Transcript Call Centers Explained

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1. Khurram Shahzad
2.
3.
4.
5.
6.
DanIsHussain
Bukhtyar Ali
Shah Mehmood
Farrukh Ali
Usman Akhtar
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Ampere’s Law
KHURRAM SHAHZAD
BS(IT)3rd
ROLL# 07-32
Presented
To:-
Dr. Tariq Bhatti
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Ampere’s Law
Andre Marie Ampere
(1775-1836)
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DEFINITION
“Ampere's Law states that for any closed
loop path, the sum of the quantities (B.ds)
for all path elements into which the complete
loop has been divided is equal to the product
of µ0 and the total current enclosed by the
loop.”
OR
“Relationship between magnetic field and
electric current.”
 B.ds   Ι
0
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EXPERIMENT
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Take a circle of radius r as the Ampere Loop.
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In the Figure:
Dot(.) represents a wire which is placed
perpendicular to the plane in the form of
circle.
Inner circle is the closed path or Amperian
path.
 “r” is the radius of closed path.
“ds” is the displacement across the closed
Path.
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Direction of magnetic field can be shown
by drawing a tangent at any point.
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Thus the dot product of B & the short vector ds is:
B  ds  B  ds cos
where
  0  cos 0  1
So
B  ds  Bds
 B  constant
 B  ds  B ds -------- (A)
Here
 ds  (ds)  (ds)  ...........  (ds)
1
2
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N
as
So
 ds  2r
 B  ds  B (2r )------ (1)
By Biot–Savart law
0 I
B
2r
By substituting the value of B in equation (1)
we get:
0 I
  B.ds 
2r
2r
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 B.ds   Ι
0
Here
…… (2)
 B.ds  ( B.ds)  ( B.ds)
1
2
 ........ ( B.ds) N
So equation (2) can be written as
 B  ds   I

By equation (2),we can say that magnetic field
depends upon current and permeability of free
space.
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Application of Ampere’s Law
DanIsHussain
BS(IT)3rd
ROLL # 07-09
Presented To:Dr. Tariq Bhatti
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Experiment :
A long straight current carrying wire is
placed In a uniform magnetic field.
“a” is the radius of wire.
“R” is the radius of circle.
“R” > “a”.
“B” is constant at each point on circular path.
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As magnitude of “B” has the same value at
each point on the circular path.
So equation A becomes:
Here
as
So
 B  ds  B ds
 ds  (ds)  (ds)  ...........  (ds)
 ds  2R
1
2
 B  ds  B (2r )------ (3)
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N
By substituting the value of B in equation
(3)
we get:
 B.ds  0  I
…… (4)
By Comparing equation (3) & equation (4)
we get:
B 2R    I
0
(R>a)
0  I
B
2R
…… (5)
This is magnetic field outside the wire.
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Now we calculate the effect of magnetic
Field Inside the surface of wire.
By equation (5):
Here
0  I
B
2r
 I  JR 2
…… (6)
…… (7)
I
J=current/cross section area of closed path=

a
2
So equation (7) becomes
R2
II
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a2
By substituting the value of  I in equation
(6) we get:
(R<a)
0 IR
B
2a 2
This is the magnitude of magnetic field
due to the current inside the surface of
wire.
Here
BάI
BάR
Bά1/a2
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The Displacement Current & Ampere's law
BUKHTYAR ALI
BS(IT)3rd
ROLL# 07-18
Presented To:Dr. Tariq Bhatti
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Displacement Current
(a) An Amperian loop
encloses a surface
through which passes a
wire carrying a current
(b) The same Amperian loop
encloses a surface that
passes between the
capacitor plates. No
conduction current
passes through the
surface.
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S
S
1
2
I
I
E
A
Q
Circular metal plates
Q
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Let ‘s1’ and ‘s2’ be two closed surface.
Figure shows a cross-section of the
capacitor and the electric field in the region
Between plates.
Here flux is
Ø =EA
E
Since the field exists only in the region
between plates.
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I  dQ / dt
----------------- (1)
 E=Charge density/permeability of free space
i.e.
E= σ/Є

σ= Q/A
----------------- (2)
0
By substituting the value of σ in equation (2)
we get:
E = Q/AЄ
 Q = Є EA
0
0
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As we know that
Electric Flux=Ø =EA
E
So
Q=Є Ø
0
E
By substituting the value of Q in equation (1)
we get:
d
I  dQ / dt
dt
I  I  Current displacement
d
I 
dt
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E
0
as
so
d
E
d
0
By Ampere law
 B  dl  o  I
By adding displacement current in Ampere law
 B  dl    I  I
o
d

By substituting the value of displacement current
we get:
d 

 B.dl    I   dt 


E
o
-------- (3)
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Equation (3) is the modified form of Ampere's
Law.
 The displacement current and the modified
form of Ampere's Law is an essential part
in study of Electromagnetic waves.
 The effect of the displacement current is
negligible in circuits with slowly varying
Currents and fields except the following
described example.
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Maxwell’s Equations
The four fundamental equations of electromagnetism,
called Maxwell’s equation.
Name
Equation
Gauss’ law for electricity
Gauss’ law for magnetism
Faraday’s law
Ampare-Maxwell law
q enc
 
 E.dA  ε o
 
 B .dA  0
 
dB
 E.d s  
dt
 
dE
 B.d s   oo dt   o i enc
These equations are the basis of many of the equations we see in
“charge particle” to “optics”.
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Force between Current
Shah Mahmood
BS(IT)3rd
ROLL# 07- 41
Presented To:Dr. Tariq Bhatti
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Force between Current
Suppose we have a two current carrying wire
placed in the magnetic field.
We can determine the magnetic force that
one current carrying wire exerts on other.
Consider the arrangement of two parallel
wires I (one) & I (two) separated by a
distance R.
The magnitude of the magnetic field is given
by.
 IR
B 0
2r
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EXPERIMENT
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Force between Current
Once the magnetic field has been
calculated, the magnetic force expression
can be used to calculate the force. The
direction is obtained from the right hand
rule. Note that two wires carrying current
in the same direction attract each other,
and they repel if the currents are opposite
in direction.
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Force between Current
The calculation below applies only to
long straight wires, but is at least
useful for estimating forces in the
ordinary circumstances of short wires.
Once you have calculated the force on
wire 2, of course the force on wire 1
must be exactly the same magnitude
and in the opposite direction
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The magnetic field of an infinitely long
straight wire can be obtained by applying
Ampere's law. Ampere's law takes the form
and for a circular path centered on the
wire, the magnetic field is everywhere
parallel to the path. The summation
then becomes just
The constant μ0 is the permeability of free space.
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Force exerted on wire 2 is
F=I LB
2
1)
1………………………(
B =µ I
1
0 1
By substituting the value of B in equation (1)
we get:
F=µ I I L /2πR
0 1 2
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Gauss’s LAW
FARRUKH ALI
Roll no 07-44
Flux
Flux in Physics is used to two distinct ways.
The first meaning is the rate of flow, such as the amount
of water flowing in a river, i.e. volume per unit area per
unit time. Or, for light, it is the amount of energy per unit
area per unit time.
Let’s look at the case for light:
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Area Vector

Represent an area as a vector DA, of length equal to the
area.
The flux of light through a hole of area DA is proportional
to the area, and the cosine of the angle between the light
direction and this area vector.


If we use a vector L to represent the light energy
per unit time,
 
then the light out of the hole is LDA cos  L  DA . In this case it
is negative (  90) which means the light flux is into the hole.
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Flux of Electric Field
Like the flow of water, or light energy, we
can think of the electric field as flowing
through a surface (although in this case
nothing is actually moving).
We represent the flux of electric field as F
(greek letter phi), so the flux of the electric
field through an element of area DA is
 
DF  E  DA  E DA cos
 
dF  E  dA  E dA cos
When we have a complicated surface, we
can divide it up into tiny elemental areas:
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Gauss’s LAW
USMAN AKHTAR
Roll no 07-17
Gauss’ Law
We are going to be most interested in
closed surfaces, in which case the outward
direction becomes self-evident.
We can ask, what is the electric flux out of
such a closed surface? Just integrate over
the closed surface:  
Flux positive => out
F   dF   E  dA

Flux negative => in
The
symbol has a little circle to indicate
that the integral is over a closed surface.
The closed surface is called a Gaussian
surface, because such surfaces are used by
Gauss’ Law, which states that:
Gauss’ Law
The flux of electric field through a closed surface
is proportional to the charge enclosed.
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Mathematical Statement of
Gauss’ Law
The constant of proportionality in Gauss’ Law
is our old friend 0.
 0 F  qenc
0 


E  dA  qenc
Coulomb’s constant is written ?
kE 
1
 
E  dA  E dA
4 0
We can see it now by integrating the electric
flux of a point charge over a spherical
gaussian surface.
 
 0  E  dA   0 E  dA  0 E 4r 2  qenc
Solving for E gives Coulomb’s
Law.
qenc
E
40 r 2
1
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qenc
r
Example of Gauss’ Law
Consider a dipole with equal positive and
negative charges.
Imagine four surfaces S1, S2, S3, S4, as
shown.
S1 encloses the positive charge. Note that
the field is everywhere outward, so the flux
is positive.
S2 encloses the negative charge. Note that
the field is everywhere inward, so the flux
through the surface is negative.
S3 encloses no charge. The flux through the
surface is negative at the upper part, and
positive at the lower part, but these cancel,
and there is no net flux through the surface.
S4 encloses both charges. Again there is no
net charge enclosed, so there is equal flux
going out and coming in—no net flux
through the surface.
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Field Inside a Conductor
We can use Gauss’ Law to show that the
inside of a conductor must have no net
charge.
Take an arbitrarily shaped conductor, and
draw a Gaussian surface just inside.
Physically, we expect that there is no electric
field inside, since otherwise the charges would
move to nullify it.
Since E = 0 everywhere inside, E must be
zero also on the Gaussian surface, hence
there can be no net charge inside.
Hence, all of the charge must be on the
surface (as discussed in the previous slide).
If we make a hole in the conductor, and
surround the hole with a Gaussian surface, by
the same argument there is no E field through
this new surface, hence there is no net charge
in the hole.
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Field Inside a Conductor
We have the remarkable fact
that if you try to deposit
charge on the inside of the
conductor...
The charges all move to the
outside and distribute
themselves so that the
electric field is everywhere
normal to the surface.
This is NOT obvious, but
Gauss’ Law allows us to
show this!
There are two ideas here
• Electric field is zero inside conductors
• Because that is true, from Gauss’
Law, cavities in conductors have E = 0
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Use Gauss’ Law to Find Out
 
 0  E  dA  qenc
Gaussian Surface
Is E = 0 in the conductor?
Yes, because as before, if there were an electric
field in the conductor, the charges would move in
response (NOT Gauss’ Law).
If we enlarge the Gaussian surface so that it is
inside the conductor, is there any net charge
enclosed?
It looks like there is, but there cannot be, because
Gauss’ Law says E = 0 implies qenc = 0!
How do we explain this?
There must be an equal and opposite charge
induced on the inner surface.
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Summary
Electric flux is the amount of electric field passing through a
closed surface.
Flux is positive when electric field is outward, and negative when
electric field is inward through the closed surface.
Gauss’ Law states that the electric flux is proportional to the net
charge enclosed by the surface, and the constant of
proportionality is 0. In symbols, it is
 0 F  qenc
 
 0  E  dA  qenc
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http://www.rpi.edu/~persap/P2F07_
persans/software/gauss/examples.ht
ml
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