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ELECTROMAGNETISM
Electric Flux:
The number of lines of force passing through a given area is known as electric flux.
The scalar product E.dS is defined as the electric flux for the surface. It is given by
s
E E^ ds
Unit coulomb
N m
2
Electric flux is a property of an electric field. Electric field lines are usually considered
to start on positive electric charges and to end on negative charges. If there is no given
net charge within a given closed surface then every field line directed into the given surface
continues through the interior and is usually directed outward elsewhere on the surface. The
negative flux just equals in magnitude the positive flux, so that the net or total, electric flux is
zero. If a net charge is contained inside a closed surface, the total flux through the surface is
proportional to the enclosed charge, positive if it is positive, negative if it is negative.
Gauss law in Electrostatics:
The mathematical relation between electric flux and the enclosed charge is known
as Gauss law for the electric field. It is one of the fundamental laws of electromagnetism. Gauss’s
law states that the total electric flux (ϕE) over a closed surface is equal to 1/ε 0 times the total
charge Q enclosed within the surface.
S
E
q EdS 0
.
^
Here S is known as Gaussian surface and ε 0 is the permittivity of the free space.
In a dielectric medium Gauss’s law is given by
S
E
q EdS
. where ε is the permittivity of free space
Proof:
Let a charge +Q is placed at O within a closed surface of irregular shape. Consider a
point P on the surface at a distance r from O. Now take a small area ds around P. The normal
to the surface ds is represented by a vector ds which makes an angle θ with the direction of
electric field E along OP. The electric flux dΦE outwards through the area ds is given by
d E E. dS EdS cos …… (1)
Dr. P. Venkata Ramana, AUCE (A)
From coulombs law, the electric intensity E at a point P distance r from a point charge
Q is given by
2 (^40)
r
q E
From eqn (1) and (2), we get
. cos 4
2 0
dS r
q d E
2 0
cos
4 r
q dS d (^) E
But
2
cos
r
dS
is the solid angle dω subtended by dS at O. Hence
d
q d (^) E (^40)
The total flux ΦE over the entire whole surface is given by
d
q E (^40)
Where
d is the solid angle subtended by whole surface at O. This is equal to 4π.
q E
q E ….. (4)
Let the closed surface encloses several charges say +q 1 , +q 2 , +q 3 ,….. +qʹ 1 , +qʹ 2 , +qʹ 3
,….. Now each charge will contribute to the total electric flux. Here it should be remembered
that for positive charges, the electric flux will be outward and hence positive while for the
Dr. P. Venkata Ramana, AUCE (A)
(b) The applicability of the law is limited to situations with simple geometrical symmetry.
Applications of Gauss's law:
- If the charge distribution has spherical symmetry, then Gauss's law can be used with
concentric spheres as Gaussian surfaces.
- If the charge distribution has cylindrical symmetry, then Gauss's law can be used with
coaxial cylinders as Gaussian surfaces.
- If the charge distribution has plane symmetry, then Gauss's law can be used with pill
boxes as Gaussian surfaces.
Electric field due to a uniformly charged sphere:
Case i: At a point outside the charged sphere:
Consider a sphere of radius R with center O. let q be the charge uniformly distributed
over it. Suppose P be an external point at a distance r from the centre of the sphere. By
symmetry, take Gaussian spherical surface with radius r and center O. The Gaussian surface
will pass through P and experience a constant electric field all around as all points is equally
distanced ‘ r ’ from the center of the sphere.
According to Gauss law, the total
electric flux over a closed surface is 1/ε 0
times the charge enclosed within the surface.
0
q E EdS
Since E and dS are in the same direction
q E EdS
q E E dS
0
2
. 4
q E E r (Area of Gaussian surface is 4πr
2 )
2 (^40)
r
q E
The electric field due to a uniformly charged thin spherical shell at a point outside the
shell is such as if the whole charge were concentrated at the centre of the shell.
Case ii: At a point on the surface:
When the point lies on the surface of the sphere, then r=R. in this case, the field
intensity is given by
2 (^40)
R
q E
The electric field due to a uniformly charged thin spherical shell at a point on the
surface of the shell is maximum.
Case iii: At a point inside the charged sphere:
Here we shall find the electric field E at a point which is inside the charged sphere at a
distance rʹ from the center.
The electric flux over the entire Gaussian surface
0
q E EdS
Since E and dS are in the same direction
q E EdS
EdS E rl
S
The total charge enclosed by the Gaussian surface is given by
Charge=volume x volume density
Charge rl
2
According to Gauss’s law
0
q E EdS
E rl rl
2
0
r
E
Expression in terms of charge per unit length of cylinder λ:
Let λ be the charge per unit length of the cylinder, then L
q
Charge= rl
2
Charge= R L
q r l 2
2
Charge= 2
2
R
lr
2
2
0
R
lr E rl
2 2 0 R
r E
Gauss’s law and Coulombs law:
Consider an isolated point charge q. construct a Gaussian
surface of radius r.
At any point on the spherical surface, the electric intensity E
will have the same magnitude and direction normal to the surface.
Both E and ds are directed outward.
According to Gauss’s law, this must be equal to 0
q or ε 0.
0
.
q Eds
As E is constant everywhere at the spherical surface, hence
0 E. ds q
E .( 4 r ) q
2
0 (Since
2 ds^4 ^ r surface area of the sphere)
2 (^40)
1
r
q E
If at the spherical surface, a second charge q 0 is placed, it experiences a force given by
F Eq 0
2
0
(^40)
r
qq F
^ This is Coulombs law.
Force on a current carrying conductor:
Consider the case of a current carrying conductor placed in magnetic field
(perpendicular to paper). We know that current is an assembly of moving charges, therefore
the magnetic field will exert a side way force on the conductor carrying current.
If the charge is infinitely small, then
dF dq ( v B )
Where v is the velocity of charge per unit time
( B )
dt
dl dF dq ( dt
dl v )
( dl B ) dt
dq dF
dF i ( dl B )
When B is uniform over the length of wire, then integrating above equation
F i ( l B )
The force is perpendicular to both l and B
iAB sin^ where A= lb = area of the coil
If the coil has N turns, then
iNAB sin
iNA B
M B where M= NiA = magnetic moment of current circuit.
Biot - Savart’s Law :
The Biot-Savart law is an equation describing the magnetic field generated by a
constant electric current. It relates the magnetic field to the magnitude, direction, length, and
proximity of the electric current. This law is fundamental to magnetostatics, playing a role
similar to that of Coulomb’s law in electrostatics. Biot-Savart observed that the magnetic
field distribution dB at any point due to small element dl of a current carrying conductor
depends upon the following factors.
- It is directly proportional to the current flowing through the conductor
dB i
- It is directly proportional to the length dl of the element considered
dB dl
- It is directly proportional to the sine of the angle θ between length of element and the
line joining the element to the point
dB sin
- It is inversely proportional to the square of the distance r of the point from the element
dl
2
dB r
Combining all these factors 2
i dl sin dB r
When the conductor is placed in vacuum or air, then
0 2
sin
i dl dB r
where
0
4
the proportionality constant and μ 0 is the magnetic permeability of the free
space.
Magnetic Induction:
The magnetic induction at any point in the magnetic field is defined as the magnetic
flux passing through the unit area at that point. It is denoted by B. it is a vector quantity. Its
S.I. unit is Wb/m
2 or tesla (T)
A
B
Where B is magnetic induction, Φ is magnetic flux; A is the area through the
magnetic flux passing.
Magnetic field due to long straight conductor carrying current:
Consider an infinitely long conductor placed in vacuum and carrying a current i. Let
the distance of P from the conductor be R. Consider a small element of length dl from the
conductor at distance l. Let be the distance of the element from the point P. suppose θ be the
angle in clockwise direction which the direction of current makes with the line joining the
element P. The magnitude of the field dB due to small element dl at point P is given by
This is the expression for the magnetic field induction near a long straight conductor.
Magnetic field on the axis of a current loop:
Let there be a circular coil of radius R and carrying current i. Let P be any point on
the axis of a coil at a distance x from the center and which we have to find the field. To
calculate the field, consider a small element d l at the top of the coil.
Let r is the distance of the element from the point P and θ is the angle which the
direction of current makes with the line joining the element to the point P.
The magnetic field dB at point P due to current element of length d l is given by
0 2
sin
i dl dB r
2
0 sin^90
4 r
idl dB
(Angle between dl and r =90°)
2
0
4 r
idl dB
Resolving dB into two components: dBsinϕ along the axis of the loop and another one
is dBcosϕ at right angles to the x - axis. Since coil is symmetrical about x - axis the contribution
dB due to the element on opposite side (along - y axis) will be equal in magnitude but
opposite in direction and cancel out. Thus we only have dBsinϕ component.
The resultant B for the complete loop is given by,
B dB sin
sin 4
2
0 dl r
i B
r
a dl r
i B 2
0
r
a
sin
dl r
ia B 3
0
4
But
dl is the circumference of the coil = 2π a and from figure
2 21 /^2 r a x
a a x
ia B
2 4 ( )
2 23 / 2
0
2 2 3 / 2
2 0
2 ( a x )
ia B
If there is N turns of the coil
2 23 / 2
2 0
2 ( a x )
Nia B
The direction of B is along the axis of the coil
- At the centre of the coil x =0. Thus at the centre of the coil
23 / 2
2 0
2 ( a )
Nia B
3
2 0
a
Nia B
a
Ni B 2
0
- At very far off from the loop x >>a and
2 23 /^23 a x x
3
2 0
2 x
Nia B
Ampere’s law:
Ampere derived a relation between current i and magnetic field B, the relationship is
known as Ampere’s law. According to Ampere’s law the line integral
B. dl for a closed
curve is equal to μ 0 times the net current i through the area bounded by the curve.
B. dl 0 i
where μ 0 is the permeability constant
Proof: Consider a long straight conductor carrying a current i. Let the conductor be
perpendicular to the page directed outward. The magnitude of the magnetic induction at a
distance R from the conductor is given by
- Field at an inside point
- Field at an axial end point
- Field at the centre of solenoid of finite length
(1) Field at an inside point: Here we shall calculate the magnetic induction B at a point P
inside the solenoid axis. For this purpose we divide the solenoid into a number of
narrow equidistance coils.
Consider one such coil of width dx. There will be ndx turns.
Let x be the distance of point P from the centre O of the coil.
The field at P due to elementary coil of width dx carrying a current i is given by
2 2 3 / 2
2 0
2 ( )
a x
ndx ia dB
From the figure ΔABC, we have
dx
rd
sin
sin
rd dx
From ΔAPO,^
2 2 2 a x r
2 23 /^23 a x r
Substituting these values in equation (1)
3
2 0
2
( /sin )
r
n rd ia dB
2 sin
2
2 0
r
n ia d dB
2 0
2 sin
r
n id a dB
sin 2 sin
n i d
dB
2
2
^ sin
r
a
0 n i d sin
dB …… (2)
The field induction B at P due to whole solenoid can be obtained by integrating the
above equation between the limits θ 1 and θ 2. The θ 1 and θ 2 are the semi vertical angles
subtended at P by first and last turn respectively of the solenoid.
2
1
2
1
0 sin
n i d
B dB
2 1 cos 2
0
n i B
1 2
0 cos cos 2
n i B ….. (3)
At any axial point P when it is well inside a very long solenoid, θ 1 =0 and θ 2 =π
cos 0 cos 2
0
n i B
1 ( 1 ) 2
0
n i B
2 2
0 n i B
B 0 n i
This can be taken as the field at the centre of a long solenoid.
(2) Field at an axial end point:
P-type semiconductor N-type semiconductor
Let us consider, a thin rectangular slab of conductor carrying current ( I ) in the
positive x - direction. If we place it in a magnetic field B which is in the y - direction, then the
electrons experience a Lorentz force given by,
FL Bevd
where ‘ e ’ is the charge of the electron and ‘ vd ’ is the drift velocity of the electron.
According to the Fleming's left hand rule, the forces exerted on the electrons are in the
negative y-direction. Hence, potential VH called Hall voltage appears between the upper and
lower surfaces of the semiconductor which establishes as electric field EH called the Hall
electric field. The Hall electric deflecting force is given by,
FH eEH
When equilibrium is reached, the magnetic deflecting force on the charge carriers are
balanced by the electric forces due to electric field.
i.e., FL FH
e v ( (^) d B ) eEH
EH ( vd B )
The relation between current density and drift velocity is d
J
v ne
, where n is the
concentration of charge carriers
EH ( vd B )
H (^ )
J
E B
ne
EH ( JB )
ne
EH RH JB
H H
E
R Hall coefficient ne JB
The coefficient can be evaluated by substituting the quantities EH, J and B. By
knowing the Hall coefficient the carrier density n can be estimated. Since the charge carriers
are holes for p-type material. The Hall coefficient is,
H^1
H
E
R
JB pe
where p is the density of holes
The Hall electric field per unit current density per unit magnetic induction is called
Hall coefficient ( RH ). If ‘ d ’ is the width of the sample across which Hall voltage VH is
measured,
H H
V
E
d
1 H
H
V
R
JB d
H H
E
R
JB
If t is the thickness of the sample, then its cross section is ( d x t ) and current density
I
J
d t
VH R JBdH
H H (^ )
I
V R B
t
H H
V t R IB
Since all the three quantities EH, J and B are measurable, the Hall coefficient RH and
hence the carrier density can be found out.
Generally for n-type material since the Hall field is developed in negative direction
compared to the field developed for a p-type material, negative sign is used while
denoting hall coefficient RH.
Applications of Hall Effect:
- The sing of charge carriers can be determined.
- The carrier density can be estimated.
- The mobility of charge carriers can be measured directly.
- It can be used to determine whether the given material is a metal, an insulator or a
semiconductor.
