Normal incidence on a Perfect Di-electric

Normal incidence on a Perfect Dielectric:-

Till now we focused on the propagation of a uniform plane wave in an unbounded medium either free space (or) dielectric.

we now consider a monochromatic uniform plane wave that travels through one medium and then enters another medium of infinite extent.

at this stage we assume that the interface between the two media is normal to the direction of propagation of the incoming wave.

The wave that is propagating in the first medium is called incident wave. Assume the direction of propagation of the incoming wave along positive z-direction.

The interface (or) the boundary is a plane z=0 in this case.

if direction of propagation was along +ve x-axis plane would be x=0 plane.
if direction of propagation was along +ve y-axis plane would be y=0 plane.

The wave reflected back into the same medium is called reflected wave and the wave that is propagating into the second medium is the transmitted wave.

incident and reflected waves are in opposite directions to each other.

Incident waves:-

\overrightarrow{E}_{i}=E_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{i}=H_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{i}=\frac{E_{io}}{\eta _{1}}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Reflected waves:-

\overrightarrow{E}_{r}=E_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{r}=-H_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{r}=-\frac{E_{ro}}{\eta _{1}}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y}.  the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow -\overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow -\overrightarrow{a}_{y}.

Transmitted waves:-

\overrightarrow{E}_{t}=E_{to}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{t}=H_{to}\ e^{\gamma _{2}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{t}=\frac{E_{to}}{\eta _{2}}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Now the Transmission and reflection coefficients are defined as follows

reflection \ coefficient\ \rho (or)\ r= \frac{reflected\ electric\ field \ strength}{incident\ electric\ field \ strength} .

\rho _{E} = \frac{E_{ro}}{E_{io}} .

similarly the transmission coefficient is

transmission \ coefficient\ \tau (or)\ T= \frac{transmitted\ electric\ field \ strength}{incident\ electric\ field \ strength}.

\tau _{E} = \frac{E_{to}}{E_{io}} .

Derivation of coefficients:-

By using the boundary conditions,

the tangential components of E are continuous

i.e, \begin{vmatrix} E_{tan1} \end{vmatrix}=\begin{vmatrix} E_{tan2} \end{vmatrix} .

E_{io}+E_{ro}=E_{to} .

E_{io}+\rho \ E_{io}=\tau \ E_{io}.

1+\rho =\tau ------EQN(1).

the tangential components of H are discontinuous

i.e, \begin{vmatrix} H_{tan1} \end{vmatrix}-\begin{vmatrix} H_{tan2} \end{vmatrix} = J_{s} .  let us assume J_{s}=0 at the Boundary.

\begin{vmatrix} H_{tan1} \end{vmatrix}=\begin{vmatrix} H_{tan2} \end{vmatrix} .

H_{io}+H_{ro}=H_{to} .

\frac{E_{io}}{\eta _{1}}-\frac{E_{ro}}{\eta _{1}}=\frac{E_{to}}{\eta _{2}}\Rightarrow \frac{E_{io}}{\eta _{1}}-\rho \ \frac{E_{io}}{\eta _{1}}=\tau \ \frac{E_{io}}{\eta _{2}}.

\frac{1}{\eta _{1}}- \frac{\rho}{\eta _{1}}= \frac{\tau}{\eta _{2}}.

1-\rho =\frac{\eta _{1}}{\eta _{2}}\ \tau -------EQN(2).

by solving the above two equations the transmission and reflection coefficients  using electric field strength are

\tau _{E}=\frac{2\eta _{2}}{\eta _{2}+\eta _{1}}     and  \rho _{E}=\frac{\eta _{2}-\eta _{1}}{\eta _{2}+\eta _{1}} .

similarly  the transmission and reflection coefficients  using magnetic field strength are

\tau _{H}=\frac{\eta _{1}-\eta _{2}}{\eta _{2}+\eta _{1}}    and  \rho _{H}=\frac{2\eta _{1}}{\eta _{2}+\eta _{1}} .

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Normal incidence on a perfect conductor

whenever an EM Wave travelling in one medium impinges second medium the wave gets partially transmitted and partially reflected depending up on the type of the second medium.

Assume the first case in Normal incidence that is Normal incidence on a Perfect conductor.

i.e an EM wave propagating in free space strikes suddenly a conducting Boundary which means the other medium is a conductor.

The figure shows a plane Wave which is incident normally upon a boundary between free space and a perfect conductor.

assume the wave is propagating in positive z-axis and the boundary is z=0 plane.

The transmitted wave E_{t}=0 since the electric field intensity inside a perfect conductor is zero.

The incident  E_{i}(t) and reflected  E_{r}(t) waves are in the medium 1  that is free space.

The energy transmitted is zero so the energy absorbed by the conductor is zero and entire wave is reflected to the same medium

now incident wave is E_{i}=E_{i}e^{-\gamma z}

\because \alpha =0  in free space \beta =\beta _{1} for medium 1

\overrightarrow{E_{i}} = E_{i}e^{-j\beta z} (-\beta _{1}z a wave propagating in positive z-direction) and the reflected wave is \overrightarrow{E_{r}} = E_{r}e^{j\beta z} (\beta _{1}z a wave propagating in positive z-direction).

\overrightarrow{E}_{total} = E_{i}e^{-j\beta z}+ E_{r}e^{j\beta z} .

by using tangential components {E}_{tan1} = E_{tan2} .

{E}_{i} +{E}_{r} = 0

{E}_{i} =-{E}_{r}

The resultant wave is  \overrightarrow{E}_{total} = E_{i}e^{-j\beta_{1} z}+ E_{r}e^{j\beta_{1} z} .

\overrightarrow{E}_{total} = E_{i}e^{-j\beta_{1} z}- E_{i}e^{j\beta_{1} z} .

\overrightarrow{E}_{total}(z) = -2E_{i}\ j \sin \beta_{1} z

the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

\widetilde{{E}_{total}}(z,t) = Re\left \{ -2E_{i}\ j \sin \beta_{1} z \ e^{j\omega t }\right \}

\widetilde{{E}_{total}}(z,t) = 2E_{i} \sin \beta_{1} z \ \sin \omega t .

This is the wave equation which represents standing wave , which is the contribution of incident and reflected waves. as this wave is stationary it does not progress.

it has maximum amplitude at odd multiples of  \frac{\lambda }{4} and minimum amplitude at multiples of \frac{\lambda }{2} .

Similarly The resultant Magnetic field is

The resultant wave is  \overrightarrow{H}_{total} = H_{i}e^{-j\beta_{1} z}+ H_{r}e^{j\beta_{1} z} .

\overrightarrow{H}_{total} = H_{i}e^{-j\beta_{1} z}+ H_{i}e^{j\beta_{1} z} .

\overrightarrow{H}_{total}(z) = 2H_{i}\ \cos \beta_{1} z

the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

\widetilde{{H}_{total}}(z,t) = Re\left \{ 2H_{i}\ \cos \beta_{1} z \ e^{j\omega t }\right \}

\widetilde{{H}_{total}}(z,t) = 2H_{i} \cos \beta_{1} z \ \cos \omega t .

this wave is  a stationary wave  it has minimum amplitude at odd multiples of  \frac{\lambda }{4} and maximum amplitude at multiples of \frac{\lambda }{2} .

 

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Nature of Magnetic materials

In order to find out the various types of materials in magnetic fields and their behaviour we use the knowledge of the action of magnetic field on a current loop with a simple model of an atom

Magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials.

a charged particle with angular momentum always contributes to the permanent contributions to the angular moment of an atom
1. orbital magnetic dipole moment.
2. electron spin moment.
3. Nuclear spin magnetic moment.

Orbital Magnetic dipole Moment:-

The simple atomic model is one which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits.

an electron in an orbit is analogous to a small current loop and as such experiences a torque in an external magnetic field, the torque tending to align the magnetic field produced by the orbiting electron with the external magnetic field.

Thus the resulting magnetic field at any point in the material would be greater than it would be at that point when the other moments were not considered.

so there are Quantum numbers which describes the orbital state of notion of electron in an atom there are n,l and ml
n-principal Quantum number, which determines the energy of an electron.
l-Orbital Quantum number which determines the angular momentum of orbit.
ml-magnetic Quantum number which determines the component of magnetic moment along the direction of an electric field.

electron spin Magnetic Moment:-

The angular momentum of an electron is called spin of the electron. as electron is a charged particle the spin of the electron produces magnetic dipole moment because electron is spinning about it’s own axis and thus generates a magnetic dipole moment.

\pm 9X 10^{-24} A-m^{2}  is the value of electron spin when we consider an atom those electrons which are in shells which are not completely filled with contribute to a magnetic moment for the atom.

Nuclear spin Magnetic Moment:-

a third contribution of the moment of an atom is caused by nuclear spin this provides a negligible effect on the overall magnetic properties of material

That is the mass of the nucleus is much larger than an electron thus the dipole moments due to nuclear spin are very small.

so the total magnetic dipole moment of an atom is nothing but the summation of all the above mentioned .

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Laplace’s and Poisson’s equations(Magnetostatics)

Laplace’s equation(Magneto statics):- 

From the equation  \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0\Rightarrow \overrightarrow{\bigtriangledown } .\overrightarrow{B}=0 .

since \overrightarrow{B}=\mu _{O}\overrightarrow{H}

\overrightarrow{\bigtriangledown }.(\mu _{O}\overrightarrow{H}) =0 .

\mu _{O}(\overrightarrow{\bigtriangledown }.\overrightarrow{H}) =0.

we know that for zero current density \overrightarrow{J}=0  , the Magnetic scalar potential  is \overrightarrow{H}=-\overrightarrow{\bigtriangledown }V_{m}

by replacing \overrightarrow{H}  in the  equation  \mu _{O}(\overrightarrow{\bigtriangledown }.\overrightarrow{H}) =0 ,

\mu _{O}(\overrightarrow{\bigtriangledown }.(-\overrightarrow{\bigtriangledown }V_{m})) =0 .

\mu _{O}(\bigtriangledown^{2}V_{m}) =0.

\bigtriangledown^{2}V_{m} =0 .   is known as Laplce’s equation in Magneto statics.

Poisson’s equation(Magneto statics):- 

In vector algebra a vector can be fully defined if it’s curl and divergence are defined.

\overrightarrow{\bigtriangledown } X\overrightarrow{H}=\overrightarrow{J}   -point form of Ampere’s law

\overrightarrow{\bigtriangledown } X\overrightarrow{B}=\mu _{o}\overrightarrow{J} because \overrightarrow{B}=\mu _{O}\overrightarrow{H}  .

from vector Magnetic potential  \overrightarrow{B}=\overrightarrow{\bigtriangledown } X \overrightarrow{A} .

\overrightarrow{\bigtriangledown } X(\overrightarrow{\bigtriangledown } X \overrightarrow{A})=\mu _{o}\overrightarrow{J}.

from the vector identity  \bigtriangledown ^{2}\overrightarrow{A}=\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{A})-\overrightarrow{\bigtriangledown } X\overrightarrow{\bigtriangledown } X \overrightarrow{A}.

\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{A})-\bigtriangledown ^{2}\overrightarrow{A}=\mu _{o}\overrightarrow{J} .

if \overrightarrow{\bigtriangledown }.\overrightarrow{A}=0 .

\bigtriangledown ^{2}\overrightarrow{A}=-\mu _{o}\overrightarrow{J} – this equation is known as Poisson’s equation for Magneto statics.

\overrightarrow{A}=A_{x}\overrightarrow{a_{x}}+A_{y}\overrightarrow{a_{y}}+A_{z}\overrightarrow{a_{z}}     and    \overrightarrow{J}=J_{x}\overrightarrow{a_{x}}+J_{y}\overrightarrow{a_{y}}+J_{z}\overrightarrow{a_{z}}

\bigtriangledown ^{2}(A_{x}\overrightarrow{a_{x}}+A_{y}\overrightarrow{a_{y}}+A_{z}\overrightarrow{a_{z}})=-\mu _{o}(J_{x}\overrightarrow{a_{x}}+J_{y}\overrightarrow{a_{y}}+J_{z}\overrightarrow{a_{z}}).

by equating the  respective components on each side

\bigtriangledown ^{2}A_{x}=-\mu _{o}J_{x} ,    \bigtriangledown ^{2}A_{y}=-\mu _{o}J_{y}    and  \bigtriangledown ^{2}A_{z}=-\mu _{o}J_{z}   are the scalar Poisson’s equations of Magneto statics.

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Relaxation time

Relaxation time (T_{r}) :-

from the equation \overrightarrow{J}=\sigma \overrightarrow{E}   and from Gauss’s law      \overrightarrow{\bigtriangledown }.\overrightarrow{E} = \frac{\rho _{v}}{\epsilon }   .

continuity equation  is      \overrightarrow{\bigtriangledown }.\overrightarrow{J} = -\frac{\partial \rho _{v}}{\partial t }.

\overrightarrow{\bigtriangledown }.(\sigma \overrightarrow{E}) = -\frac{\partial \rho _{v}}{\partial t } .

\sigma( \overrightarrow{\bigtriangledown }. \overrightarrow{E}) = -\frac{\partial \rho _{v}}{\partial t } .

\sigma(\frac{\rho _{v}}{\epsilon }) = -\frac{\partial \rho _{v}}{\partial t } .

\therefore \frac{\partial \rho _{v}}{\partial t }+(\frac{\sigma}{\epsilon })\rho _{v} = 0.

The solution to the above equation is of the form  \rho _{v} = \rho _{vo} \ e ^{-\frac{t}{T_{r}}}  .

where T_{r} = \frac{\epsilon }{\sigma }  is known as relaxation time and defined as the time it takes a charge placed in the interior of a material to drop to e^{-1}= 36.8 percent of it’s initial value.

\rho _{vo} is the initial charge density (i.e,  \rho _{v}  at t=0) the equation   \rho _{v} = \rho _{vo} \ e ^{-\frac{t}{T_{r}}}    shows that as a result of introducing  charge at some interior point of the material there is a decay of volume charge density  \rho _{v}  this decay is associated with the charge movement from the interior point at which it was introduced to the surface of the material.

T_{r}  –  is the time constant known as the relaxation time (or) rearrangement time.

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Boundary conditions static electric fields

Boundary conditions static Electric fields:-

So far, we have considered the existence of the electric field in a Homogeneous medium. If the field exists in a region consisting of two different media, the conditions that the field must satisfy at the interface separating the media are called “Boundary conditions”.

These conditions are helpful in determining the field on one side of the boundary if the field on the other side is known.

The conditions will be dictated by the types of material the media are made of.

We shall consider the Boundary conditions at an interface separating

  • Di-electric (\epsilon _{1}) and Di-electric (\epsilon _{2}).
  • Conductor (\sigma ) and Di-electric(\epsilon ).
  • Conductor and free space.

To determine the boundary conditions, we need to use two Maxwell’s equations.

  1. \oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q_{enclosed} –  Gauss’s law in Electrostatics.
  2. \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0.

The Electric field intensity could be considered as result of two components tangential and normal components.

\overrightarrow{E}=\overrightarrow{E_{t}}+\overrightarrow{E_{n}}.

Di-electric (\epsilon _{1}) and Di-electric (\epsilon _{2}):-

In order to find out the \overrightarrow{E} , \overrightarrow{D}   at the interface between two different magnetic materials boundary conditions are required.

Consider two Di-electric materials having permeabilities  \epsilon _{1}   and  \epsilon _{2} as shown in the figure

To apply    \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0  a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda  on the surface of the boundary

Then \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0 .

E_{1t}\Delta w -E_{1n}\frac{\Delta h}{2}-E_{2n}\frac{\Delta h}{2}-E_{2t}\Delta w+E_{2n}\frac{\Delta h}{2}+E_{1n}\frac{\Delta h}{2}=0.

(since \Delta h=0   On the boundary)

E_{1t}\Delta w-E_{2t}\Delta w=0 .

E_{1t}=E_{2t}.

\frac{D_{1t}}{\epsilon _{1}}=\frac{D_{2t}}{\epsilon _{2}} , So the tangential components of  \overrightarrow{D}    are discontinuous where as \overrightarrow{E}   are continuous.

In order to apply \oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q , a surface is required which is nothing but Gaussian surface (or) a Pillbox enclosing some charge.

\oint_{s} \overrightarrow{D} .\overrightarrow{ds} = \oint_{side} \overrightarrow{D} .\overrightarrow{ds} + \oint_{top} \overrightarrow{D} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{D} .\overrightarrow{ds} .

(\oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q  -Because on the boundary  \Delta h=0, so there exists no side of surface).

D_{1n}\Delta S -D_{2n}\Delta S =\Delta Q.

D_{1n}-D_{2n}=\frac{\Delta Q}{\Delta S} .

D_{1n}-D_{2n}=\rho _{s}-----(1).

since \overrightarrow{D} = \epsilon \overrightarrow{E} ,

\epsilon _{1}E_{1n}-\epsilon _{2}E_{2n}=\rho _{s}-----(2)

From equations (1) and (2)  the normal components of \overrightarrow{D} and \overrightarrow{E}   are  discontinuous at the boundary .

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Magnetic Boundary conditions

Magnetic Boundary conditions:-

In order to find out the \overrightarrow{B} , \overrightarrow{H}  and  \overrightarrow{M}  at the interface between two different magnetic materials boundary conditions are required.

Consider two magnetic materials having permeabilities  \mu _{1}   and  \mu _{2} as shown in the figure

We will find out these boundary conditions by using

  1. \oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0  –  Gauss’s law in Magneto statics.
  2. \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I–  Ampere’s law.

In order to apply \oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0 , a surface is required which is nothing but Gaussian surface (or) a Pillbox.

\oint_{s} \overrightarrow{B} .\overrightarrow{ds} = \oint_{side} \overrightarrow{B} .\overrightarrow{ds} + \oint_{top} \overrightarrow{B} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{B} .\overrightarrow{ds} .

(\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0  -Because on the boundary  \Delta h=0, so there exists no side of surface).

B_{1n}\Delta S -B_{2n}\Delta S =0.

B_{1n}=B_{2n}-----------(1)

since \overrightarrow{B} = \mu \overrightarrow{H} ,      \mu _{1} H_{1n}= \mu _{2} H_{2n}.

H_{1n}= \frac{\mu _{2}}{\mu _{1}} H_{2n} ---------(2).

From equations (1) and (2)  the normal component of \overrightarrow{B}    is continuous at the boundary and the normal components of  \overrightarrow{H}   is discontinuous at the boundary.

we know that \overrightarrow{B} = \mu _{o}(1+ \chi _{m})\overrightarrow{H} .

\overrightarrow{B} = \mu _{o}\overrightarrow{H}+ \mu _{o}\overrightarrow{M} .

\overrightarrow{M} = \chi _{m}\overrightarrow{H} .

 M_{2n} = \chi _{m2} H_{2n} .

M_{1n} = \chi _{m1} H_{1n}.

M_{2n} = \chi _{m2} H_{2n}  since  H_{2n} = \frac{\mu _{1}}{\mu _{2}} H_{1n} .

M_{2n}=\chi _{m2} \frac{\mu _{1}}{\mu _{2}} H_{1n} .

M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} \frac{\mu _{1}}{\mu _{2}} M_{1n}.

The magnetisation normal components are also discontinuous.

To apply    \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I  a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda , which encloses a surface current density k on the surface of the boundary

Then \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I .

H_{1t}\Delta w -H_{1n}\frac{\Delta h}{2}-H_{2n}\frac{\Delta h}{2}-H_{2t}\Delta w+H_{2n}\frac{\Delta h}{2}+H_{1n}\frac{\Delta h}{2}=I.

(since \Delta h=0   On the boundary)

H_{1t}\Delta w-H_{2t}\Delta w=k \Delta w .

H_{1t}-H_{2t}=k.

\frac{B_{1t}}{\mu _{1}}-\frac{B_{2t}}{\mu _{2}}=k , So the tangential components of  \overrightarrow{B}   and  \overrightarrow{H}  are discontinuous.

(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}})X \overrightarrow{a_{n12}} = \overrightarrow{k}.

(or) (\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}}) = \overrightarrow{k}X \overrightarrow{a_{n12}}.

When  \overrightarrow{k}=0\overrightarrow{H_{1t}}=\overrightarrow{H_{2t}}   and  \frac{\overrightarrow{B_{1t}}}{\mu _{1}} =\frac{\overrightarrow{B_{2t}}}{\mu _{2}} .

Similar to normal components of magnetisation the tangential components are

M_{2t} = \chi _{m2} H_{2t} .

M_{1t} = \chi _{m1} H_{1t}.

M_{2t} = \chi _{m2} H_{2t} since  H_{2t} = H_{1t}-k .

M_{2t}=\chi _{m2} (H_{1t}-k) .

M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} M_{1t}-\chi _{m2} k.

Obtain the results for magnetisation by using the same procedure as that of  normal components.

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Biot-savart’s law

It states that the magnetic field intensity dH produced, at the point P by the differential current element I dl

  1. is proportional to the product I dl and the \sin \alpha  the angle between the element and the line joining the point P to the element.
  2. and is inversely proportional to the square of the distance  R between P and the current element.

then the direction of \overrightarrow{dH} can be determined by right hand rule with the right hand thumb pointing in the direction of the current and the fingers encircling the wire in the direction of  \overrightarrow{dH} .

i.e,  dH \propto \frac{I \ dl \ \sin \alpha }{R^{2}} .

dH = \frac{ k\ I \ dl \ \sin \alpha }{R^{2}} .

where k is the constant of proportionality , k=\frac{1}{4\pi } .

\overrightarrow{dH} = \frac{ \ I \ dl \ \sin \alpha }{4\pi \ R^{2}}\ \overrightarrow{a_{R}}  A/m.

\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} A/m.

\overrightarrow{dH}   is perpendicular to the plane that contains \overrightarrow{dl}   and \overrightarrow{a_{R}}.

\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{R} }{4\pi \ R^{3}}  A/m.

then the total magnetic field strength  measured at a point P is given by

\overrightarrow{H} = \oint \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} A/m.

closed path is taken since the current can flow only in closed path and this is called as integral form of Biot-Savart’s law.

as similar to  different charge distributions in electro-statics , there exists different current elements like line, surface and volume in the study of  static magnetic fields.

\overrightarrow{H} = \int_{l} \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} A/m.  —-for a line current element.

\overrightarrow{H} = \int_{s} \frac{ \overrightarrow{k} \ ds X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}  A/m. —-for a surface current element.

\overrightarrow{H} = \int_{v} \frac{ \overrightarrow{J} \ dv \ X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}  A/m. —-for a volume current element.

the dot and cross products between dl and I represents either H is out of  (or) into the page(plane) .

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Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density  \overrightarrow{k} = k_{y}\overrightarrow{a_{y}}    A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

\overrightarrow{H} = H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y  and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

\overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0\\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

from Ampere’s Circuit law  \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right ) .

the component \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right )

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= \overrightarrow{H}_{1-0}. \overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}. \overrightarrow{dl}_{0-2}

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})- H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}}) .

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}=0 .

similarly \overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4} =0 .

\therefore \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right) .

\oint \overrightarrow{H}. \overrightarrow{dl} = -(H_{o}\ \overrightarrow{a_{x}}).(b \ -\overrightarrow{a_{x}})+(H_{o}\ \overrightarrow{a_{x}}).(b \ \overrightarrow{a_{x}}) .

\oint \overrightarrow{H}. \overrightarrow{dl} = 2H_{o}\ b .

I= 2H_{o} \ b .

k_{y}\ b = 2H_{o} \ b .

H_{o} =\frac{1}{2} k_{y} .

 as   \overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

this will be changed to \overrightarrow{H} = \left\{\begin{matrix} \frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -\frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

In general for a finite sheet of current density  \overrightarrow{k}  A/m  Magnetic field is generalised as     \overrightarrow{H} = \frac{1}{2} (\overrightarrow{k} X \overrightarrow{a_{n}}) .

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Magnetic Forces

Magnetic forces are required to study the force , a magnetic field exerts on charged particles, current elements and loops which is used in electrical devices in ammeters, volt meters, Galvano meters.

There are 3 ways in which force due to magnetic fields can be experienced.

  1. The force can be due to a  moving charged particle in a  Magnetic field.
  2. on a current element in an external B  field.
  3. between two current elements.

Force on a charged particle:-

we know that \overrightarrow{E} = \frac{\overrightarrow{F}}{Q}  .

\overrightarrow{F_{e}} = Q \overrightarrow{E}-----EQN(1) .

where  \overrightarrow{F_{e}}  is the electric force on a stationary (or) moving electric charge in an electric field and is related to  \overrightarrow{E} .  where  \overrightarrow{F_{e}}  and   \overrightarrow{E} are in the same direction.

a magnetic field can exert force only on a moving charge , suppose a charge Q is moving with velocity u  (or) v in a magnetic field (B) is

\overrightarrow{F_{m}} = (Q\ \overrightarrow{u} \ X \overrightarrow{B})-----EQN(2) .

from the  equations    \overrightarrow{F_{e}}   is independent of velocity of the charge and performs work on the charge which changes its kinetic energy but  \overrightarrow{F_{m}}  depends on the charge velocity and is normal to it so work done \overrightarrow{F_{m}}.\overrightarrow{dl} =0 it does not cause increase in the kinetic energy of the charge.

 \overrightarrow{F_{m}}  is small compared to    \overrightarrow{F_{e}}  except at high velocities.

so a charge which is in movement has both electric and magnetic fields.

Then \overrightarrow{F} = \overrightarrow{F_{e}}+\overrightarrow{F_{m}} .

\overrightarrow{F} = Q(\overrightarrow{E}+\overrightarrow{u}X\overrightarrow{B}) .

This is known as Lorentz’s force equation. It relates mechanical force to electrical force.

if the mass of the charges particle is m

Then \overrightarrow{F} = m\overrightarrow{a} .

\overrightarrow{F} = m\frac{d\overrightarrow{u}}{dt} .

\overrightarrow{F} = Q(\overrightarrow{u}X\overrightarrow{B}) .

m\frac{d\overrightarrow{u}}{dt}= Q(\overrightarrow{u}X\overrightarrow{B}) .

\frac{d\overrightarrow{u}}{dt}-\frac{1}{m} (\overrightarrow{u}X\overrightarrow{B})=Qm\overrightarrow{E} .

The solution to this equation is  important in determining the motion of charged particles in in such cases the energy transfer is only by means of electric field.

Force on a current element:-

Consider a current carrying conductor I\overrightarrow{dl} , in order to find out the force acting on the current carrying element by the  magnetic field \overrightarrow{B} .

I\overrightarrow{dl}\approx \overrightarrow{k}ds\approx \overrightarrow{J}dv .

I\overrightarrow{dl}=\frac{dQ}{dt}\overrightarrow{dl} .

I\overrightarrow{dl}=dQ\frac{\overrightarrow{dl}}{dt} .

I\overrightarrow{dl}=dQ\overrightarrow{u} .

I\overrightarrow{dl} is nothing but a elemental charge dQ moving with the velocity \overrightarrow{u} .

\overrightarrow{F}=Q\overrightarrow{u}X\overrightarrow{B} .

d\overrightarrow{F}=dQ\overrightarrow{u}X\overrightarrow{B}.

d\overrightarrow{F}=I\overrightarrow{dl}X\overrightarrow{B} .

\overrightarrow{F}=\oint_{l}I\overrightarrow{dl}X\overrightarrow{B} .

The line integral is for the current is along the closed path.

i.e,

The magnetic field produced by the current element  I\overrightarrow{dl}  does not exert force on the element itself just as a point charge does not exert force on itself.

So the  magnetic field  \overrightarrow{B}  that exerts force on I\overrightarrow{dl} must be from the another element in other words the magnetic field  \overrightarrow{B}is external to the current element I\overrightarrow{dl}.

Similarly we have  force equations for other current elements  \overrightarrow{k}ds  and \overrightarrow{J}dv as follows

\overrightarrow{F}=\oint_{s}\overrightarrow{k} dsX\overrightarrow{B}       and        \overrightarrow{F}=\oint_{v}\overrightarrow{J} dvX\overrightarrow{B} .

So the magnetic field is defined as the force per unit current element

i.e,  \frac{d\overrightarrow{F}}{\overrightarrow{k}ds}=\overrightarrow{B}  .(or)    \overrightarrow{F_{m}} = Q\overrightarrow{u} X \overrightarrow{B}\Rightarrow \frac{\overrightarrow{F_{m}}}{Q}=\overrightarrow{u} X \overrightarrow{B}    similar to \overrightarrow{E} = \frac{\overrightarrow{F_{e}}}{Q} .

so the    \overrightarrow{B}  describes the force properties of a magnetic field.

Force between two current elements (Ampere’s force law):-

Consider two current loops  I_{1}\overrightarrow{dl_{1}}  and  I_{2}\overrightarrow{dl_{2}} then by Biot- Savart’s law  both current elements produces respective magnetic fields so we may find the force on element sI_{1}\overrightarrow{dl_{1}} due to the field produced by I_{2}\overrightarrow{dl_{2}}.

Field produced by current element  is   I_{2}\overrightarrow{dl_{2}}  is  d\overrightarrow{B_{2}} .

So the force applied on  the element   I_{1}\overrightarrow{dl_{1}} is    d\overrightarrow{F_{1}}  by the field  d\overrightarrow{B_{2}} .

d(d\overrightarrow{F_{1}}) = I_{1}\overrightarrow{dl_{1}}Xd\overrightarrow{B_{2}} ..

d\overrightarrow{B_{2}} = \frac{\mu _{o}I_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}} .

d(d\overrightarrow{F_{1}}) = \frac{\mu _{o}I_{1}\overrightarrow{dl_{1}}XI_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}}.

This is similar to coulomb’s law in electrostatics. Here it is law of force between two current elements and is analogous to coulomb’s law

\overrightarrow{F_{1}} =\frac{\mu _{o}I_{1}I_{2}}{4\pi}\oint_{l_{1}} \oint_{l_{2}}\frac{\ \overrightarrow{dl_{1}}X\ \overrightarrow{dl_{2}}X \ \widehat{{a_{R21}}}}{ R_{21}^{2}} .

Then the force  \overrightarrow{F_{2}}  acting on loop 2 due to the field produced by the current element  t I_{1}\overrightarrow{dl_{1}}   is nothing but \overrightarrow{F_{2}}=-\overrightarrow{F_{1}}

Note:- This is nothing but the Ampere’s force law that is the force between two current carrying conductors is given by it.

 

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Capacitance of a spherical conductor

choose two spherical conductor systems as shown in the figure with the inner conductor – M_{2}  -radius-a   and outer conductor – M_{1}  -radius-b .

Now induced field is directed from M_{2}   to  M_{1} .  then the potential difference between the two conductors is 

V= -\int_{1}^{2}\overrightarrow{E} .\overrightarrow{dl} .

by assuming a point P between the two conductors such that P is out of inner spherical conductor ( M_{2} ) an inside the outer conductor ( M_{1} ).

\therefore \overrightarrow{E}_{at P} = \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r} .

\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl} .

V = -\int_{1}^{2} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta }) .

V = -\int_{r=b}^{a} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta }) .

V = - \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{a}-\frac{1}{b}) .

V = \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{b}-\frac{1}{a}) .

\therefore C_{spherical} =\frac{Q}{V} = \frac{4\pi \epsilon _{o}}{(\frac{1}{b}-\frac{1}{a}) } .

b-radius of the outer conductor.

a- radius of the inner conductor.

\epsilon – permittivity of the medium.

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Capacitance of a Co-axial cable

A Co-axial cable is a Transmission line, in which two conductors are placed co-axially and are separated by some dielectric material with dielectric constant (or) permittivity  (\epsilon ).

a conductor is in the form of a cylinder with some radius, let the radius of inner conductor is ‘a’ meters and that of outer conductor be ‘b’ meters.

Now connect this co-axial conductor to a supply of ‘V’ volts , after applying ‘V’ assume positive charges are distributed on M_{2} and negative charges on M_{1} .

Now, a field is induced \overrightarrow{E} between M_{2}  and M_{1} because of flux lines, to find out \overrightarrow{E} at any point P  between these two conductors

location of P is out of the conductor M_{2} an inside the conductor M_{1}.

\therefore \overrightarrow{E}_{at P} = \overrightarrow{E}_{\ due \ to \ inner \ conductor \ M_{1}} .

assume a cylindrical co-ordinate system \rho ,\ \phi , \ z  and axis of cable coincides with z-axis this is similar to a line charge distribution \rho_{L} placed along the z-axis.

\rho _{L}=\frac{Q}{L} .

\therefore \overrightarrow{E}_{at P} = \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho } .

\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl} .

V = -\int_{1}^{2} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }+d\phi \overrightarrow{a}_{\phi }+dz \overrightarrow{a}_{z }) .

V = -\int_{b}^{a} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }) .

V = - \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln a-\ln b) .

V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln b-\ln a) .

V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }\ln (\frac{b}{a}) .

V = \frac{Q}{2\pi \epsilon _{o}L }\ln (\frac{b}{a}) \ \because \ \rho _{L} = \frac{Q}{L} .

\therefore C_{co-axial} =\frac{Q}{V} = \frac{2\pi \epsilon_{o}L}{\ln (\frac{b}{a})} .

L- length of the conductors.

b-radius of the outer conductor.

a- radius of the inner conductor.

\epsilon – permittivity of the medium.

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Capacitance of Parallel Plate Capacitor

Choose two parallel conducting plates with charge densities separated by a distance ‘d’ meters as shown in the figure.

Assume charges are distributed uniformly on the plates.

Now apply a voltage source ‘V’ to these plates, then all positive charges are accumulated on conductor M_{2} similarly negative charges are accumulated on conductor  M_{1} .

\therefore the charges give rise to a field \overrightarrow{E}  in between them called as induced electric field.

To find out capacitance, choose a co-ordinate system x, y and z as shown in the figure

\therefore V =-\int \overrightarrow{E}.\overrightarrow{dl}.

Noe the potential difference is V=V_{2}-V_{1}.

V_{12}=V_{2}-V_{1}.

\therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl}.

\overrightarrow{E}  is directed from 2 to 1 work has to be done in opposite direction from 1 to 2.

Now \rho _{s}=\frac{Q}{S} . assume two conducting plates has equal surface area S

\therefore  to find out \overrightarrow{E} at any point P in between the ‘2’ plates,  use the concept of infinite sheet of charge distributions with densities  \rho _{s}  and   -\rho _{s} .

E_{\rho _{s}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}}) — from the positive charge distribution.

E_{-\rho _{s}} = \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}}) –with the negative charge distribution.

\therefore  Electric field intensity at P is the sum of electric field intensities due to two infinite charge distributions

\overrightarrow{E_{total}} = \overrightarrow{E_{\rho _{s}}}+\overrightarrow{E_{-\rho _{s}}} .

\overrightarrow{E_{total}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}})+ \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}}) .

\overrightarrow{E_{total} }= \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}}) .

Now the potential difference between the two conductors is \therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl} .

\therefore V =-\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}}).\overrightarrow{dl} .

we know that \overrightarrow{dl} = dx\ \overrightarrow{a_{x}}+ dy\ \overrightarrow{a_{y}}+ dz\ \overrightarrow{a_{z}} .

\therefore V =\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(\overrightarrow{a_{z}}).\ dz \ \overrightarrow{a_{z}} .

V =\int_{0}^{d} \frac{\rho _{s}}{\epsilon _{o}}.\ dz .

V = \frac{\rho _{s}d }{\epsilon _{o}} .

V = \frac{Q\ d }{S\ \epsilon _{o}} \ \because \rho _{s}=\frac{Q}{S} .

\therefore C= \frac{Q}{V}=\frac{S\ \epsilon _{o} }{d} .

If the medium between two parallel plates is air  (or) free space (or) Vaccum  use \epsilon _{o}  or else use \epsilon .

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Capacitance (C)- Farads

When two conductors are embedded in a homogeneous dielectric medium as shown in the figure

 

M_{2} – conductor 2 carries a positive charge (+Q) and M_{1} -conductor 1 carries a negative charge (-Q). Assume the two charges are equal.

\therefore The resultant charge is zero.

we know that charge is carried on the surface as a surface charge density \rho _{s} and \overrightarrow{E}  is normal to the conducting surface.

since M_{2}  carries positive charge the flux is directed from M_{2}  to  M_{1} . flux starts from positive charge and ends at negative charge.

i.e, flux lines leaving one conductor must terminate at the surface of the other conductor.

\therefore  a field is induced between the two conductors  M_{2}  and  M_{1}  and the medium between these two conductors is a dielectric material.

Now, work must be done to carry a positive charge from M_{1} to  M_{2} , that is opposite to the induced field.

V_{21}= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}

V_{21}=V_{1}-V_{2} .

V= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}.

Now, a system as shown in the above figure will have a special physical property called capacitance.

Definition of Capacitance of a conductor:-

The capacitance of a conductor is defined as the physical property of the conductor which is the ability to store Electrical Energy

i.e, C=\frac{Q}{V}  Farads

(or)

Capacitance is the ration of charge of the conductor (either m M_{1} (or)  M_{2} ) to the voltage applied between the two conductors.

Procedure to find Capacitance:-

  • Assume Q  and determine V in terms of Q , or else assume V and determine Q in terms of V.
  • choose a suitable Co-ordinate system.
  • assume the two conducting plates are carrying charges +Q and -Q.
  • Determine  using Gauss’s law (or) by using Coloumb’s law  and find V= - \int \overrightarrow{E}.\overrightarrow{dl} .
  • Finally calculate  C=\frac{Q}{V} .

 

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Non uniform Quantization in PCM

Quantization has been divided into two types

  1. Uniform Quantization.
  2. Non-Uniform Quantization.
Uniform Quantization Non-Uniform Quantization
\Delta is constant through out Quantization process. \Deltais variable that is the step size is variable.

characteristics of the Quantizer are linear.

ex:- Mid-tread type Quantizer. Mid-rise type Quantizer.

characteristics of the Quantizer are  Non-linear.

we know that the sampled version of message signal is Quantized. The Quantization process may follow a Quantization law , which is uniform in general.

But in tele – communication applications like PCM telephony however it is preferable to use a variable separation between the representation levels.

i.e, Non-Uniform Quantization law is to be followed.

for example the range of voltages covered by voice signals from the peaks of loud talk to the weak passages of weak talk is on the order of 1000 to 1.

weak passages (weak talk) which need more protection can be Quantized with smaller step size.

and loud talk will be Quantized with large step size.

Need for Non-Uniform Quantization for speech signal (or) Need for companding in PCM:-

as said above weak talk can be protected more than that of loud talk of a speech signal.

 Since at the receiving end of a telephone system a loud talk can be received with tolerable reduced voltage but a weak talk can not be received if the voltage level is reduced during Quantization process.

we know that speech and music signals are characterized by large crest factor.

for a signal with large crest factor, the signal to noise ratio is given by 

SQR=\frac{P}{N_{q}}=\frac{Signal Power}{Noise Power} .

if we keep  \Delta as constant that is Uniform Quantization is applied then noise power  N_{q} = \frac{\Delta ^{2}}{12}  is almost constant.

SQR\propto P .

SQR is proportional to signal power .

if signal power is low  then SQR also becomes low which is inevitable at the receiver.

if signal power is high then SQR also becomes high which is desirable at the receiver.

to maintain constant SQR at the receiving end companding is preferred in a PCM system (or) for the transmission of voice / speech / Music signals.

i.e, for low power levels of input signal  \Delta is reduced

\therefore N_{q} =\frac{\Delta ^{2}}{12} becomes less and \because \frac{S}{N_{q}} =\frac{P}{N_{q}} is improved.

similarly for high power level \Delta  is increased ,

P is more Noise power increases and SQR is reduced.

In this way , SQR is maintained uniform throughout the Quantization process.

This Uniform Quantization can be achieved through companding.

companding in PCM :-

  1. is required to improve SQR of weak signals.
  2. is also known as Non-Uniform Quantization.

In practical it is difficult to implement the Non-Uniform Quantization because it is not known in advance hoe the signal level is varying?

we follow a particular method as mentioned below

  1. Before the application of the signal to a Uniform Quantizer in the Transmitter weak signals are being amplified and strong signals are being attenuated.
  2. then uniform Quantization is used on the modified signal.
  3. At the receiver a reverse process is to be done.

in general compander is a combination of two devices a compressor(at the Transmitter) and an expander (at the receiver).

the use of Non-Uniform Quantizer is equivalent to passing the base band signal through a compressor passing through a compressor and then applying the compressed signal to a uniform Quantizer.

a particular compression law is used in practice the so called -law and A-law .

to restore the signal samples to their correct relative values we must use a device in the receiver with a characteristic  complementary  to the compressor such a device is called as an expander.

ideally, the compression and expansion laws are exactly inverse.

 

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Single stub matching

Stub matching in Transmission lines:-

A section of Transmission line can be inserted between Load and source besides it is also possible to connect sections of open (or) short circuited lines as stub (or) tuning stubs in shunt with the main Transmission lines at a certain point to effect the matching.

Matching with the help of tuning stub (or) stub is called as stub matching and has the following advantages

  1. Length  l,\ Z_{o} are  unchanged.
  2. It is possible to add adjustable susceptance in shunt with the line.

Stub matching can be possible by

  1. Single stub matching.
  2. Double stub matching.

Single stub matching:-

A type of transmission line frequently used in single stub matching is a short circuited section of transmission line , which is connected in parallel to the main transmission line at a particular distance from the load. By using such stub anti resonance is achieved providing impedance at resonance equal to  R_{o} .

The conductance at that point is equal to   Y_{o}  and the stub length is adjusted to provide a susceptance which is equal in value but opposite in sign to the input susceptance of the main line at that point so that the total susceptance at that point is zero.

The combination of stub and line will thus represent a conductance which is equal to Y_{o} of the line.

We are connecting the stub in parallel to the main Tx line , we will work out with admittance’s instead of impedance’s.

Y_{s}=G_{o}\pm jB

This is the admittance at a point A before stub was connected. The point A is now connected to a stub which provides a susceptance of  \mp jB.

Y_{s}=G_{o}\pm jB\mp jB.

Y_{s}=G_{o}.

Y_{s}=\frac{1}{R_{o}}.

Z_{s}=R_{o}.

Thus the input impedance of the line is  R_{o} itself up to the point A, after A to load the reflection and hence standing waves occurs but by making this distance less than the wave length the losses can be minimized.

For the single stub, it is important to know where the stub is to be connected exactly and the length of the stub for these two measurements must be made on the line it is easy to measure S and  V_{min}  nearest to load.

The measurement is accurate in case of    V_{min}  rather than   V_{max}

From the expression of  Z_{s} in terms of reflection coefficient 

Z_{s}=Z_{o}(\frac{e^{\gamma l}+ke^{-\gamma l}}{e^{\gamma l}-ke^{-\gamma l}}) .

Z_{s}=Z_{o}(\frac{1+ke^{-2\gamma l}}{1-ke^{-2\gamma l}}) .

 

Thus at point from load, input impedance is resistive and its value is minimum equal to this is point from the load

At distance

 

Infinite Line equivalent circuit

Consider the basic form of Transmission line with some impedance Z_{o} at the Load end.

an infinite line can be approximated by an equivalent finite line with load impedance  Z_{o} as shown in the above figure , then the input impedance can be calculated from the voltage and current equations

now at x= l , V=V_{R} \ and \ I=I_{R} .

V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1).

I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l----EQN(2).

The load voltage is given by the equation V_{R}=I_{R}Z_{o}.

V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l = Z_{o}(I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l)

V_{s}Z_{O}\cos h\gamma l-I_{s}Z_{o}^{2}\sin h\gamma l = Z_{o}(I_{s}Z_{o}\cos h\gamma l-V_{S}\sin h\gamma l)

V_{s}(Z_{O}\cos h\gamma l+Z_{R}\sin h\gamma l) = I_{s}Z_{o}(Z_{o}\sin h\gamma l+Z_{R}\cos h\gamma l)

Z_{S}=\frac{V_{S}}{I_{S}}=Z_{o} \frac{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}

Z_{S} \ (or) \ Z_{in}=Z_{o}.

represents the source (or) input impedance of an infinite Transmission Line.

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input impedance of Transmission Line

Consider the basic form of Transmission line with some impedance Z_{R} at the Load end.

In order to find out the impedance at the input terminals of a Transmission line choose the Basic Transmission Line equations

V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x.

I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x.

now at x= l , V=V_{R} \ and \ I=I_{R} .

V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1).

I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l----EQN(2).

The load voltage is given by the equation V_{R}=I_{R}Z_{R}.

V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l = Z_{R}(I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l)

V_{s}Z_{O}\cos h\gamma l-I_{s}Z_{o}^{2}\sin h\gamma l = Z_{R}(I_{s}Z_{o}\cos h\gamma l-V_{S}\sin h\gamma l)

V_{s}(Z_{O}\cos h\gamma l+Z_{R}\sin h\gamma l) = I_{s}Z_{o}(Z_{o}\sin h\gamma l+Z_{R}\cos h\gamma l)

Z_{S}=\frac{V_{S}}{I_{S}}=Z_{o} \frac{(Z_{R}\cos h\gamma l+Z_{O}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{R}\sin h\gamma l)}

Z_{S} \ (or) \ Z_{in}=Z_{o} \frac{(Z_{R}\cos h\gamma l+Z_{O}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{R}\sin h\gamma l)}.

Z_{S}=Z_{o} \frac{(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{o}+Z_{R}\tan h\gamma l)}   represents the source (or) input impedance of a basic Transmission Line.

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open circuit line and Short Circuit line

open circuit line:-

In order to observe the properties of an open circuited Transmission Line, chose a Transmission line of length l and is open circuited at the load end.

from the figure

V_{S}  –  Source voltage (at sending end (or) source end).

I_{S} –  Source current (at sending end (or) source end).

V_{R} –  Load voltage (at receiving end (or) Load end).

I_{R} –  Load current (at receiving end (or) Load end).

at Load end I_{R} =0  ,  V_{R} -  maximum voltage  and Z_{R}=\infty .

We knew  that  the input impedance of a Transmission line is given by  Z_{S} = Z_{O} \frac{(Z_{R}+Z_{O} \tan h\gamma l)}{(Z_{O}+Z_{R} \tan h\gamma l)}.

after substituting the conditions of a open circuit conditions  that is Z_{R}=\infty

Z_{S} =\frac{Z_{O}}{\tan h\gamma l}.

Z_{S} =Z_{O}\cot h\gamma l.

if the Transmission line is a loss-less line then \alpha =0   and  \gamma = j\beta .

Z_{S} =Z_{O}\cot (h j\beta l) .

Z_{S} =-jZ_{O}\cot \beta l.     since \cot (hj\beta l)=-j\cot \beta l .

alternative method:- 

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x.

I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x.

at x=l   ,  V=V_{R}    and  I=I_{R} then the equations changes to 

V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1).

I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l-----EQN(2).

by substituting I_{R} =0  in equation(2)  and also choosing the line as loss-less line

0= I_{s}\cos (j\beta l)-\frac{V_{s}}{Z_{o}}\sin (j\beta l) .    

I_{s}\cos \beta l=\frac{V_{s}}{Z_{o}} j\sin \beta l .     since  \cos (hj\beta l) = \cos \beta l   and  \sin (hj\beta l) = j\sin \beta l .

Z_{S} =\frac{V_{s}}{I_{s}}=Z_{o} \frac{\cos \beta l}{j\sin \beta l}.

Z_{S}=Z_{OC} =-jZ_{o} \cot \beta l.

short circuit line:-

In order to observe the properties of an short circuited Transmission Line, chose a Transmission line of length l and is short circuited at the load end.

from the figure

V_{S}  –  Source voltage (at sending end (or) source end).

I_{S} –  Source current (at sending end (or) source end).

V_{R} –  Load voltage (at receiving end (or) Load end).

I_{R} –  Load current (at receiving end (or) Load end).

at Load end V_{R} =0  ,  I_{R} -  maximum current and Z_{R}=0 .

We knew  that  the input impedance of a Transmission line is given by  Z_{S} = Z_{O} \frac{(Z_{R}+Z_{O} \tan h\gamma l)}{(Z_{O}+Z_{R} \tan h\gamma l)}.

after substituting the conditions of a short circuit conditions  that is Z_{R}=0

Z_{S} =Z_{O}\tan h\gamma l.

if the Transmission line is a loss-less line then \alpha =0   and  \gamma = j\beta .

Z_{S} =Z_{O}\tan (h j\beta l) .

Z_{S} =jZ_{O}\tan \beta l.     since \tan (hj\beta l)=j\tan \beta l .

alternative method:- 

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x.

I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x.

at x=l   ,  V=V_{R}    and  I=I_{R} then the equations changes to 

V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1).

I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l-----EQN(2).

by substituting V_{R} =0  in equation(1)  and also choosing the line as loss-less line

0= V_{s}\cos (j\beta l)-I_{s}Z_{o}\sin (j\beta l) .    

V_{s}\cos \beta l=I_{s}Z_{o} j\sin \beta l .     since  \cos (hj\beta l) = \cos \beta l   and  \sin (hj\beta l) = j\sin \beta l .

Z_{S} =\frac{V_{s}}{I_{s}}=Z_{o} \frac{j\sin \beta l}{\cos \beta l}.

Z_{S}=Z_{SC} =jZ_{o} \tan \beta l.

 

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Half,Quarter and Eight-wave lines

Eight-wave line (or) \frac{\lambda }{8}-line:-

Consider a eight-wave line with length of the wave length as \frac{\lambda }{8} .

from the input impedance of a Transmission line

Z_{s}  (or) Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}

for a loss-less line \alpha =0  , \gamma = j\beta.

then the input impedance changes to Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}.

Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}.  since \tan (j\beta l) =j \tan \beta l.

now by substituting \beta l=\frac{2\pi }{\lambda }.\frac{\lambda }{8}\Rightarrow \beta l=\frac{\pi }{4}  .

Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \frac{\pi }{4} )}{(Z_{O}+jZ_{R}\tan \frac{\pi }{4} )}.

\therefore Z_{input} = Z_{O}.

i.e, The eight wave line , the input impedance is the Characteristic impedance (Z_{O}) .

Half-wave line (or) \frac{\lambda }{2}-line:-

Consider a Half-wave line with length of the wave length as \frac{\lambda }{2} .

from the input impedance of a Transmission line

Z_{s}  (or) Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}

for a loss-less line \alpha =0  , \gamma = j\beta.

then the input impedance changes to Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}.

Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}.  since \tan (j\beta l) =j \tan \beta l.

now by substituting \beta l=\frac{2\pi }{\lambda }.\frac{\lambda }{2} \Rightarrow \beta l=\pi .

Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \pi )}{(Z_{O}+jZ_{R}\tan \pi )}.

\therefore Z_{input} = Z_{R}.

i.e, The Half wave line , the input impedance is the load impedance (Z_{R}) , this line repeats it’s terminating impedance, hence it is called as one-to-one transformer.

the main application of \frac{\lambda }{2} line is to connect a load to source where both of them can’t made adjacent in such case , we may connect a parallel Half-wave line at the load end.

Quarter-wave line (or) \frac{\lambda }{4}-line:-

Consider a Quarter-wave line with length of the wave length as \frac{\lambda }{4} .

from the input impedance of a Transmission line

Z_{s}  (or) Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}

for a loss-less line \alpha =0  , \gamma = j\beta.

then the input impedance changes to Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}.

Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}.  since \tan (j\beta l) =j \tan \beta l.

now by substituting \beta l=\frac{2\pi }{\lambda }.\frac{\lambda }{4}\Rightarrow \Rightarrow \beta l=\frac{\pi }{2}  .

Z_{input} = \frac{Z_{O}(\frac{Z_{R}}{\tan \beta l}+jZ_{O})}{(\frac{Z_{O}}{\tan \beta l}+jZ_{R})}.

Z_{input} = \frac{Z_{O}(\frac{Z_{R}}{\tan \frac{\pi }{2}}+jZ_{O})}{(\frac{Z_{O}}{\tan \frac{\pi }{2}}+jZ_{R})}.

Z_{input} = \frac{Z_{O}(jZ_{O})}{(jZ_{R})}.

Z_{input} = \frac{Z_{O}^{2}}{Z_{R}}.

\therefore Z_{O} = \sqrt{Z_{input}Z_{R}} .

if Z_{O} is constant then Z_{s}\propto \frac{1}{Z_{R}} .

i.e, The Quarter wave transformer transforms a load impedance Z_{R}< Z_{O} into a value Z_{S} which is larger than Z_{O} and vice-versa.

sometimes it is called as impedance inverter  if Z_{R} is pure resistive Z_{S} also becomes resistive.

This Quarter wave transformer is useful when it is desired to transform a resistance into a different resistance value either larger (or) smaller.

The desired value of Z_{O} can be obtained by choosing proper value of the ratio of spacing of the line conductors to their diameters  because the physical dimensions can not practically have an infinite range of values where as the transformer ratio of the \frac{\lambda }{4} transformer is subject to the practical limitation.

it is very useful device because of its simplicity and the ease with which it’s behavior is calculated . 

if the length is an odd multiple of \frac{\lambda }{4} will have same transformation properties but when the length is made larger the sensitivity to a small change of frequency becomes larger.

 since input impedance is inversely proportional to Z_{R}. If Z_{R} is high , Z_{S} will be low and vice-versa.

if  Z_{R}  is capacitive , Z_{S} will be inductive and vice-versa. if  Z_{R}  is resistive , Z_{S} will be resistive and vice-versa.

Thus depending up on Z_{R} , a \frac{\lambda }{4} transformer acts as step-up (or) step-down impedance transformer and that is why it is being called as impedance inverter.

\frac{\lambda }{4}  transformer is disadvantageous , as it is sensitive to change in frequency because a new frequency section will no longer be   \frac{\lambda }{4} in length.

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condition for minimum attenuation

The attenuation constant \alpha of a transmission line  is

\alpha =\sqrt{\frac{(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}{2}}-----EQN(1)

\alpha depends up on the  4 primary constants  along with frequency \omega. In order to achieve the minimum attenuation , these primary constants should be varied in turn.

case 1:-

consider the  case where the attenuation (\alpha )  variation depends only on Inductance(L) .

i.e by considering R, G , C and \omega as constants 

from EQN(1), 2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}

differentiating the above equation w.r. to L

\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}C + \frac{2L(\omega ^{2}G^{2}+\omega ^{4}C^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

\frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

now      \frac{d\alpha }{dL}=0.

-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})=0.

\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}L(G^{2}+\omega ^{2}C^{2}).

C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}.

C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2}).

\Rightarrow LG=RC.

\therefore L=\frac{RC}{G}  H/m.

The condition for L to get minimum attenuation is   \frac{RC}{G}  H/m but in general L should be maintained at  less than this value.

case 2:-

consider the  case where the attenuation (\alpha )  variation depends only on Capacitance(C) .

i.e by considering R, L, G and \omega as constants 

from EQN(1), 2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}

differentiating the above equation w.r. to L

\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}L + \frac{2C(\omega ^{2}R^{2}+\omega ^{4}L^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

\frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}.

now      \frac{d\alpha }{dL}=0.

-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})=0.

\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}C(R^{2}+\omega ^{2}L^{2}).

C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}.

C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2}).

\Rightarrow LG=RC.

\therefore C=\frac{LG}{R} F/m.

The condition for C to get minimum attenuation is  \frac{LG}{R}  F/m but in general C should be maintained at  more  than this value.

Transmission line equations

The transmission line will be analyzed in terms of voltage and current and also the relation between V and I of a line can be obtained by the simplest type of Transmission line.

i.e, a pair of parallel wires of uniform size which are spaced a small distance (S) apart in air.

Consider a short section of Transmission line AA^{'}BB^{'} of length \Delta x (i.e an infinitesimal section with a very small length).

at AA^{'} – voltage is V and current is I and at BB^{'} voltage is V+\Delta V and current is I+\Delta I .

the voltage drop between AA^{'} and BB^{'} is 

V-(V+\Delta V) = I(R+j\omega L) \Delta x---------EQN(I)

similarly, the current difference between AA^{'} and BB^{'} is 

I-(I+\Delta I) = V(G+j\omega C) \Delta X-------------EQN(II)

from EQNS (I) and(II) 

\therefore -\frac{\Delta V}{\Delta x}=I (R+j\omega L)     and         -\frac{\Delta I}{\Delta x}=V (G+j\omega C).

to neglect the transit time effect the condition to be applied is  \Delta x\rightarrow 0, then the two equations will become

\lim_{\Delta x\rightarrow 0} -\frac{\Delta V}{\Delta x}=\lim_{\Delta x\rightarrow 0}I (R+j\omega L)    \Rightarrow -\frac{d V}{d x}=I (R+j\omega L).

\lim_{\Delta x\rightarrow 0} -\frac{\Delta I}{\Delta x}=\lim_{\Delta x\rightarrow 0}V (G+j\omega C)  \Rightarrow -\frac{d I}{d x}=V (G+j\omega C).

choose -\frac{d V}{d x}=I (R+j\omega L)

differentiating the above equation w.r.to x  

\Rightarrow -\frac{d^{2} V}{d x^{2}}=\frac{dI}{dx} (R+j\omega L)

replacing \frac{dI}{dx} in the above equation will result

\frac{d^{2} V}{d x^{2}}= (R+j\omega L)(G+j\omega C)V.

\frac{d^{2} V}{d x^{2}}=\gamma ^{2} V----------EQN(1) ,   Let \gamma ^{2} =(R+j\omega L)(G+j\omega C).

where \gamma is the propagation constant which is a complex number.

Similarly  from -\frac{d I}{d x}=V (G+j\omega C)

differentiating the above equation w.r.to x  

\Rightarrow -\frac{d^{2} I}{d x^{2}}=\frac{dV}{dx} (G+j\omega C)

replacing \frac{dV}{dx} in the above equation will result

\frac{d^{2} I}{d x^{2}}= (R+j\omega L)(G+j\omega C)I.

\frac{d^{2} I}{d x^{2}}=\gamma ^{2} I------EQN(2)

the solutions of  Equations(1) and (2) are 

V=A e^{-\gamma x}+Be^{\gamma x}.

I=Ce^{-\gamma X}+D e^{\gamma X}.

where A, B, C and D are arbitrary constants in which A and B have dimensions of voltage and C and D have current dimensions.

since \gamma is complex that is by replacing \gamma = \alpha +j\beta in the V and I equations

V=A e^{-\alpha x} e^{-j\beta x}+Be^{\alpha x} e^{j\beta x} .

I=C e^{-\alpha x} e^{-j\beta x}+D e^{\alpha x} e^{j\beta x}.

the  term e^{-\alpha x} e^{-j\beta x} represents waves travelling from source end  to load end and are called as incident waves . 

Similarly the term e^{\alpha x} e^{j\beta x} represents reflected waves when a transmission line is terminated with any load impedance Z_{R} at the output end.

V=A e^{-\gamma x}+Be^{\gamma x}.

differentiating V w.r.to x

\frac{dV}{dx}=-A\gamma e^{-\gamma x}+B\gamma e^{\gamma x}.

but \frac{dV}{dx}=-I(R+j\omega L)

\therefore -I(R+j\omega L)=-A\gamma e^{-\gamma x}+B\gamma e^{\gamma x}.

I=\frac{\gamma }{(R+j\omega L)}(Ae^{-\gamma x}-B e^{\gamma x}).

I=\frac{\sqrt{(R+j\omega L)(G+j\omega C)}}{(R+j\omega L)}(Ae^{-\gamma x}-B e^{\gamma x})

since \gamma ={\sqrt{(R+j\omega L)(G+j\omega C)}}.

I=\sqrt{\frac{(G+j\omega C)}{(R+j\omega L)}}(Ae^{-\gamma x}-B e^{\gamma x}).

\therefore I=\frac{1}{Z_{o}}(Ae^{-\gamma x}-B e^{\gamma x}),   where Z_{o} =\sqrt{\frac{(R+j\omega L)}{(G+j\omega C)}} .

AS  \cos h\gamma x = \frac{e^{\gamma x}+e^{-\gamma x}}{2}    and \sin h\gamma x = \frac{e^{\gamma x}-e^{-\gamma x}}{2} .

this implies \cos h\gamma x-\sin h\gamma x = e^{-\gamma x}      ,    \cos h\gamma x+\sin h\gamma x = e^{\gamma x}.

by substituting e^{-\gamma x}  and e^{\gamma x} in the Voltage and current equations , V and I results to be

V= (A+B)\cos h\gamma x-(A-B)\sin h\gamma x---EQN(3).

I= \frac{1}{Z_{o}}((A-B)\cos h\gamma x-(A+B)\sin h\gamma x)---EQN(4)

at the input terminals x=0 , V = V_{s}  and I = I_{s} , after substituting this condition in EQNs (1) and (2) 

V_{s} =(A+B)   and  I_{s} =\frac{1}{Z_{o}}(A-B).

then the EQNs (3) and (4)  

V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x.

I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x.

The above equations are known as general equations of a transmission line for voltage and current at any point  which is located at x from the sending end.

 

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Brewster’s angle

In parallel polarization the incident angle at which there is no reflection is called Brewster’s angle.

\rho _{parallel} = 0.

As   \rho _{parallel} = \frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}.

\frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=0.

(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})=0.

\eta _{2}\cos \theta _{t}=\eta _{1}\cos \theta _{i}.

by squaring on both sides  \eta^{2} _{2}\cos^{2} \theta _{t}=\eta^{2} _{1}\cos^{2} \theta _{i}.

\eta^{2}_{2}(1-\sin^{2} \theta _{t})=\eta^{2} _{1}(1-\sin^{2} \theta _{i})....EQN(I)

By using Snell’s law  \frac{\sin \theta _{i}}{\sin \theta _{t}} =\sqrt{\frac{\mu _{2}\epsilon _{2}}{\mu _{1}\epsilon _{1}}} =\frac{r_{2}}{r_{1}}.

using the above equation  \sin^{2} \theta _{t} =\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i}. substituting this in EQN (I).

\eta^{2}_{2}(1-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=\eta^{2} _{1}(1-\sin^{2} \theta _{i}).

(\eta^{2}_{2}-\eta^{2}_{2}\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=(\eta^{2} _{1}- \eta^{2} _{1}\sin^{2} \theta _{i}).

after simplification    \sin^{2} \theta _{i}(\eta^{2} _{1}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\eta^{2} _{2} )=(\eta^{2} _{1}-\eta^{2} _{2 }).

as \eta _{1} = \sqrt{\frac{\mu _{1}}{\epsilon _{1}}}  and  \eta _{2} = \sqrt{\frac{\mu _{2}}{\epsilon _{2}}}.

\sin^{2} \theta _{i}(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\frac{\mu _{2}}{\epsilon _{2}} )=(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{2}}{\epsilon _{2}})..

by simplification  \sin^{2} \theta _{i}= \frac{(1-\frac{\mu _{2}\epsilon _{1}}{\mu _{1}\epsilon _{2}} )}{(1-\frac{\epsilon^{2} _{1}}{\epsilon^{2} _{2}})}.

here \theta _{i} is called as Brewster’s angle.

Let us assume two mediums are lossless dielectrics and are non-magnetic  then

\mu _{1} =\mu _{2}=\mu _{0}.

\sin^{2} \theta _{Brewster}= \frac{1}{(1+\frac{\epsilon _{1}}{\epsilon _{2}})}.

\sin \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{2}+\epsilon _{1}}}.

\tan \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{1}}} = \frac{r_{2}}{r_{1}}.

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Parallel Polarization

Parallel Polarizaton:-

Parallel polarization means \overrightarrow{E} field lies in the XZ-plane (y=0) that is the plane of incidence, the figure illustrates the case of parallel polarization

then the incident and reflected fields is given by

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(\overrightarrow{k}.\overrightarrow{r})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}}).(k_{i}\cos \theta _{i}\overrightarrow{a_{z}}+k_{i}\sin \theta _{i}\overrightarrow{a_{x}})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(k_{i}\cos \theta _{i}z+xk_{i}\sin \theta _{i})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}

\overrightarrow{H_{is}} = \frac{E_{i}}{\eta _{1}}e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}\overrightarrow{a_{y}}

now the reflected wave is 

\overrightarrow{E_{rs}} = E_{r}(\sin \theta_{r }\overrightarrow{a_{z}}+ \cos \theta_{r }\overrightarrow{a_{x}} )e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}

\overrightarrow{H_{rs}} =\frac{ H_{r}}{\eta _{1}}e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}\overrightarrow{a_{y}}

since k_{i}=k_{r}=\beta _{1} =\omega \sqrt{\mu_{1} \epsilon_{1} }

let’s find out  \overrightarrow{k}  then by using the equation \overrightarrow{k}.\overrightarrow{E_{s}} =0\overrightarrow{H_{s}} = \frac{k}{\omega \mu }X \overrightarrow{E_{s}} = \overrightarrow{a_{k}}X \frac{E_{s}}{\eta } .

and the transmitted fields in the second  medium is  

\overrightarrow{E_{ts}} = E_{t}(-\sin \theta_{t }\overrightarrow{a_{z}}+ \cos \theta_{t}\overrightarrow{a_{x}} )e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}

\overrightarrow{H_{ts}} = \frac{H_{t}}{\eta _{2}}e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}\overrightarrow{a_{y}}

where \beta _{2} = \omega \sqrt{\mu _{2}\epsilon _{2}}.

Transmission coefficient:-

as \theta _{r} = \theta _{i} and also that the tangential components of \overrightarrow{E} and \overrightarrow{H} are continuous at the boundary z=0.

E_{tan1}=E_{tan2} 

(E_{i}+E_{r}) \cos \theta _{i} = E_{t}\cos \theta _{t}

H_{tan1}-H_{tan2} = \overrightarrow{J_{s}} , H_{tan1} = H_{tan2}  since \overrightarrow{J_{s}} =0

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

E_{i}+E_{r}=E_{t}\frac{\cos \theta _{t}}{\cos \theta _{i}}------EQN(2)

(1)+(2) implies 

2E_{i}=E_{t}(\frac{\eta _{1}}{\eta _{2}}+\frac{\cos \theta _{t}}{\cos \theta _{i}})

2E_{i}=E_{t}\frac{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}{\eta _{2}\cos \theta _{i}}

\tau _{parallel} = \frac{E_{t}}{E_{i}} = \frac{2\eta _{2}\cos \theta _{i}}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}.

Reflection coefficient:-

from EQN (1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}= \frac{2E_{i}\eta _{2}\cos \theta _{i}}{\eta _{2}(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

after simplification

E_{r}\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=E_{i}

\rho _{parallel} = \frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

1+\rho _{parallel} = \tau _{parallel}(\frac{\cos \theta _{t}}{\cos \theta _{i}}).

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Surface impedance

At high frequencies, the current is almost confined to a very thin sheet at the surface of the conductor which is used in many applications.

The  surface impedance may be defined as the ratio of the tangential component of the electric field \overrightarrow{E_{tan}} at the surface of the conductor to the current density (linear) \overrightarrow{J_{s}} which flows due to this electric field.

given as Z_{s} (or) \eta _{s} = \frac{\overrightarrow{E_{tan}}}{\overrightarrow{J_{s}}}.

\overrightarrow{E_{tan}}   is the Electric field strength parallel to and at the surface of the conductor.

and \overrightarrow{J}  is the total linear current density which flows due to \overrightarrow{E_{tan}}.

The \overrightarrow{J_{s}} represents the total conduction per meter width flowing in this sheet.

Let us consider a conductor of the type plate, is placed at the surface y=0 and the current distribution in the y-direction is given by 

 

Assume that the depth of penetration (\delta) is very much less compared with the thickness of the conductor.

J_{s}= \int_{0}^{\infty } \overrightarrow{J}.\overrightarrow{dy}

J_{s}= \int_{0}^{\infty } J_{o}e^{-\gamma y}dy

J_{s}= J_{o}(e^{-\gamma y})_{0}^{\infty }

J_{s}= \frac{J_{o}}{\gamma }

from ohm’s law \overrightarrow{J_{o}} = \sigma \overrightarrow{E_{tan}}  

E = \frac{J_{o}}{\sigma } .

then  \eta _{s} = \frac{\gamma }{\sigma } .

Z_{s}  (or)  \eta _{s} = \frac{\gamma }{\sigma } .

we know that \gamma = \sqrt{j\omega \mu (\sigma +j\omega \epsilon )}  

for good conductors \sigma > > \omega \epsilon .

then \gamma \approx \sqrt{j\omega \mu \sigma } 

\eta _{s} = \frac{\gamma }{\sigma } = \sqrt{\frac{j\omega \mu }{\sigma }} .

therefore the surface impedance of a plane good conductor which is very much thicker than the skin depth is equal to the characteristic impedance of the conductor.

This impedance is also known s input impedance of the conductor when viewed as transmission line conducting energy into the interior of metal.

when the thickness of the plane conductor is not greater compared to the depth of penetration , reflection of wave occurs at the back surface of the conductor.

 

 

 

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oblique incidence

when a uniform plane wave  is incident obliquely (making an angle \theta _{i} other than 90^{o}) to the boundary between the two media then it is known as oblique incidence.

Now consider the situation that is more general case  that is the oblique incidence.

In this case the EM wave (incident wave) not strikes normally the boundary. i.e,  the incident wave is not propagating  along any standard axes (like x,y and z).

Therefore EM wave is moving in a random direction then the general form is    \overrightarrow{E} = E_{o} \cos (\omega t-\overrightarrow{k}.\overrightarrow{r})  

it is also in the form \overrightarrow{E} = E_{o} \cos (\overrightarrow{k}.\overrightarrow{r}-\omega t).

then \overrightarrow{k} = k_{x}\overrightarrow{a_{x}}+k_{y}\overrightarrow{a_{y}}+k_{z}\overrightarrow{a_{z}} is called the wave number vector (or) the propagation vector.

and \overrightarrow{r} = x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}} is called the position vector (from origin to any point on the plane of incidence) , then the magnitude of \overrightarrow{k} is related to \omega according to the dispersion.

k^{2} = k_{x}^{2}+k_{y}^{2}+k_{z}^{2} = \omega ^{2}\mu \epsilon

\overrightarrow{k} X \overrightarrow{E} = \omega \mu \overrightarrow{H}.

\overrightarrow{k} X \overrightarrow{H} = -\omega \epsilon \overrightarrow{E}.

\overrightarrow{k} . \overrightarrow{H} =0.

\overrightarrow{k} . \overrightarrow{E} =0.

i.    \overrightarrow{E},\overrightarrow{H} and \overrightarrow{k} are mutually orthogonal.

ii.  \overrightarrow{E}  and  \overrightarrow{H} lie on the plane \overrightarrow{k} . \overrightarrow{r} = k_{x}x+k_{y}y+k_{z}z=constant.

then the \overrightarrow{H} field corresponding to \overrightarrow{E} field is \overrightarrow{H} = \frac{1}{\omega \mu } (\overrightarrow{k} X \overrightarrow{E}) = \frac{\overrightarrow{a_{k}} X \overrightarrow{E}}{\eta }.

Now choose oblique incidence of a uniform plane wave at a plane boundary.

the plane defined by the propagation vector \overrightarrow{k} and a unit normal vector \overrightarrow{a_{n}} to the boundary is called the plane of incidence.

the angle \theta _{i} between \overrightarrow{k} and \overrightarrow{a_{n}} is the angle of incidence.

both the incident and reflected waves are in medium 1 while the transmitted wave is in medium 2 .

Now, 

\overrightarrow{E_{i}} =E_{i}\cos (k_{ix}x+k_{iy}y+k_{iz}z-\omega _{i}t)

\overrightarrow{E_{r}} =E_{i}\cos (k_{rx}x+k_{ry}y+k_{rz}z-\omega _{r}t)

\overrightarrow{E_{t}} =E_{t}\cos (k_{tx}x+k_{ty}y+k_{tz}z-\omega _{t}t).

the wave propagates 

  1.     \omega _{i}=\omega _{r}=\omega _{t}=\omega.
  2.     k_{ix} = k_{rx}=k_{tx}=k_{x}.
  3.    k_{iy} = k_{ry}=k_{ty}=k_{y}.

(1) indicates that all waves are propagating with same frequency. (2) and (3) shows that the tangential components of propagation vectors be continuous.

k_{i} \sin \theta _{i}=k_{r} \sin \theta _{r}  implies  k_{i} = k_{r} =\beta _{1} =\omega \sqrt{\mu _{1}\epsilon _{1}}    since \theta _{r}=\theta _{i}.

k_{i} \sin \theta _{i}=k_{t} \sin \theta _{t}   implies k_{t} =\beta _{2} =\omega \sqrt{\mu _{2}\epsilon _{2}}.

\frac{\sin \theta _{t}}{sin \theta _{i}}=\sqrt{\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}}

now velocity u=\frac{\omega }{k}.

then from Snell’s law       r_{1}\sin \theta _{i} = r_{2}\sin \theta _{t},  where r_{1} and r_{2}  are the refractive indices of the two media.

 

 

 

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Poynting theorem

Introduction:-

Poynting theorem is used to get an expression for propagation of energy in a medium.

It gives the relation between energy stored in a time-varying magnetic field and the energy stored in time-varying electric field and the instantaneous power flow out of a given region.

EM waves propagates through space from source to destination. In order to find out power in a uniform plane wave it is necessary to develop a power theorem for the EM field known as poynting theorem.

The direction of power flow is perpendicular to \vec{E} and \overrightarrow{H} in the direction of plane containing \overrightarrow{E} and \overrightarrow{H}.

i.e, it gives the direction of propagation .

\overrightarrow{P} = \overrightarrow{E}X\overrightarrow{H}  Watts/m2  (or)   VA/m2.

Proof:-

from Maxwell’s  equations \overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{d}}

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{J}=\overrightarrow{\bigtriangledown } X \overrightarrow{H}-\frac{\partial \overrightarrow{D}}{\partial t}

the above equation has units of the form current density Amp/m2. When it gets multiplied by \overrightarrow{E} V/m. The total units  will  have of the form power per unit volume.

\overrightarrow{J}\rightarrow Amp/m2\overrightarrow{E}\rightarrow Volts/m.

EJ\rightarrow Amp. Volt/m3 \rightarrow Watts/m3 \rightarrow Power/volume.

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

by using the vector identity 

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{E}.(\overrightarrow{\bigtriangledown }X \overrightarrow{H} )

then 

\overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}.

from the equation of Maxwell’s \overrightarrow{\bigtriangledown } X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} = -\frac{\partial (\mu \overrightarrow{ H})}{\partial t}

\overrightarrow{H}.(-\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\overrightarrow{J}-\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}-\overrightarrow{H}.(\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )------EQN(I)

by using the vector identity  \frac{\partial (\overrightarrow{A}.\overrightarrow{B})}{\partial t}=\overrightarrow{A}.\frac{\partial \overrightarrow{B}}{\partial t}+\overrightarrow{B}.\frac{\partial \overrightarrow{A}}{\partial t}

if \overrightarrow{A} = \overrightarrow{B}    \Rightarrow \frac{\partial (\overrightarrow{A}.\overrightarrow{A})}{\partial t}=2 \overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}

\overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}=\frac{1}{2}\frac{\partial A^{2}}{\partial t}

from EQN(I)   ,  \overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\sigma\overrightarrow{E }-\frac{1}{2}\epsilon\frac{\partial E^{2}}{\partial t}-\frac{1}{2}\mu \frac{\partial H^{2}}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\sigma E^{2}-\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})

by integrating the above equation by over  a volume 

\int_{v}\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) dv=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv

by converting the volume integral to surface integral

\oint_{s}(\overrightarrow{E} X \overrightarrow{H}). \overrightarrow{ds}=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv.

the above equation gives the statement of Poynting theorem.

Poynting theorem:-  

It states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored with in that volume V and the ohmic losses.

 

 

 

 

 

Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

Consider a coaxial cable with inner conductor having radius a and outer conductor of radius b and the current is flowing in the cable along z-axis  in which the cable is placed such that the axis of rotation of the cable co-incides with z-axis.

the length of the co-axial cable be d meters.

we know that the \overrightarrow{H_{\phi }} between the region a< \rho < b  is \overrightarrow{H_{\phi }}=\frac{I}{2\pi \rho }\overrightarrow{a_{\phi }}

and \overrightarrow{B_{\phi }}=\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }}   since \overrightarrow{B_{\phi }}=\mu _{o}\overrightarrow{H_{\phi }}

as L = \frac{\lambda }{I}, here the flux linkage \lambda =1\phi

where \phi-Total flux coming out of the surface

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

Since the magnetic flux will be radial plane extending from \rho =a   to   \rho =b and z=0 to z=d.

\overrightarrow{ds_{\phi }} = d\rho dz \overrightarrow{a_{\phi }}

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

\phi = \oint_{s}\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }} .d\rho dz \overrightarrow{a_{\phi }}

\phi = \int_{\rho =a}^{b}\int_{z =0}^{d}\frac{\mu _{o}I}{2\pi \rho } d\rho dz

\phi = \frac{\mu _{o}I}{2\pi } ln d

L=\frac{\phi }{I} =\frac{\mu _{o}d}{2\pi } ln  Henries.

 

Inductance of a Solenoid:-

Consider a Solenoid of N turns and  let the current flowing inside it is  ‘I’ Amperes. The length of the solenoid is ‘l’ meters and ‘A’ is its cross sectional area.

\phi – Total flux coming out of solenoid.

flux linkage \lambda = N\phi      \Rightarrow \lambda = NBA----------EQN(I)        \because \frac{\phi }{A} = B

L = \frac{\lambda }{I}, from the definition

As B is the Magnetic flux density given B= \frac{flux}{unit area} =\frac{\phi }{A}

from EQN(I) ,

\lambda = N\mu _{o}HA-------EQN(II)  because B = \mu _{o}H

The field strength H of a solenoid is H = \frac{NI}{l} A/m

EQN (II) becomes  \lambda = N \mu _{o}\frac{NI}{l}A

\lambda = \frac{N^{2} \mu _{o}IA}{l}

from the inductance definition L = \frac{\lambda }{I}

L= \frac{N^{2} \mu _{o}A}{l}   Henries.

Inductance of a Toroid:-

Consider a toroidal ring with N-turns and carrying current I.

let the radius of the toroid be ‘R’ and the total flux emerging be \phi

then flux linkage  \lambda = N\phi

the magnetic flux density inside a toroid is given by B = \frac{\mu _{o}NI}{2\pi R}

\lambda = NBA

where A is the cross sectional area of the toroid then \lambda = N \frac{\mu_{o} NI}{2\pi R}A

\lambda = \frac{\mu_{o} N^{2}I}{2\pi R}A

L=\frac{\lambda }{I}

L=\frac{\mu_{o} N^{2}A}{2\pi R}  Henries.

if the toroid has a height ‘h’ , inner radius \rightarrow r_{1} and outer radius \rightarrow r_{2} then its Inductance is L=\frac{\mu_{o} N^{2}h}{2\pi }ln   Henries.

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continuity equation

The continuity equation of the current is based on the principle of conservation of charge that is  the charge can neither be created not destroyed.

consider a closed surface S with a current density \overrightarrow{J}, then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is I_{out} = \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

since the direction of current is in the direction of positive charges, positive charges also move out of the surface because of the current I.

According to principle of conversation of charge, there must be decrease of an equal amount of positive charge inside the closed surface.

therefore the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the closed surface of the volume.

I_{out} = \frac{-dQ_{in}}{dt}= \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

By Divergence theorem \oint_{s} \overrightarrow{J}.\overrightarrow{ds} = \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

\frac{-dQ_{in}}{dt}= \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

Q_{in}=\int_{v}\rho _{v}dv    implies  -\frac{dQ_{in}}{dt}=-\frac{d}{dt}\int_{v}\rho _{v}dv

for a constant surface the derivative becomes the partial derivative 

\oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\int_{v}\frac{\partial \rho _{v}}{\partial t}dv   -this is for the whole volume.

for a differential volume \overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\frac{\partial \rho _{v}}{\partial t}dv

\overrightarrow{\bigtriangledown }.\overrightarrow{J} =-\frac{\partial \rho _{v}}{\partial t} , which is called as continuity of current equation (or) Point form (or) differential form of the continuity equation.

This equation is derived based on the principle of conservation charge states that there can be no accumulation of charge at any point.

for steady  (dc) currents     \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0 \Rightarrow -\frac{\partial \rho _{v}}{\partial t}=0

from \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0. The total charge leaving a volume is the same as total charge entering it. Kirchhoff’s law follows this equation.

This continuity equation states that the current (or) the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.

 

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E due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density \rho _{L} C/m and lies on Z-axis from -\infty to +\infty.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are (0,\rho ,0) ( a point on y-axis) and assume dQ is a small differential charge confirmed to a point  M (0,0,Z) as co-ordinates.

\therefore dQ produces a differential field \overrightarrow{dE} 

\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}

the position vector \overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} and the corresponding unit vector \widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}

\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

then the Electric field strength \overrightarrow{E} produced by the infinite line charge distribution \rho _{L} is 

\overrightarrow{E} = \int \overrightarrow{dE}

\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

to solve this integral  let Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta

as Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}

Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}

\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})

\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })

\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }

\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb.

\overrightarrow{E} is a function of \rho   only, there is no \overrightarrow{a_{z}} component and \rho is the perpendicular distance from the point P to line charge distribution \rho _{L}.

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Energy Density in Electrostatic Fields

To determine the Energy present in an assembly of charges (or) group of charges one must first determine the amount of work necessary to assemble them.

It is seen that , when a unit positive charge is moved from infinity to a point in a field, the work is done by the external source and energy is expended.

If the external source is removed then the unit positive charge will be subjected to a force exerted by the field and will be moved in the direction of force.

Thus to hold the charge at a point in an electrostatic field, an external source has to do work , this energy gets stored in the form of Potential Energy when the test charge is hold at a point in a field.

when external source is removed , the Potential Energy gets converted to a Kinetic Energy.

In order to derive the expression for energy stored in electrostatics (i.e, the expression of such a Potential Energy)

Consider an empty space where there is no electric field at all, the Charge Q_{1} is moved from infinity to a point in the space ,let us say the point as P_{1}, this requires no work to be done to place a charge Q_{1} from infinity to a point P_{1} in empty space.

i.e, work done = 0 for placing a charge Q_{1} from infinity to a point P_{1} in empty space.

now another charge Q_{2} has to be placed from infinity to another point P_{2} . Now there has to do some work to place Q_{2} at P_{2} because there is an electric field , which is produced by the charge Q_{1} and Q_{2} is required to move against the field of Q_{1}.

Hence the work required to be done is  Potential=\frac{work done}{unit charge}

i.e, V = \frac{W}{Q} \Rightarrow W = V X Q .

\therefore Work done to position Q_{2} at P_{2} = V_{21} X Q_{2}.

Now the charge Q_{3} to be moved from infinity to P_{3} , there are electric fields due to Q_{1} and Q_{2}, Hence total work done is due to potential at P_{3} due to charge at P_{1} and Potential at P_{3}