Teerthanker Mahaveer University,Moradabad
College of Engineering
III Class Test (Summer) 2015 – 16
For Ist Year/ Ist Semester (
M.Sc. Physics
) Sub. Name : Electromagnetic Theory Max Marks: 50Subject Code : MPH104 MaxTime: 1hr 45 min. Course/Branch/Section: M.Sc./Physics/Ist
(First–15 Min.are for distribution and reading of the paper & paper writing time 1 Hr 30 Min.) Note: Attempt all questions. Question 1 is compulsory.
1. (A) Derive the expression for total power radiated by a dipole.
OR (20)
(B) Define radiation resistance. Obtain its value for a dipole antenna. Justify the selection of 𝜆/2 antenna on its basis.
2. (A) Find the solution of Laplace equation in Cartesian co – ordinate system, using the method of separation of variables.
OR (15)
(B) Develop the magnetic boundary condition at the interface between two magnetic medium. 3. (A) Deduce poynting theorem for the flow of energy in an electromagnetic field.
OR (15)
(B) Determine the solution of electric and magnetic fields of TM waves guided along rectangular wave – guides?
1. (A) Energy radiated by the dipole
Recall our old buddy the complex Poynting vector for harmonic fields:
(11.127)
The factor of ½ comes from time averaging the fields. This is the energy per unit area per unit time that passes a point in space. To find the time average power per solid angle, we must relate the normal area through which the energy flux passes to the solid angle:
(11.128)
and project out the appropriate piece of S, i. e. -- . We get (with µ=1)
(11.129)
where we must plug in E and H from the expressions above for the far field. After a bunch of algebra that I'm sure you will enjoy doing, you will obtain:
The polarization of the radiation is determined by the vector inside the absolute value signs. By this one means that one can project out each component of p (and hence the radiation) before evaluating the square independently, if so desired. Note that the different components of p need not have the same phase (elliptical polarization, etc.).
If all the components of p (in some coordinate system) have the same phase, then p necessarily lies along a line and the typical angular distribution is that of (linearly polarized) dipole radiation:
(11.131)
where θ is measured between p and n. When you integrate over the entire solid angle (as part of your assignment) you obtain the total power radiated:
(11.132)
The most important feature of this is the dependence. In this antenna, d<< λ and
(11.133)
From the continuity equation,
(11.134)
and we find that the linear charge density (participating in the oscillation, with a presumed neutral background) is independent of z:
(11.135)
where the ± sign indicates the upper/lower branch of the antenna and the means that we are really treating ρ/(dxdy) (which cancels the related terms in the volume integral below). We can then evaluate the dipole moment of the entire antenna for this frequency:
(11.136)
The electric and magnetic fields for r > d in the electric dipole approximation are now given by the previously derived expressions. The angular distribution of radiated power is
(11.137)
and the total radiated power is
(11.138)
Remarks. For fixed current the power radiated increases as the square of the frequency (at least when kd<<1, i. e. - long wavelengths relative to the size of the antenna). The total power radiated by the antenna appears as a
``loss'' in ``Ohm's Law'' for the antenna. Factoring out 𝐼02/2 , the remainder must have the units of resistance and is called the radiation resistance of the antenna:
(11.139)
where we do the latter multiplication to convert the resulting units to ohms. Note that this resistance is there for harmonic currents even if the conductivity of the metal is perfect. Note further that by hypothesis this expression will only be valid for small values of 𝑅𝑟𝑎𝑑.
1. (B)
Radiation resistance is that part of an antenna's feedpoint resistance that is caused by the radiation of electromagnetic waves from the antenna, as opposed to loss resistance (also called ohmic resistance) which generally causes the antenna to heat up. The total of radiation resistance and loss resistance is the electrical resistance of the antenna.
The radiation resistance is determined by the geometry of the antenna, where loss resistance is primarily determined by the materials of which it is made. While the energy lost by ohmic resistance is converted to heat, the energy lost by radiation resistance is converted to electromagnetic radiation.
Radiation resistance is caused by the radiation reaction of the conduction electrons in the antenna.
When electrons are accelerated, as occurs when an AC electrical field is impressed on an antenna, they will radiate electromagnetic waves. These waves carry energy that is taken from the electrons. The loss of energy of the electrons appears as an effective resistance to the movement of the electrons, analogous to the ohmic resistance caused by scattering of the electrons in the crystal lattice of the metallic conductor.
Power is calculated as P=I2R
Where I is the electric current flowing into the feeds of the antenna and is the power in the resulting electromagnetic field. The result is a virtual, effective resistance:
R = P/I2.
This effective resistance is called the radiation resistance.
The idea at λ/2 is the reactive portion of the antenna falls to zero making the transfer of maximum power easier. This is often termed a "resonant antenna" because it consists of only real impedance at the operating frequency. (Note Impedance does not fall to zero, only the reactive portion falls to zero). This also occurs at 1.5λ, 2.5λ, etc. λ/2 length is used because it is the shortest length by which high radiation could be obtained for the particular freq, bcoz the in an antenna current distribution will not be uniform, only time varying field can give rise to radiation. the electrons get accelerated for quater of the cycle and then suddenly chages its direction and moves in the opp direction, the same action is performed for quater cycle (i.e. λ/2 length) thats why the minimum length required is λ/2.
2. (A) Three-Dimensional Solutions to Laplace's Equation: In three-dimensions, Laplace's equation is
We look for solutions that are expressible as products of a function of x alone, X(x), a function of y alone, Y(y), and a function of z alone, Z(z).
A function of x alone, added to one of y alone and one of of z alone, gives zero. Because x, y, and z are independent variables, the zero sum is possible only if each of these three "functions" is in fact equal to a constant. The sum of these constants must then be zero.
Note that if two of these three separation constants are positive, it is then necessary that the third be negative. We anticipated this by writing (4) accordingly. The solutions of (4) are
where
It is possible to choose the constants and the solutions from (6) so that zero potential boundary conditions are met on five of the six boundaries. To make the potential zero in planes x = a and z = w, it is necessary that
Solution of these eigenvalue equations gives kx = m /a, kz = n /w, and hence
where m and n are integers.
To make the potential zero on the fifth boundary, say where y = 0, the hyperbolic sine function is used to represent the y dependence. Thus, a set of solutions, each meeting a zero potential condition on five boundaries, is
where in view of (5)
These can be used to satisfy an arbitrary potential constraint on the "last" boundary, where y = b.
In Fig, a \ll w. If the cross-sectional dimensions a and w are comparable, as shown in Fig., the field can be represented by the modal superposition given by (9).
Region bounded by zero potentials at x = 0, x = a, z = 0, z = w, and y = 0. Electrode constrains plane y = b to have potential v.
In the five planes x = 0, x = a, y = 0, z = 0, and z = w the potential is zero. In the plane y = b, it is constrained to be v by an electrode connected to a voltage source.
Evaluation of (10) at the electrode surface must give v.
where
The steps that now lead to an expression for any given coefficient Amn. Both sides of (11) are multiplied by the eigenfunction Xi Zj and then both sides are integrated over the surface at y = b.
Because of the product form of each term, the integrations can be carried out on x and z separately. In view of the orthogonality conditions, (12), the only none-zero term on the right comes in the summation with m = i and n = j. This makes it possible to solve the equation for the coefficient Aij. Then, by replacing i m and j
\rightarrow n, we obtain
The integral can be carried out for any given distribution of potential. In this particular situation, the potential of the surface at y = b is uniform. Thus, integration gives
The desired potential, satisfying the boundary conditions on all six surfaces, is given by (10) and (15).
2. (B) To determine the boundary condition for the normal components of the magnetic flux density at the interface between the two regions, let us construct a Gaussian pillbox with vanishingly small thickness, as shown in Figure. Since the magnetic flux lines are continuous, we have
S S d B. 0
where S is the entire surface of the pillbox. Neglecting the flux that flows through the vanishingly small thickness of the pillbox, this equation becomes
1
2 1
2
0 .
.
S S S
n n
S
ds B ds B S d B S d
B Bn1 Bn2 1Hn1 2Hn2
Thus normal component of B is continuous while for H is discontinuous across the boundary.
To obtain the boundary condition for the tangential components of the H field, consider the closed path shown in Figure. Applying Ampere’s law to the closed path, we obtain
C
I l d H.
Applying Stoke’s law
0 ). (
.
.
S S
S d J H S
d J S d
H
where I is the total current enclosed by the closed path c. The path c2 and c4 are each of vanishingly small thickness, w0, and their contributions to the total mmf drop can be neglected. Thus, dropping these integrals, we have
1 2 1 2 . .C S S
t t C I dl H dl H l d H l d
H
2 2 1 1 2
1
t t t t B B H
H
which states that the tangential components of the field at the boundary are discontinuous.
3. (A) Poynting theorem: Electric and magnetic fields store energy. This, energy can also be carried by the electromagnetic waves, which consist of both fields. The rate of flow of energy per unit area in a plane electromagnetic wave can be described by a vector S, called the Poynting vector, which is represented by
H E
S The Unit of Sin MKS is watt/m2.
Let an electromagnetic field interacts with a particle of charge q traveling at a velocity v. The Lorentz force on this particle
) ( )
( mv
dt d E B v q
FLorentz
To obtain the energy relation, multiply this equation by. mv qvE dt
d
) 2 1
( 2
Here v.(vB)is zero as the magnetic field does not contribute to the particle's energy. Multiply by the particle density n and introduce the current density J nqv, we obtain J E
dt
dT
.
where T is the kinetic energy of the ensemble of particles. Using fourth Maxwell's equations to express J in terms of the magnetic and electric fields. J(B)/0 0E
2 0 0 2 1 / ) .( . E dt d B E E
J
Using vector identity (AB)B.(A)A.(B) above equation can be written
0 2
0
0 .( )/
2 1 /
) .(
. E B E
dt d B
E E
J
The last term in the equation above is actually the time derivative of the magnetic field energy density.
This can be shown by using Faraday's law to substitute dt
B d
for the curl of E. The first term on the R.H.S contains the Poynting vector S.
2 0 2 0 1 2 1 .
. E B
dt d S E J
The electromagnetic field energy density U is given by
2 0 2 0 1 2 1 B E U
We get the Poynting theorem for the case of an ensemble of free particles in an electromagnetic field in its most compact form.
S J E t
U
total energy rate mechanical work with the Poynting vector SEH
This theorem states that the work done on the charge by an electromagnetic force is equal to the decrease in energy stored in the field, less than the energy, which flowed out through the surface. It is also called the energy conservation law in electrodynamics.
Above relation for pointing vector show the instantaneous rate of flow of energy per unit area. Average value of S can be obtained for more than one cycle of the wave. The average values of E and B for the cycle are E0/2 and B0/2 respectively. Therefore the average value of the Poynting vector
2 0 0 0 0 0
0 2 2
1 2 1 o oE c c E E B E S
3. (B) We consider the conditions under which EM waves can propagate when confined to the interior of some kind of “hollow” pipe – also known as a wave guide. We consider first the simplest type of a wave guide – a perfect conductor (ρ=1/σ=0) such that inside the walls of the perfect conductor: 𝐸 = 0 &𝐵 = 0
The boundary conditions at the inner walls of a perfect conductor are:
𝐸 𝐶 . 𝑑𝑙 = − 𝐵𝑆 . 𝑑𝑎 = 0 (1) ⇒Tangential 𝐸 continuous: E = 0 𝐵𝑆 . 𝑑𝑎 = 0 (2) ⇒Normal B continuous: B⊥=0
We are interested in/seek monochromatic/single-frequency plane traveling wave-type solutions - that propagate down the inside of the wave guide, e.g. in the +zˆ direction.
Using phasors& assuming waveguide filled with (i) lossless dielectric material and (ii) walls of perfect conductor,
the wave inside should obey…
c
k2 2
where
Then applying on the z-component…
Fields inside the waveguide
0 0 2 2 2 2 H k H E k E 2 2 2 2 2 2 2 2 : obtain we where from ) ( ) ( ) ( ) , , ( : Variables of Separation of method by Solving 0 k Z Z Y Y X X z Z y Y x X z y x E E k z E y E x E '' '' '' z z z z z 0 2
2
Ez k Ez
z z '' y y y '' x x x '' y x '' '' '' e c e c z Z Z Z y k c y k c Y(y) Y k Y x k c x k c X(x) X k X k k k k Z Z Y Y X X 6 5 2 4 3 2 2 1 2 2 2 2 2 2 ) ( 0 sin cos 0 sin cos 0 : s expression in the results which 2 2 2 2 2 y x k k k
Substituting
Other components
From Faraday and Ampere Laws we can find the remaining four components:
So once we know Ezand Hz, we can find all the other fields.
TM Mode:
Boundary conditions: From these, we conclude:
X(x) is in the form of sin kxx, where kx=mp/a, m=1,2,3,… Y(y) is in the form of sin kyy, where ky=np/b,n=1,2,3,… So the solution for Ez(x,y,z) is
zy y x x z z y y x x z z z y y x x z e y k B y k B x k B x k B H e y k A y k A x k A x k A E z e c e c y k c y k c x k c x k c E sin cos sin cos , field magnetic for the Similarly sin cos sin cos : direction -in traveling wave at the looking only If sin cos sin cos 4 3 2 1 4 3 2 1 6 5 4 3 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 y x z z y z z x z z y z z x k k k h where y H h x E h j H x H h y E h j H x H h j y E h E y H h j x E h E
zy y
x x
z A k x A k x A k y A k ye
E 1cos 2sin 3cos 4sin
,a x E ,b y E z z 0 at 0 0 at 0
j zy x
z A A k x k ye E 2 4 sin sin
