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1.1. Khurram ShahzadKhurram Shahzad
2.2. DanIsDanIsHHussainussain
3.3. Bukhtyar AliBukhtyar Ali
4.4. Shah MehmoodShah Mehmood
5.5. Farrukh Ali Farrukh Ali
6.6. Usman AkhtarUsman Akhtar
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KHURRAM SHAHZAD BS(IT)3rd
ROLL# 07-32
Ampere’s LawAmpere’s Law
Presented Presented To:-To:-
Dr. Tariq Bhatti
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Ampere’s Law Ampere’s Law Andre Marie Ampere Andre Marie Ampere
(1775-1836)(1775-1836)
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“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.”
DEFINITIONDEFINITION
“Relationship between magnetic field and electric current.”
OR
ΙdsB 0.
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EXPERIMENTEXPERIMENT
www.BZUpages.com 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 shownby drawing a tangent at any point.
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cosdsBdsB
10cos0
dsBdsB
Bds dsB
constant B
Thus the dot product of B & the short vector ds is:
So
where
Ndsdsdsds )(...........)()( 21Here
-------- (A)
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By Biot–Savart law
rds 2)2( rBdsB So
as
r
IB
2
0
By substituting the value of B in equation (1) we get:
rr
IdsB
22
. 0
------ (1)
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ΙdsB 0.
I dsB
So equation (2) can be written as
NdsBdsBdsBdsB ).(........).().(. 21 Here
…… (2)
By equation (2),we can say that magnetic field depends upon current and permeability of free space.
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DanIsHussain BS(IT)3rd
ROLL # 07-09
Presented To:-Presented To:- Dr. Tariq Bhatti
Application of Ampere’s Application of Ampere’s LawLaw
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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.
Experiment :
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As magnitude of “B” has the same value ateach point on the circular path.
So equation A becomes:
dsBdsB
Ndsdsdsds )(...........)()( 21Here
Rds 2as
)2( rBdsB So ------ (3)
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IdsB0
.
…… (5)
By substituting the value of B in equation (3) we get:
By Comparing equation (3) & equation (4) we get: IRB
02
R
IB
20(R>a)
This is magnetic field outside the wire.
…… (4)
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Now we calculate the effect of magnetic Field Inside the surface of wire.
By equation (5):
r
IB
20
2RJI Here
J=current/cross section area of closed path= 2a
I
So equation (7) becomes
…… (6)
…… (7)
2
2
a
RII
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By substituting the value of in equation
I
(6) we get:
2
0
2 a
IRB
(R<a)
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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BUKHTYAR ALI BS(IT)3rd
ROLL# 07-18Presented Presented
To:-To:- Dr. Tariq Bhatti
The Displacement Current & Ampere's lawThe Displacement Current & Ampere's law
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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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I I
1S 2S
A E
Q Q
Circular metal plates
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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 ØE=EA
Since the field exists only in the region between plates.
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dtdQI /
E=Charge density/permeability of free space
----------------- (2)E= σ/Є0
σ= Q/A
E = Q/AЄ0
By substituting the value of σ in equation (2) we get:
i.e.
Q = Є0EA
----------------- (1)
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Electric Flux=ØE=EA
As we know that
So Q=Є0ØE
By substituting the value of Q in equation (1) we get:
dtdQI /dt
d E0
dt
dI
E
d
0
dIIas
so
Current displacement
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By Ampere law
IdlB o
do IIdlB
By adding displacement current in Ampere law
By substituting the value of displacement current we get:
dt
dI
E
odlB . -------- (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 enciooosd.Bdt
d E
dt
d Bsd.E
0 AB.d o
encqAE.d
ε
These equations are the basis of many of the equations we see in “charge particle” to “optics”.
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Shah Mahmood BS(IT)3rd
ROLL# 07- 41Presented Presented
To:-To:- Dr. Tariq Bhatti
Force between CurrentForce between Current
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Force between CurrentForce between CurrentSuppose we have a two current carrying wire Suppose we have a two current carrying wire placed in the magnetic field.placed in the magnetic field.
We can determine the magnetic force that We can determine the magnetic force that one current carrying wire exerts on other.one current carrying wire exerts on other.
Consider the arrangement of two parallel Consider the arrangement of two parallel wires I (one) & I (two) separated by a distance wires I (one) & I (two) separated by a distance R.R.
The magnitude of the magnetic field is given The magnitude of the magnetic field is given by.by.
r
IRB
20
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EXPERIMENT
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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.
Force between CurrentForce between Current
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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
Force between CurrentForce between Current
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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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F=I2LB1………………………(1)
B1=µ0I1
F=µ0I1I2 L /2πR
Force exerted on wire 2 is
By substituting the value of B in equation (1) we get:
FARRUKH ALIFARRUKH ALI
Roll no 07-44Roll no 07-44
Gauss’s LAWGauss’s LAW
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FluxFluxFlux in Physics is used to two distinct ways. Flux in Physics is used to two distinct ways. The first meaning is The first meaning is the the rate of flowrate of flow, such as the amount , such as the amount of water flowing in a river, i.e. of water flowing in a river, i.e. volume per unit area pervolume per unit area per unit timeunit time. Or, for light, it is the amount of energy per unit . Or, for light, it is the amount of energy per unit area per unit time.area per unit time.Let’s look at the case for light:Let’s look at the case for light:
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Area VectorArea VectorRepresent an area as a vector , of length equal to the Represent an area as a vector , of length equal to the area.area.The flux of light through a hole of area The flux of light through a hole of area AA is proportional is proportional to the area, and the cosine of the angle between the light to the area, and the cosine of the angle between the light direction and this area vector. direction and this area vector.
A
If we use a vector If we use a vector to represent the light energy per unit time, then the light out of the hole is to represent the light energy per unit time, then the light out of the hole is . In this case it is . In this case it is negativenegative which means the light flux is into the hole. which means the light flux is into the hole.L
ALAL
cos
)90(
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Flux of Electric FieldFlux of Electric FieldLike the flow of water, or light energy, we Like the flow of water, or light energy, we can think of the electric field as flowing can think of the electric field as flowing through a surface (although in this case through a surface (although in this case nothing is actually moving).nothing is actually moving).We represent the flux of electric field as We represent the flux of electric field as (greek letter phi), so the flux of the electric (greek letter phi), so the flux of the electric field through an element of area field through an element of area AA is is
When we have a complicated surface, we When we have a complicated surface, we can divide it up into tiny elemental areas:can divide it up into tiny elemental areas:
cos dAEAdEd
cos AEAE
USMAN AKHTARUSMAN AKHTAR
Roll no 07-17Roll no 07-17
Gauss’s LAWGauss’s LAW
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Gauss’ LawGauss’ LawWe are going to be most interested in We are going to be most interested in closedclosed surfaces, in which case the outward surfaces, in which case the outward direction becomes self-evident.direction becomes self-evident.We can ask, what is the electric flux out of We can ask, what is the electric flux out of such a closed surface? Just integrate over such a closed surface? Just integrate over the closed surface:the closed surface:
The symbol has a little circle to indicate The symbol has a little circle to indicate that the integral is over a closed surface.that the integral is over a closed surface.
The closed surface is called a The closed surface is called a Gaussian Gaussian surfacesurface, because such surfaces are used by , because such surfaces are used by Gauss’ Law, which states that:Gauss’ Law, which states that:
AdEd
Gauss’ LawThe flux of electric field through a closed surface
is proportional to the charge enclosed.
Flux positive => out Flux negative => in
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Mathematical Statement of Mathematical Statement of Gauss’ LawGauss’ Law
The constant of proportionality in Gauss’ Law is our The constant of proportionality in Gauss’ Law is our old friend old friend ..
Coulomb’s constant is written ?Coulomb’s constant is written ?
We can see it now by integrating the electric flux of We can see it now by integrating the electric flux of a point charge over a spherical gaussian surface. a point charge over a spherical gaussian surface.
enc
enc
qAdE
q
0
0
04
1
Ek
r
qenc
encqrEdAEAdE 2000 4
204
1
r
qE enc
dAEAdE
Solving for Solving for EE gives Coulomb’s Law. gives Coulomb’s Law.
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Example of Gauss’ LawExample of Gauss’ LawConsider a dipole with equal positive and Consider a dipole with equal positive and negative charges.negative charges.Imagine four surfaces Imagine four surfaces SS11, , SS22, , SS33, , SS44, as shown., as shown.SS11 encloses the positive charge. Note that encloses the positive charge. Note that the field is everywhere outward, so the flux the field is everywhere outward, so the flux is positive.is positive.SS22 encloses the negative charge. Note that encloses the negative charge. Note that the field is everywhere inward, so the flux the field is everywhere inward, so the flux through the surface is negative.through the surface is negative.SS33 encloses no charge. The flux through the encloses no charge. The flux through the surface is negative at the upper part, and surface is negative at the upper part, and positive at the lower part, but these cancel, positive at the lower part, but these cancel, and there is no net flux through the surface.and there is no net flux through the surface.SS44 encloses both charges. Again there is no encloses both charges. Again there is no net charge enclosed, so there is equal flux net charge enclosed, so there is equal flux going out and coming in—no net flux going out and coming in—no net flux through the surface.through the surface.
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Field Inside a ConductorField Inside a ConductorWe can use Gauss’ Law to show that the We can use Gauss’ Law to show that the inside of a conductor must have no net inside of a conductor must have no net charge.charge.Take an arbitrarily shaped conductor, and Take an arbitrarily shaped conductor, and draw a Gaussian surface just inside.draw a Gaussian surface just inside.Physically, we expect that there is no electric Physically, we expect that there is no electric field inside, since otherwise the charges would field inside, since otherwise the charges would move to nullify it.move to nullify it.Since E = 0 everywhere inside, E must be zero Since E = 0 everywhere inside, E must be zero also on the Gaussian surface, hence there can also on the Gaussian surface, hence there can be no net charge inside.be no net charge inside.Hence, all of the charge must be on the Hence, all of the charge must be on the surface (as discussed in the previous slide).surface (as discussed in the previous slide).If we make a hole in the conductor, and If we make a hole in the conductor, and surround the hole with a Gaussian surface, by surround the hole with a Gaussian surface, by the same argument there is no E field through the same argument there is no E field through this new surface, hence there is no net charge this new surface, hence there is no net charge in the hole.in the hole.
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Field Inside a ConductorField Inside a ConductorWe have the remarkable fact We have the remarkable fact that if you try to deposit that if you try to deposit charge on the inside of the charge on the inside of the conductor...conductor...
The charges all move to the The charges all move to the outside and distribute outside and distribute themselves so that the themselves so that the electric field is everywhere electric field is everywhere normal to the surface.normal to the surface.
This is NOT obvious, but This is NOT obvious, but Gauss’ Law allows us to show Gauss’ Law allows us to show this!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 OutUse Gauss’ Law to Find Out encqAdE
0
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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SummarySummaryElectric flux is the amount of electric field passing through a Electric flux is the amount of electric field passing through a closed surface.closed surface.
Flux is positive when electric field is outward, and negative when Flux is positive when electric field is outward, and negative when electric field is inward through the closed surface.electric field is inward through the closed surface.
Gauss’ Law states that the electric flux is proportional to the net Gauss’ Law states that the electric flux is proportional to the net charge enclosed by the surface, and the constant of charge enclosed by the surface, and the constant of proportionality is proportionality is . In symbols, it is. In symbols, it is
enc
enc
qAdE
q
0
0
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http://www.rpi.edu/~persap/P2F07_persans/software/gauss/examples.htmlWWW.WIKIPEDIA.COM
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