In Ssssy: Dept: ECE
In Ssssy: Dept: ECE
In Ssssy: Dept: ECE
1. A positive point charge Q is at the center of a spherical conducting shell of an inner radius Ri and an outer
radius Ro. Determine E and V as functions of the radial distance R.
2. R > Ro
a. Use Gauss law,
D.ds Q
Q
Er
4 0 R 2
R
Q
V1 E r dr
4 0 R
b. Ri < R < Ro
D.ds Q en closed
Q
Er
4 0 R 2
Ro Ri R
R
Q Q 1 1 Q 1 1 1
V E.dl E.dl E.dl E.dl
Ro Ri 4 o R o 4 o R R i 4 o R o R R i
2. A positive point charge Q is at the center of a spherical dielectric shell of inner radius R i and outer radius
R . The dielectric constant of the shell is r . Determine E , V, D and P as functions of radial distance R.
o
Sol: i) R > Ro (S1)
D.ds Q
S1
-1-
Q
E aR
4 o R 2
Q
D aR
4 R 2
R
Q
V E.dl
4 o R
D o E P
P D o E o r E o E o E r 1 .
If r 1 , P 0 .
ii) Ri < R < Ro (S2)
D.ds Q
S2
en closed
Q
ER
4 o R 2
Q
E aR
4 R 2
Q
D
4 R 2
Ro R1
Q Q 1 1
V E.dl E.dl
Ro 4 o R o 4 o r Ri Ro
Q 1 1 1 Q 1 1 1
1
4 o R o r R i r R o 4 o r R o r R i
iii) R < Ri, S3
D.ds Q
S3
Q
E ar
4 o R 2
Q
D ar
4 R 2
P 0
Ro Ri R
V E.dl Q Q 1 1 Q 1 1
2 Ro Ri 4 o R o 4 R i R o 4 o R R i
Q 1 1 1 1 1
V
4 o R o r R i r R o R R i
Q 1 1 1 1 1
1 1
4 o r R
o r R i R
V
-2-
3. Two dielectric spherical conductors with radii b1 and b2 (b2 > b1) that are connected by a conducting
wires. The distance between them is very large in comparison to b1 to b2 , so that the charges on the spherical
conductors may be considered as uniformly distributed. A total charge Q is deposited on the spheres. Find (a) the
charges on the two spheres (b) Find the electric field intensities on sphere surfaces.
Sol: a) Two spherical conductors are on same potential i.e., V1 = V2.
Q1 Q2 Q b
1 1
4 o b1 4 o b 2 Q2 b2
But total charge Q = Q1 + Q2.
b1 b2
Q1 Q Q2 Q
b1 b 2 b1 b 2
, .
b) The electric field intensities at the surfaces of the two conducting spheres are
Q1 Q2
E 1n E 2n
4 o b1 2
4 o b 2 2 .
2
E 1n Q1 b 2 b2
E 2n Q 2 b1 b1
.
Sol: Eo Eo a x
D o o E x a x
D i D o a x o E o a x
Di 1 E a x E o
Ei o o ax
E o r 3.2
Di 1 1
Di 1 o E i 1 o E o a x
P i D i o E i
o
o r r 32 0.6875 o E o c m 2
5. When a coaxial cable is used to carry electric power the radius of inner conductor is determined by the load
current, and the overall size by the voltage and the type of insulating material used. Assume that the radius of the
, polystyrene =
20 × 10 6
V m ).
L 1
MAX E r 0.25 25 10 6 2 o 3.2 r i
Sol: In rubber,
-3-
1
MAX E r 0.25 20 10 6 .
2.6r p
In polystyrene,
r p 1.54r i 0.616m
.
1. Find the energy required to assemble a uniform sphere of charge of radius b and volume
charge density .
Q
VR
4 o R
Sol:
4
Q R 3
If Q is the total charge contained in a sphere of radii R, 3 .
4
dQ 3 R 2 dR 4 R 2 dR.
The differential charge in a spherical layer of thickness dR is, s .
4 R 2 dR. 4 R3 dR
4 2
Q
dW VdQ
4 o R
and the work done in moving the dQ is o .
Hence the total work done in moving charge to form a sphere of radius b,
b
4 2 4 4 2 b 5 4 2 b 5
W dW R dR
3 o 0 3 o 5 15 o
.
4 3 Q b 2 2
3Q 2
Q b
3 5 o 20 o b
In terms of charge, .
b
We VdV V4 R 2 dR
2V 20
By using practical formula,
Q
b 3
E a R aR Rb
4 o R 2 3 o R 2
QR
R
E aR aR 0Rb
4 o R 2
3 o
R
b R
b b 3 R
R
V E.dR E dR E dR dR dR
b 3 o R
2
b
3 o
2 b 2
R
2
3 2 R 2
b b
3 o 2 2 3o 2 2
V b 3 2 R2 4 2 b 5
2 0 3 o
We b 4 R dR
2
2 2 15 o
.
2. A conducting material of uniform thickness h and conductivity has a shape of a quarter of a flat circular
washer, with inner radius a and outer radius b as shown in figure. Determine resistance between the end traces.
-4-
Sol: Assume Vo is the potential difference between the ends at and 2.
i.e., V = 0 at 0 ; V = Vo at 2 ;
d2V
0
Apply Laplace’s equation, d
2
2Vo
Vo C1 C1
V C1 C 2 ; 0 C 2 ; 2 ;
2Vo
V
1 V 2Vo 2 Vo
J E V . .
Current density, e
2 Vo b d 2 h Vo b
I J .dS h ln
a a
Vo
R
I b
2 h ln
a
Consider
1 j11 o r j r .
1 11
The magnitude response is usually very weak compared to the dielectric response in most materials, so
o .
j j j
Now propagation constant,
j11
o 1 1 1
0 in dielectric medium, j j.j o j
1 11
11
o 1 1 j 1
.
The presence of
, cause is a complex, and so losses occur, which are qualified through
11
attenuation
constant
,
1
2 2
11
1
1 1 1
2
Re j .
-5-
1
2 2
1
11
Im j 1 1 1
2
11
A non-zero value of
, indicates loss in medium. This is due to presence of
11 1
. The ratio
is called
loss tangent.
In loss-less medium (free-space),
2
p
; ; ;
c c
p
; 1 1
r r 1
r 1
;
2 2 o
o o
1
r 1 o r 1
r 1
; ;
1
.
j11
1
1 11
1 j 1
If 0 , will become complex quantity,
11
.
……………..(1)
J c E s , J D j E s ;
1
We know,
JC ES
JD j E S j1
.
11 JC
1
1
JD
Loss tangent, .
11
tan
Loss tangent, 1 1 .
-6-
j j 1 1 j
1
1
2
j j 1 1 j ......... j
2 8 1
1
Re j j 1 j 2 1
For a good dielectric, 2 1
1
2
1 2
Im j 1 1 1
1
8 1 8 1
They can compare with and values of a distortion less line. In that,
3
2
1 1 1 j 1 1 j
8 21 21
Intrinsic Impedance, .
1. A 1MHZ plane wave propagating in fresh water. At this frequency, losses in water are negligible, which
means that we can assume that ε = 0 . In water μ r = 1 , and at 1MHZ,
11 ε r 1 = 81 . Calculate
i) ii) iii) p iv) v) Expression for Ex and Hy, if Eo = 0.1V.
2 10 6
1 o . r 1 2 f o o 81 9 0.19 r a d m
Sol: (i) 2 10 8
2 2
3.3 10 7 m sec
(ii) 0.19
2 10 6
p 3.3 10 7 m sec
(iii) 0.19
o 377 377
42
r1 81 9
Intrinsic impedance,
Expression for Ex and Hy having maximum amplitude of 0.1 V m , then
E x 0.1 cos 2 10 6 t 0.19z V m
E
2.4 10 3 cos 2 10 6 t 0.19z A m
Hy
Wave reflections from multiple interfaces: -
-7-
H 1 H 1 H 1 H 2 Z l
.
Reflection at interface (1) is
Z 1
W
W Z 1 Z l
2
2 l = .
2
cos 2 l = cos 1
sin 2 l = sin 0
in 2 . 3 3
2
1
3
3 1
, if 3 1 , 0 no reflection i.e., half wave matching.
2
2 l = .
4 2
cos 2 l = cos 0
2
sin 2 l = sin 1
2
j 2
in 2 . 2 2
j3 3
-8-
in ,a 1
in ,a 1
2
% of power reflected is .
2
% of power transmitted is 1- .
Plane waves
IES-87-10M
1. A 100 MHz plane wave is normally incident on a plane copper interface from air. It the incident E field
has rms value of 1 v/m. How large the reflected and transmitted waves?
Ans: For air, 1 0 ; copper, 2 2
On copper conductor, reflection of EM wave takes place. It is just like a mirror for EM wave.
j
2
jE
2 0
2 1
2 1
1
1
1
Er
1 E r E i
Ei
E r 1V,
1 1 1 0
Et
0 E t 0.
Ei
E r 1, E t 0
for given Ei = 1 V/m.
IES-88-12M
2. Show that in a UPW propagating in Z-direction stored. Energy density at each point and each instant is
equally divided between electric and magnetic field energy densities, and
n p ro p a g a tin g
Ans: Consider an unpolarized wave propagating in Z-direction
E E x a x E y a y E E x 2 E y2
Take
H H x a x H y a y H H x2 H y2
-9-
1 *
P a vg Re E × H
2
1
*
Re E x a x E y a y × H x a x H y a y
2
1
2
Re E x H y * E y H x * a z
After substituting (i) in above equation
1 1
Pa vg Pa vg E x 2 E y 2 or H x 2 H y 2
2 2
1 2 1
Pa vg E or H 2
2 2
1 2 1
Pa vg E or H 2
2 2
IES-88-10M
3. Staring from MAXWELL’S equations, obtain the equations for a uniform plane wave in an unbounded,
isotropic, homogeneous medium.
Sol: MAXWELL’S equations are,
. D Pv
.B0
H
× E
t
D
×H J
t
Unbounded means no boundaries, i.e, no reflections.
Isotropic means, , t are scalars.
Homogenous means , t are constants.
Generally A free space is unbounded, isotropic and Homogeneous. So it also charge free i.e Pv = 0.
MAXWELL’S equations are
. D 0 . E 0.
. B 0 . H 0.
H
× E
t
E
× H E t
t
E
× ( × E ) ×
t
Take
× H
t
E
E t
t t
E
2 E
( . E ) E 2
t
t t 2
E 2 E
E
2
t
t t 2
- 10 -
H
2 H
H
2
t
t t 2
Similarly,
IES-88-14M
4. Evaluate, using pointing theorem, the power that is conveyed along. A co-axial cable carrying a
current (d.c) in the lnner conductor and sheath forming the return path for the current. Voltage between inner
conductor and the sheath is V. Neglect resistance of conductor.
Ans:
E E a
H H a
L
E a
We know 2
fin a l
V E . dt
Potential between inner conductor and outer conductor is in it
Poynting vector (P ) E × H
Power flow P
E × H ds.
2 b
V I
2 b .
b 2
..d .d
E H ..d .d 0 a
ln
E H a z . d d a z 0 a a
V.I 1 b
. ln .2
2 b a
ln
a =V.I.
P = V.I watts.
IES-ECE-89-5M
5. State the solution of uniform plane EM wave and show that the solution has E in x- direction, H in Y direction
and wave propagating in z-direction.
t
0
Wave is in +Z direction, EZ = HZ = 0, x y .
2 E 2 E
z 2 t 2
E Ex a x Ey a y
Here, .
a
a a a
2 L
ln L ln b
V E a .d .d E a .d .a d
2 2 2 o a
b b b o o b .
L
E a
2 o
- 11 -
b
V E ln
a
V
E
b
ln
a .
H . a ..d . a I
H 2 I
I
H
2
I
H a
2
2 E x 2 E x
Comparing x and y components, z 2 t 2
The solution of this two-dimensional partial differential equation is,
E x z, t E x * cos t z E x cos t z
for wa rd wa ve rever se wa ve .
H y z,t H y cos t z H y cos t z
Similarly, .
IES-89-5M
6. State the properties of linearly polarized uniform plane wave.
Sol: For a uniform linear polarized waves,
E
H y x cos t z
E x E x cos t z
, .
Properties:
1. Both Ex and Hy are perpendicular to each other.
2. Both Ex and Hy are perpendicular to direction of propagation.
3. Power carries normal to the plane containing E and H. i.e., P E H .
IES-90-12M
7. Starting from MAXWELL’s equations derive wave equations in free space?
- 12 -
E 100 100
H 0.593 A m
168.6 .
IES-90-4M
9. The magnitude of magnetic field intensity H of a plane wave in free space is 0.20 A m and is in Y-direction.
If the wave is propagating in z-direction with a frequency of 3 GHz, find the wavelength, amplitude and direction of
-7 -12
E vector, μ o = 4π × 10 H m , ε o = 8.825 × 10 F m .
H H y e jZ a y 0.2 e jZ a y
Sol:
f 3 10 9 Hz
c 3 1010
10 cm 0.1m
In free space, f 3 10 9
+ Z direction,
Ex E y
o
Hy Hx
.
E x o H y 120 0.2 24
E E x e jZ a x 24 e jZ a x
E 75.36 e jZ a x V m
.
IES-90-16M
10. The approximation radiation fields of certain antenna expressed in spherical co-ordinate system are
k
k
E = s in θc o s ωt - βr a θ H = s i n θc o s ωt - βr a
r , ηr . Determine instantaneous power and average power
flowing out of volume surrounded by the spherical surface of radius ‘r’ with center at origin.
Sol: Instantaneous power, P E H
k
k
sin cos t r a sin cos t r a
r r
k2
sin 2 cos 2 t r a r
P
r 2
k2 4
2 cos 2 t r .
3
k2 4
2 cos 2 t r .
120 3
0.022k 2 cos 2 t r wa t t s
.
1 *
1
2
k2
2
Re E H ds 0 0 r 2 sin 2
cos 2
t r .sin .d .d . a r
2
Average power =
1 k2 4 k2
. . cos 2 t r 2 . cos 2 t - r 0.011k 2 cos 2 t r wa t t
2 120 3 90 .
IES-91-12M
ε ε
11. Two isotropic dielectric media (1) and (2) with r1 and r2 are separated by a charge-free boundary with
σ = 0 . Two field lines in the 2 media made angles θ1 and θ 2 with normal to charge free boundary. Prove that
- 13 -
ta n θ1 ε r1
=
ta n θ 2 ε r2
. Show that static electric field line at dielectric conductor boundary is always perpendicular to the
conductor surface when no current is present.
Sol:
E.dl 0
2 3 4 1
E.dl 0
1 2 3 4
0 E t 2 l + 0 - E t 1 l 0
E t2
E t 1 l = 0
E t 2 E t1
E t 1 E1 sin 1
E t 2 E 2 sin 2
From figure, ……………..(1)
Apply Gauss law for electric fields to pill Box.
D.ds 0 (Charge–free)
D.ds 0
sides t op bot t om
0 D n ds D n ds 0
1 2
Dn1 Dn 2
1 E 1 E 2
1 E 1 cos 1 2 E 2 cos 2 ………………(2)
1 and 2
1 E sin 1 E 2 sin 2
1
2 1 E1 cos 1 2 E 2 cos 2
t a n 1 r1
t a n 2 r2
IES-92-12M
12. Distinguish the linearly, elliptically and circular polarized waves. An elliptically polarized wave in air has X
and Y components.
E x = 4s in t - βZ V m
,
E y = 8s i n t - βZ + 75 o
V m . Find poynting vector for air η o = 377Ω
.
Linearly polarized, i.e., Ex = 0 or Ey = 0 and = 0
o
Sol:
E 0
Ex = 0, y Vertically polarized.
x 0 ,E = 0 Horizontally polarized.
y
- 14 -
Ex = Ey = Eo, 90
o
Circularly polarized,
1 2
P E
Poynting vector, 2
2 2 2
E Ex Ey 4 2 8 2 16 64 80
.
1 1
80 0.106 wa t t
Average (P) = 2 120 3 .
Instantaneous P = 0.106 2 0.212 wa t t .
IES-93-12M
E = E o c o s ωt - βZ a x +E o s i n ωt - βz a y
13. Electric field of UPW propagation in +Z-directionis . Find i)
Corresponding H ii) Poynting vector if E o = 10 V m .
E E o cos t Z a x E o sin t Z a y E x a x E y a y
Sol:
+Z directions.
Ex Ey
o
Hy Hx
Ey Ex
Hx ; Hy
o o
.
Ey Ex
H cos t Z a x sin t Z a y
o o
2
E
1 * 1 100 5
P E H 0.132 W m 2
2 2 2 120 12
Poynting vector,
-6
14. Find the skin depth of penetration in copper at 10,000 MHz, resistivity of copper is 1.7 × 10 ohm-cm.
8
f = 10,000 MHz, ρ = 1.7 ×10 oh m cm 1.7 10 oh m m .
-6
Sol:
1 1
5.8 10 7 S m
Conductivity, 1.7 10 8
1 1
0.066 10 5 0.66 m
Skin depth, f 6 7
3.14 10000 10 4 10 5.8 10 7
.
IES-93-12M
15. Obtain the three dimensional wave equation for an absorbing medium assumed to be both magnetically and
electrically homogeneous and isotropic. The charge density in the field may be assumed to be zero. Hence
determine the wave equation if the field is varying harmonically with time.
Sol: Look at Q 3 and 5.
IES-94-12M
16. The electric field intensity of electromagnetic wave in the free space is given by
z
Ex = Eoc os ω t - a x
Ey = 0, Ez =0, V . Determine the expression for respective magnetic field intensity.
Z
E x E o cos t a x
V
Sol: Given,
+Z direction of propagation.
Ex E
o y
Hy Hx
Eo Z
Hy cos t a y . A m
o V
.
IES-94-12M
17. A uniform plane EM wave is incident at an angle θ i at the surface of discontinuing between two
homogeneous isotropic, dielectric with permittivity ε 1 and ε 2 , ε 2 being permittivity of the dielectric into which the
- 15 -
wave gets refracted at an angle θ t . Ei, Er and Et are the electric field intensities of incident, reflected, and
transmitted waves. Show that the reflection co-efficient for parallel polarization is given by
Ex ta n θ i - θ t
=
Ei ta n θ i + θ t
.
1 cos i 2 cos t
11
1 cos i 2 cos t
Sol:
1 o 1 o
r1 r2
, ;
r1
cos i cos t
r2
r2 cos i r1 cos t r1
11 cos i cos t
r2 cos i r1 cos t r2
1 1
Refractive Index, .
sin i r2
1 sin i 2 sin t sin t r1
.
sin t
cos i cos t
sin i
11
sin t sin i cos i sin t cos t 2 sin 2 i sin 2 t 2 sin i t cos i t
cos i cos t
sin i sin i cos i sin t cos t 2 sin 2 i sin 2 t 2 sin i t cos i t
t a n i t
t a n i t
Er t a n i t
11
Ei t a n i t
.
IES-95-12M
18. What do you understand by i) uniform plane wave ii) linearly polarized wave, and iii) elliptically
polarized wave?
Sol: i) Uniform plane wave is nothing but TEM, transverse electric and magnetic wave. In this, both E and H are
perpendicular and also perpendicular to direction of propagation.
ii) Linear polarization means either E = 0 (or) E = 0 and 0 .
x y
Standard electric field equation is
E E x cos t Z a x E y cos t Z p a y p Polarized angle.
p 0
If and Ey = 0, horizontal polarization.
p 0
If and Ex = 0, vertical polarization.
- 16 -
Ex Ey
and P 90 .
o
iii) Elliptical polarization,
P 90 o
P 90 o
IES-95-12M
19. Define in relation to travelling waves, the following i) Reflection coefficients ii) Transmission coefficients
iii) Standing wave ratio
E 1
r 2
E i 2 1
Sol: i) Reflection coefficient ,
E 2 2
t 1
E i 2 1
ii) Transmission coefficient,
1
E max Ei Er
SWR
E m in Ei Er
iii) Standing wave ratio,
1
SWR
1
IES-95-12M
20. An electric field wave travelling in air is incident normally on a boundary between air and dielectric having
permittivity ε r = 4 , and permeability μ o .Prove that one ninth of the incident power is reflected and eight ninth
of it transmitted in the second medium.
Sol: Air
1 , 1
1
1 120
1
Dielectric
2 , 2
2 120 120
2 60
2 r 4
2 1 60 120 60 1
2 1 60 120 180 3
2 1
100 100 11.11%
% power reflected is 9
2 2 2 60 2
2 1 180 3
- 17 -
2 1 8
1 1 % 88.88%
% power is transmitted is 9 9 .
2 E 2 E
E 2 = 0 2
21. Derive the Helmholtz equation for E in the form t t . The electric field intensity
associated with a plane wave travelling in perfect dielectric medium is given by
E x z, t = 12c o s 2π × 10 7 t - 0.1π z V m
.Find i) Velocity of propagation ii) Intrinsic impedance.
Sol: Look at Q 3 and 5.
E x z,t 12 cos 2 10 7 t 0.1 z V m
2 10 7
V 2 10 8 m sec
0.1
120
r
1
c 0 0
r
1
Refractive index,
3 10 8 3
r 1.5
2 10 8 2
120
80
1.5 251.2 .
22. Explain what do you understand by perpendicular polarization. Given two dielectric media, medium
(1) is a free space and medium (2) has and .Determine reflection co-efficient for oblique incident
i 30 for
a) perpendicular polarization, b) parallel polarization.
Sol: If incident electric field component is perpendicular to plane of the figure (or) horizontal to earth surface
then it is called perpendicular polarization (or) horizontal (or) s-polarization.
2 cos i 1 cos t
perpen dicu la r
2 cos i 1 cos t
Reflection coefficient,
2 cos t 1 cos i
pa r a llel
2 cos t 1 cos i
1 0 , 2 4 0 , , i 30 . o
sin i 2
sin t 1
;(Snell’s law).
0 1 1 1
sin i 1 sin t sin 30 0.25
2 4 0 2 2 4
.
i sin 1
0.25 14.47 .
-j
1
5 β
E r = -2 3 - j a x + 2 - j 3 a y + j2 3 a z e
2 . Find (i) direction of propagation (ii) frequency and
wavelength(iii) Apparent wavelengthand phase velocities along three axes.
Sol: (i) Direction of propagation,
3 a x 3 a y 2 a z
3 ax 3ay 2az
4
.
All positive x, y,z directions.
2
(ii)
x 2 y 2 z 2 4 0.8
5 5 .
2 1
2.5m
0.8 0.4 .
C 3 10 8
f 1.2 10 8 Hz
2.5 .
3 2
x 3 ; y ; z ;
(iii) 5 5 5
2 2 5 10 2 10 2
x m ; y m; z 5m;
x 3 3 y 3 z
2 1.2 10 8
VPx 6.92 10 8 m s
x
3
5 ;
2 1.2 10 8
Vp y 4 10 8 m sec
y 3
5 ;
2 1.2 10 8
Vp z 6 10 8 m sec
z 2
5 ;
IES-98-10M
- 19 -
25. State MAXWELL’s equation for harmonically varying fields and deduce the wav e equation in conducting
medium.
.D PV 0
Sol:
. B 0
H
E
t
E
H J
t
H H
E
t t
E
. E E
2
E
t t
E 2
E
2
2 E
t t .
IES-99-12M
26. Discuss the wave equation in (i) a lossy dielectric (ii) a conductor. Derive relatent equations.
2
1 1
2
2
1 1
2
1
, for good conductor.
1
, for good dielectric.
j j j.j 1
j
j.j. 1
j
(For good conductor j ).
j j
1
j 2 j
1 j
e 2 e 4 1 j
2 2
2
j j
45
j
.
and are complex quantities. i.e., 1 j11 , 1 j11 .
In a lossy dielectric,
Propagation constant is a complex variable. i.e., j .
j j j1 j j 1 j11 .
- 20 -
27. (i) Using MAXWELL’s equation, derive equations to demonstrate the propagation of uniform plane
waves in a perfect dielectric medium.
(ii) The magnetic field intensity H of a plane wave in free space is 0.20 A m and is in Y- direction if the
wave is in Y-direction. If the wave is propagating in Z-direction, find wavelength, amplitude and direction of E-
vector.
Sol: Look at Q.9.
IES-2000-12M
28. From MAXWELL’s curl equations, derive the wave equation in E for a plane wave travelling in positive y-
direction in an isotropic, homogeneous, lossless medium. The electric field is in z-direction. Assuming harmonic
variation, state a solution of this equation and prove it is a solution.
Sol: Look at Q.3.
IES-01-12M
- 21 -
ax ay az
0 0
z
0 H x a x H y a y
Ex Ey
t .
E y E x H y
ax ay H x a x ay
z z t t
E H x E x H y
y ; ;
By comparing, z t z t
j,
We know, t z ;
Ey j Ex j
E y jH x E x jH y
Hx Hy
; ;
j j
But,
Ex j
Hy j j
.
Ex j
Hy j
(ii)
E is E i e j1z a x
in ciden t fields
H is H i e j1z
a y
E r s E r e j1z a x
r eflect ed fields
H rs H r e j1z
a y
E t s E t e j1z a x
t r a n sm it t ed fields
H ts H t e j1z
a y
34. A uniform plane wave with the field components Ex and Hy has an electric field amplitude of 10
V m
σ
= 0.5
μ = μ , ε = 9ε 0 and ωε
and propagates at f = 2 MHz in a conductive region having parameter 0 .
(i) Find the values of attenuation constant, phase constant, phase velocity, wave length, skin depth and
intrinsic impedance of the wave.
(ii) Express electric and magnetic field in both their complex and real time forms, with the numerical values
of (i) inserted.
0 , 9 0 , 0.5
Sol:
E
i) x
100 30 , f = 2MHz, .
In conductors medium,
2 9 3 1
1 1 2 2 10 6 0 0 1 5 2 1 2 2 10 6 0.3078 1 NP m
2 2
2 2 2 3 10 8
0.03078 NP m .
2 9
2
1
2 2
1 2 10 6 2 0 0
2
1 0.25 1 0.1293 r a d m
.
6
2 2 10
Vp 97.18 10 6 m sec
Phase velocity, 0.1293 .
2 2
48.59 m
Wave length, 0.1293
1 1
32.49 m
Skin depth 0.03078 .
e jn
Intrinsic impedance, .
0
j 9 0
118.84
j 1 1
2 2 1 0.25 4
1
2 2
.
- 23 -
1 1
t a n 1 2 t a n 0.5 13.28
1
n
2 .
(ii) Ex and Hy are travelling in same direction.
E x (Z,t ) E x0 e z cos t z
E x z,t 100e 0.03708z e
j 4 10 6 t 0.129z 30 axV m
o
100e j30 o
H y (z,t ) 0.841e j16.72
j13.28 o
118.8e .
H y (z,t ) 0.84e 0.03078z e
j 4 10 6 t 0.129z 16.72 o ayA m
.
Electric and magnetic fields in real time is given by,
E x (z,t ) 100e 0.03078z cos 4 10 6 t 0.129z 30 o a x V m .
H y (z,t ) 0.84e 0.03078z cos 4 10 6 t 0.129z 16.72 o a y A / m .
IES-07-10M
Region 1(Z <0) is free space, where as region z(z>0) is a material medium characterized by σ = 10 S m , = 5 0 ,
-4
35.
5 -3
μ = μ 0 . For uniform plane wave having the electric field, E i = E 0 c o s 3π × 10 t - 10 π z a x V m incident normally
on the interface z=0, from region 1, obtain the expression for the transmitted wave electric field.
Sol:
j 3 10 5 4 10 7
2 104.52 33.68 o 86.97 j57.96
1
10 4 5
j 3 10 9
10 5
36 .
Et 2 2 2 104.52 33.68 o 209.04 33.68 o
0.446 26.56 o
E i 2 1 463.97 j57.96 467.58 7.12 o
Transmission coefficient, .
E t 0.446 26.56 o E i 0.445 0 cos 3 10 5 t 10 3 z 26.56 o a x V m
36. Same as problem 34.
37. Same as problem 35.
38. A uniform plane wave travels in free space in z-direction and is described by,
H = 0.8c o s ωt - βz a x + 0.80s in ωt - βz a y
. Find the corresponding electric field and the poynting
vector. Also obtain intrinsic impedance of the medium.
H 0.8 cos t z a x 0.80 sin t z a y
Sol:
Ex Ey
0
+ve z-direction Hy Hx
E x 0 H y E y 0 H x
; ;
E x 0 0.80 sin t z E y 0 0.80 cos t z
, ;
E 0 0.80 sin t z a x 0.80 cos t z a y
- 24 -
Poynting vector, P E H .
*
1
Pa vg Re E H
2
Average poynting vector, .
j z j z
2 jz j z 2
E 0.80 0 e a x 0.80 0 e a y ; H 0.80.e ax 0.80e ; a y
j z j z
1 2 j z j z 2
P a vg Re 0.80 0 e a x 0.80.0 e a y 0.80.e a x 0.80e a y
2
.
1
Re 0.8 0 a z 0.80 a z 0.8 0 a z
2
P a vg 241.28 a z .
IES-11-10M
39. Explain the following:
(i) Poynting vector and its significance.
(ii) Loss tangent of dielectrics as used in wave propagation.
(iii) Intrinsic impedance of a wave medium.
Sol: (i) Poynting vector: -
P E H V m A m wa t t m .
2
Poynting vector always gives the direction of power flow. It gives power per unit area i.e., power density.
1 *
P E H
2
Complex poynting vector,
1 *
Pa vg Re E H
2
Average poynting vector,
(ii) Loss tangent(tan )
Loss tangent is defined as, the ratio of magnitude of conduction current density to magnitude of
displacement current density, when they are in phasor form.
JC ES
tan =
JD
j E s
is called loss tangent.
>>1, good conductor.
<<1, good dielectric.
= 1, dissipation factor(D).
- 25 -
1
2
2 4
1
1
.
1
n t a n 1
2 .
IES-12-10M
Sol:
E 4 sin 2 10 7 t 0.8x a z V m
2 10 7
V 7.85 10 7 m sec
0.8
C 3 10 8
r
= 3.82 r 14.6 .
V 7.85 10 7
(i) Refractive index
120 120
98.69
r 3.82
.
1 E 2 1 16 81.06 m W m 2
Pa vg
(ii) Time-average power, 2 2 98.69 .
IES-12-12M
41. Define intrinsic wave impedance for a medium and derive the equations for intrinsic impedance for a
- y
E x = 0, E y = 0, E z = E 0 e
lossy dielectric medium (Consider, ).
H
E
Sol: t
ax ay az
0 0 Hx a x
y t
0 0 EZ
E z
a x H x a x
y t
, j
y t .
Ez j j j
E z j x x j j j
.
E j
z
Hx j
.
42. Define intrinsic impedance. A uniform plane wave in free space is given by
E s = 200 30 o e -j250z a x V m .Find i) Phase constant ii) Angular frequency iii) Wavelength
iv) Intrinsic impedance
v) magnetic field (Hs).
- 26 -
ii) 2 f
c 2 2
f ,
250
3 10 8
f 250 75 10 9 r a d sec
2 .
2
0.025 m
iii) 250
0
120 (or ) 377
0
iv)
200 30 o j250z
H e .a y j 7510 9 t 250z 30 o
120 0.53e ay.
v)
IES-14-8M
π
j ωt - z
20
43. For a uniform plane wave in air, the magnetic field is given by H = 2e a x .Calculate i)
1
t= μsec
Wavelength ii) Frequency iii) the value of E at 15 , Z= 5m.
j t z
20
Sol: H 2e , a x 20 .
2 2
40m
i) 20
C 3 10 8
f 0.75 10 7 Hz
ii) 40
iii) +z directions.
Ex Ey
Hy Hx
.
j t z j t z
E y 0 H x 120 . 2 .e 20
a y 754.e 20
1
t sec
at 15 , Z = 5m.
1 3
j 2 0.75107 10 6 5 j
Ey 754e 15 20
754e 4 754 135 o V m
.
IES-14-12M
44. Differentiate between linear, homogeneous and isotropic dielectric material. At the center of a hollow
dielectric sphere
0 r is placed a point charge Q. If two spheres has inner radius a and outer radius b,
calculate D , E and P .
Sol: Linear: - A material is said to be linear if D varies linearly with applied E , otherwise nonlinear.
Homogeneous: - A homogeneous medium is one in which and are constants. i.e., they are independent of
spatial co-ordinates.
Isotropic medium: - and are scalar i.e., and are independent of magnitude and direction of applied
electric field.
For anisotropic medium,
D x 11 E x 12 E y 13 E z D y 21 E x 22 E y 23 E z D z 31 E x 32 E y 33 E z
, , .
D x 11 12 13 E x
D y 21 22 23 E y
D
z 31 32 33 E z
Consider a hallow dielectric sphere,
- 27 -
i) r< a
By Gauss law,
D.ds Q en c
Q
D r . 4 r 2 Q D r 4 r 2
Q
E ar
4 0 r 2
r a
P e 0 E r 1 0 E 0
.
ii) a < r < b
D.ds Q
Q
Dr
D r .4 r 2 Q 4 r 2 .
Q
D ar
4 r 2
Q
E ar
4 0 r r 2
Q Q r 1
P e 0 E r 1 0 a r ;P ar;
4 0 r r 2
4 r r 2
(iii) r > b
D.ds Q
Q
D ar
4 r 2
Q
E ar
4 0 r 2
P e 0 E r 1 0 E 0
IES-14-10M
The magnetic field intensity is m in a material when B = 2 w b m , When H is reduced to 400
2
45.
A m , B = 1.4 w b m 2 . Calculate the change in magnetization M ?
Sol: B1 2 Wb m 2 , H 1 1200 A m
B1 1 H 1
B1 2
r1 1326.96
0 H 1 1200 4 10 7
m1 r1 1 1325.96
.
M1 m1 H 1 1325.96 1200 1591152
.
B 2 1.4 Wb m 2 ,H 2 400 A m .
1.4
r2 2786.624
400 4 10 7 .
M 2 m 2 H 2 2785.624 400 1114249.6
.
M M M 476.9024 KA m .
Change in magnetization, 1 2
IES-15-10M
- 28 -
46. What is the term is called in study of EM waves? Find the general expression of attenuation constant
and relate with skin depth
. Prove that the skin depth
is independent of frequency when <<1(poor
conductor) and decreases with frequency when >>1(good conductor).
Sol: is called dissipation factor D.
>>> 1 good conductor.
<<<1 good dielectric
=0 perfect dielectric.
Perfect conductor
is also called loss tangent.
J C
Es
tan
JD E s
.
j j j j j
2 2
1 2 2 1 1 2 2 1
2
, 2
.
For good conductor, >>> 1
. f
2 2
1 1
f .
Skin depth,
decreases with increasing frequency.
For good dielectric, <<<1
2
1 2 2 1
2 2 2 , .
1 1 2
Skin depth
2 .
is independent of frequency.
- 29 -
- 30 -