Based on the R-19 Regulation

3. 107
Introduction
Magnetism
Magnetism is defined as an attractive and repulsive
phenomenon produced by a moving electric charge.
the affected region around a moving charge consist
of both an electric and magnetic field.
ex:- bar magnet
It is attracted to magnetic field and can attract or
repel other magnets
Magnetic dipole
A magnet consist of a both north and south poles.
any two opposite sides separated by a finite
distance constitute magnetic dipole
Magnetic dipole moment(or)magnetic
moment
The strength of the magnetic dipole is called the
magnetic dipole moment
Let us consider a m be the magnetic pole strength
and '2l' be the length of the magnet.this magnetic
moment is given by the
μₘ=m*2l

4. 108
==> Whenever current 'I' ampere flows through a
circular wire of one turn having an area of
cross-section Am²,it is said to be have a magnetic
moment
μₘ=IA
Units:-Am²
==>It is vector quantity pointing from S to N poles
Magnetic field
A magnetic field is a vector field that describes the
magnetic influence of the electric charges in relative
motion and magnetic materials
ex:-This can be seen in the permanent magnets
Magnetic field strength(H)
The force experianced by a unit north pole placed
at a given point in a magnetic field.
==>It is expressed in A/m
Magnetisation or intensity of
magnetisation(M)
==>it is denoted by the 'M'
==>The magnetic moment per unit volume
Units:-A/m

5. 109
Magnetic suceptibility(χ)
The ratio of the magnetisation produced in a
sample to the magnetic field strength
χ=M/H
Magnetic lines of forces
The magnetic field is characterised by magnetic
lines of forces
Magnetic flux density(or)magnetic
induction field strength
The ratio of the no.of magnetic lines of force
passing through the unit area of cross section
B=φ/A
Units:- wb/m² (or) Tesla
Magnetic permeability(μ)
The ratio of the magnetic flux density to the
applied magnetic field density
μ=B/H

6. Magnetic permeability of free space(μ₀)
The ratio of the magnetic induction field space
to the applied magnetic field strength H
μ₀=B₀/H
μ₀=4π*10⁻⁷H/m
Units:-Henry/meter
Relative permeability(μᵣ)
The ratio between the magnetic permeability to
the magnetic permeability of free space is called
Relative permeability
μᵣ=μ/μ₀
Relation between the B,H,M
B=μH μ=B/H
=μ₀μᵣH [μᵣ=μ/μ₀ ==> μ=μ₀μᵣ]
B=μ₀μᵣH+μ₀H-μ₀H
B=(μᵣ-1)μ₀H+μ₀H
B=μ₀H(μᵣ-1+1)
B=μ₀μᵣH ----> 1
we know that
M=(μᵣ-1)H
110

7. M=μᵣH-H
μᵣH=M+H ----> 2
Now substitute 2 in 1
B=μ₀(H+M)
Relation between the "χ" and "μᵣ"
B=μ₀(H+M)
μ₀=B/H+M
w.k.t, μᵣ=μ/μ₀
=B/H/B/H+M
μᵣ=H+M/H
μᵣ=1+M/H
μᵣ=1+χ [χ=M/H]
Origin of magnetic moment-Bohr
magneton
111
==>The orbital motion of electrons establish a
magnetic field and hence we get "Orbital magnetic
moment of the electrons"
==>while revolving around the nucleus,the electrons
spin about their own axis which results in a magnetic
field"spin magnetic moment of nucleus"
==>the nucleons in the nucleus also spin,resulting in
magnetic

8. 112
a)Orbital magnetic moment of
electron
^
electron orbit
μ(orbital)
^
ω
m
Let us consider a "m" be the mass of an electron
and "e" be the charge revolving around the nucleus
in a circular orbit of radius 'r'.we can take the clock
wise (or) Anticlock wise to revolve an electron.Now
the electron is revolving around the nucleus in an
anticlock wise direction.Let v be the linear velocity of

9. 113
an electron and "ω" be the angular velocity of a
revolving electron
The frequency of the revolving electron is given by
the
frequency(F)=w/2π
As we know that,time period(T)=1/F
=2π/ω
T=2π/ω ----> 1
Whenever the electron is revolve in a circular orbit it
establishes the current
I=charge/time period
I=e/T
I=e*1/T
I=e*ω/2π [from 1]
The area of the circular orbit is given by
A=πr²
the magnetic moment is given by the current and area
of cross section is
μ
ᵒʳᵇⁱᵗᵃˡ =IA
=e*ω/2π*πr²
=eωr²/2
the orbital magnetic moment is pointing from
s-pole to north pole i.e, in downward direction
the angular momentum(L) of the revolving
electron is given by the

10. 114
L=mvr
L=m(rω)(r) [ w.k.t, the relation between the v and
ω is v=rω]
L=mωr²
from 1
μ =eωr²/2
ᵒʳᵇ
=e/2m[mωr²]
=-e/2m[orbital angular momentum]
"-" sign indicates the orbital angular momentum and
magnetic moment are in opposite direction
orbital gyromagnetic ratio
(γ)=magnetic moment/orbital angular momentum
γ=e/2m
As we know thatangular moment of electrons
associated with the orbital quantum number 'l' is given
by the lh/2π
μ
ᵒʳᵇⁱᵗᵃˡ
=-e/2m(orbital angular momentum)
=-e/2m[lh/2π]
μ
ᴮ
=ehl/4πm is called the bohr magneton
μ
ᴮ =9.27*10⁻²⁴ Am²
μ =μ l
ᵒʳᵇ ᴮ
l=1,2,3,....

11. 115
μ =-μ ,-2μ ,-3μ ,...
ᵒʳᵇ ᴮ ᴮᴮ
b)Spin magnetic moment of electron
^
^ spin angular momentum
spinning electron in a
orbit

12. 116
The magnetic moment due to electron
spin
The magnetic moment associated with spinning of
the electron is called the spin magnetic moment(μₛₚᵢₙ).
magnetic moment due to the rotation of the electronic
charge about one of the diameters of the electron is
similar to the earths spinning motion around its north
south axis
μₛₚᵢₙ=γ(e/2m)s
where, γ=gyromagnetic ratio
s=spin angular momentum
for an electron,s=h/4π
μₛₚᵢₙ=γ(e/2m)(h/4π)
=-9.4*10⁻²⁴ Am²
c)Magnetic moment due to Nuclear
spin
Another contribution may arise from the nuclear
magnetic moment.by analogy with bohr magneton,
the nucleus magneton arises due to the spin of the
nucleus
The nuclear spin magnetic moment is given by

13. 117
μₙₛ=eh/4πmₚ
where, mₚ=mass of proton
μₙₛ=5.05*10⁻²⁷Am²
Classification of magnetic materials
The magnetic materials have been classified into
five types based on the influence of electricfield on
them.there were given below
a)Diamagnetic materials
b)Paramagnetic materials
c)Ferromagnetic materials
d)Anti-ferromagnetic materials
e)Ferrimagnetic Materials
Magnetic materials
a)Diamagnetic
materials
b)Paramagnetic
materials
c)Ferromagnetic
Materials
d)Anti-ferromagn
etic materials
e)Ferrimagnetic
materials

14. 118
a)Diamagnetic materials
Diamagnetic materials are repelled by the magnets
due to the fact that they produce negative
magnetization
Before presence of magnetic
field
---->
H
==>In the presence of external magnetic field the
atoms move in randomly in all direction
==>The net magnetization is zero
==>In the presence of an external magnetic field tha
atoms move opposite to the magnetic field si 'm' is
-ve
m="-ve"
==>Relative permeability(μᵣ) is less than 1
μᵣ < 1
o
o
o
o
o
o
o
o
o
o
o o
o o o
o o
o o
o o o o o
o o o o o
o o o o o
o o o o o
o o o o o

15. 119
==>the suceptibility(χ) is smal and negative
==>the suceptibility is independent on temperature
ex:- organic materials
b)Paramagnetic materials
==>it allows the magnetic lines of force to pass
through it H=0
---->
H
==>As we know that in the absence of magnetic field
they move randomly in all direction
==>whenever the paramagnetic materials placed in
the magnetic field there is some of atoms alligning
opposite to the magnetic field
==>then net magnetion is positive
==>μᵣ > 1 (relative permeability)
==>suceptibility(χ) is low and positive vaues
==>χ∝1/T
o o

16. 120
χ=c/T [c=curies constant]
this is also called as curies law
c=Nμ₀μ² /k
ᴮ ᴮ
> >T 1/T
l
χ
c)Ferromagnetic materials
(weiss theory)
==>it allows a large number of magnetic lines of
forces to pass through it

17. 121
==>In the case of absence of magnetic field the
atoms more randomly in all direction.whenever the
ferromagnetic materials are placed in the presence
of magnetic field all the atoms moves parallel to the
magnetic field
==>magnetization is positive and high
==>suceptibility(χ) is positive and high
==>Relative permeability μᵣ>>1
χ=T/T+̅T
ᶜ
^
>Tᶜ
l
ferromagnetic
paramagnetic
variation of χ
with T
---->HH=0

18. 122
T is called curies temperatureᶜ
Curies temperature(or)curies weiss
law(T )
At the constant temperature,the ferromagnetic
materials is converted into the paramagnetic
materials is called the curies temperature
ᶜ
ex:- transition and rare earth elements
d)Anti-ferromagnetic materials
^ ^ ^ ^
Alignment of antiferromagnetic moments
In the process of magnetic field,the magnetic
moments are antiparallel is called
antiferromagnetic materials
==>it allows the magnetic lines of forces to pass
through it
==>μᵣ>1
==>m is +ve
==>χ is low and +ve
==>χ=T/T+T
ᴺ

19. 123
where, T =neel's temperature
ᴺ
^
>
|
|
|
|
|
|
|
^
> T
χ
ex:-salts of transition element
==>The magnetic point or the χ is decreases is
called the Anti ferromagnetic materials
e)Ferrimagnetic materials
^ ^^ ^ ^
Low magnitude

20. 124
==>The substances which consist of the antiparallel
magnetic moments of different magnitudes are
known as ferrimagnetic materials
^
>
|
|
|
|
|
|
|
|
^
>
χ
T
ᶜ
==>μᵣ>>1
==>m is +ve and high
==>χ is positive and high
ex:-ferrites

21. 125
Hysteresis
when an ferromagnetic material is placed in a
magnetic field,the curve is drawn in between the
magnetic field(B) and magnetic intensity"H".this is
called the hysteresis of curve(or)hysteresis loop(or)
as shown in graph the B is lagging behind it is called
the hysteresis loop(or)B-H curve
In the hysteresis curve it exhibits the two properties
are residual magnetism and cohersive forces
from the figure as the H is decreased,B also
decreased but following a path ab.In the B lags
behind it whenever the H becomes zero,B is still has
a value equal to ob.this magnetic flux density
remaining in the speciment in the absence of

22. 126
any external field is called the "residual magnetism"
of the speciment.the power of retaining this
magnetism is called the "retentivity(or)remenence
of the speciemen
Whenever,if the magnetic intensity H is now
increased in the reversed direction,the value of B
further decreased,still lagging behind H and
becomes zero when H has a value equal to oc
this value of the magnetic intensity is called the
coercive forces(or) coercitivity.
Coercitivity:-
it is a measure of magnetic intensity required to
destroy the residual magnetism
When the H increases beyond oc,the speciment is
increasingly magnetised in the opposite direction
and a poibt such d is reached
thus we can found that flux density B always lags
behind the magnetic intensity H.the lagging of B
behind H is called hysteresis and the curve in
between B and H as shown in abcde is called
"Hysteresis loop(or)curve

23. Based on the area of the hysteresis loop,the
ferromagnetic materials are classified into soft and
hard magnetic materials
a)soft magnetic materials
b)hard magnetic materials
Soft magnetic
materials
hard magnetic
materials
==>The ferromagnetic
materials which are
easily magnetised and
demagnetised are
known as the soft
magnetic materials
==>The ferromagnetic
materials which are hard
to magnetise and
demagnetise are known
as hard magnetic
materials
==>The Retentivity and
coercitivity is small(or)
low
==>The Retentivity and
coercitivity is high
127

24. ==>The BH curve has a
small area
==>The BH curve has a
large area
==>The suceptibility and
permeability is very high
==>The suceptibility
and permeability is low
==>it exhibits the high
resistivity(ρ)
==>it exhibits the low
resistivity(ρ)
==>They required low
magnetic field for
magnetisation
==>They required high
magnetic field for
magnetisation and
demagnetization
==>Loss of wastage is
small
ex:-iron-silicon alloys
==>Loss of wastage are
high
ex:-carbon steels,
tungesten steel,alnico
etc.,
Uses:- Uses:-
==>in motors,
relays&sensors
==>microwave isolators
==>in making of
permanent magnets
==>dc motors and
measuring devices
128

25. Applications of Magnetic materials
==>Ferromagnetic materials are used in magnetic
recording devices,such as for cassette tapes,floppy
discs for computers,and the magnetic strine on the
back of credit cards
==>Diamagnetic materials are used for magnetic
levitation,where an object will be made to float in
are above a strong magnet
==>Magnetic soft materials are used in making
electromagnets and these electromagnets are used
in telephone receiver,bells,loud speakers etc.,
==Magnetic hard materials are usedd in making
prmanent magnets
Solved problems
1)The magnetic field in the interior of a certain
solenoid has the value of 6.5*10⁻⁴T.when the solenoid
is empty.when it is filled with iron,the field becomes 1.
4T.find the relative permeability of iron
Ans:-
Given that
B₀=6.5*10⁻⁴Tesla
B=1.4Tesla
w.k.t, μᵣ=B/B₀
=1.4/6.5*10⁻⁴
μᵣ=2153.85
129

26. =1.4/6.5*10⁻⁴
μᵣ=2153.85
2)Find the relative permeability of a
ferromagnetic material if a field of strength
220amp/meter products a magnetisation
3300A/m in it
Ans:-
Given that
M=3300A/m
H=220A/m
w.k.t, the relative permeability,we have
μᵣ=m/H +1
=3300/220 +1
= 15+1
μᵣ=16
3)A magnetic material has a magnetisation
3300A/m and flux density of 0.0044wb/m².
calculate the magnetising force and the relative
permeability of the material
Ans:-
Given that
M=3300A/m
B=0.0044wb/m²
130

27. w.k.t, B=μ₀[H+M]
H=B/μ₀ - M
μ₀=4π*10⁻⁷H/m
H=0.0044/4*3.14*10⁻⁷ - 3300
H=203A/m
μᵣ=M/H + 1
=3300/200 + 1
=16.5+1
μᵣ=17.5
4)A circular loop of a copper having a diameter
of 10cm carries a current of 500mA.calculate the
magnetic moment associated with the loop
Ans:-
Given that
d=10cm
r=d/2
r=5cm
=0.05m
I=500mA
=0.5A
w.k.t, magnetic moment is
μ=Al
=πr²I
=3.14*(0.05)²*0.5
μ=3.93*10⁻³Am²
131

28. 5)An electron in a hydrogen atom circulates
with a radius of 0.052nm.calculate the change
in its magnetic moment if a magnetic
induction(B)=3wb/m² acts at right angles to
the plane of orbit
Ans:-
Given that
r=0.052nm
=0.052*10⁻¹⁰m
B=3wb/m²
Δμ=e²r²B/4m
=[1.6*10⁻¹⁹]²*[0.52*10⁻¹⁰]²*3/4*9.1*10⁻³¹
=2.56*02704*3*10⁻⁵⁸/4*9.1*10⁻³¹
=5.705*10⁻²⁹ Am²Δμ
6)Calculate the change in magnetic moment of a
circulating elctron in an applied field of 2Tesla acting
perpendicular to the plane of the orbit.given r=5.
29*10⁻¹¹m for the radius of the orbit
Ans:-
Given that
B=2Tesla
r=5.29*10⁻¹¹m
w.k.t, Δμ=e²r²B/4m
=(1.6*10⁻²⁹)²*(5.29*10⁻¹¹)²*2/4*9.1*10⁻³¹
132

29. 133
Δμ=3.9362*10⁻²⁹Am²
7)A paramagnetic material has 10²⁸ atoms per m³.its
suceptibility at 350k is 2.8*10⁻⁴.calculate the
suceptibility at 300k
Ans:-
Given that
N=10²⁸ atoms/m³
χ₁ at 350k, χ₁=2.8*10⁻⁴
T₁=350k
T₂=300k
At 300k,χ₂=?
w.k.t, χ=C/T
==>c=χ
χ₁T₁=χ₂T₂
χ₂=χ₁T₁/T₂
=2.8*10⁻⁴*350/300
χ₂=3.267*10⁻⁴
8)When a magnetic material is subjected to a
magnetic field of intensity 250A/m,its relative
permeability is 15.calculate its magnetisation
and magnetic flux density
Ans:-

30. 134
Given that
H=250A/m
μᵣ=15
w.k.t, M=H[μᵣ-1]
=250[15-1]
=3500A/m
magnetic flux density
B=μ₀[H+M]
=4π*10⁻⁷[250+3500]
B=4.71*10⁻³wb/m²
9)Calculate magnetic moment per unit volume and
flux density of a magnetic material placed in a
magnetic field of intensity 1000A/m.the magnetic
suceptibility is -0.42*10⁻³
Ans:-
Given that
H=1000A/m
χ=-0.42*10⁻³*1000
=-0.42 A/m
flux density D=μ₀(H+M)
=4π*10⁻⁷*[1000-0.42]
D=1.257*10⁻³ wb/m²

36. 135
Dell operator
The elctromagnetic field theory deals with the field
vectors i.e, electric field and magnetic field.In
general the vector field may be a functions of space
and time .the differential vector ∇ called
Dell(or)nabla is a factor space functions
==>it can be expressed as
∇=i∂/∂x+j∂/∂y+k∂/∂z
Gradient of a scalor field(grad)
Gradient is a differential operator(∇) which can
be performed on a scalar field to change it into a
corresponding vector field
ex:-Let φ(x,y,z) be a scalar function of three
coordinate axis x,y&z.the partial derrivates of along
three coordinate axis ∂φ/∂x ,∂φ/∂y ,∂φ/∂z
==>Therefore gradient of a scalar field can be
expressed as
grad(φ)=(i∂/∂x +j∂/∂y +k∂/∂z)φ(x,y,z)
∇φ=i∂φ/∂x +j∂φ/∂y +k∂φ/∂z
here,φ is a scalar&∇ is a vector
∇φ is a vector quantity and also having both
magnitude and direction

37. 136
Divergence of a vector field
The differential operator(∇) when operator on a
vector field change it into a scalar field.the scalar
(or) dot product of a operator with a vector A is
called divergence of vector field
==>It is represented by "∇.A"
ex:- Let us consider a vector
A=iAₓ+jA +kA
ʸ ᶻ
then the divergence of A
div.A=(i∂/∂x + j∂/∂y +k∂/∂z)(iAₓ+jA +kA )ʸ ᶻ
∇.A = ∂Aₓ/∂x+∂A /∂y+∂A /∂zʸ ᶻ
The amount of "flux per unit volume" diverging
from that point is reffered as "divergence of a vector
field"
Curl of a vector field
The curl of a vector field is defined as the maximum
line integral of a vector per unit area
If A is a vector function
A=iAₓ+jA +kAʸ ᶻ
therefore the curl can also be expressed as
curl(or) "∇*A"

38. 137
∇*A= i j k
∂/∂x ∂/∂y ∂/∂z
Aₓ A A
ʸ ᶻ
∇*A=i(∂A /∂y - ∂A /∂z) - j(∂A /∂x - ∂Aₓ/∂z)
+k(∂A /∂x -∂Aₓ/∂y)
ᶻ ʸ ᶻ
ʸ
∇*A=i∂A /∂y - i∂A /∂z - j∂A /∂x+j∂Aₓ/∂z+k∂A
/∂x - k∂Aₓ/∂y
ᶻ ʸ ᶻ ʸ
Fundamental laws of
Electromagnetism(E&B)
a)Gauss law in electric field
The total normal electric flux φₑ over a closed surface
is equal to the 1/ε₀ times of the total charge within the
surface
φₑ=q/ε₀ ----> 1
But φₑ=∫E.ds ---->2
from 1&2
∫E.ds=q/ε₀

39. 138
b)Gauss law in magnetic field
The net magnetic flux "φ " through the surface
normal is zero i.e, φ =0 ----> 1
ᴮ
ᴮ
But φ =∫B.ds ----> 2
from 1&2
∫B.ds=q/ε₀
c)Faraday laws of electromagnetic
induction
The induced emf in a closed loop is equal to
negative rate of change of magnetic flux is called
faradays laws
e=-∂φ /∂t ----> 1
e=∫E.dl ----> 2
from 1&2
∫E.dl=-∂φ /∂t
ᴮ
ᴮ
d)Amperes law
It is gives the relation between magnetic induction
and current(i) of conductor in a closed loop
This law states that "the line integral of magnetic
field over a closed path (or) loop is equa to the (μ₀)
times of the current i.e,

40. 139
∫B.dQ = μ₀i
Gauss Theorem
statement:-
This theorem states that the surface integral of a
vector A over the closed surfaces ie equal to the
volume integral of the divergence of the vector field
over the volume 'v' enclosed by the surface
∫∫A.ds = ∫∫∫(divA)dv =∫(∇.A)dvᵥ
Proof:-
^
>
s
r
ds
x
y
z

41. 140
Let us consider a closed surface 's' of any irregular
shape drawn in a vector field 'A' as shown in the
figure
Let us consider a surface 's' enclosed a volume 'v'.
Now the volume is divided into a large number of z
cubical volume elements of "dv".w.k.t, the
divergence represents the amount of flux diverging
per unit volume and hence a small volume of
element dv will be
divA.dv
therefore,the total flux from the entire volume is
given by
∫∫∫divA.dv ----> 1
consider a small element of area "ds" on the
surface has shown in the figure.let n̂ be the unit
positive vector outward normal of area ds.if the
field vector A and outward normal n̂ makes an
angle "θ" then the component of A is given by the
Acosθ=A.n̂
where n̂=1 [AB=|A||B|cosθ]
the total flux "A.ds" through the entire surface 's' is
given by
__
∫∫A.ds ----> 2

42. 141
This must be equal to the total flux diverging from
the whole volume v enclosed by the surfaces
from eq(1)&(2)
∫∫A.ds = ∫∫∫(div A)dv =∫(∇*A)dv
Stokes law
Statement:-
This theorem states that the line integral of a vector
field around a closed curve is equal to the surface
integral of curl A over the surfaces "s" surrounded by
the closed curve
∫A.dl =∫(curl A)ds =∫(∇*A)ds
ˡ ˢ ˢ
Proof:-
^
>
n̂
curl A
ds
s
B
dl
A

43. 142
Let us consider a surface "s" enclosed in a vector
field "A" as shown in figure.The line integral around
the surface is
∫A.dl ----> 1
Let us consider the surface be divided into larger
no.of square loops of small surface "ds" as shown in
figure B.Let n̂ be the unit positive vector outward
normal on "ds".Thus the vector area of the element
is
n̂.ds=ds [ n̂=1]
we know that the curl of a vector field at any point
is the maximum line integral per unit area along the
boundary of small surfaces ds.hence the line
integral of A around the boundary of the surface ds
is
curl A.ds
hence,the total line integral of A around the
boundary of all area elements is given by the
∫∫curl A.ds ----> 2
This is the surface integral of A
The eq(2) must be equal to the sum of the line
integral on the boundary lines of the surface by
eq(2) hence

44. 143
∫A.dl = ∫curl*A ds = ∫(∇*A).ds
l s s
Hence,the stokes law is proved
"This theorem states that the it convert the surface
integral into line integral and vice versa"
Whenever the curl A is zero then the line integral
of A over a closed path is also zero.then the field is
called as conservative field.
whenever the curl A is zero then the line integral
of A over a closed path is also zero.then the field is
called as consertive field
Maxwell equations
Maxwell formulated the bask laws of electric field
and the magnetic field are the four fundamental
equations are known as "Maxwell equations"
The integral form of maxwell equations is given by
the
∫E.ds =q/ε₀
∫B.ds=0
∫E.dl=-dφ /dt
∫B.dl=μ₀i
ᴮ

45. 144
The maxwell equations can also stated in
diferentiate form
∇.E=ρ/ε₀
∇.B=0
∇*E=-∂φ /∂t
∇*B=μ₀(J+ε₀ ∂E/∂t)
ᴮ
Maxwell 1ˢᵗ equation:-
The total electric flux is equal to the 1/ε₀ times to the
enclosed by the charge
∫E.ds = q/ε₀ ----> 1
Let q be the charge density of a small volume dv
q=∫ρdv ----> 2
Now substitute 2 in 1
∫E.ds=1/ε₀∫ρdv
On applying the Gauss-divergence
v
∫A.ds = ∫(∇.A)dv
s ᵛ
∫(∇.E)dv = 1/ε₀∫ρdv
ᵛ
∇.E=ρ/ε₀ ----> 3

46. 145
This is the differential form of the maxwell first
equation
w.k.t, D=ε₀E
D=elctric displacement vector
ε₀(∇.E)=ρ
∇(ε₀E)=ρ
∇.D=ρ
Maxwell 2ⁿᵈ equation:-
The total magnetic flux is zero in a normal surface
∫B.ds=0
On applying the Gauss divergence theorem
∫A.ds = ∫(∇.A)dv
∫(∇.B)dv=∫0 dv
∇.B=0 ----> 4
ˢ ᵛ
ᵛ
This is the differential form of maxwell 2ⁿᵈ
equation

47. 146
Maxwell 3ʳᵈ equation:-
The induced emf is equal to the negative rate of
change of magnetic flux(φ )
ᴮ
i.e, ∫E.dl = -∂φ /∂t
ᴮ
w.k.t, φ =∫B.ds
∫E.dl = -∫ ∂φ/∂t B.ds
∫E.dl = -∂/∂t ∫B.ds
l
ᴮ
By applying the stokes law
∫A.dl = ∫(∇*E)ds
∫(∇*E)ds = -∂/∂t ∫B.ds
∇*E = - ∂B/∂t ----> 5
ˢₗ
ₛ
The above equation represents the differential
form of the 3ʳᵈ equations
Maxwell 4ᵗʰ equation:-
The line integral of the magnetic field is equal to the μ₀
times of the current(i)
∫B.dl = μ₀i

48. 147
w.k.t, J=i/A
i=Jds
i value in above equation
∫B.dl = μ₀ ∫Jds
ₗ ₛ
∫A.dl = ∫(∇*A)ds [By stokes law]
ₗ ₛ
∫(∇*B)ds = μ₀∫Jdsₛ ₛ
∇*B=μ₀J ----> 6
This equation gives that amperes law is
incomplete and it is not valid for both steady state
and time varying field
The maxwell suggested the Amperes law must be
modified by the adding displacement current(J )
and it is valid for both steady state time varying field.
This is known as displacement current
vector(or)modified amperes law
∇*B = μ₀(J+J ) ----> 7
ᵈ
ᵈ
By taking the divergence on both side
∇.(∇*B) = ∇.(μ₀(J+J )
w.k.t, the divergence of any curl vector is zero
∇.(μ₀(J+J )=0
ᵈ
ᵈ

49. 148
μ₀(∇.J + ∇.J )=0
∇.J=-∇.J -----> 8
ᵈ
ᵈ
w.k.t, According to the equation of continity
∇.J = -∂ρ/∂t
eq(8) becomes
-∇.J =-∂ρ/∂t
∇.J =∂ρ/∂t ----> 9
ᵈ
ᵈ
w.k.t, ρ=∇.D
∇.J =∂ρ/∂t
∇.J =∂/∂t(∇.D)
∇.J =∇ ∂D/∂t
ᵈ
ᵈ
ᵈ
J =∂D/∂t=ε₀∂E/∂t
ᵈ
[D=ε₀E]
J =ε₀∂E/∂t
ᵈ
----> 10
Now substitute 10 in 7
∇*B=μ₀(J+ε₀∂E/∂t)
This is the differential form of the maxwell 4ᵗʰ
equation

50. 149
Propagation of electromagnetic waves
through conducting and
non-conducting media
According to the maxwell equations are given
by the
∇.E=P/ε₀
∇.B=0
∇*E=-∂Β /∂t
∇*B=μ₀(J+ε₀ ∂E/∂t)
----> 1
Let us apply the maxwell equations to
homogeneous,isotropic and conducting media of
permitivity(ε) and permeability(μ)
J=σE
B=μH
D=ε₀E
As in the case of conducting media there is no net
charge within a conductor over the surface i.e,
ρ=0
∇.E=0 ----> 2
∇.B=0 ----> 3

51. 150
∇*E=-∂Β/∂t ----> 4
∇*B=μ(J+ε ∂E/∂t)
∇*B=μ(σΕ+ε ∂E/∂t) ---- 5
On taking the curl on both sides of equations 4 & 5
we get the equations of maxwell equations for
conducting media from 5
∇*B=μ(σΕ+ε ∂E/∂t)
Now take "curl" on both side
∇*∇*B=∇*(μ(σΕ+ε ∂E/∂t)
=μσ(∇*E)+με ∂/∂t(∇*E)
from eq(1) we know that
∇*E=-∂Β/∂t
∇*∇*B=∇*(μ(σΕ+ε ∂E/∂t)
∇*∇*B=μσ(-∂Β/∂t)-με ∂/∂t(∂Β/∂t)
∇*∇*B=-μσ ∂B/∂t-με ∂²B/∂t² ----> 6
w.k.t ∇*∇*B=(∇.B)∇ -(∇.∇)B
=(∇.B)∇ -∇²B [∇.B=0]
∇*(∇*B)=-∇²B ----> 7
from eq (6) and (7)
-∇².B=-μσ∂B/∂t - με ∂²B/∂t²
∇².B=μσ∂B/∂t + με∂²B/∂t² ----> 8

52. 151
This equation gives that maxwell equation for
conducting media in terms of magnetic field vector
==>In the case of non-conducting media which
offers the infinite resistance then conductivity is
zero
σ=0
then eq(8) becomes
∇²B=με ∂²B/∂t² ----> 9
this equation gives that maxwell equation for
non-conducting interms of magnetic field vector
Now again by taking
∇*E=-∂B/∂t
Now take "curl" on both side
∇*(∇*E)=-∇*∂B/∂t
=-∂(∇*B)/∂t
=-∂[μ(J+ε ∂E/∂t)]/∂t [ from 5]
=-∂[(μ(σE+ε ∂E/∂t)]
∇*(∇*E)=-μσ ∂E/∂t -με ∂²E/∂t²
∇*(∇*E)=(∇.E) -(∇.∇)E
=(∇.E)∇-∇²E
=-∇²E
[ ∇.E=0]
-∇²E=-μσ∂E/∂t - με ∂²E/∂t²
∇²E=μσ∂E/∂t + με ∂²E/∂t² ----> 10

53. 152
This equation gives that the maxwell equation for
conducting media interms of electric field vector
In the case of an non conducting media which
offers the infinite the conductivity is zero
σ=0
then eq(10) becomes
∇²E=με ∂²E/∂t²
from eq(8) and (10) are the general equations for
electromagnetic waves in conducting medium
with 1ˢᵗ and 2ⁿᵈ order derivativesPropagation of electromagnetic
waves through dielectric medium (or)
equation of electromagnetic waves in
free space
∇.E=P/ε₀
∇.B=0
∇*E=-∂Β /∂t
∇*B=μ₀(J+ε₀ ∂E/∂t)
According to the equations of the maxwell is given
by the
1

54. 153
Let us apply the maxwell equations to
homogeneous isotropic and dielectric media
w.k.t J=σE
B=μH
D=ε₀E
In the case of dielectric medium which offers the
infinite resistance then conductivity is zero.Also
there is no distribution of charges near the volume
σ=0 (or) J=0 & ρ=0
∇.E=ρ/ε₀=0 ----> 2
∇.B=0 ----> 3
∇*E=-∂B/∂t ----> 4
∇*B=με∂E/∂t ----> 5
By taking the "curl" on both side 5 & 4
we get the equation for electromagnetic waves in
dielectric medium eq(4)
∇*E=-∂B/∂t
Take curl on both side
∇*(∇*E)=-∇*∂B/∂t
=-∂(∇*B) /∂t
∇*(∇*E)=-∂(μεE∂E/∂t)/∂t [ From 5]
∇*(∇*E)=-με∂²E/∂t²
(∇.E)∇-∇²E=-με∂²E/∂t²

55. 154
-∇²E=-μεE ∂²E/∂t² [ ∇.E=0]
∇²E=με∂²E/∂t² ----> 6
This equation gives that the maxwell equation in
dielectric medium in terms of electric field vector
for eq(5) take "curl" on both side
∇*(∇*B)=∇*(με∂E/∂t
=με∂(∇*E)/∂t
=με∂/∂t(-∂B/∂t) [ from 4]
(∇.B)∇-∇²B=-με ∂²B/∂t²
-∇²B=-με ∂²B/∂t² [ from 3]
∇²B=με ∂²B/∂t² ----> 7
From the eq(6)&(7) shows that variation of electric
field and magnetic field.this equations are
compared to the general wave equation.the
differential equation for general wave equation is
given by
∇²y=1/v² ∂²y/∂t² ----> 8
from (7) & (8)
1/v² =με
v=1/√με

56. 155
In free space,
v = 1/√μ₀ε₀
μ₀=4π*10⁻⁷ ; ε₀=1/4π*10⁹*9
v =1/√4π*10⁻⁷*1/4π*10⁹*9
v=3/√1/10¹⁶
v=3*10⁸ m/s
hence,electromagnetic waves of velocity travels
with the speed of velocity of light

57. 156
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62. 157
Optical fibre:-
The guiding medium or material which guides the
information carrying light waves is called the optical
fibre
==>it is a thin and transparent
==>it is used in the communication sysytem to
transfer the data
Transmitter information
channel
Recieve> >
"electrical signal
to optical signal"
"Optical signal to the
electrical signal"
Structure of the optical fibre:-
n₂ cladding
n₁>n₂

63. 158
An optical fibre has three co-axial regions.
they are:
a)core
b)cladding
c)sheet
a)Core:-
The innermost region of an optical is called the core.it
is made up with glass possessing high refractive
index(n₁).the core carries light
b)Cladding:-
==>The core surrounds by cylindrical shell called
cladding
==>The refractive index of cladding(n₂) is always lower
than the that of the core i.e n₂<n₁ (or) n₁>n₂
==>The cladding helps to kee the light within the core
through the phenomenon of the total internal reflection
c)Sheet:-
==>The outermost region of an optical fibre is called
the sheet.
==>The sheet protects the cladding and core from
obiasions contamination an the harmful influence of
moistructred.

64. 159
==>Sheet increases the mechanical strength of
the fibre
Cross sectional
area of optical
fibre
Working of the optical fibre:-
---------------------------------------------------------------
---
n₂(cladding)
n₁(core)
(n₁>n₂)
r

65. 160
The optical fibre based on the principle of the total
internal reflection.when the light is incident on the
one end of the fibre at small angle,it passes through
the fibre as explained above
Let i and r be the angles of incidence and angle of
refraction of the light with the acis.Let "θ" be the angle
of incident on the interference of core and cladding
(θ=90-r).Let n₁ and n₂ are the refractive index of the
fibre of core and cladding respectively(n₁>n₂)
from the snell's law
sini/sinr = n₂/n₁ ----> 1
At the interference of core and cladding if θ>θᶜ
sinθ /sin90 = n₂/n₁
ᶜ
sinθ = n₂/n₁
θ =sin⁻¹(n₂/n₁)
ᶜ
ᶜ
==>To undergoes the total internal reflection the
angle of incident of core(θ) should be greater than
the critical angle(θ ) i.e (θ>θ ) undergoes the total
internal reflection.thus,the light ray is guided
through the fibre from one end without any energy
being lost due to refraction.
ᶜᶜ

66. 161
Acceptance angle & Numerical
apearture
Acceptance angle(θₐ):-
The angle of incident(θₐ) for which refracted ray
at the interference of core and cladding reflect back
to the same medium is called acceptance angle.the
zone or medium at which light enters into the fibre
is called the Launching zone
------------------------------------------------------------------
n₂ cladding
n₀
ᵢ
axis of the fibre (or) Launching zone
Total internal
reflection
n₁ core
μ=1 (refractive
index of air)
Let us consider the refractive index of the core be n₁
and that of cladding n₂(n₁>n₂).Let n₀ be the refractive
index of the launching zone.The light ray enters the
fibre at angle i to the axis of the fibre.the ray refracts
at an angle θ.if θ>θ the ray undergoes total internal
reflection
ᶜ

67. 162
According to the snell's law
μ=sini/sinr
n₁/n₀ =sini/sinr
n₀sini=n₁sinr ----> 1
from ΔABC, r=90-θ
sinr=sin(90-θ)
sinr=cosθ ----> 2
Now substitute 2 in 1
n₀sini = n₁cosθ ----> 3
w.k.t, i=iₘₐₓ ; θ=θ
n₀siniₘₐₓ = n₁cosθ
ᶜ
ᶜ
[θ =critical angle]
ᶜ
siniₘₐₓ = n₁/n₀ cosθ
since, sin(θ )=n₂/n₁ᶜ
ᶜ
siniₘₐₓ=n₁/n₀ √1-n₂²/n₁²
siniₘₐₓ=√n₁²-n₂²/n₁ (n₁/n₀)
siniₘₐₓ=√n₁²-n₂²/n₀
where, n₀=refractive index of axis of fibre in air
n₀=1
siniₘₐₓ=√n₁²-n₂²
iₘₐₓ=sin⁻¹(√n₁²-n₂²)
In the otherwords we can say that,the maximum
angle of light which enters into the axis of fibre is
acceptance angle(θ )
ᶜ

68. 163
Numerical apearture:-
The sine of the acceptance angle is called the
numerical apearture
(or)
The light gathering of the optical fibre is called NA
NA=sin(iₘₐₓ)
w.k.t, acceptance angle iₘₐₓ=sin⁻¹(√n₁²-n₂²)
NA=sin(sin⁻¹(√n₁²-n₂²))
NA=√n₁²-n₂²
==>The numerical apearture for short communication
is 0.4 to 0.5
ex:- LED's
==>The NA for long communication is 0.1-0.3
ex:- LASER etc.,
The fractional change of refractive index can be
defined as the
Δ=n₁-n₂/n₁
Δn₁=n₁-n₂ ----->1
NA=√n₁²-n₂²
=√(n₁+n₂)(n₁-n₂)
=√(n₁+n₂)Δn₁
=√2n₁ Δn₁ [ n₁=n₂]
=√2n₁²Δ

69. 164
NA=n₁ √2Δ
Classification of optical fibre
==>These are classified based on the no.of modes
are two types.they are:
a)single mode optical fibre
b)multi mode optical fibre
a)Single mode optical fibre
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core(5μm)
cladding
(70μm)

70. 165
==>In an single mode fibre(smf),core diameter is
5μm-100μm and cladding diameter is 70μm
==>In between the core and cladding the refractive
index is very small
==>In an smf,only one single ray-mode propagate in
optical fibre
==>It is used in long communication
ex:-Laser,telephone lines etc.,
b)Multi mode optical fibre
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core(40μm)
claddin
(70μm)

71. 166
==>In an multi mode optical fibre(mmf),core
diameter is 40μm and cladding is 70μm
==>In between the core&cladding the refractive
index is very large
==>In an mmf,the no.of rays can propagate in
optical fibre
==>It is used in short communication with
disortion(or)Dispersion
ex:- LED,LAN Network etch
On the basis of Refractive index
==>On the basis of Refractive index,these are
classified into two types.they are:
a)step indes fibre
b)grade index fibre
a)Step index fibre
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core
cladding
a
b

72. 167
--------------------------------------------------------------
axis of the fibre
Low state
input
signal
meridional ray
output signal
==The step index optical fibre is also called as the "multi
mode step index fibre"
==>In step index optical fibre,the core has a uniform
refractive index(n₁) and the cladding has also a uniform
refractive index(n₂) such that n₁>n₂
==>The diameter of the core in step index fibre is 50μm
and the diameter of the cladding in step index fibre is
110μm
==>In the step index fibre,we get the signal distortion
carried by the distorted output signal
==>LED's are used in short communication
==>The meridonial rays are used in step index fibre
==>The rays are difference angle of incidence,path
difference and with different time period

73. 168
==>The no.of possible propagation modes in the
core is given by the v-number as
v=2π/λ a(NA)
where, λ=wavelength of light
a=radius of the core
NA=Numerical apearture
==>NA is more for multimode step index fibre
==>It is of reflective type
==>The no.of modes through the step index fibre
=v²/2
b)Grade index optical fibre
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core
cladding
a
b

74. 169
==>In the grade index fibre,the core has the
non-uniform refractive index and the cladding has
the uniform refractive index
==>In the grade index fibre,the diameter of the core
is 50μm and diameter of the cladding is 110μm
==>when a light ray enters into the core and moves
towards the cladding interface there is a decrease
of refractive index.then the light ray bends away
from the normal path and finally bends towards the
axis and move towards core-cladding interface as
shown in figure
==>thus,we can say that "this fibre is of refractive
type".
==>there is very less distortion in grade index fibre
==>the no.of possible modes through graded index
fibre is v²/4
==>NA is less for grade index fibre

75. 170
Fibre optic communication
system
An efficient optical fibre communication system
requires high frequency and high information
carrying capacity,fast operating speed over along
distances without loss of energy.An optical fibre
communication sysytem main consist of the
following classification as shown below
Encoder
transmitter
Drive
circuit
optical
signal
photo
detector
signal
restores
Decodes
>
Analog i/p
Analog o/pAmplifiers
Connector

76. 171
a)Encoder
b)Transmitter
c)Wave guide
d)Receiver
e)Decoder
a)Encoder:-
It is an electronic device which converts the analog
informaion like voice,images,pictures and vedious
etc., into the binary codes 0's and 1's is called
encoder
b)Transmitter:-
In the case of transmitter,there were two parts
drive circuit and optical signal.the drive circuit
supplied into the light signals from the encoder.the
light signal converts electrical signal into the optical
signal.
By using the specially made connector the optical
signal will be propagate into the wave guide from
the transmitter.
==>In this case of transmitter,the electrical signal
is converted into the optical signal

77. 172
c)Wave guide:-
It is an optical fibre which carries the information
in the form of optical signal over along diatances.
By using of the specially made connector again the
optical signal received by the receiver
d)Receiver:-
In the Receiver,there were three parts.they are:
1)Photo detector
2)Amplifier
3)Signal restorer
1)Photo detector:-
The photo detector converts the optical signal into
the electrical signal and it supplies to the Amplifier
2)Amplifier:-
The Amplifier ampilfies the electrical signals as
they become weak during a long distances through
the wave guide
3)Signal restorer:-
The signal restorer helps the electrical signal in a
sequential order and supplies to the decoder

78. 173
e)Decoder:-
It converts the electrical signals into the analog
information
Advantages of optical fibres in
communication
==>potential of delivering signals at low cast
==>much safer than copper cables
==>it has longer life span
==>high temperature resistance
==>more reliable and easier to maintain than copper
cables
In medicines:-
==>They are used in photo dynamic therapy for
cancer
==>They are used in the treatment of lung disorders
==>They are used in the treatment of bleeding ulcers
==>They are used in the investigation of the heart,
respiratory system and pancreas
Applications of optical fibres in
communication:-
==>optical fibres are extensively used in optical
communication systems

79. 174
==>Nearly 10,000 information carries signals can be
transmitted simultaneously through optical fibres
==>Due to higher band-widths,optical fibres carry
more information
==>As the optical fibre is highly immune to
temperature moisture etc., the information can be
delivered without any environmental effects
==>without any crosstalk,the information can be
safely delivered
==>during wars,they are used for secret
communication
==>they are used for guiding weapons and
submarine communication systems
Solved Problems
1)An optical fibre has a core material of
refractive index 1.55 and the cladding
material of refractive index 1.50 and light is
launched into it in air.calculate its numerical
apearture.
Ans:-
Given that
n₁=1.55
n₂=1.50
The numerical apearture of the optical fibre
NA=√n₁²-n₂²

80. =√1.55²-1.50²
=√2.40-2.25
=√0.15
NA=0.39
2)Calculate the angle of acceptance of a given
optical fibre if the refractive indices of the core
and cladding are 1.563 and 1.498 respectively
Ans:-
Given that
n₁=1.563
n₂=1.493
w.k.t, θₐ=sin⁻¹(NA)
NA=√n₁²-n₂²
=√1.563²-1.493²
=√0.1989
NA=0.446
Now, θₐ=sin⁻¹(0.446)
θₐ=26°29'
175
3)Calculate the fractional index change for a
given optical fibre if the refractive indices of
the core and cladding are 1.563 and 1.498
respectively
Ans:-

81. 176
Given that
n₁=1.563
n₃=1.498
the fractional refractive indices change
Δ=n₁-n₂/n₁
=1.563-1.498/1.563
=0.065/1.563
Δ=0.0416
4)The refractive indices of the core and cladding
materials of a step index fibre are 1.48 and 1.45
respectively,calculate
1)numerical apearture
2)acceptance angle
3)the critical angle at the core-cladding interface
and
4)fractional refractive indices change
Ans:-
Given that
n₁=1.48
n₂=1.45
1)Numerical apearture
NA=√n₁²-n₂²
=√1.48²-1.45²

82. 177
=√2.1904-2.1025
=√0.0879
NA=0.2965
2)Acceptance angle(θₐ)=sin⁻¹(NA)
=sin⁻¹(0.2965)
θₐ=17°14'
3)Critical angle(θ )=sin⁻¹(n₂/n₁)
=sin⁻¹(1.45/1.48)
=sin⁻¹(0.9797)
θ =78°26'
5)The numerical apearture of an optical fibre is 0.
39.if the refractive indices of the material of its
core and the cladding is 0.05,calculate the
refractive index of the material of the core
Ans:-
Given that
NA=0.39
n₁-n₂=0.05 ----> 1
w.k.t, NA=√n₁²-n₂²
NA=√(n₁-n₂)(n₁+n₂)
0.39=√0.05*(n₁+n₂)
n₁+n₂=(0.39)²/0.05
=0.1521/0.05
n₁+n₂=3.042 ----> 2

83. 178
Now adding the 1 + 2 ==>
n₁-n₂=0.05
n₁+n₂=3.042______________
2n₁=3.092
n₁=3.092/2
n₁=1.546
The refractive index of the core is 1.546
6)Calculate the refractive indices of core and
cladding of an optical fibre with a numerical
apearture of 0.33 and their fractional difference
of refractive indices being 0.02.
Ans:-
Given that
Numerical apearture NA=0.33
Δ=0.02
Δ=n₁-n₂/n₁ =0.02
w.k.t, NA=n₁√2Δ
n₁=NA/√2Δ
=0.33/√2*0.02
n₁=1.1
from the above equation
0.02=1.1-n₂/1.1
n₂=1.1-0.02
n₂=1.078

84. 179
7)An optical fibre has a numerical
apearture of 0.20 and a cladding refractive
index of 1.59.find the acceptance angle for
the fibre in water which has a refractive
index of 1.33
Ans:-
Given that
NA=0.20
n₂=1.59
n₀=1.33
w.k.t, NA=√n₁²-n₂²
0.20=√n₁²-(1.59)²
(0.20)²=n₁²-(1.59)²
n₁²=(0.20)²+(1.59)²
n₁=√2.57
n₁=1.6031
sinθₐ = √n₁²-n₂²/n₀
θₐ =sin⁻¹(√n₁²-n₂²/n₀)
Now
√n₁²-n₂²/n₀ = √(1.6031)²-(1.59)²/1.33
=√2.5699-2.5281/1.33
=√0.0418/1.33
=0.2045/1.33
=0.1538

85. 180
θₐ=sin⁻¹(0.1538)
θₐ=8°50'
8)An optical fibre has a diameter of 60μm and
numerical apearture of 0.45.calculate the
number of modes for an operating wavelength of
10μm
Ans:-
Given that
d=60μm
a=d/2
=30μm
=30*10⁻⁶m
NA=0.45
λ=10μm
=10*10⁻⁶m
w.k.t v=2π/λ a(NA)
=2π*30*10⁻⁶*0.45/10*10⁻⁶
v=9.21
Number of possible modes=9
9)The refractive indices of the core and cladding
of a fibre are 1.48 and 1.45 respectively.the
diameter of the fibre is 60μm.calculate
a)NA
b)Number of possible modes at an operating
wavelength of 1.5μm and

86. 181
c)velocity of light rays in the core and cladding
of the fibre.
Ans:-
Given that
n₁=1.48
n₂=1.45
d=60μm
a=d/2
=30μm
=30*10⁻⁶m
λ=1.5μm
=1.5*10⁻⁶m
a)NA=√n₁²-n₂²
=√(1.48)²-(1.45)²
=√(1.48+1.45)+(1.48-1.45)
NA=0.2964
b)Number of possible nodes
v=2π/λ a(NA)
=2π*30*10⁻⁶*0.2964/1.5*10⁻⁶
=37.25
=37
c)Velocity of light in core =Velocity of light in
air/refractive index of core
=3*10⁻⁸/1.48
=2.02*10⁸ m/s

87. 182
d)Velocity of light in cladding=velocity of light in
air/refractive index of cladding
=3*10⁸/1.45
=2.06*10⁸m/s
10)An optical fibre of 1mw is guided into the
optical fibre of length 100m.if the output power
a the other end is 0.3mw,calculate the fibre
attenuation.
Ans:-
Given that
Pᵢₙₚ=1mw
Pₒᵤₜ=0.3mw
length of the fibre L=100m
=0.1km
w.k.t, fibre attenuation
α=10/L log(Pᵢₙₚ/Pₒᵤₜ)
=10/0.1 log(1/0.3)
α=52.29 dB/km
11)The optical power launched into an optical
fibre is 1mw.the fibre has the attenuation of 0.5
dB/km.if the power output is 0.5mw,then
calculate the fibre length.
Ans:-

88. 183
Given that
Pᵢₙ=1mw
Pₒᵤₜ=0.5mw
α=0.5 dB/km
w.k.t, α=10/L log(Pᵢₙ/Pₒᵤₜ )
0.5=10/L log(1/0.5)
L=10/0.5 log(1/0.5)
L=13.8629m
12)A multimode step index fibre with refractive
indices of core and cladding as 1.53 and 1.50,
respectively.if the wavelength of light signal is
1μm,and the radius of the core is 50μm,then
find the number of possible modes through it.
Ans:-
Given that
n₁=1.53
n₂=1.50
λ=1μm
=1*10⁻⁶m
a=50μm
=50*10⁻⁶m
w.k.t, v=2π/λ a(NA)
=2πa√n₁²-n₂²/λ
=2*3.15*50*10⁻⁶*√(1.53)²-(1.50)²/1*10⁻⁶

89. 184
v=94.72
Possible number of modes=v²/2
=4486

94. 185
Semiconductor:-
The whose conductivity is lies in between the
conductors and insulators are known as
semiconductor.
==>The semiconductors are bipolar and current is
transported by the two charge carriers(electrons
and holes)
==>At 0k temperature the semiconductor act as a
insulator
==>Around the temperature the semoliconductor
act as a conductor
Energy Band:-
The closely packed energy levels is known as energy
band.these are classified into two types.they are:
1)conduction band
2)valance band
conduction band
valance band
E
ᵍ

95. 186
1)Conduction band:-
The energy band which is occupied by free
electrons in a solid is called the conduction band.
==>it may be partially filled (or) empty
==>it has lowest unfilled band
==>electrons available for conduction band
2)Valance band:-
The energy band which is occupied by valance
electrons of an atom is called the valance band.
==>it is filled completely(or)partially
==>it has highest filled band
==>non availability of the electrons in valance band
Forbidden energy gap:-
The energy difference between the lowest level of
conduction band and the highest level of valance
band is called the forbidden energy gap
(or)
The portion of the energy band structure if a
semiconductor (or) insulator between the valance
band and conduction band is energy gap.

96. 187
Classification of solids based on the
energy bands
conductor semiconductor insulator
C.B
V.B
C.B
V.B
E =0
ᵍ
E (small)
ᵍ
C.B
V.B
E (large)
ᵍ
==>the energy
band gap is
zero(E =0)
ᵍ
==>the energy
band gap is
small
==>the
energy band
gap is large
==>due to the
overlapping of
bands,the
electrons and
freely moves
from V.B to the
C.B
==>to move the
electrons from V.
B to the C.B
sufficient amount
of energy is
required.
==>the
electrons
cannot move
from valance
band to the
conduction
band.

97. 188
==>due to the
absence of
band gap free
electrons are
available in
the
conduction
band at room
temperature
==>due to the
small band
gap,only few
electrons are
available in
conduction
band at room
temprature
==>due to the
high band,no
electrons are
available in
the C.B at
room
temperature
==>the
electrical
conductivity is
due to the
electrons only
==>electrical
conductivity
may be either
electrons or
holes
depending
upon the type
of
semiconductor
==>there is no
charge carriers
in insulator
==>The semiconductors are classified into the two
types.they are:
a)intrinsic semiconductor
b)extrinsic semiconductor

98. 189
Semiconductor
intrinsic semiconductor extrinsic semiconductor
a)intrinsic semiconductor:-
• • • •
• si • • si • • si • • si •
• • • •
• • • •
• si • • si • • si • • si •
• • • •
• • • •
• si • • si • • si • • si •
• • • •
• • • •
• si • • si • • si • • si •
• • • •

99. 190
==>the pure form of the semiconductors is known as
the intrinsic semiconductor
==>the most important semiconductors
aregermanium(Ge) and silicon(si).each
semiconductor has four valance electrons in its
valance shell.to get stability,each of these atoms
makes the four covalent bonds with the
neighbouring atoms in the semiconductor crystal.At
low temperature all valance electrons are strongs
bound to their atoms and are actively participating in
the covalent bond formation.As we know that the At

100. 191
0k the semiconductor act as a insulator.thus there
is a non availability of electrons in the conduction
band."E " represents the lowest portion of the
conduction band and "E " represents the highest
portion of the valance band.the "E " is the energy
gap between the conduction band and valance
band the E for Ge is 0.07ev and si is 1.2ev
ᵍ
ᶜ
ᵛ
Intrinsic carrier concentration and
fermi energy level
-----------------------------------------------------E
ᶠ
C.B
V.B
E
ᵍ
Eₒ
Eᵥ

101. 192
In an intrinsic semiconductor,due to the each
broken bond leads to the generation of two charge
carriers are electrons and holes.At any temperature T,
the no.of electrons generatedbis equal to the no.of
holes generated.
Let us consider "n" be the no.of electrons and "p"
be the number of holes,then from an intrinsic
semiconductor.
we have
n=p=nᵢ ----> 1
where,nᵢ is called the intrinsic carrier concentration
the electron concentration C.B is
n=N e
ᶜ
(E -E )/k T
ᶜ ᶠ
-
the hole concentration in V.B is
----> 2
p=Nᵥe
ᴮ
-(E -E )/k T
ᶠ ᵛ ᴮ
----> 3
where,N ,N are pseudo-constrants depending on
temperature
ᶜ ᵛ
k =Boltzman constant=1.38*10⁻²³ j/k
ᴮ
T=absolute temperature

102. 193
from 1 ==>
nᵢ²=np
nᵢ²=[N e
ᶜ
-(E -E )/k T
ᶜ ᶠ ᴮ ][N e-(E -E )/k T
ᶠ ᵛ ᴮ
]
nᵢ²=N N e
ᶜ ᵛ
(E +E -E +E )/k Tᶜ ᶠ ᶠ ᵛ ᴮ
nᵢ²=N N e
ᶜ ᵛ
(E -E )/k T
ᵛ ᶜ ᴮ ----> 4
As we know that
E =Eᵥ-E ----> 5ᵍ ᶜ
Now substitute 5 in 4
nᵢ²=N N e
ᶜ ᵛ
-E /k T
ᵍ ᴮ
nᵢ=(N N )
ᶜ ᵛ
1/2
e-E /2k Tᵍ ᴮ
from the above equation we know that
==>the intrinsic carrier concentration is
independent of the fermi energy level
==>the intrinsic carrier concentration is function of
the band gap E
==>the intrinsic carrier concentration depends on
the temperature T

103. 194
b)Fermi energy level:-
At 0k temperatur,the probability of occupation of
energy levels in conduction band and valance band
is called the fermi energy level
==>the fermi energy level lies in the middle of the
energy gap "E "
==>in an intrinsic semiconductor,the no.of electrons
present in conduction band is equal to the no.of
holes present in the valance band
n=p
N e
ᶜ
-(E -E )/k T
ᶜ ᶠ ᴮ
= N e
ⱽ
-(E -E )/k T
ᶠ ᵛ
Nᵥ/Nᶜ
= e
(-E +E )/k T
e
-(E -E )/k T
ᶜ ᶠ
ᶠ ᵛ
ᴮ
ᴮ
ᴮ
Nᵥ/N =
ᶜ
e-E +E +E -E /k T
ᶜ ᶠ ᶠ ᵛ ᴮ
Nᵥ/N =
ᶜ
e
2E -E -E /k T
ᶠ ᶜ ᵛ ᴮ
Now take log on both side

104. 195
log(Nᵥ/N ) =2E -(E +E )/k T
ᶜ ᶠ ᶜ ᵛ ᴮ
k T log(Nᵥ/N )=2E -(E +E )
ᴮ ᶜ ᶠ ᶜ ᵛ
2E =(E +E )/2 +k T log (Nᵥ/N )
ᶠ ᶜ ᵛ ᶜᴮ
we know that
E =E +E /2 +1/2 k T log 1
ᶠ ᶜ ᵛ ᴮ
E =E +E /2ᶜ ᵛ
[log 1=0]
thus,the fermi energy level in intrinsic
semiconductor is always lies between the valance
band and conduction band
==>the fermi energy level is independent of
temperature
Intrinsic conductivity
• • • • •
o o o o o
+ -
electron flow
hole flow
v

105. 196
Let us consider an intrinsinc semiconductor with
the potential "v".it establishes the electric field E
and the charge carriers are moving from opposition
in direction.the drift velocity acquired by the charge
carrier is given by
v =μE
ᵈ
----> 1
where, μ=mobility
Let n be the concentration of electrons in the
semisonductor.then the current density due to an
electron is given by the
Jₙ=nev
Jₙ=neμₙE [from 1] ----> 2
where, μₙ=mobility of an electron
ᵈ
As same as the current density due to an hole is
given by the
Jₚ=peμₚE ----- 3
p=hole concentration
μₚ=mobility of hole
the total current density is given by the
J=Jₙ+Jₚ
J=neμₙE + peμₚE
J=eE[nμₙ+pμₚ] ----> 4

106. 197
But we know that,total current density
J=σE ---- 5
σ=J/E
σ=conductivity
Now substitute 5 in 4
σ=eE[nμₙ+pμₚ]/E
σ=e[nμₙ+pμₚ]
According to the intrinsic semiconductor,we have
n=p=nᵢ
σ=enᵢ[μₙ+μₚ] -----> 6
we have
nᵢ=(N N )
ᶜ ᵛ
1/2
e
-E /2k T
ᵍ ᴮ
----> 7
Now substitute 7 in 6
σ=e(N N ) (μₙ+μₚ) e
ᶜ ᵛ
1/2 -E /2k T
ᵍ ᴮ
where, A=(N N ) (μₙ+μₚ)ᶜ ᵛ
1/2
=constant
σ=Ae-E /2k Tᵍ ᴮ

107. 198
Energy gap of a semiconductor
dx
dy
1/T
logρ
lnA
>
^The energy gap
between the
conduction band and
valance band is
called the energy
band gap(E )(or)
forbidden energy gap
from the fermi energy level,we have
σ=Ae
-E /2k T
ᵍ ᴮ
w.k.t, ρ=1/σ
ρ=1/Ae
-E /2k T
ᵍ ᴮ
ρ=1/A e
E /2k T
ᵍ ᴮ

108. 199
B=1/A =constant
ρ=Be
E /2k T
ᵍ ᴮ
Take log on both side
log ρ = log B+E /2k T
ᵍ ᴮ
log of the resistivity when plotted with 1/T
produces straight line as shown in above figure
from the above figure
E /2k T = dy/dx
ᵍ ᴮ
2)Extrinsic semiconductor
When the impurities are added to an intrinsic
semiconductor then it becomes an extrinsic
semiconductor
Depending upon the type of impurity added to the
intrinsic semiconductor are two types.they are:
a)n-type semiconductor
b)p-type semiconductor

109. 200
a)n-type semiconductor:-
• • • •
• Ge • • Ge • • Ge • • Ge•
• • • •
• • • •
• Ge • • P • • Ge • • Ge•
• • • •
• • • •
• Ge • • Ge • • Ge • • Ge•
• • • •
• • • •
• Ge • • Ge • • Ge • • Ge •
• • • •
•
When the pure semiconductor "Ge" (or) "si" is
doped with the 5ᵗʰA group elements like "P" or "N"
is known as the n-type semiconductor
==>In an n-type semiconductor,the majority charge
carriers are electrons and minority charge carriers
are holes

110. 201
------------------------------------------------------------
C.B
V.B
E (donar energy
level)
ᵈ
o o o o o o o o o o o o o o o
• • • • • • • •
• • • • • • • • • • • • • •
E
ᵍ
E
ᶜ
E
ᵛ
Enerygy band diagram
==>when an impurity atom like phosphorous is
forms the four covalent bands with the four "Ge"
atoms.But we know that 5ᵗʰA group elements has
5 valance electrons.so the fifth electron is left free.
At 0k.this electron is bound to phosphorous with 0.
045ev.the corresponding energy band with the
separation of energy gap.the donar energy level is

111. formed below the conduction band.
202
==>when the temperature(T>0k) is increased the
bound electrons becomes free and move to the
conduction.As a result,the donar energy level get
ionised
==>As the temperature still increases,the breakage
of covalent band releases an electron hole pair i.e,
the electrons moves from valance band to the
conduction band by leaving holes in valance band.
==>Thus,the concentration of electrons in the
conduction band rather than holes.thus we can say
that majority charge carriers are electrons and
minority charge carriers are holes
b)p-type semiconductor:-
When the pure semiconductor "si" is doped with
the trivalent impurities like B,Al,etc.,is known as the
p-type semiconductor
==>In an p-type semiconductor,the majority charge
carriers are holes and minority charge carriers are
electrons

112. 203
• • • •
• si • • si • • si • • si •
• • • •
• • • •
• si • • B • si • • si •
• • • •
• • • •
• si • • si • • si • • si •
• • • •
• • • •
• si • • si • • si • • si •
• • • •
O
hole
----------------------------------------Eₐ
C.B
V.B
Acceptors have
accepted
electrons fron
the V.B
E
ᵍ
• • • • • • • • •
o o o o o o o
E
ᶜ
Eᵛ

113. 204
==>when an impurity atom like Boron(B) is doped
with the "si",then the boran form the three
covalent with the adjacent neighbouring si atoms
and one bond is left a positive hole is created at
this place.since boran is in a position to accept an
electron.it acts as a acceptor.As you seen in the
above figure there is a acceptor energy level(Eₐ)
just above the valance band
==>As the temperature is raised(T>0) the elctrons
from the valance band try to occupy the acceptor
and it gets ionised creating holes in the valance band.
As the temperature is still increases the breakage of
covalent bonds,electrons moves into the C.B leaving
more holes in the V.B
.As a result holes becomes majority charge carriers
and electrons becomes minority charge carriers.
Fermi energy level in extrinsic
semiconductor:-
==>for the n-type semiconductor,the fermi energy
level is given by
E =E -k T log(N /N )
ᶠ ᶜ ᴮ ᶜ ᴰ
where, N =concentration of donar atoms
The fermi lies below the C.B

114. 205
==>for the p-type,the fermi energy level is given by
the
E =E +k T log(Nᵥ/N )
ᶠ ᵛ ᴬ
N =concentration of acceptor atoms
ᴬ
fermi lies above the valance band
Effects of temperature on E
ᶠ
In an n-type semiconductor,as temperature T
increases,more number of electron-hole pairs are
formed.At a very high temperature T,the
concentration of thermally generated electrons in
the conduction band will be far greater than the
concentration of donar elements.In such a case,as
concentration of electrons and holes become equal,
the semiconductor becomes essentially intrinsic
and E returns to the middle of the forbidden energy
gap.Hence,it is concluded that as the temperature
of the p-type and n-type semiconductor increses,E
progressively moves towards the middle of the
forbidden energy gap.
ᶠ
ᶠ
Law of mass action
According to the law of mass action "the product
of the majority and minority charge carrier
concentration at a particular temperature is equal

115. 206
to the square of the intrinsic carrier concentration at
a particular temperature
nₚpₚ=nᵢ²
In intrinsic semiconductor,expression for the
electron and hole concentration are
n=N e
ᶜ
-(E -E )/k T
ᶜ ᶠ ᴮ
p=N e
-(E -E )/k T
ᶠ ᵛ ᴮ
from the intrinsic carrier concentration,electron
concentration is equal to the hole concentration
ᵛ
n=p=nᵢ
np=nᵢ²
nᵢ²= [N e
ᶜ
-(E -E )/k T
ᶜ ᶠ ᴮ ] [ N e
ᵛ
-(E -E )/k T
ᶠ ᵛ ᴮ
]
nᵢ²= N N e
ᶜ ᵛ
-(E -E )/k T
ᶜ ᵛ ᴮ
for n-type semiconductor,the law of mass action
relation can be written as
nₙpₙ=nᵢ² ----> 1
for p-type semiconductor,the law of mass action
can be written as
nₚnₚ=nᵢ² ----> 2

116. 207
from 1 & 2 we observe that,the product of the
majority and minority charge carrier concentration
at a particular temperature is equal to the square
of the intrinsic carrier concentration
Charge densities in n and p-type
semiconductor:-
A semiconductor,whether intrinsic or extrinsic,is an
electrically neutral body in its equillibrium conditio
n-type semiconductor:-
---------------------------------
C.B
V.B
• • • • •
o o o o o
E
ᵈ
ions forms of carrier
concentration
+ve hole forms
In an n-type semiconductor,the total number of
electrons in the conduction band must be equal to
the sum of electrons originated from the donar
atoms and electrons excited from the valance band.
In the donar energy level the ions are formed by

117. 208
leaving the holes in the valance band.
The charge neutrality condition applied to the
n-type semiconductor implies that the total negative
charge of mobile electrons is equal to the total
positive charges created in the crystal,thus
nₙ=N + pₙ
ᴰ
if nₙ>>>pₙ
nₙ≈N
ᴰ
But w.k.t,
nₙpₙ=nᵢ²
pₙ=nᵢ²/nₙ
pₙ=nᵢ²/N
ᴰ
p-type semiconductor:-
pₚ=N +nₚ
ᴬ
if pₚ>>>nₚ
pₚ≈N
ᴬ
According to the law of mass action
pₚnₚ=nᵢ²
nₚ=nᵢ²/Nᴬ

118. 209
Drift and Diffusion
Drift:-
o o o o
o o o o
o o o o
o o o o
e⁻ e⁻ e⁻ e⁻
e⁻ e⁻ e⁻ e⁻
e⁻ e⁻ e⁻ e⁻
e⁻ e⁻ e⁻ e⁻
e⁻
e⁻
e⁻
e⁻
E
-
+
Under the influence of an electric field,the
charge carriers are forced to move in a particular
direction constituting the electric current.this
phenomenons is known as drift.the current in the
semiconductor is called the drift current.

119. 210
Let us consider n be the electrons in a
semiconductor under the electric field E and v be
the different velocity.
ᵈ
the current density is given by the
J=nev ----> 1
ᵈ
where, e=charge of an electron
the conductivity σ=J/E ---->2
=nev /E
ᵈ
According to the drift velocity
v =μₙE ----> 3
ᵈ
μₙ=mobility of electron
Jₙ=neμₙE [ from 3]
σ=J/E
σ=neμₙE/E
σ=neμₙ
ρ=1/σ
ρ=1/neμₙ
In case of a semiconductor,the drift current density
due to electrons is given by the
Jₙ=neμₙE
the drift current density due to holes
Jₚ=peμₚE

120. 211
the total current density is given by the
J=Jₙ+Jₚ
=neμₙE + peμₚE
J=eE[nμₙ+pμₚ]
w.k.t, σ=J/E
=eE[nμₙ+pμₚ]/E
σ=e[nμₙ+pμₚ]
σ=enᵢ [μₙ+μₚ] [ nᵢ=n=p]
Diffusion:- E Diffusion of charge
carriers
>Drifting of
charge
carriers

121. 212
Due to non-uniform carrier concentration in a
semiconductor,the charge carriers moves from a
region of higher concentration region to the low
concentration region,this process is known as
diffusion of charge carriers.
According to the fick's law,the rate of diffusion of
electrons ∝ -∂(Δn)/∂x
=-Dₙ ∂(Δn)/∂x
where, Dₙ=diffusion constant
the diffusion current density due to electrons is
Jₙ=-e[-Dₙ ∂(Δn)/∂x)]
Jₙ=eDₙ∂(Δn)/∂x
the diffusion current density due to holes
Jₚ=e[-Dₚ ∂(Δp)/∂x]
the total current density due to electrons is the
sum of the current densities due to drift and
diffusion
Jₙ=Jₙ(drift) +Jₙ(diffusion)
Jₙ=neμₙE+eDₙ ∂(Δn)/∂x
similarlly,for the holes
Jₚ=peμₚE-eDₙ∂(Δp)/∂x

122. 213
Diffusion
distance in
semiconductor
Δn excess electron density
Hall Effect
z
y
hall volatge(v )
ᴴ+ + + + + + + + +
- - - - - - - - - - -
v
ᴴ Hall electric
field(E )ᴴ

123. 214
when a magnetic field is applied perpendicular to
the current carrying semiconductor,a potential
difference of electric field is developed between the
two points on the opposite sides of the
semiconductor in direction perpendicular to both
current and magnetic field.this is called the Hall
effect.
Let us consider a current "i" is pssed in the
conductor along x-axis and magnetic field B is
applied along y-axis.Due to the magnetic field to
the charge carriers experience a force of F is given
by the flemming left hand rule.
there were two forces acts on it
F =eE
F =Beυ
ᴱ
ᴮ
The electric field force is equal to the magnetic field
force F = F
ᴱ ᴮ
eE =Beυ
ᴴ
ᴴ
v /d =Bυ
E =Bυ
ᴴ
ᴴ
[E =v /d]
ᴴ ᴴ
v =Bυd ----> 1
ᴴ

124. 215
As we know that
J=neυ
i/A=neυ
υ=i/neA ----> 2
Now substitute 2 in 1
v =Bid/neA
ᴴ
v =Bid/nedw
ᴴ [ A=dw]
v =Bi/new
ᴴ
Hall coefficient=1/ne
Hall voltage
v =BiR /w
ᴴ ᴴ
R =v w/Bi
ᴴ ᴴ
the conductivity in a semiconductor due to electrons
is given by
σ=neμ
μ=σ/ne
μ=R σ
ᴴ
where, R =Hall coefficient
σ=conductivity
ᴴ

125. 216
Applications:-
==>Knowing the values of R and σ,mobility 'μ' of
the charge carrier determined.
σ=J/E
=neμₙE/E
σ=neμₙ
R =μ/σ
ᴴ
[R =1/ne]
ᴴ
μ=R σ
ᴴ
==>the hall effect gives the information about the
sign of the charge carriers
==>the carrier concentration can be determined by
knowing R
ᴴ
==>we can measure the conductivity of the given
specimen knowing μ and R
ᴴ
==>we can measure the magnetic field strength B by
measuring the relative quantities

126. 217
Direct and Indirect band-gap
semiconductors:-
==>As shown in the above diagram,the minimum
energy of conduction band and maximum energy of
valance band are have the same value of wave
vector
==>An electron from the conduction band can
recombine with a hole in the valance band directly
emitting a light photon of energy 'hυ'
==>Life time(i.e, recombination time) of charge
carriers is very less
Direct band gap semiconductors:-

127. 218
==>Due to the emission of light photon during
recombination of charge carriers,these are used to
fabricate LEDs and laser diodes
==>these are mostly from the compound
semiconductors
ex:-InP,GaAs etc.,
Indirect band gap semiconductors:-

128. 219
==>As shown in the band diagram,the minimum
energy of the conduction band and maximum
energy of the valance band having different values
of wave vector.
==>An electron from the conduction band can
recombine with a hole in the valance band indirectly
through taps.here,there is emission of photon along
with phonon.the emission of phonon leading to the
rise of temprature of the material
==>Lifetime of charge carriers is more
==>Due to longer life time of charge,these are used
to amplify the signals as in the case of diodes and
transistors.
==>These are mostly from the elemental
semiconductors
ex:-si,Ge etc.,
Applications of semiconductor:-
==>they are used to convert large amounts of power
into electricity for electric rail roads
==>they are used in research laboratories of
electronic instruments to perform tests,
measurements and numerous other experimental
tasks
==>they are used in industrial control systems and
automatic telephone exchanges

129. 220
==>semiconductors are used in pocket calculators,
televisions and portable radius
==>they are used in military equipments,data
display systems,computers and data processing
units
==>they are used in the consumer applications of
space systems,computers and data processing
equipments
Solved Problems
1)The following data are given for intrinsic Ge at 300k,
nᵢ=2.4*10⁻¹⁹m⁻³,μₑ=0.39m²v⁻¹s⁻¹,μₚ=0.19m²⁻¹s⁻¹
calculate the resistivity of the sample
Ans:-
Given that
e=1.6*10⁻¹⁹c
nᵢ=2.4*10⁻¹⁹m⁻³
μₑ=0.39m²v⁻¹s⁻¹
μₚ=0.19m²v⁻¹s⁻¹
T=300k
w.k.t, σ=nᵢe(μₑ+μₚ)
ρ=1/σ
ρ=1/nᵢe(μₑ+μₚ)
ρ=1/2.4*10⁻¹⁹*1.6*10⁻¹⁹(0.39+0.19)

130. 221
ρ=0.448Ωm
2)The electron and hole mobilities of si sample are
0.135 and 0.048m²/vs respectively.Determine the
conductivity of intrinsic si at 300k.the sample is
then doped with 10²³ phosphorous atom/m³.
Determine the equillibrium hole concentration and
conductivity.Given nᵢ=1.5*10¹⁶m⁻³.
Ans:-
Given that
μₑ=0.135m²/vs
μₚ=0.048m²/vs
nᵢ=1.5*10¹⁶/m³
N =10²³ phosphorous atoms/m³
ᴰ
ρ= ?
σ= ?
w.k.t, σ=nᵢe(μₑ+μₚ)
=1.5*10¹⁶*1.6*10⁻¹⁹[0.135+0.048]
=1.5*1.6*0.183*10⁻³
=0.439*10⁻³
=0.439*10⁻³/Ω-mσ
Hole concentration
ρ=nᵢ²/N
=(1.5*10¹⁶)²/10²³
=2.25*10⁹/m³ρ

131. 222
3)The R of a specimen is 3.66*10⁻⁴m³c⁻¹.its
resistivity is 8.93*10⁻³Ω-m.find μ and n
Ans:-
Given that
R =3.66*10⁻⁴m³c⁻¹
ρ=8.93*10⁻³m
μ= ?
n=?
ᴴ
w.k.t, μ=σR
ᴴ
μ=R /ρ [σ=1/ρ]
ᴴ
μ=3.66*10⁻⁴/8.93*10⁻³
μ=0.4116m²/vs
n=1/R e
ᴴ
n=1/3.66*10⁻⁴*1.6*10⁻¹⁹
n=1.7*1022/m³
4)Find the diffusion coefficient of electron in
silicon at 300k if μₑ is 0.19m²/vs
Ans:-
Given that
T=300k
μₑ=0.19m²/vs
e=1.6*10⁻¹⁹
k =1.38*10⁻²³
Dₙ=?
ᴮ

132. 223
w.k.t, Dₙ=μₑk T/e
ᴮ
=0.19*1.38*10⁻²³*300/1.6*10⁻¹⁹
Dₙ=4.9*10⁻³m²/sec
5)The resistivity of an intrinsic semiconductor
is 4.5Ω-m at 20°c and2.0Ω-m at 32°c.what is the
energy band gap?
Ans:-
Given that
ρ₁=4.5
ρ₂=2.0
T₁=20°c=293k
T₂=32°c=305k
k =1.38*10⁻²³j/k
w.k.t, ρ=A exp[E /2k T]ᵍ ᴮ
ρ₁/ρ₂ = A exp[E /2k T₁]/ A exp[E /2k T₂]
ᵍ ᵍᴮ ᴮ
=exp[E /2k (1/T₁ - 1/T₂)]ρ₁/ρ₂
Now take 'log' on both side
log(ρ₁/ρ₂)=E /2k (1/T₁ - 1/T₂)
ᵍ ᴮ
E =2k /(1/T₁-1/T₂) log(4.5/2.0)
ᵍ ᴮ
=2*1.38*10⁻²³/1/293 -1/305 log(4.5/2.0)
=1.6669*10⁻¹⁹J

133. 224
E =1.04ev
ᵍ
6)The following data are given for an intrinsic Ge at
300k.calculate the conductivity of the sample.(Given
nᵢ=2.4*10 ¹⁹m⁻³,μₑ=0.39m²v⁻¹s⁻¹,μₚ=0.19m²v⁻¹s⁻¹)
Ans:-
Given that
nᵢ=2.4*10¹⁹m⁻³
μₑ= 0.39m²v⁻¹s⁻¹
μₚ=0.19m²v⁻¹s⁻¹
w.k.t,
σᵢ=nᵢe(μₑ+μₚ)
=2.4*10¹⁹*1.6*10⁻¹⁹(0.39+0.19)
=2.22Ω⁻¹m⁻¹σᵢ
the conductivity of Ge at 300k is 2.22Ω⁻¹m⁻¹
7)A current of 50A is established in a slab of copper
that is 0.5cm thick and 2cm wide.the slab is placed in a
magnetic field B of 1.5T.the magnetic field is
perpendicular to the plane of the slab and to the
current.the free electron concentration in cu is 8.
4*10²⁸m⁻³.what will be the magnitude of the
Hall-voltage across the width of the slab?
Ans:-

134. 225
Given that
i=50A
B=1.5T
thickness of slab=0.5cm
d=2cm
N=8.4*10²⁸m⁻³
w.k.t, Hall voltage(v )=Bi/neA
=1.5*50/8.4*10²⁸*1.6*10⁻¹⁹*2*10⁻²
v =2.79*10⁻⁷
ᴴ
ᴴ
8)Mobilities of electron and holes in an intrinsic
germanium at 300k are 0.36m²/vs and 0.17m²/vs
respectively.if the resistivity is 2.12Ω-m,calculate
the intrinsic concentration.
Ans:-
Given that
μₑ=0.36
μₚ=0.17
σ=2.12
w.k.t, σ=1/ρ
=nᵢ(μₙ+μₚ)
1/2.12 =nᵢ*1.6*10⁻¹⁹[0.36+0.17]
nᵢ=556.25*10¹⁶/m³

135. 226
9)For the conductivity of p-type Ge which is doped
with acceptor atoms of concentration 2*10²³/m³ and
all acceptor atoms are active.Given μₚ=0.19m²/vs and
e=1.6*10⁻¹⁹c
Ans:-
Given that
N =2*10²³m³
μₚ=0.19m²/vs
e=1.6*10⁻¹⁹c
w.k.t, for a p-type Ge
pₚ=N
conductivity σₚ=nₚμₚe
=2*10²³*0.19*10⁻¹⁹
σₚ=6100Ω-m
ᴬ
ᴬ
10)In an intrinsic semiconductor the energy gap
is 1.2ev.what is the ratio between conductivity at
600k and at 300k?
Ans:-
Given that
E =1.2ev
=1.2*1.6*10⁻¹⁹J
T₁=600k
T₂=300k
k =1.38*10⁻²³J/k
ᵍ
ᴮ

136. 227
w.k.t, conductivity
σ=Ae
-E /2k T
ᵍ ᴮ
σ₁/σ₂ =
e-E /2k T₁
ᵍ ᴮ
e
-E /2k T₂
ᵍ ᴮ
= e
-E /2k [1/T₁ -1/T₂]
ᵍ ᴮ
= e-1.2*1.6*10⁻¹⁹/2*1.38*10⁻²³ [1/600 - 1/300]
= e
-0.6956*10⁴ (-0.3316)
= e
0.2306*10⁴
=1.25*10⁴ρ₁/ρ₂