B) are parallel to the cross sectional area
(B) perpendicular to geographic axis of of the magnet
Earth (C) are in the direction from N pole to S
(C) passing through the magnetic axis of pole
Earth (D) are in the direction from S pole to N
(D) passing through the geographic axis pole
Earth
227
v) A place where the vertical components of ii) A magnet makes an angle of 45o with the
Earth's magnetic field is zero has the angle horizontal in a plane making an angle of
of dip equal to 30o with the magnetic meridian. Find the
true value of the dip angle at the place.
(A) 0o (B) 45o
[Ans: tan-1 (0.866)]
(C) 60o (D) 90o
iii) Two small and similar bar magnets have
vi) A place where the horizontal component magnetic dipole moment of 1.0 Am2
of Earth's magnetic field is zero lies at each. They are kept in a plane in such a
way that their axes are perpendicular to
(A) geographic equator each other. A line drawn through the axis
of one magnet passes through the center
(B) geomagnetic equator of other magnet. If the distance between
their centers is 2 m, find the magnitude of
(C) one of the geographic poles magnetic field at the mid point of the line
joining their centers.
(D) one of the geomagnetic poles
vii) A magnetic needle kept nonparallel to the
magnetic field in a nonuniform magnetic
field experiences
(A) a force but not a torque [Ans: 5 u10 7 T ]
(B) a torque but not a force iv) A circular magnet is made with its north
pole at the centre, separated from the
(C) both a force and a torque surrounding circular south pole by an air
a gap. Draw the magnetic field lines in the
(D) neither force nor a torque gap. [The magnet is hypothetical magnet].
2. Answer the following questions in brief.
i) What happens if a bar magnet is cut into Draw a diagram to illustrate the magnetic
two pieces transverse to its length/ along lines of force between the south poles of
its length? two such magnets.
ii) What could be the equation for Gauss' v) Two bar magnets are placed on a straight
law of magnetism, if a monopole of pole line with their north poles facing each other
strength p is enclosed by a surface? on a horizontal surface. Draw magnetic
lines around them. Mark the position of
3. Answer the following questions in detail. any neutral points (points where there is no
resultant magnetic field) on your diagram.
i) Explain the Gauss' law for magnetic fields.
ii) What is a geographic meridian. How does ***
the declination vary with latitude? Where
is it minimum?
iii) Define the Angle of Dip. What happens to Internet my friend
angle of dip as we move towards magnetic https://www.ngdc.noaa.gov
pole from magnetic equator?
4. Solve the following Problems.
i) A magnetic pole of bar magnet with pole
strength of 100 A m is 20 cm away from
the centre of a bar magnet. Bar magnet has
pole strength of 200 A m and has a length
5 cm. If the magnetic pole is on the axis
of the bar magnet, find the force on the
magnetic pole.
[Ans: 2.5×10-2N]
228
13. Electromagnetic Waves and Communication System
Can you recall?
1. What is a wave? 4. What are Lenz's law, Ampere's law and
2. What is the difference between longitudinal Faraday's law?
and transverse waves? 5. By which mechanism heat is lost by hot
3. What are electric and magnetic fields and bodies ?
what are their sources? with the circuit changes.
(4) Ampere’s law gives the relation between
13.1 Introduction :
The information age in which we live the induced magnetic field associated with
a loop and the current flowing through
is based almost entirely on the physics of the loop. Maxwell (1831-1879) noticed
electromagnetic (EM) waves. We are now a major flaw in the Ampere’s law for
globally connected by TV, cellphone and time dependant fields. He noticed that the
internet. All these gadgets use EM waves as magnetic field can be generated not only
carriers for transmission of signals. Energy by electric current but also by changing
from the Sun, an essential requirement for life electric field. Therefore in the year 1861,
on Earth, reaches us by means of EM waves he added one more term to the equation
that travel through nearly 150 million km of describing this law. This term is called
empty space. There are EM waves from light the displacement current. This term is
bulbs, heated engine blocks of automobiles, extremely important and the EM waves
x-ray machines, lightning flashes, and some which are an outcome of these equations
radioactive materials. Stars, other objects in our would not have been possible in absence
milky way galaxy and other galaxies are known of this term.
to emit EM waves. Hence, it is important for us As a result, the set of four equations
to make a careful study of the properties of EM describing the above four laws is called
waves. Maxwell’s equations.
In 1888, H. Hertz (1857-1894) succeeded
13.2 EM wave: in producing and detecting the existence of EM
There are four basic laws which describe waves. He also demonstrated their properties
namely reflection, refraction and interference.
the behaviour of electric and magnetic fields, In 1895, an Indian physicist Sir Jagdish
the relation between them and their generation Chandra Bose (1858-1937) produced EM
by charges and currents. These laws are as waves ranging in wavelengths from 5 mm to 25
follows. nm. His work, however, remained confined to
(1) Gauss' law for electrostatics, which is laboratory only.
In 1896, an Italian physicist G. Marconi
essentially the Coulomb’s law, describes (1874-1937) became pioneer in establishing
the relationship between static electric wireless communication. He was awarded the
charges and the electric field produced by Nobel prize in physics in 1909 for his work in
them. developing wireless telegraphy, telephony and
(2) Gauss' law for magnetism, which is broadcasting.
similar to the Gauss' law for electrostatics
mentioned above, states that "magnetic 13.2.1 Sources of EM waves:
monopoles which are thought to be According to Maxwell’s theory, "accelerated
magnetic charges equivalent to the electric
charges, do not exist". Magnetic poles charges radiate EM waves". Consider a charge
always occur in pairs. oscillating with some frequency. This produces
(3) Faraday’s law which gives the relation
between electromotive force (emf) induced
in a circuit when the magnetic flux linked
229
an oscillating electric field in space, which 3)
produces an oscillating magnetic field which in
turn is a source of oscillating electric field. Thus (Faraday’s law with Lenz’s law)
varying electric and magnetic fields regenerate Here φm is the magnetic flux and the
each other.
integral is over a closed loop. Time varying
Waves that are caused by the acceleration magnetic field induces an electromotive
of charged particles and consist of electric force (emf) and hence, an electric field. The
and magnetic fields vibrating sinusoidally at direction of the induced emf is such that the
right angles to each other and to the direction change is opposed.
of propagation are called EM waves or EM
radiation. Figure 13.1 shows an EM wave 4)
propagating along z-axis. The time varying
electric field is along the x-axis and time varying (Ampere-Maxwell law)
magnetic field is along the y-axis.
and tHheereinµte0girsatlhies permeability of vacuum
over a closed loop, I is
the current flowing through the loop. φE
is the electric flux linked with the circuit.
Magnetic field is generated by moving
Fig. 13.1: EM wave propagating along z-axis. charges and also by varying electric fields.
13.2.2 Characteristics of EM waves:
Do you know ? 1) The electric and magnetic fields, E and
B are always perpendicular to each other
In 1865, Maxwell proposed that an and also to the direction of propagation
oscillating electric charge radiates energy
in the form of EM wave. EM waves are of the EM wave. Thus the EM waves are
periodic changes in electric and magnetic
fields, which propagate through space. transverse waves.
Thus, energy can be transported in the form
of EM waves. 2) The cross product E ×B gives the direction
Maxwell’s Equations for Charges and in which the EM wave travels. E ×B also
Currents in Vacuum gives the energy carried by EM wave.
3) The E and B fields vary sinusoidally and
are in phase.
1) (Gauss’ law) 4) EM waves are produced by accelerated
electric charges.
Here E is the electric Tfiheledinatnedgrεa0l is 5) EM waves can travel through free space
permittivity of vacuum. is
the as well as through solids, liquids and
over a closed surface S. The law states that gases.
electric flux through any closed surface S is 6) In free space, EM waves travel with
bedqyeusctahrlietboestsuhretfhateocteareldleailvteiicodtneridcbcebhtywareεge0e.nQGaianneunsescle’loclstarewidc velocity c, equal to that of light in free
space.
charge and electric field it produces. c 1 3u108 m / s ,
P0H 0
2) (Gauss’ law for magnetism). where µ0 (4π×10-7 Tm/A) is permeability
and ε0 (8.85×10-12 C2/Nm2) is permittivity
Here B is the magnetic field. The of free space.
integral is over a closed surface S. The law
states that magnetic flux through a closed 7) In a given material medium, the velocity
surface is always zero, i.e., the magnetic
field lines are continuous closed curves, (vm) of EM waves is given by vm 1
having neither beginning nor end. PH
where µ is the permeability and ε is the
230
permittivity of the given medium. Do you know ?
8) The EM waves obey the principle of According to quantum theory, an
electron, while orbiting around the nucleus
superposition. in a stable orbit does not emit EM radiation
9) The ratio of the amplitudes of electric and even though it undergoes acceleration. It
will emit an EM radiation only when it falls
magnetic fields is constant at any point and from an orbit of higher energy to one of
lower energy.
is equal to the velocity of the EM wave.
EM waves (such as X-rays) are
| E0 |= c | B0 | or | E0 | = 1 --- (13.1) produced when fast moving electrons hit
| B0 | P0 H 0 a target of high atomic number (such as
molybdenum, copper, etc.).
E0 and B0 are the amplitudes of E and An electric charge at rest has an
B respectively. electric field in the region around it but has
no magnetic field. When the charge moves,
10) As the electric field vector ( E0 ) is more it produces both electric and magnetic
prominent than the magnetic field vector fields. If the charge moves with a constant
velocity, the magnetic field will not change
( B0 ), it is responsible for optical effects with time and as such it cannot produce an
EM wave. But if the charge is accelerated,
due to EM waves. For this reason, electric both the magnetic and electric fields change
with space and time and an EM wave is
vector is called light vector. produced. Thus an oscillating charge emits
an EM wave which has the same frequency
11) The intensity of a wave is proportional to as that of the oscillation of the charge.
the square of its amplitude and is given by
the equations
IE 1 H E0 2 , IB 1 B02 --- (13.2)
2 2 P0
0
12) The energy of EM waves is distributed
equally between the electric and magnetic
fields. IE = IB both x- and y-axes. As per property (2) of EM
Example 13.1: Calculate the velocity of EM
waves, E ×B should be along the direction of
waves in vacuum. propagation which is along the x- axis
Solution: The velocity of EM wave in free Since (+ j ) ×(+ k ) = i , B is along the k ,
space is given by i.e., along the z-direction.
1 1 Thus, the amplitude of B = 3.2×10-8 T and
c its direction is along the z-axis.
(8.85 u 10 12 C2 )(4S u10 7 T.m )
P0H 0 Nm 2 A
c 3.00 u10 8 m / s Example 13.3: A beam of red light has an
amplitude 2.5 times the amplitude of second
Example 13.2: In free space, an EM wave of beam of the same colour. Calculate the ratio of
frequency 28 MHz travels along the x-direction. the intensities of the two waves.
The amplitude of the electric field is E = 9.6
V/m and its direction is along the y-axis. What Solution: Intensity ∝ (Amplitude)2
is amplitude and direction of magnetic field B?
I2 ∝ (a)2 and I1 ∝ (2.5a)2
I1 (2.5a )2
Solution : We have, ? I2 a2 (2.5) 2 6.25
.
|B| |E| 9.6 V / m
c 3u108 m / s
In an EM wave, the magnetic field and
B 3.2 u10 8 T
electric field both vary sinusoidally with x. For
It is given that E is along y-direction and ay-waxaivseatnrdavBellailnognagltohnegzxa-axxisi,swhiatvhirnegfeErenacloentog
Chapter 8, we can write Ey and Bz as
the wave propagates along x-axis. The magnetic
field B should be in a direction perpendicular to
231
Ey= E0 sin (kx-ωt) --- (13.2) Ey E0 sin 2S ªx t »¼º
¬« O
and Bz= B0 sin (kx-ωt), --- (13.3) Q
where E0 is the amplitude of the electric field Ey E0 sin 2S ªx 3 u 1010 t º
«¬10 2 »¼
Ey and B0 is the amplitude of the magnetic field
2S is the propagation constant and λ
B. k O Ey E0 sin 2S ¬ª100x 3u1010 tº¼ V / m
z
is the wavelength of the wave. ω = 2πυ is the Example 13.6: The magnetic field of
angular frequency of oscillations, υ being the
frequency of the wave. an = EM wave travelling along x-axis t is
B k 4×10-4 sin (ωt - kx). Here B is in tesla, is
Both the electric and magnetic fields in second and x is in m. Calculate the peak value
attain their maximum (and minimum) values at
the same time and at the same point in space, of electric force acting on a particle of charge 5
i.e., E and B oscillate in phase with the same µC travelling with a velocity of 5×105 m/s along
frequency. the y-axis.
Solution :
Example 13.4: An EM wave of frequency B0= 4×10-4 T, q = 5 µC = 5×10-6 C
50 MHz travels in vacuum along the positive v = 5×105 m/s
x-axis and E at a particular point, x and at a E0= cB0=(3×108) × (4×10-4)
particular instant of time t is 9.6 j V/m. Find =12×104 N/C
the magnitude and direction of B at this point Maximum electric force = qE0
= (5×10-6) (12×104)
x and at time t. = 60×10-2
= 0.6 N
Solution : B=E 9.6 3.2 u10 8 T 13.3 Electromagnetic Spectrum:
c 3u108 The orderly distribution (sequential
arrangement) of EM waves according to their
As the wave propagates along +x axis and wavelengths (or frequencies) in the form of
distinct groups having different properties
E is along +y axis, direction of B will be along is called the EM spectrum (Fig. 13.2). The
+z-axis i.e. B = 3.2×10-8 k T. properties of different types of EM waves are
given in Table 13.1.
Example 13.5: For an EM wave propagating
along x direction, the magnetic field oscillates
along the z-direction at a frequency of 3×1010
Hz and has amplitude of 10-9 T.
a) What is the wavelength of the wave?
b) Write the expression representing the
corresponding electric field.
Solution :
a) O = c 3u108 m / s 10 2 m
X 3u1010 / s
b) E0= cB0= (3×108 m/s) × (10-9 T) = 0.3 V/m. Fig. 13.2: Electromagnetic spectrum.
Since B acts along z-axis, E acts along y-axis. We briefly describe different types of EM
Expression representing the oscillating electric waves in the order of decreasing wavelength (or
field is increasing frequency).
13.3.1 Radio waves :
Ey E0 sin (kx -Zt) Radio waves are produced by accelerated
motion of charges in a conducting wire. The
Ey E0 sin ª§ 2S · x (2SQ )t º
«¬¨© O ¸¹ ¼»
232
frequency of waves produced by the circuit Notation used for high frequencies
depends upon the magnitudes of the inductance 1 kHz = one kilo Hertz =1000 Hz = 103 Hz
and the capacitance (This will be discussed 1 MHz = one mega Hertz = 106 Hz
in XIIth standard). Thus, by choosing suitable 1 GHz = one giga Hertz =109 Hz
values of the inductance and the capacitance, Notation used for small wavelengths
radio waves of desired frequency can be 1 µm = one micrometer = 10-6 m
produced. 1 Å= one angstrom = 10-10 m= 10-8 cm
Properties : 1nm = one nanometer = 10-9 m
1) They have very long wavelengths ranging
Uses :
from a few centimetres to a few hundreds 1) Radio waves are used for wireless
of kilometres.
2) The frequency range of AM band is 530 communication purpose.
kHz to 1710 kHz. Frequency of the waves 2) They are used for radio broadcasting and
used for TV-transmission range from 54
MHz to 890 MHz, while those for FM radio transmission of TV signals.
band range from 88 MHz to 108MHz. 3) Cellular phones use radio waves to
transmit voice communication in the ultra
high frequency (UHF) band.
Table 13.1: Properties of different types of EM waves
Name Wavelength Frequency Generated By
range in m range in Hz
Gamma 6×10-13 to 1×10-10 5×1020 to 3×1018 a) Transitions of nuclear energy levels
rays 1×10-11 to 3×10-8 3×1019 to 1×1016 b) Radioactive substances
X-rays
3×10-8 to 4×10-7 1×1016 to 8×1014 a) Bombardment of high energy
Ultraviolet 8×1014 to 4×1014 electrons (keV) on a high atomic
(UV waves) 4×10-7 to 8×10-7 4×1014 to 1×1012 number target (Cu, Mg, Co)
8×10-7 to 3×10-4
Visible light 3×10-4 to 6×10-2 b) Energy level transitions of
6×10-4 to 1×105 innermost orbital electrons
Infrared (IR)
radiations Rearrangement of orbital electrons of
Microwaves atom between energy levels. As in high
Radio waves voltage gas discharge tube, the Sun and
mercury vapour lamp, etc.
Rearrangement of outer orbital
electrons in atoms and molecules e.g.,
gas discharge tube
Hot objects
1×1012 to 5×109 Special electronic devices such as
5×1011 to 8×1010 klystron tube
Acceleration of electrons in circuits
13.3.2 Microwaves : they are incident.
These waves were discovered of by H. 2) They can be detected by crystal detectors.
Uses
Hertz in 1888. Microwaves are produced by 1) Used for the transmission of TV signals.
oscillator electric circuits containing a capacitor 2) Used for long distance telephone
and an inductor. They can be produced by
special vacuum tubes. communication.
Properties 3) Microwave ovens are used for cooking.
1) They heat certain substances on which 4) Used in radar systems for the location of
233
distant objects like ships, aeroplanes etc, Do you know ?
5) They are used in the study of atomic and
Stars and galaxies emit different types
molecular structure. of waves. Radio waves and visible light can
13.3.3 Infrared waves pass through the Earth’s atmosphere and
reach the ground without getting absorbed
These waves were discovered by William significantly. Thus the radio telescopes
Herschel (1737-1822) in 1800. All hot bodies and optical telescopes can be placed on the
are sources of infrared rays. About 60% of ground. All other type of waves get absorbed
the solar radiations are infrared in nature. by the atmospheric gases and dust particles.
Thermocouples, thermopile and bolometers are Hence, the γ-ray, X-ray, ultraviolet, infrared,
used to detect infrared rays. and microwave telescopes are kept aboard
Properties artificial satellites and are operated remotely
1) When infrared rays are incident on any from the Earth. Even though the visible
radiation reaches the surface of the Earth,
object, the object gets heated. its intensity decreases to some extent due
2) These rays are strongly absorbed by glass. to absorption and scattering by atmospheric
3) They can penetrate through thick columns gases and dust particles. Optical telescopes
are therefore located at higher altitudes.
of fog, mist and cloud cover.
Uses The Indian Giant Metrewave Radio
1) Used in remote sensing. Telescope (GMRT) near Pune is an important
2) Used in diagnosis of superficial tumours milestone in the field of Radio-astronomy.
Also, Indian Astronomical Observatory
and varicose veins. houses the Himalayan Chandra Telescope
3) Used to cure infantile paralysis and to (HCT), the 2 m optical-IR Telescope, which
is situated at Hanle, Ladakh, at an altitude of
treat sprains, dislocations and fractures. 4500 m.
4) They are used in Solar water heaters and
Table 13.2: Wavelengths of colours in
cookers. visible light
5) Special infrared photographs of the body
Colour Wavelength
called thermograms, can reveal diseased
organs because these parts radiate less violet 380-450 nm
heat than the healthy organs.
6) Infrared binoculars and thermal imaging blue 450-495 nm
cameras are used in military applications
for night vision. green 495-570 nm
7) Used to keep green house warm.
8) Used in remote controls of TV, VCR, etc yellow 570-590 nm
13.3.4 Visible light :
It is the most familiar form of EM waves. orange 590-620 nm
These waves are detected by human eye.
Therefore this wavelength range is called the red 620-750 nm
visible light. The visible light is emitted due to
atomic excitations. 13.3.5 Ultraviolet rays :
Properties : Ultraviolet rays were discovered by
J. Ritter (1776-1810) in 1801. They can be
1) Different wavelengths give rise to different produced by the mercury vapour lamp, electric
colours. These are given in Table 13.2. spark and carbon arc lamp. They can also be
obtained by striking electrical discharge in
2) Visible light emitted or reflected from hydrogen and xenon gas tubes. The Sun is the
objects around us provides us information most important natural source of ultraviolet
about those objects and hence about the rays, most of which are absorbed by the ozone
surroundings. layer in the Earth’s atmosphere.
234
Properties : Properties
1) They are high energy EM waves.
1) They produce fluorescence in certain 2) They are not deflected by electric and
materials, such as 'phosphors'.
magnetic fields.
2) They cause photoelectric effect. 3) X-rays ionize the gases through which
3) They cannot pass through glass but pass
they pass.
through quartz, fluorite, rock salt etc. 4) They have high penetrating power.
4) They possess the property of synthesizing 5) Their over dose can kill living plant and
vitamin D, when skin is exposed to them. animal overdose tissues and hence are
Uses : harmful.
Uses
1) Ultraviolet rays destroy germs and bacteria 1) Useful in the study of the structure of
and hence they are used for sterilizing crystals.
surgical instruments and for purification 2) X-ray photographs are useful to detect
of water. bone fracture. X-rays have many other
medical uses such as CT scan.
2) Used in burglar alarms and security 3) X-rays are used to detect flaws or cracks in
systems. metals.
4) These are used for detection of explosives,
3) Used to distinguish real and fake gems. opium etc.
4) Used in analysis of chemical compounds. 13.3.7 Gamma Rays (γ-rays)
5) Used to detect forgery. Discovered by P. Villard (1860-1934) in
1900. Gamma rays are emitted from the nuclei
Do you know ? of some radioactive elements such as uranium,
radium etc.
1. A fluorescent light bulb is coated from Properties
with a powder inside and contains a gas; 1) They are highest energy EM waves.
electricity causes the gas to emit ultraviolet (energy range keV - GeV)
radiation, which then stimulates the tube 2) They are highly penetrating.
coating to emit light. 3) They have a small ionising power.
2. The pixels of a television or computer 4) They kill living cells.
screen fluoresce when electrons from an Uses
electron gun strike them. 1) Used as insecticide disinfection for wheat
3. What we call 'visible light' is just the part and flour.
of the EM spectrum that human eyes see. 2) Used for food preservation.
Many other animals would define 'visible' 3) Used in radiotherapy for the treatment of
somewhat differently. For instance, many cancer and tumour.
animals including insects and birds, see 4) They are used to produce nuclear
in the UV region. Natural world is full of reactions.
signals that animals see and humans cannot.
Many birds including bluebirds, budgies, 13.4 Propagation of EM Waves:
parrots and even peacocks have ultraviolet You must have seen a TV antenna used to
patterns that make them even more vivid to
each other than they are to us. receive the TV signals from the transmitting
tower or from a satellite. In communication
13.3.6 X-rays: using radio waves, an antenna in the transmitter
German physicist W. C. Rontgen (1845- radiates the EM waves, which travel through
space and reach the receiving antenna at the
1923) discovered X-rays in 1895 while studying other end. As the EM wave travels away from
cathode rays (which is a stream of electrons the transmitter; the strength of the wave keeps
emitted by the cathode in a vacuum tube).
X-rays are also called Rontgen rays. X-rays
are produced when cathode rays are suddenly
stopped by an obstacle.
235
on decreasing. Several factors influence the
propagation of EM waves and the path they
follow. It is also important to understand the
composition of the Earth’s atmosphere as
it plays a vital role in the propagation of EM
waves. Different layers of Earth’s atmosphere
are shown in Fig. 13.3.
Do you know ?
Ionizing radiations : Fig. 13.4: Propagation of EM waves.
Ultraviolet, X-ray and gamma rays 13.4.1 Ground (surface) wave:
have sufficient energy to cause ionization When a radio wave from a transmitting
i.e. they strip electrons from atoms and antenna propagates near surface of the Earth
molecules lying along their path. The atoms so as to reach the receiving antenna, the wave
lose their electrons and are then known as propagation is called ground wave or surface
ions. Ionization is harmful to human beings wave propagation.
because it can kill or damage living cells,
or make them grow abnormally as cancers. In this mode, radio waves travel close to
Fluorescent lamps are based on ionization of the surface of the Earth and move along its
gas. Ionizing radiation is also used in various curved surface from transmitter to receiver.
equipments in laboratory and industry.
The radio waves induce currents in the
Fig 13.3: Earth and atmospheric layers. ground and lose their energy by absorption.
Different modes of propagation of EM Therefore, the signal cannot be transmitted over
large distances. Radio waves having frequency
waves are described below and are shown in less than 2 MHz (in the medium frequency band)
Fig. 13.4. are transmitted by ground wave propagation.
This is suitable for local broadcasting only.
Do you know ? For TV or FM signals (very high frequency),
X-rays have many practical applications ground wave propagation cannot be used.
in medicine and industry. Because X-ray 13.4.2 Space wave:
photons are of such high energy, they can
penetrate several centimetres of solid matter When the radio waves from the transmitting
and can be used to visualize the interiors of antenna reach the receiving antenna either
materials that are opaque to ordinary light. directly along a straight line (line of sight) or
after reflection from the ground or satellite or
after reflection from troposphere, the wave
propagation is called space wave propagation.
The radio waves reflected from troposphere are
called tropospheric waves. Radio waves with
frequency greater than 30 MHz can pass through
the ionosphere (60 km - 1000 km) after suffering
a small deviation. Hence, these waves cannot be
transmitted by space wave propagation except
by using a satellite. Also, for TV signals which
have high frequency, transmission over long
distance is not possible by means of space wave
propagation.
The maximum distance over which a signal
can reach is called its range. For larger TV
236
coverage, the height of the transmitting antenna Example 13.8: If the height of a TV transmitting
should be as large as possible. This is the reason antenna is 128 m, how much square area can be
why the transmitting and receiving antennas are covered by the transmitted signal if the receiving
mounted on top of high rise buildings. antenna is at the ground level? (Radius of the
Earth = 6400 km)
Range is the straight line distance from the Solution:
point of transmission (the top of the antenna)
to the point on Earth where the wave will hit Range = d 2Rh
while travelling along a straight line. Range is
shown by d in Fig. 13.5. Let the height of the = 2(6400u103 )(128)
transmitting antenna (AA') situated at A be h. B
represents the point on the surface of the Earth = 16.384 u105 u103
at which the space wave hits the Earth. The
triangle OA'B is a right angled triangle. From = 16.384 u108
∆ OA' B we can write
= 4.047u104
OA'2 = A'B2 + OB2
(R+h)2 = d2 + R2 40.470 km
or R2 + h2 + 2Rh = d2 + R2
As h << R, we can ignore h2 and write Area covered = S d 2 3.14 u (40)2
d ≅ 2Rh
The range can be increased by mounting = 5144 .58 k4m2
the receiver at a height h' say at a point C on the
surface of the Earth. The range increases to d + Example 13.9: The height of a transmitting
d' where d' is 2Rh ' Thus antenna is 68 m and the receiving antenna is
Total range = d d ' 2Rh 2Rh ' at the top of a tower of height 34 m. Calculate
the maximum distance between them for
Fig. 13.5: Range of the signal (not to scale). satisfactory transmission in line of sight mode.
Example 13.7: A radar has a power of 10 kW (radius of Earth = 6400 km)
and is operating at a frequency of 20 GHz. It
is located on the top of a hill of height 500 m. ht = 68 m, hr = 34m, R = 6400 km = 6.4×106m
Calculate the maximum distance upto which it Solution:
can detect object located on the surface of the
Earth . (Radius of Earth = 6.4×106 m) dmax 2Rht 2Rhr
Solution:
= 2u 6.4 u106 u 68 2u 6.4 u106 u 34
Maximum distance (range) =
d = 2Rh = 870 u103 435 u103
= 2u (6.4 u106 ) u 500 m = 29.5u103 20.9u103
= 8u104 80 km,
50.4 u103 m
where R is radius of the Earth and h is the
height of the radar above Earth’s surface. = 50.4 km
13.4.3 Sky wave propagation:
When radio waves from a transmitting
antenna reach the receiving antenna after
reflection in the ionosphere, the wave
propagation is called sky wave propagation.
The sky waves include waves of frequency
between 3 MHz and 30 MHz. These waves
can suffer multiple reflections between the
ionosphere and the Earth. Therefore, they can
be transmitted over large distances.
Critical frequency : It is the maximum value
of the frequency of radio wave which can be
reflected back to the Earth from the ionosphere
when the waves are directed normally to
ionosphere.
237
Skip distance (zone) : It is the shortest distance place and the receiver at another place. The
from a transmitter measured along the surface communication channel is a passage through
of the Earth at which a sky wave of fixed which signals transfer in between a transmitter
frequency (if grater than critical frequency) will and a receiver. This channel may be in the form
be returned to the Earth so that no sky waves of wires or cables, or may also be wireless,
can be received within the skip distance. depending on the types of communication
system.
13.5 Introduction to Communication System:
Communicationisexchangeofinformation. There are two basic modes of
communication: (i) point to point communication
Since ancient times it is practiced in various and (ii) broadcast.
ways e.g., through speaking, writing, singing,
using body language etc. After the discovery In point to point communication mode,
of electricity in the late 19th century, human communication takes place over a link between a
communication systems changed dramatically. single transmitter and a receiver e.g. Telephony.
Modern communication is based upon the In the broadcast mode there are large number of
discoveries and inventions by a number of receivers corresponding to the single transmitter
scientists like J. C. Bose (1858-1937), S. F. B. e.g., Radio and Television transmission.
Morse (1791-1872), G. Marconi (1874-1937) 13.5.2 C ommonly used terms in electronic
and Alexander Graham Bell (1847-1922) in the
19th and 20th centuries. communication system:
Following terms are useful to understand
In the 20th century we could send messages any communication system:
over large distances using analogue signals, 1) Signal :- The information converted into
cables and radio waves. With the advancements electrical form that is suitable for transmission
of digitization technologies, we can now is called a signal. In a radio station, music and
communicate with the entire world almost in speech are converted into electrical form by a
real time. microphone for transmission into space. This
electrical form of sound is the signal. A signal
The ability to communicate is an important can be analog or digital as shown in Fig. 13.7.
feature of modern life. We can speak directly to
others all around the world and generate vast (a) (b)
amount of information every day.
Fig 13.7: (a) Analog signal. (b) Digital signal.
Here we will briefly discuss how (i) Analog signal: A continuously varying
communication systems work.Acommunication
system is a device or set up used in transmission signal (voltage or current) is called
and reception of information from one place to an analog signal. Since a wave is a
another. fundamental analog signal, sound and
13.5.1 Elements of a communication system: picture signals in TV are analog in nature
(Fig 13.7 a)
There are three basic (essential) elements (ii) Digital signal :Asignal (voltage or current)
of every communication system: a) Transmitter, that can have only two discrete values
b) Communication channel and c) Receiver. is called a digital signal. For example, a
square wave is a digital signal. It has two
Fig. 13.6: Block diagram of the basic elements values viz, +5 V and 0 V. (Fig- 13.7 b)
of a communication system. 2) Transmitter :- A transmitter converts the
signal produced by a source of information
In a communication system, as shown into a form suitable for transmission through a
in Fig. 13.6, the transmitter is located at one channel and subsequent reception.
238
3) Transducer :- A device that converts one used to increase the range of a communication
form of energy into another form of energy is system. These are shown in Fig. 13.8.
called a transducer. For example, a microphone
converts sound energy into electrical energy. Fig.13.8: Use of repeater station to increase
Therefore, a microphone is a transducer. the range of communication.
Similarly, a loudspeaker is a transducer which
converts electrical energy into sound energy. Do you know ?
4) Receiver :- The receiver receives the message
signal at the channel output, reconstructs it in To transmit a signal we need an antenna
recognizable form of the original message for or an aerial. For efficient transmission and
delivering it to the user of information. reception, the transmitting and receiving
5) Noise :- A random unwanted signal is called antennas must have a length at least λ/4
noise. The source generating the noise may be where λ is the wavelength of the signal.
located inside or outside the system. Efforts
should be made to minimise the noise level in a For an audio signal of 15kHz, the
communication system. required length of the antenna is λ/4 which
6) Attenuation :- The loss of strength of the can be seen to be equal to 5 km.
signal while propagating through the channel
is known as attenuation. It occurs because the The highest TV tower in Rameshwaram,
channel distorts, reflects and refracts the signals Tamilnadu, is the tallest tower in India and
as it passes through it. is ranked 32nd in the world with pinnacle
7) Amplification :- Amplification is the height of 323 metre. It is used for television
process of raising the strength of a signal, using broadcast by the Doordarshan.
an electronic circuit called amplifier.
8) Range :- The maximum (largest) distance 13.6 Modulation:
between a source and a destination up to which As mentioned earlier, an audio signal has
the signal can be received with sufficient
strength is termed as range. low frequency (< 20 KHz). Low frequency
9) Bandwidth :- The bandwidth of an electronic signals can not be transmitted over large
circuit is the range of frequencies over which it distances. Because of this, a high frequency
operates efficiently. wave, called a carrier wave, is used. Some
10) Modulation :- The signals in communication characteristic (e.g. amplitude, frequency or
system (e.g. music, speech etc.) are low phase) of this wave is changed in accordance
frequency signals and cannot be transmitted with the amplitude of the signal. This process
over large distances. In order to transmit the is known as modulation. Modulation also
signal to large distances, it is superimposed on helps avoid mixing up of signals from different
a high frequency wave (called carrier wave). transmitters as different carrier wave frequencies
This process is called modulation. Modulation can be allotted to different transmitters. Without
is done at the transmitter and is an important the use of these waves, the audio signals, if
part of a communication system. transmitted directly by different transmitters,
11) Demodulation :- The process of regaining would have got mixed up.
signal from a modulated wave is called
demodulation. This is the reverse process of Modulation can be done by modifying
modulation. the (i) amplitude (amplitude modulation)
12) Repeater :- It is a combination of a (ii) frequency (frequency modulation), and (iii)
transmitter and a receiver. The receiver receives phase (phase modulation) of the carrier wave in
the signal from the transmitter, amplifies it and proportion to the amplitude or intensity of the
transmits it to the next repeater. Repeaters are
239
signal wave keeping the other two properties susceptible to noise. This modulation is used forDisplacement
same. Figure 13.9 (a) shows a carrier wave high quality broadcast transmission.
and (b) shows the signal. The carrier wave is
a high frequency wave while the signal is a Phase modulation (PM) is easier than
low frequency wave. Amplitude modulation, frequency modulation. It is used in determining
frequency modulation and phase modulation of the velocity of a moving target which cannot be
carrier waves are shown in Fig. 13.9 (c), (d) and done using frequency modulation.
(e) respectively.
a
Amplitude modulation (AM) is simple
to implement and has large range. It is also b
cheaper. Its disadvantages are that (i) it is not
very efficient as far as power usage is concerned c
(ii) it is prone to noise and (iii) the reproduced
signal may not exactly match the original signal. d
In spite of this, these are used for commercial
broadcasting in the long, medium and short e
wave bands. Time
Frequency modulation (FM) is more Fig. 13.9: (a) Carrier wave, (b) signal (c)
complex as compared to amplitude modulation AM (d) FM and (e) PM.
and, therefore is more difficult to implement.
However, its main advantage is that it Internet my friend
reproduces the original signal closely and is less https://www.iiap.res.in/centers/iao
ExercisesExercises
1. Choose the correct option. vi) The waves used by artificial satellites for
i) The EM wave emitted by the Sun and communication purposes are
responsible for heating the Earth’s (A) Microwave
atmosphere due to green house effect is (B) AM radio waves
(A) Infra-red radiation (B) X ray (C) FM radio waves
(C) Microwave (D) Visible light (D) X-rays
ii) Earth ’s atmosphere is richest in vii) If a TV telecast is to cover a radius of
(A) UV (B) IR 640 km, what should be the height of
(C) X-ray (D) Microwaves transmitting antenna?
iii) How does the frequency of a beam of (A) 32000 m (B) 53000 m
ultraviolet light change when it travels (C) 42000 m (D) 55000 m
from air into glass? 2. Answer briefly.
(A) No change (B) increases i) State two characteristics of an EM wave.
(C) decreases (D) remains same ii) Why are microwaves used in radar?
iv) The direction of EM wave is given by iii) What are EM waves?
(B) E . B iv) How are EM waves produced?
(A) E ×B v) Can we produce a pure electric or
(C) along E (D) along B
v) The maximum distance upto which TV magnetic wave in space? Why?
transmission from a TV tower of height h vi) Does an ordinary electric lamp emit EM
can be received is proportional to waves?
(A) h1/2 (B) h vii) Why do light waves travel in vacuum
(C) h3/2 (D) h2 whereas sound wave cannot?
240
viii) What are ultraviolet rays? Give two uses. iii) The speed of light is 3×108 m/s. Calculate
the frequency of red light of wavelength of
ix) What are radio waves? Give its two
uses. 6.5×10-7 m.
x) Name the most harmful radiation [Ans: υ = 4.6×1014 Hz]
entering the Earth's atmosphere from the iv) Calculate the wavelength of a microwave
outer space. of frequency 8.0 GHz.
xi) Give reasons for the following: [Ans: 3.75 cm]
(i) Long distance radio broadcast uses v) In a EM wave the electric field oscillates
short wave bands. sinusoidally at a frequency of 2×1010 Hz.
(ii) Satellites are used for long distance What is the wavelength of the wave?
TV transmission. [Ans: 1.5×10-2 m]
xii) Name the three basic units of any vi) The amplitude of the magnetic field part
communication system. of a harmonic EM wave in vacuum is
xiii) What is a carrier wave? B0= 5×10-7 T. What is the amplitude of the
electric field part of the wave?
xiv) Why high frequency carrier waves are
used for transmission of audio signals? [Ans: 150V/m]
xv) What is modulation? vii) A TV tower has a height of 200 m.
xvi) What is meant by amplitude modulation? How much population is covered by TV
xvii) What is meant by noise? transmission if the average population
xviii) What is meant by bandwidth? density around the tower is 1000/km2?
xix) What is demodulation? (Radius of the Earth = 6.4×106 m)
xx) What type of modulation is required for [Ans: 8×106]
television broadcast? viii) Height of a TV tower is 600 m at a given
xxi) How does the effective power radiated place. Calculate its coverage range if the
by an antenna vary with wavelength? radius of the Earth is 6400 km. What
xxii) Why should broadcasting programs use should be the height to get the double
different frequencies? coverage area?
xxiii) Explain the necessity of a carrier wave [Ans: 87.6 km, 1200 m]
in communication. ix) Atransmitting antenna at the top of a tower
xxiv) Why does amplitude modulation give has a height 32 m and that of the receiving
noisy reception? antenna is 50 m. What is the maximum
xxv) Explain why is modulation needed. distance between them for satisfactory
2. Solve the numerical problem. communication in line of sight mode ?
i) Calculate the frequency in MHz of a radio Given radius of Earth is 6.4×106 m.
wave of wavelength 250 m. Remember [Ans: 45.537 km]
that the speed of all EM waves in vacuum ***
is 3.0×108 m/s. [Ans: 1.2 MHz]
ii) Calculate the wavelength in nm of an
X-ray wave of frequency 2.0×1018 Hz.
[Ans: 0.15 nm]
241
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