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Title: Electromagnetic Field Theory
Description: It contains all the topics of Electromagnetic fields with important objective questions and answers.
Description: It contains all the topics of Electromagnetic fields with important objective questions and answers.
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ELECTROMAGNETIC FIELDS
ELECTRICAL ENGINEERING
CONTENTS
01
...
ELECTRIC FIELD INTENSITY
03
...
DIELECTRICS
05
...
CURRENT DENSITY & CONTINUITY EQUATION
07
...
AMPERE’S LAW
09
...
INDUCTANCE
OF SIMPLE GEOMETRIES
TOPIC – 1 : VECTOR ANALYSIS
1
...
Circuit theory deals with only two variables that is voltage and current whereas
Electromagnetic theory deals with many variables like electric field intensity, magnetic field
intensity etc
...
Hence the
solution becomes complex
...
Maxwell has applied vectors to Gauss’s law, Biot Savart’s law, Ampere’s Law and
Faraday’s Law
...
2
...
Let ‘θ’ be the angle between vectors
A and B
...
B = | A | | B | cosθ
The result of dot product is a scalar
...
an
b) Cross product: is also called vector product
...
i
...
perpendicular to the plane containing the vectors A and B
...
Operator Del ( ∇ ):
Del is a vector three dimensional partial differential operator
...
There are 3 possible operations with del
...
4
...
Consider a scalar function ‘t’
...
∇t
=
(Grad t)
∂ i +
∂x
∇t =
∂ k
∂z
+ ∂t j
∂y
∂t i
∂x
∂ j +
∂y
+
t
∂t k
∂z
Vector
Gradient of scalar function is a vector function
...
In a cable, potential is scalar
...
5
...
Considering a vector function A = Axi + Ayj + Azk
The divergence of vector A mathematically and symbolically expressed as shown below
...
A =
(Div A)
∂ i+
∂x
∇
...
Axi +
Ayj + Azk
∂Az
∂z
Scalar
Divergence of vector function is a scalar function
...
ds = flux through the surface ds
The flux through the entire surface is ∫∫ s D
...
∴
∇
...
ds
0
∆V
6
...
∇xA
=
(Curl A)
∂ i+
∂x
=
∂
∂y
i
∂
∂x
Ax
vector
j
j
∂
∂y
Ay
∂Az _ ∂Ay
∂y
∂z
=
∂
∂z
+
k
Axi +
x
Ayj + Azk
k
∂
∂z
Az
∂Az _ ∂Ax
∂x
∂z
i _
j +
∂Ay _ ∂Ax
∂x
∂y
k
Curl of a vector function is a vector function
...
If the curl of a vector field vanishes, it is called Irrotational field
...
Curl v
Curl v
=
circulation
UnitArea
Lt ∫ v
...
Laplacian of a Scalar function (t) :Double operation
∇
...
8
...
9
...
If associated function is a scalar then it is a scalar field and if the associated function is a
vector function then it is a vector field
...
Basic types of vector fields:
a) Solenoidal vector field (∇
...
Fundamental theorem of Gradient:
Statement: consider an open path from ‘a’ to ‘b’ in a scalar field as shown
...
Z
If ‘t’ is the associated scalar function,
then according to the fundamental theorem of gradient
Scalar field
∇t b
∇tcosθ
b
∫ (∇t)
...
dl
=0
a
Corollary-2:
b
A line integral ∫ (∇t)
...
a
12
...
The volume integral of the divergence of the
associated vector function carried within a enclosed volume is equal to the surface integral of
the normal component of the associated vector function carried over an enclosing surface
...
A )dv = ∫∫ A
...
Fundamental theorem of Curl:- (Stokes theorem)
Statement: Considering an open surface placed in a vector field, the surface integral of the normal
component of the curl of the associated vector function carried over the open surface is equal to the
line integral of the tangential component of the associated vector function along the boundary of the
open surface
...
da =
∫
A
...
da = 0
Since there is no boundary and hence
∫
A
...
dl is constant for a fixed boundary
...
da is independent
of the type of open surface
...
Vector Identities:
a) ∇ x ∇ φ = 0
b) ∇
...
φ A = ∇ φ
...
A)
d) ∇ x φ A = ∇ φ x A + φ (∇ x A)
e) ∇ x ∇ x A = ∇ (∇
...
∇ φ = ∇2 φ
g) ∇( φF) = φ(∇
...
B x C = B
...
A x B
j) ∇
...
∇ x A – A
...
A) - ∇ x (∇ x A)
15
...
a) Cartesian co-ordinate system :- (x,y,z)
Z
dz
p2(x+dx,y+dy,z+dz)
p1(x,y,z)
z
y
dy
Y
x
dx
X
Differential length,
dl = dx i + dy j + dz k
15
...
c) Cylindrical co-ordinate system: ( r, φ, z )
Z
r
z
P(r,φ,z)
Y
φ
X
Cartesian to Cylindrical
r = √ x2 + y2
φ = tan-1 (y/x)
z = z
Cylindrical to Cartesian
x = r cosφ
y = r sinφ
z= z
Note: In cylindrical system unit vectors are r, φ , z
Differential Length Vector:
Ranges:
r = 0
φ = 0
z =-∝
∝
2∏
+∝
Differential length, dl = (dr) r + (rdφ) φ + (dz) z
16
...
Differential volumes: (dv)
a) Cartesian system:
dl = dx i + dy j +dz k
dv = dx dy dz
b) Spherical system:
dl = (dr) r + (rdθ ) θ + (r sinθ dφ) φ
dv = r2 sinθ dr dθ dφ
c) Cylindrical system:
dl = (dr) r + (rdφ) φ + (dz) z
dv = rdr dφ dz
18
...
Cartesian
i
j
k
Spherical
r
sinθ cosφ sinθ sinφ
θ
cosθ cosφ
φ
- sinφ
cosθ sinφ
cosθ
- sinθ
cosφ
0
19
...
Cartesian
i
j
k
Cylindrical
r
cosφ
sinφ
0
φ
- sinφ
cosφ
0
z
0
0
1
20
...
A = h1h2h3
∂t e2
∂u2 +
∂ ( A1h2h3)
∂u1
+
3) ∇ x A = 1
h1h2h3
h1e1
∂
∂u1
A1h1
1
4) ∇2t = h1h2h3
h2h3
h1
∂t e3
∂u3
∂ ( A2h3h1)
∂ ( A3h1h2)
∂u2
+ ∂u3
h2e2
∂
∂u2
A2h2
∂
∂u1
1
h3
h3e3
∂
∂u3
A3h3
∂t
∂u1
+ ∂
∂u2
h3h1
h2
Let: A = A1 i + A2 j + A3 k
Cartesian system
= A1 er + A2 eθ + A3 eφ
Spherical system
= A1 er + A2 eφ + A3 ez
∂t
∂u2
Cylindrical
In Cartesian system:
1) ∇t = ∂t i + ∂t j
∂x
∂y
∂t k
∂z
2) ∇
...
A = r sinθ
1
3) ∇ x A = r2 sinθ
r
∂
∂r
A1
rθ
∂
∂θ
rA2
r sinθφ
∂
∂φ
r sinθ A3
+ ∂
∂u3
h1h2
h3
∂t
∂u3
4) ∇2 t =
1
r sinθ
r2 sin θ ∂t
∂r
∂
∂r
2
+∂
∂θ
r sinθ
r
∂t
∂θ
+∂
∂φ
r
r sinθ
∂t
∂φ
In Cylindrical system:
1) ∇t = ∂t
∂r
2) ∇
...
l along a circular radius of 2 units is
a) zero
b) 2∏
c) 4∏
d) 8∏
3)
which of the following relations is correct?
b) ∇
...
B + A
...
∇B
c) ∇ (AB) = A
...
∇A
d) all the three
4)
∇
...
74
b) 4
...
7
d) 6
...
Find ∫c A
...
where ‘c’ is the curve y = 2x2 in
the x-y plane from (0,0) to (1,2)
a) -9/2
b) 7/6
c) -7/6
d) 2/3
9)
Find the laplacian of the scalar function v = (cosφ)/r (cylindrical system)
...
Z
(0,2,0)
D
( 0,0,0) A
Y
(2,0,0) B
C (2,2,0)
X
a) 0
b) 10
c) -1
d) 8
12) For the vector function A = xy2 i + yz2 j + 2 xz k, calculate ∫c A
...
If
A = 3y2 i + 4z j + 6y k, find the line integral ∫c A
...
Where C is the circumference of the circle
...
a) ( 1, π/3, π)
b) (√ 2 , π/4, π)
c) (√ 2 , -π/4, π)
d) (√ 2 , π/4, -π )
15) Find ∫∫ s (∇x A )
...
Static charges produce electric
field
...
Let us start our study with an introduction of coulomb’s law
Coulomb’s Law:
Q2
Q1
F12
A
B
d
F21
This law states that considering two point charges separated by a distance, the force of
attraction (or) repulsion is directly proportional to the product of the magnitudes and inversely
proportional to the square of the distance between them
...
Michel Faraday gives a satisfactory explanation of coulomb’s law by introducing the
concept of electric field
...
And Q2 experiences a force because it is placed in the electric field of Q1
...
If ‘q’ is the test charge and F is the force experienced by the test charge, then the force per
unit test charge is known as Electric field intensity
...
In order to find electric field intensity at point of
observation P, consider a Unit test Charge ‘q’ c at P
...
In Cartesian system:
Q
F =
4π∈0r2
Q
F =
2
...
2
(x i + y j + z k)
2 3/2
4π∈0(x +y +z )
ELECTRIC FIELD DUE TO A POINT CHARGE LOCATED AT ANY GENERAL
POSITION:
+ QC
∴E =
A
- QC
P
A
P
∴E =
Q
4π∈0(AP)2
AP
Q
4π∈0(PA)2
PA
Electric field is always directed away from the point charge towards the point of
observation(P), if it is a positive charge
...
PRINCIPLE OF SUPERPOSITION:
The principle of superposition says that electric field due to any charge is unaffected by the
presence of other charges
...
E1
E2
E3
Net electric field intensity E = E1 + E2 + E3 +……
...
a) Line charge distribution:
If the charge is continuously distributed along the line with line charge density “ρL” c/m, it is
called line charge distribution
...
c) Volume Charge Distribution:
If the charge is continuously distributed over a volume with volume charge density “ρv “
3
c/m , it is called volume charge distribution
...
Let the
point of observation ‘P’ be on x-y plane
...
ρL
i) E at P = 2π∈ NP
0
ρL
ii) E at P = 2π∈ NP
0
ρL c/m
N
NP,
if it is a +ve line charge
...
P
Electric field due to a finite Line charge(2L) along the perpendicular bisector
...
ρL
...
BN
ii) E at P =
2
2
2π∈0NP √BN +NP
i)
E at P =
A
if it is a -ve line charge
...
PN,
B
α
dq
x
θ
(90-θ)
iii) Net electric field intensity, E = √E2H + E2V
o
dEH
P
(90-θ)
d
dE
iv) If ‘O’ is the mid point, β=(180-α)
...
dEBC
dEAB
dECD
dEDA
Z
P
α
D
ρL c/m
α
β
β
Q
C
d
N
R
2a
a
A
Y
B
M
2b
X
Eat P = EAB + ECD + EBC + EDA
= 2 [ EAB + EBC ]
E=
ρLd
...
2
2
2
2
πεo√2a +d
...
e
...
Let point of
observation ‘P’ be along ‘Z’ axis
...
z
2
2 3/2
2εo(a +z )
Eat P =
Electric field due to an infinite charge sheet:
dEA
rb
dEB
Z
P
ra
B
Z
ρs c/m2
Y
r dφ
A
x
da
Consider two diametrically opposite elementary surface charges located at A & B
...
ρs
...
It has a constant magnitude equal to ρs/2ε0 and has a
direction normal to the surface charge sheet
...
Electric field due to a circular disc along its axis:dEA
dEB
Z
P
Ba
z
ρS c/m2
Y
o dφ
da
A
X
Consider two diametrically opposite elemental surface charges located at A & B
...
Eat p = ρs (1 – z / √a2 + z2) z
2ε0
Z
Gauss’s Law:
r
X
+Qc
Y
Let us consider a point charge of ‘+Q’C at origin
...
The electric field at any point over the closed surface
E = (Q / 4πε0r2)
...
da = Q
...
da = (Q/ε0)
s
Though the above result is deduced with respect to a spherical closed surface enclosed a point
charge, it is a general result applicable for any closed surface enclosing any charge in any form
...
da = (1/ε0) × Qenclosed = (1/ε0) ∫ ρv dv
s
v
Gauss law in integral form (or) Maxwell’s 1st equation
Using divergence theorem,
∫ (∇
...
E = (ρv / ε0)
point form of Gauss law
substitute D = εE
∫∫ D
...
D = ρv
Maxwell’s 1st Equation
Statement:Surface integral of normal component of electric field Vector is equal to (1/εo) times charge
enclosed
...
Gaussian Surface:
Gauss’s law is very useful to find out electric field intensity
...
It must be normal to the
surface considered
...
d) Electric field changes with distance from the center of the sphere
2) A metal sphere with 1m radius and a surface charge density of 10 c/m2 is
enclosed in a curve of 10m side
...
The associated electric field strength is
a) 30 i V/m
b) 30 j V/m
c) 30 kv/m
d) 60 J v/m
5) The electric field strength at a distance point, P due to a point charge, +q, located at the origin, is
100μV/m
...
–
7) Given the potential function in free space to be v(x) = 50x2+ 50y2 + 50z2 volts, the
magnitude (in v/m) and the direction of electric field at point (1, -1,1), where the dimensions are
in meters, are
a) 100; (i+j+k)
b) 100/√3;
(i-j+k)
c) 100/√3; [(-i +j-k)/√3]
d) 100/√3; [(-i –j –k)/√3]
8) In a uniform electric field, field lines and equipotentials
a) are parallel to one another
b) intersect at 45°
c) intersect at 30°
d) are orthogonal
9) When a charge is given to a conductor
a) It distributes uniforming all over the surface
b) It distributes uniformly all over the volume
b) It distributes on the surface, inversely proportional to the radius of curvature
c) It stays where it was placed
...
The electric field in the gap b/w the plates is
a) The same as that produced by one plate
b) Double of the field produced by one plate
b) Dependent on coordinates of the field point d) Zero
13) Three concentric spherical shells of Radii R1, R2,R3(R1
...
a) 0 and 4
b) 3 and 1
c) –3 and 7
d) –2 and 6
14) A positive charge of ‘Q’ coulombs is located at point A(0,0,3) and a negative charge &
magnitude Q coulomb is located at point B (0,0,-3)
...
If the charges are increased by
10%; to get the same force b/w them, their separation must be
a) increased by 21%
b) increased by 10%
c) decreased by 10%
d) non of the above is correct
Two mark Questions
Common data for Q
...
17, 18 & 19
A small isolated conducting sphere of radius r1 is charged with +Qc
...
The inner radius of
the shell is r2, and outer radius r3
...
17) The electric field distribution from 0 to r1 will be
a)zero
b) same
c)increases
d)decreases
18) The electric field from r1 to r2 will be
a) zero
b) same
c)decreases
d) increases
19) The electric field from r2 to r3 will be
a) same
b) zero
c) decreasing
d) increasing
Common data for Q
...
20 & 21
Infinite surface charge sheets are placed along the Y-axis with surface charge density +ρs
c/m and -ρs c/m2 respectively
...
No
...
∞
Z
X
0
1
2
3
4 …………∞
Y
22) An infinite number of charges, each equal to ‘Q’ c, the electric field at the point x = 0 due to
these charges will be
a) Q
b) 2Q / 3
c) 4Q/3
d) 4Q/5
23) The electric field at x = 0, when the alternate charges are of opposite in nature, will be
a) 4Q/3
b) 4Q/5
c) 1
...
r =1
r=2
r=3
24) How much electric flux leaves the surface at r = 5 ?
a) 2π × 10-3
b) 8π
c) 3π × 10-9
d) 8π × 10-9
25) Find electric flux density at P(1, -1, 2)
a) 8
...
3 × 10-10 r
d) 40 × 10-9 r
c) 3
...
a
2
...
b
4
...
c
6
...
c
8
...
a
10
...
b
12
...
b
15
...
b
17
...
c
19
...
d
21
...
c
23
...
d
25
...
b
TOPIC – 3: ELECTRIC POTENTIAL, WORK & ENERGY
Electric Potential:
Z
b
r(b)
r
+Qc
p
Y
r(a)
X
a
Consider +QC of charge at origin
...
We know,
Eat p = (Q / 4πε0r2)
...
dl = (Q /4πε0)[1/r(a) – 1/r(b)]
a
The integral E
...
Now let the starting point be replaced by a reference point (θ) and the ending point be
replaced by the point of observation (p)
...
dl attached with a ‘negative’ sign is known as electric potential at the
θ
point of observation p
...
dl
θ
Note: For finite charge distribution, ‘infinity’ is recommended as the reference point and for “infinite
charge distribution” other than infinity can be assumed as the reference point
...
dl = ∫ E
...
dl
--(1)
A
The fundamental theorem of gradient,
B
V(B) – V(A) = ∫ (∇V)
...
dl --(2)
A
Compare (1) & (2)
E = - ∇V
i) Taking ‘curl’ on both sides
∇ × E = ∇ × (-∇V)
∴
Maxwell’s 3rd Equation
∇×E =0
ii) Taking ‘divergence’ on both sides
∇
...
(-∇V)
= ∇2V ≠ 0
∴
∇
...
Z
p
Electric potential due to a point charge (Absolute potential):
We know,
p
V(p) = - ∫ E
...
r
V(p) = - ∫ (Q / 4πε0r2)
...
r1
r2
Q3
p
r3
Electric Potential due to a continuous charge distribution:
V(p) = ∫ (ρL dl) / 4πε0r
for line charge distribution
= ∫∫ (ρs da) / 4πε0r
s
for surface charge distribution
= ∫ (ρv dv) / 4πε0r
v
for volume charge distribution
Electric potential due to an infinite line charge distribution:
Z
Consider an infinite line charge placed along the Z – axis
...
E = ρ / ε0 --(1)
But,
E = -∇V
--(2)
Substitute,
∇
...
e
...
Solution to Laplace’s equation in Cartesian Co – Ordinates:
Laplace equation, ∇2V = 0
⇒ (∂2V / ∂x2) + (∂2V / ∂y2) + ∂2V / ∂Z2) = 0
Case1: ‘V’ is a function of only ‘x’
∴
V = Ax + B
Case2: ‘V’ is a function of only ‘y’
∴
V = Ay + B
Case3: ‘V’ is a function of only ‘z’
∴
V = Az + B
p
Solution of Laplace equation in spherical co – ordinates:
∇2V = 0
⇒ 1/r2Sinθ [ ∂/∂r(r2sinθ ∂V/∂r) + ∂/∂θ(Sinθ ∂V/∂θ) + ∂/∂φ[(1/Sinθ) ∂V/∂φ)] = 0
Case1: ‘V’ is a function of ‘r’ only
∴
V = -A / r + B
Case2: ‘V’ is a function of ‘θ’ only
∴
V = A ln tan(θ/2) + B
Case3: ‘V’ is a function of ‘φ’ only
∴
V = Aφ + B
Solution of Laplace equation in Cylindrical Co – ordinates:
∇2 V = 0
⇒ 1/r[1/∂r(r ∂V/∂r) + ∂/∂φ(1/r
...
Work Done:
q
Fa
+
+
+
b
a
F
–
–
–
A charge ‘q’ kept in the electric field experiences a force in the direction of electric field
...
Fa is the force applied in opposite direction
...
If Fa is slightly greater than F, the charge can
be moved from point a to point b
...
dl
...
Work done = ∫ Fa
...
dl = -q ∫ E
...
dl
a
Energy: If point ‘a’ is replaced by the reference point ‘θ’ and point ‘b’ is replaced by point of
observation (p), then
p
W = -q ∫ E
...
dl
θ
The above expression represents the energy because this amount of work done is stored in the
form of electrostatic energy
...
Energy stored in this system = ½ (V1Q1 + V2Q2 + …… + VnQn)
In compact form
n
W = 1/2 ∑ Qi Vi
i =1
OBJECTIVES
One Mark Questions
1
...
Taking
potential to be zero at infinity, the potential any point with in the shell (r1 < r < r2) will be
a) q / 4πε0r
p
r
b) q / 4πε0a
q
c) q / 4πε0r2
a
r2
d) q / 4πε0r1
r1
2
...
A point charge of +1nc is placed in a space with a permittivity of 8
...
The
potential difference VPQ between two points P and Q at distance of 40mm and 20mm respectively
from the point charge is
Q
a) 0
...
24 KV
d) 15 V
1nc
40mm
4
...
Equation ∇2V = -ρ/∈ is called the
a) Poisson’s equation b) Laplace equation
c) One Coulomb / Joule
d) None
c) Continuity equation
d) None
6
...
If the
potential at the corner A is taken as 1V, then the potential at B, the centre of the square will be
a) zero
Q
A
b) 1/√2 V
B
c) 1 V
d) √2 V
-Q
7
...
Two spheres of radii ‘r1’ and ‘r2’ are connected by a conducting wire
...
Now,
a) larger sphere will have greater potential
b) larger sphere will have smaller potential
c) both the spheres will have same potential
d) smaller sphere will have zero potential
9
...
A sphere of radii 1m can attain a maximum potential of
a) 3 × 106 V
b) 30 KV
c) 1000 V
11
...
-Q
1
0
+Q
4
An infinite number of concentric rings carry a charge Q each alternately positive and
negative
...
metres in geometric progression as shown
...
Find the work involved in moving a charge of 1C from (6,8,-10) to (3,4,-5) along a straight line in
the field E = -xi + yj - zk
...
5 Joules
b) 25
...
Find the work done in moving a point charge 3 μc from (2,π, 0) to (4, π,0) in the field E = 105/r r
+ 105 z z
...
207 Joules
b) 1
...
8 Joules
d) zero
15
...
Find the potential at the origin
...
A line charge of 10-9/2 c/m lies on the Z – axis
...
24 V
c) 6
...
24 V
17
...
4 nc is located at (2,3,3) in Cartesian system
...
a) 2
...
6 V
c) 4
...
1 V
18
...
a) 12
...
8 V
d) 13
...
3 point charges of 1C, 2C and 3C are located at the corner of an equilateral triangle of 1m side
each
...
a) 9 / 4π∈0 Joules
b) 4π∈0 / 3 Joules
c) 11 / 4π∈0 Joules d) 30 × 109 Joules
20
...
How much charge is located
with in a sphere of 1m radius centered at the origin
...
21
...
Find the potential outside the spherical shell
a) Q02 / 4π
b) Q0 / 4πε0a
c) zero
d) zero
d) Q0 / 4πε0r
Linked Question
Two parallel infinite conducting plates separated by a distance ‘d’ along the X – axis have a potential
V0 and zero respectively as shown
...
Find the expression for voltage distribution
a) V = V0(1 + d/x)
b) V = V0(1 – x/d)
24
...
i
d) 0
d) (x / V0)
...
a
2
...
b
4
...
a
6
...
b
8
...
a
10
...
b
14
...
d
16
...
a
18
...
c
20
...
c
22
...
b
24
...
d
13
...
It is capable of storing energy for a short duration
...
Electric Dipole: Two equal and opposite charges separated by a small distance is called a dipole
...
+q
Dipole moment p = q s
s
–q
E Applied
–+
–+
Fig(1)
Polar Dielectric without
applied electric field
–+
–+
fig(2)
Polar Dielectric with
applied Electric field
Polar Dielectrics: The charges in the molecules of polar type have permanent displacement from
each other
...
They are randomly oriented as shown in
fig(1)
...
When an electric field is applied, the dipoles orient in a particular direction such that the
induced electric field is in a direction opposite to the applied electric field
...
Non – Polar Dielectrics: In non – polar dielectrics, the centres of positive and negative charges
coincide each other
...
Potential due to a dipole: Let us consider a physical dipole located on Z – axis and the point of
observation P(r, θ, φ)
...
It is easy to handle this problem using spherical co – ordinates
...
Therefore, the potential at ‘p’ due to the physical dipole is given by
V(p) = V1 + V2
= q / (4πε0ra) + (-q) / (4πε0rb)
∴
V(p) = q / (4πε0)[1/ra - 1/rb]
∴
V(p) = q/(4πε0)[(rb – ra) / rarb]
Case: When the point of observation is at a very large distance α = β = θ and ra = rb = r
rb – ra = BC
ra
∴ V(p) = q / (4πε0) [BC / rarb]
α
r
= q / (4πε0) [Scosθ / r2]
[Q BC = S cosθ]
+q A
θ
rb
s
2
β
V(p) = (p cosθ) / (4πε0r )
∴
[Q p = q S]
C
-q B
V(p) α 1/r2
Electric field intensity due to a Dipole:
We know
E=-∇V
E = p / (4πε0 r3)[2cosθ r + sinθ θ]
∴
∴
in spherical system
E ∝ (1/r3)
Observations:
i) Potential due to an electric dipole V(p) ∝ 1 / r2
ii) Electric field intensity due to an electric dipole E ∝ 1/r3
Polarization (P)
Some materials already contain the internal electric dipoles
...
Many materials do not contain any internal electric dipoles
...
Qualitatively defined as production and / or alignment of internal electric dipoles
...
∴
P = p / dv
units for polarization is coulomb / m2
...
χ = εr – 1
Displacement density is directly proportional to electric field intensity
...
Polarization density (p) is directly
proportional to the applied electric field
...
D
[∵ χ = εr – 1]
Gauss’s Law for Dielectrics:
We know that differential form of Gauss law in free space
...
E = ρf / ε 0
Where ρf
free volume charge density
...
–+ –+ –+ –+ –+
The positive charge is nullified by the negative charge near by (or) the head and tail gets
cancelled throughout except at the beginning and end
...
This charge is called ‘Bound charges’
...
∇
...
∫∫ E
...
D = ρf + ρb
Dielectric Boundary Conditions:
When flux lines flow through a single medium, they are continuous
...
This can be studied by
using boundary conditions
...
Surface of porcelain
insulator is a porcelain air boundary
...
An infinite charged
sheet with charge density ρs c/m2 is placed at the boundary
...
θ1 is the angle of incidence
...
Dn1 and
Dn2 are the normal components of flux density vectors
...
Apply Gauss’s law to
the pill box under limiting condition Δh 0
...
da = Qf enclosed
S
Dn2 ∫da – Dn1∫da = ρsf × A
Dn2A – Dn1A = ρsf A
∴
Dn2 – Dn1 = ρsf
Statement: Normal component of flux density vector is discontinuous by an amount equal to the
charge density of the sheet
...
∴
Dn1 = Dn2
Statement: Normal components of flux density vectors are equal
...
Boundary condition for Electric field intensity vector(E):
Second boundary condition deals with tangential
component of electric field
...
(2) εr2
Et1 and Et2 are the tangential components of the electric
(1) εr1
field in media 1 and 2 respectively
...
A
B
θ2
θ1
D
Δl
E2
Et2
Δh
Et1
C
E1
We know that static electric field is a conservative field
...
dl = 0
Apply this equation to the contour ABCDA under limiting condition Δh
∫ E
...
dl + ∫ E
...
dl = 0
...
Et2 ∫ dl – Et1∫dl = 0
Et2 Δl – Et1 Δl = 0
∴
0
...
Relation between angle of incidence (θ1) and angle of emergence(θ2):
Dn2 = D2 Cosθ2
Assume interface does not contain any surface charge
E2
D2
Apply boundary condition for D,
θ2
Dn2 = Dn1
[Q ρsf = 0]
(2) εr2
D2cosθ2 = D1 cosθ1
(1) εr1 E1t= E1sinθ1
E2t = E2Sinθ2
θ1
ε2E2cosθ2 = ε1E1cosθ1 ---- (1)
Apply boundary condition for E,
D1
E2t = E1t
E1
E2sinθ2 = E1sinθ1
---(2)
Dn1 = D1 Cosθ1
(2) ÷ (1)
=
E1sinθ1
E2sinθ2
ε2E2cosθ2
ε1E1cosθ1
∴
Tanθ1 = εr1
Tanθ2
εr2
If εr1, εr2 and angle of incidence are given, angle of emergence can be calculated using the above
equation
...
Capacitance is the property of a dielectric to store electrical energy
...
The dielectric is subjected to
electric stress and strain
...
Capacitance is similar
to inertia
...
Similarly voltage across
capacitor cannot change suddenly
...
dl |
=
=
---------
ρs (Area)
∫ ρs i
...
Apply Gauss’s law
D
...
4πr2 = Q
∴ E = Q/ 4πεor
Q
+ +
+
+
+
a
+
+
+r + +
2 r
b
electric field exists only in the direction of r
a
a
we know, v = - ∫ E
...
dr
b
= Q/4πε [1/a - 1/b]
Q = 4πε v ab
b–a
substitute Q = CV
CV = 4πε v ab
b–a
C = 4πε ab
b–a
Capacitance of cylindrical capacitor (or) cable:
Consider a Gaussian cylinder (G)
...
da = Qenclosed
D
...
dr
b
[∵ l = 1mt]
b
g
a
= -∫
b
Q dr
2πεr
b
= Q [ ln r]
a
2πε
= Q
2πε
[ lnb – lna]
V = Q ln (b/a)
2πε
Substitute Q = CV
C = 2πε
ln(b/a)
Farad/m
Note: To calculate the total capacitance, multiply with total length
Capacitance of a 2-wire transmission line:
Single phase transmission line is shown
...
Radius of
each conductor is ‘r’ and spacing is ‘d’ meters
...
The distance between the wire B
and the point P is (d-x)
...
Direction of electric field is away from
the positive charge or towards the negative charge
...
C
C = c| x c| = c|
c| + c|
2
c| = 2c
c| = 2πε
ln ((d-r)/r)
F/m/conductor
Energy stored in capacitor:
we know that energy stored in ‘n’ point system,
n
W = 1/2 Σ q v(pi)
i=1
= 1/2 ∫ (ρsda) v
s
+
+
+
+
+
C
V
= 1/2 ρs v ∫ da
s
= 1/2 x Q x V x a
a
= 1/2 QV
[∵ Q = CV]
V
-
W = 1/2 cv2
Joules
Energy density = Energy
Volume
= 1/2 cv2
Axd
= 1/2 x 1 x εA x v2
Axd
d
= (1/2) ε (v/d)2
= 1/2 εE2
= 1/2 (εE) E
= 1/2 DE
[∵v/d = E]
D and E can be written as D
...
E
Energy = 1/2 D
...
E dv
v
Force of Attraction between plates:
Between the oppositely charged plates there is a force of
attraction
...
The work done is stored in the form of energy in the
additional volume Adx
...
Every line current is
associated with a mobile line charge density λ c/m
...
The amount of mobile charge contained at any instant within the elementary segment is
λ(VΔt)
...
All these mobile charges coming out of segment in Δt seconds is called current
...
Every surface current is
associated with a mobile surface charge density σ c/m2
...
The amount of mobile charges contained at any instant within the elementary rectangle is
“σ Δl⊥(VΔt)”
...
ΔI = σΔl⊥(VΔt)
Δt
ΔI = σ V Δl⊥
ΔI = σV = K
Δl⊥
K = σV
, A/m
where K = surface current density, A/m
ρ c/m3
Volume Currents:
Δa⊥
vΔt
Flow of electric charges over a volume represents volume currents
...
Considering an elementary cylinder within
the volume current region, the amount of mobile charges contained at any instant is “ρ Δa⊥(V Δ t)”
...
ΔI = ρΔa⊥ (VΔt)
Δt
ΔI = ρVΔa⊥
ΔI = ρV = J
Δa⊥
∴
J = ρV
A/m2
Where J = Volume current density, A/m2
Continuity Equation:
ρ c/m3
dv
Enclosing
surface
Let us consider a region carrying volume currents
...
The net outward current through the enclosing surface can be obtained as
...
da --s
(1)
[ from volume currents]
And also, the rate of reduction of electric charges within the encloser
...
da = - d/dt ∫ ρ dv
s
V
According to the fundamental theorem of divergence
∫ (∇
...
Therefore ∂/∂t can be taken inside the integral since the variables are different
...
J) dv = - ∫ (∂ρ / ∂t) dv
V
V
∇
...
The above equation is called continuity equation or Fifth Maxwell’s equation
...
Net overflow of current per unit volume is negative of time rate of charge per unit volume
...
The above equation explains continuity of current
...
Some charge keeps flowing in the circuit
...
E
Ohm’s Law:
J
A
B
l
V
Current flowing through a conductor is directly proportional to the potential difference across
it, provided temperature is kept constant
...
Joule’s Law:
According to Joules law, whenever current flows through a conductor, heat energy is
produced
...
Heat Energy ∝ I2Rt
Energy ∝ (I2Rt) / J1
Where J1 is called Joule’s constant
...
Al / Al
Rearrange the terms,
P = V × l (Al)
l
A
Substitute E = V/l, J = I/A and volume = Al
∴ P = EJ volume
EJ can be written as E
...
J) volume
∴
P = ∫ (E
...
Relaxation Time:
To study relaxation time we start with ohm’s law and equation of continuity
...
J = - (∂ρ / ∂t)
∇
...
ε σE = - (∂ρ / ∂t)
ε
σ
- ∂ρ
ε ∇
...
The charge density decays exponentially as time passes with time constant equal to ∈/σ seconds
...
Conductance – Capacitance Theorem:
G = Conductance, C = Capacitance
σ = Conductivity, ε = permittivity ρ = resistivity
According to conductance theorem, conductance of an insulated medium is equal to σ/∈
times the capacitance of the insulation provided between two conducting media
...
Conductance can be obtained by multiplying
capacitance expression with σ/∈
...
da
s
Substitute ohm’s law J = σE
I = ∫∫ σE
...
da = Q
s
∫∫ E
...
(Q / ε)
Substitute I = V/R and Q = CV
V = σ
...
Duality means that it is possible to pass from one equation to another equation by suitable
interchanges of dual quantities
...
Capacitance is
εA/l
...
For example capacitance of cylindrical capacitor is
C = [(2πε) / log(b/a)]
Conductance can be obtained by replacing ‘ε’ with ‘σ’
...
∴ G = (4πσab) / (b – a)
Therefore, ε and σ are dual quantities
...
If there is a field inside, the charges experience a force and
they move outwards
...
Q = 0, D = 0 and E = 0
2) The charges can only reside on the surface of the conductor and not inside a conductor
...
4) Electric field intensity at all points on the surface of a conductor must be normal to the surface
...
OBJECTIVES
One Mark Questions
1
...
5mm diameter
...
a) capacitance
b) charge
c) dielectric breakdown
d) tan δ
2
...
ds = ∫∫∫ ρ dV
b) V × H = D
c) ∇
...
E = ρ/ε
3
...
When a charge is given to a conductor
a) it distributes uniformly all over the surface
b) it distributes uniformly all over the volume
c) it distributes on the surface, inversely proportional to the radius of curvature
d) it stays where it was placed
5
...
c source is
a) the same as that due to a d
...
c source of equivalent magnitude
c) zero
d) none
6
...
The potential difference between the forces A and B of a uniformly polarized infinite slab shown
in figure
...
If n is the polarization vector and K is the direction of propagation of a plane electromagnetic
wave, then
a) n = K
b) n = - K
c) n
...
Consider the following statements regarding field boundary conditions:
1
...
2
...
The discontinuity in the normal component of the flux density at a dielectric conductor
boundary is equal to the surface charge density on the conductor
...
The normal component of the flux density is continuous across the charge free boundary
between two dielectrics
...
Consider the following statements associated with a parallel plate capacitor
...
Capacitor is proportional to area of plates
2
...
The dielectric material is in a state of compression
...
Two electric dipoles aligned parallel to each other and having the same axis exert a force F on
each other, when a distance ‘d’ apart
...
When a lossy capacitor with a dielectric of permittivity ε and conductivity σ operates at a
frequency ω, the loss tangent for the capacitor is given by
b) ωε / σ
c) σ / ωε
d) σωε
a) ωσ / ε
13
...
The characteristic impedance of a co – axial cable depends on
1
...
length of the cable
3
...
logarithmic ratio of outer and inner diameter and inversely as the square root of dielectric
constant
...
The unit of μ0ε0 is
a) Farad Henry b) m2 / sec2 c) amp sec / volt sec
d) Newton metre2/coulomb2
16
...
B = 0
d) ∇ × H = J + ∂D/∂t
a) ∇
...
n ds = 0
17
...
(ε∇V) = - ρ
d) ∇
...
A material is described by the following electrical parameters as a frequency of 10 GHz
...
The material at this frequency is considered to be
(σ0 = 1/36π × 10-9 F/m)
a) a good conductor
b) a good dielectric
c) neither a good conductor nor a good dielectric
d) a good magnetic material
19
...
For a dipole antenna
a) The radiation intensity is maximum along the normal to the dipole axis
b) The current distribution along the length is uniform irrespective of the length
c) The effective length equals its physical length
d) The input impedance is independent of the location of the feed – point
21
...
An antenna, when radiating, has a highly directional radiation pattern
...
A composite parallel capacitor is made up of two different materials with different thickness (t1
and t2) as shown
...
The
voltage of the conductivity foil is
...
5mm
εr2 = 4; t2 = 1mm
100V
F
c) 67 V
0V
d) 33 V
2
...
1 mm between the
electrodes
...
85 × 10-12 F/m
...
The stored energy in the capacitor is
a) 8
...
1 nJ
d) 44
...
A circular ring carrying a uniformly distributed charge Q and a point charges –Q on the axis of the
ring are shown
...
Find the polarization in a dielectric material with εr = 2
...
-2
2
d) 0
a) 1
...
602 × 10 c/m
5
...
5 and P is 2
...
a) 7
...
1 × 10-3
c) 74
...
Calculate the emerging angle by which the vector E changes its direction as it passes from a
medium with εr = 100 into air making an angle of 45° with the interface as it enters
a) 90°
b) 0
...
89°
d) 45°
7
...
The electric
field in the insulator is 1000V/m
...
46°, 400 V/cm
b) 2
...
2°, 4925 V/cm
d) 9
...
No 8 to 11
A parallel plate capacitor consists of two square metal plates of side 500 mm and separated by
a 10 mm slab of Teflon with εr = 2 and 6mm thickness is placed on the lower plate leaving an air gap
of 4mm thick between it and upper plate
...
8
...
16 × 10-10 F
a) 2
...
26 × 10-6 F
9
...
12 μc/m2, 0
...
35 μc/m2, 0
...
11 μc/m2, 0
...
Find the electric field intensity of Teflon and air
a) 12555 V/m, 6776 V/m
b) 13553 V/m, 6776 V/m
c) 0, 5826 V/m
d) 38265 V/m, 38265 V/m
11
...
21 V, 40
...
11 V, 34
...
1 V, 2
...
Two conducting planes are located at Z equal to ‘0’ and 6 mm
...
Find the capacitance per
square meter of surface if the region for 5 < Z < 6 mm is filled with air
...
8 nF/m2
b) 3
...
1 nF/m2
d) 2
...
A 2 μF capacitor is charged by connecting it across 100V D
...
The supply is now
disconnected and the capacitor is connected in parallel with another uncharged 2μF capacitor
...
a) 0
...
005 Joules
c) 1
...
5 Joules
14
...
It is charged to 100V by connecting it across a battery
...
6 × 106 Joules
b) 0
...
23 × 104 Joules
d) 1 × 10-6 Joules
15
...
The maximum voltage gradient should not exceed
300KV per cm
...
Take the relative permittivity of gas to be
1
...
a) 3
...
2 cm, l = 1m
c) 8
...
7 cm, l = 7m
Key:
One Marks:
1
...
a
3
...
a
5
...
a
7
...
c
15
...
b
17
...
a
19
...
a
21
...
c
3
...
a
5
...
b
7
...
b
9
...
a
11
...
c
13
...
d
9
...
b
11
...
b
13
...
d
Two Marks:
1
...
d
2
...
Magnetic field:
A static magnetic field can be produced from a permanent magnet or a current carrying
conductor
...
The field exists as concentric circles having centres at the axis of conductor
...
The unit of magnetic
flux is weber
...
Magnetic flux density (B):
The magnetic flux per unit area is called magnetic flux density (or) magnetic induction vector
...
The magnetic flux through any surface is the surface integral of the normal component of B
...
B = dφ /da
dφ = B
...
da
S
Magnetomotive force ( M
...
F)
M
...
F is produced when an electric current flows through a coil of several turns
...
M
...
Therefore, the unit for M
...
F is ampere turns
...
Reluctance (s):
Reluctance is the opposition to the establishment of magnetic flux and can be defined as the
ratio of M
...
F to the flux produced
...
The reciprocal of reluctance is called “PERMEANCE”
...
According to this law considering a current element of length ‘dl’ carrying a current
‘I’, the magnetic flux density at a point of observation ‘p’ is elementary field intensity
...
H = ∫ Idl x r
4π | r3|
B = μ /4πr3 ∫ Idl x r
Taking divergence on both sides,
div
...
( u x v ) = v
...
curl v
div (Idl x r ) = r
...
curl r
∇
...
curl Idl – Idl
...
The current element vector and distance vector have no rotation
...
∇
...
∇
...
It is also called second Maxwell’s equation
...
Let ‘p’ be the point of observation on the x-y plane at a distance ‘r’ from the zaxis
...
Let ‘p’ be the
point of observation on XY plane
...
e α = 0, β = 180o
∴
B = μ0I
...
φ
4πr
Corollary-3:
Magnetic field due to finite along the perpendicular bisector
i
...
cosα φ
2πr
Magnetic field due to a circular current carrying loop along its axis:
Z
dBB
dBA
p
d
d
B
r
a
Y
φ
A
I
X
Consider a circular loop of radius ‘a’ lying on a x-y plane with centre at origin and carrying a
current I as shown
...
Considering
two diametrically opposite current element located at A & B
...
Resolving
dBA & dBB vectors horizontal and vertical components, we find that horizontal components get
cancelled and vertical components added up
...
z
2(a2+d2)3/2
Corollary-1:
Magnetic field due to circular current carrying loop at its centre i
...
∴
B = μ0I z
2a
Corollary-2:
Magnetic field due to a semicircular current carrying loop at its centre
∴
B = μ0I
4a
z
Corollary-3:
Magnetic field due to a thin circular coil of ‘N’ turns along the axis
B = μ0NIa2
...
of turns per unit length
(n=N/l) and carrying a current I
...
Consider an elemental thickness ‘dx’ at a distance ‘x’ from the origin
...
The direction of the magnetic field depends on the sense of current carrying by the solenoid
and the right hand screw rule
...
let ‘n’ be the no
...
Assume that the solenoid axis coincides with the xaxis and the origin coincides with centre
...
∴
B = μ0nI (cosβ+cosα)
2
Corollary-1:
Magnetic field due to an infinite circular solenoid
i
...
e α = β
B = μ0nI cosα
Corollary-3:
Magnetic field at the end of a finite circular solenoid
β = 90o, α
B = μ0nI cosα
2
Magnetic field due to an infinite surface current sheet:
Z
dBB
α
dBA
α
α
P
α α
ra
rb
dy
dy
d xy-plane
0
Y
y
y
k
B
X
A
Let us consider an infinite current sheet lying on x-y plane carrying a surface current along
the positive x-direction ( K ) with a surface current density K
...
Net magnetic field
B = μ0k (-j)
2
It the point of observation is below the surface current sheet, then
B = μ0k j
2
Note:
The magnetic field due to an infinite surface current sheet is independent of the distance of the point
of observation from the sheet
...
Where n is a unit vector normal to the
sheet directed away from the sheet towards the point of observation
...
Z
Consider an infinite straight conductor lying along the
Z-axis carrying a current I along the +ve Z-axis
...
Considering any point
‘P’ on the closed path, the magnetic flux density at the point ‘P’ is
given by
b
P
I
μI
B= 0 φ
2πr
dl = (dr) r + (r dφ) φ + (dz) z
B
...
r dφ
2πr
μ0I 2π
∫ B
...
dl = μ0 Ienclosed
Ampere’s law in integral form
C
Statement :
Considering any closed path in a magnetic field the line integral of tangential
component of the magnetic field around the closed path is equal to μo times current enclosed
...
dl =
∫∫
μoIenclosed
((∇ x B)
...
da
∇ x B = μoJ
(or)
∇xH = J
Point from (or) Maxwell’s 4th equation
2
...
Inside (r < a) :
Ai r A0
Considering the ampere loop A and applying Ampere’s Law,
Bi =
μoIr φ
2πa2
∴Bαr
Outside ( r > a) :
Considering an ampere Loop Ao and applying
Ampere’s Law,
μ0I
2πa
μoI φ
2πr
Bo =
B
B α 1/r
r
r=a
3
...
Apply Ampere’s Law
a
b
Bi = 0
Case (ii) : (a < r < b)
B
Construct an ampere’s loop and apply Ampere’s Law
B =
μoI
2πr
r
(r2-a2) φ
(b2-a2)
Case(iii) : (r > b)
Construct an ampere’s loop and apply Ampere’s Law
B =
μoI φ
2πr
4
...
Maxwell’s Equations for time varying fields:
Differential Form
Integral Form
1
...
∫ D
...
Div B = 0
2
...
da = 0
s
3
...
∫ E
...
da
∂t s
4
...
∫ H
...
da + ∫ Jd
...
Div J = - ∂ρ
∂t
5
...
da = - ∂ ∫ ρdv
s
∂t v
2
...
2
...
4
...
Div D = ρ
Div B = 0
Curl E = 0
Curl H = J
Div J = 0
3
...
2
...
4
...
Div D = 0
Div B = 0
Curl E = - (∂B/∂t)
Curl H = (∂D/∂t)
Div J = 0
4
...
2
...
3
...
Div D = 0
Div B = 0
Curl E = - (∂B/∂t)
Curl H = J
Div J = 0
5
...
2
...
4
...
Div D = 0
Div B = 0
Curl E = - (∂B/∂t)
Curl H = (∂D/∂t)
Div J = 0
6
...
2
...
4
...
Substitute (∂D/∂t) = jωD; (∂B/∂t) = jωB
Div D = ρ
Div B = 0
Curl E = -jωB
Curl H = J + jωD = σE + jωεE = (σ+jεω)E
Div J = - (∂ρ/∂t) = -jωρ
7
...
(1)
Curl H = J +(∂D/∂t)
∇ x H = (∂D/∂t) since J = 0 in free space
∴ Curl H = ε (∂E/∂t)
……
...
H) - ∇2 H = ε ∂/∂t (∇ x E)
……
...
H = 0 and substitute (1) in (3)
- ∇2 H = ε ∂/∂t (-μ ∂H/∂t)
∇2 H = με ∂2H/∂t2
∴
in free space εr = 1, μr = 1
∇2 H = μ0 ε0 ∂2H/∂t2……
...
From equation 1:
∇ x E = - μ (∂H/∂t)
taking curl on both sides and substitute ∇ x H = ∂D/∂t
∇ (∇
...
E = 0
- ∇2 E = -μ ∂/∂t (ε ∂E/∂t)
∴
∇2 E = μ0 ε0 ∂2E/∂t2 …
…
...
OBJECTIVES
One Mark Questions
1
...
Maxwell’s divergence equation for the magnetic field is given by
(a) ∇ × B = 0 (b) ∇
...
B = ρ
3
...
For free space ∇ × H = (σ + jωε)E
2
...
B = ρ
3
...
For static electric field ∇
...
When an iron core is placed between the poles of a permanent magnet as shown below, the
magnetic field pattern is:
(a)
S
N
(b)
S
N
(c)
S
N
(d)
S
N
5
...
K
...
The reflection coefficient, characteristic impedance and load impedance of a transmission line are
connected together by the relation
(a) Kr =
ZL + Z0
Z0 – ZL
(b) Kr =
Z0 ZL
Z0 – ZL
(c) Kr =
ZL - Z0
ZL + Z0
(d) Kr =
ZL - Z0
Z0 ZL
7
...
Poynting vector signifies
(a) current density vector producing electrostatic field
(b) power density vector producing electromagnetic field
(c) current density vector producing electromagnetic field
(d) power density vector producing electrostatic field
9
...
The velocity of a traveling wave on the transmission line is:
(a) Z0C
(b) 1 / (Z0C)
(c) Z0 / C
(d) C / Z0
10
...
To attain it, in a line,
(a) of all the parameters, it is best to increase L for distortionless transmission
(b) Keeping R, and L constant it is preferable to increase or decrease G, and C
(c) the inductance can be added at any interval
(d) the inductance can be of any value
11
...
Which one of the following statements DOES NOT pertain to the equation ∇
...
For incidence from dielectric medium (ε1) into dielectric medium 2(ε2) the browster angle θp and
the corresponding angle of transmission θt for ε2/ε1 = 3 will be respectively
(c) 60° and 30°
(d) 60°and 60°
(b) 30° and 60°
(a)30° and 30°
14
...
A very lossy, λ/4 long, 50 ohms transmission line is open circuited and the load end
...
The intrinsic impedance of copper at high frequencies is
(b) purely inductive
(a) purely resistive
(c) complex with a capacitive component
(d) complex with an inductive component
17
...
The equation ∇
...
A slab of uniform magnetic field deflects a moving charged particle by 45° as shown in figure
...
In the figure shown below, the force acting on the conductor PQ is in the direction of
(a) PQ
(b) QP
I
I
n
(c) – n
(d) n
Q
P I
-n
21
...
If R
is the resistance per unit length of the wire, the poynting vector at the surface of the wire will be
(b) RI2
...
n
I
2πr
2πr
n
r
(c) RI2
...
(- n )
2π
2π
22
...
Due
to polarization mismatch, the power transfer efficiency from the wave to the antenna is reduced to
about
(a) 50%
(b) 33
...
Following equations hold for the time – varying fields:
i ) ∇ × E = - (∂B/∂t)
ii) E = - ∇V – (∂A/∂t)
iii) ∇2V + ∂/∂t(∇
...
Match List – I with List – II and select the correct answer using the codes given below the lists:
List – I
List – II
A ) ∫ (J + ∂D/∂t)
...
n ds
2) ∫ dv
s
v
C) ∫ D
...
dl
s
c
D) ∫ B
...
dl
s
c
5) ∫ B dv
v
Codes:
(a) A-4,B-3,C-2,D-1
(b) A-3,B-4,C-2,D-1
(c) A-2,B-5,C-4,D-1
(d) A-4,B-2,C-3,D-1
25
...
015 Joule (b) 0
...
5 Joule
(d) 1
...
Match List – I with List – II and select the correct answer using the codes given below the lists:
List – I(Maxwell’s equation)
List – II
A) ∇ × H = J + ∂D/∂t
1) Faraday’s law
B) ∇ × E = -(∂B/∂t)
2) Gauss’s Law
C) ∇
...
Match List – I with List – II and select the correct answer using the codes given below the lists:
List – I
List – II
A) ∇ × E = 0
1) ∫ H
...
dA
B) ∇
...
dl = 0
C) ∇ × B = μ0J
3) ∫ B
...
B = 0
4) ∫ E
...
Match List – I with List – II and select the correct answer using the codes given below the lists:
List – II
List – I
A) Electric field E
1) amp/metre2
B) Magnetic flux density B
2) coulomb/metre2
C) Current density J
3) amp/metre
D) Magnetic field strength H
4) Volt/metre
5) Tesla
Codes:
(a) A-5,B-4,C-1,D-2
(b) A-4,B-3,C-2,D-1
(c) A-1,B-4,C-2,D-5
(d) A-4,B-5,C-1,D-3
29
...
The transmission coefficient is
(b) 1
...
68°
(c) 1
...
43°
(d) 0
...
45°
(a)1
...
68°
30
...
The characteristic impedance of the line is
(b)
164Ω
(c)
36Ω
(d) 64Ω
(a) 80Ω
31
...
D = ρ
B) ∇
...
Which of the following pairs of parameters and expressions is/are correctly matched?
1
...
Power flow density ……………
...
Displacement current in
non - conducting medium ………
...
Codes:
(a) 1 alone
(b) 2 and 3
(c) 1 and 3
(d) 1 and 2
33
...
1te-1ax and ε = 4ε0, then the displacement current crossing an area of
0
...
04 ε0
(c) 0
...
The wave length of a wave with propagation constant (0
...
2π)m-1 is
(a) 2/√0
...
The polarization of wave with electric field vector E = E0 ej(ωt + βz) (ax + ay) is
(a) Linear
(b) elliptical
(c) left hand circular
(d) right hand circular
36
...
H = 0 and ∇ × H = 0
(b) ∇
...
H = 0 and ∇ × H ≠ 0
(d) ∇
...
c
2
...
c
4
...
b
6
...
b
8
...
b
10
...
c
15
...
d
17
...
b
19
...
c
21
...
a
23
...
d
28
...
c
30
...
b
32
...
b
34
...
a
36
...
a
24
...
b
25
...
c
26
...
INDUCTANCE OF A TOROIDAL COIL:
I
R is Mean radius and N is No
...
INDUCTANCE OF A COAXIAL CABLE:
μo ln(b/a)
2π
∴L=
a
H/m
Total inductance of the cable can be obtained by multiplying the above
equation with the length of the cable
...
INDUCTANCE OF SOLENOID:
∴
L = N2Aμ
l
L
4
...
Total inductance is the product of inductance
per meter length and the length of the line
...
Loop inductance of single phase line is 2Ll
...
5 N a
9a+10l
2 2
6
...
6 N2r1
6r1+ 9l + 10(r2 – r1)
(d-x)
(d-r)
d
Total inductance =L l
5
Title: Electromagnetic Field Theory
Description: It contains all the topics of Electromagnetic fields with important objective questions and answers.
Description: It contains all the topics of Electromagnetic fields with important objective questions and answers.