Teerthanker Mahaveer University,Moradabad
College of Engineering
Second Class Test (Odd Sem) 2015 – 16
For Ist Year/ Ist Semester (
M.Sc.
Physics
)
Sub. Name
: Electromagnetic Theory Max
Marks: 50
Subject 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. Attempt any four: (5 × 4 = 20)
Unit III
(a) Explain the concept of displacement current.
(b) Using the Maxwell’s third equation show that𝑑𝑖𝑣 𝑩 = 0.
(c) Using the Maxwell’s equation drive the Poisson’s equation∇2𝑉 = 𝜌
𝜖0.
Unit IV
(d) Define uniform plane wave. What do you mean by polarization of a wave?
(e) Write the characteristics and cut off frequency of TE wave?
(f) A rectangular wave – guides has the following values a = 6 cm, b = 4 cm. The frequency of impressed signal is 3 GHz. For TE1,0 mode, compute the cut off frequency ?
2. (A) Derive the general expression for the scalar and vector potentials for electrodynamics. (15)
OR
(B)Deduce poynting theorem for the flow of energy in an electromagnetic field. (15) 3. (A)Starting from Maxwell’s equation derive the plan
electromagnetic solution for E in conducting media. Also point out the attenuation constant
and phase velocity. (15)
OR
(B) Determine the solution of electric and magnetic fields of TM waves guided along rectangular wave –
guides? (15)
1 (a) Maxwell postulated that a changing electric field is equivalent to a current, which flows as long as the electric field is changing. This equivalent current produces the same magnetic effect as an ordinary current in a conductor. This equivalent current is known as displacement current. In electromagnetism, displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of
electric displacement field. Displacement current has the units of
electric current density, and it has an associated magnetic field just as actual currents do.
The magnitude of displacement current is given by,
0d
E/
dt
, where E is the flux of the electric field through an area bounded by the closed curve. Therefore for changing electric field along with electric current, the modified Ampere’s law is given by,
1
0(
.
S
I l d
B 0dE/dt)
1(b) Taking Divergence of Maxwell’s third equation 𝑑𝑖𝑣. 𝑐𝑢𝑟𝑙 𝑬 = − 𝜕
𝜕𝑡 𝑑𝑖𝑣 𝑩 ⇒ 0 = − 𝜕
𝜕𝑡 𝑑𝑖𝑣 𝑩 ⇒ 𝑑𝑖𝑣 𝑩 = 𝟎 1(c) 𝑑𝑖𝑣 𝑫 = 𝜌 ⇒ 𝑑𝑖𝑣 𝜖 𝐸 = 𝜌 ⇒ 𝑑𝑖𝑣 𝜖(−∇𝑉) = 𝜌 ⇒ ∇ 2V = −ρ
ϵ
1(d)Plane wave in which the electric and magnetic intensities have constant amplitude over the equi-phase surfaces; such a wave can only be found in free space at an infinite distance from the source.
1(e) Cutoff The cutoff frequency is the same expression
But the lowest attainable frequencies are lowest because here n or m can be zero.
Dominant Mode
The dominant mode is the mode with lowest cutoff frequency. It’s always TE10
The order of the next modes change depending on the dimensions of the guide. 1(f) 𝑓𝑐=
𝑣0 2𝑎 =
3×108
2×0.06= 2.5 × 10 9
2 (A) A and φ potential fields
Introducing the electric potentialφ (a scalar potential) and the magnetic potential A (a vector potential) defined from the E and B fields by:
the four Maxwell's equations in a vacuum with charge ρ and current J sources reduce to two equations, Gauss' law for electricity is:
and the Ampère-Maxwell law is:
The source terms are now much simpler, but the wave terms are less obvious. Since the potentials are not unique, but have
gauge freedom, these equations can be simplified by gauge fixing. A common choice is the Lorenz gauge condition:
Then the nonhomogeneous wave equations become uncoupled and symmetric in the potentials:
For reference, in cgs units these equations are
with the Lorenz gauge condition
2 (B) 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 particle2 2
2 '
b n a
m u fcmn
)
(
)
(
m
v
dt
d
E
B
v
q
F
Lorentz
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 thePoynting vector S.
20 2 0
1
2
1
.
.
E
B
dt
d
S
E
J
The electromagnetic field energy density U is given by
20 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
S
E
H
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 0 0 2
0 0
0
0
2
2
1
2
1
o o
E
c
c
E
E
B
E
S
3(A)
The curl equation for H has two source terms, the conduction current and the displacement current. So far we have only considered the displacement current, which makes travelling electromagnetic waves possible. It is easy to include the conduction current if we assume a plane sinusoidal wave. Then Maxwell's equations become algebraic equations, and we can find a wave equation just as we did above, but now there are new terms. The dispersion relationbecomes k2 = ω2κμ/c2 + j(4πμσω/c2). What happens is that k becomes complex, k = k' + jk", so that when we put this in the wave expression, we find e-k"zej(ωt - k'z), so that k" represents an exponential absorption of the wave, while k' gives the usual wavelength.
The dispersion relation yields (k')2 - (k")2 = ω2κμ/c2 and k'k" = 2πμσω/c2 = (1/δ2) when we equate real and imaginary parts. The new parameter δ, with the dimensions of cm, is called the skin depth, a name we shall soon justify. If v/ω = λ/2π, and r = (λ/2π)/δ, then (k')2
= (2π/&lambda)2[√(1 + r2) + 1]/2 and k" = (2π/λ)2[√(1 + r2) - 1]/2. Note that λ is not quite the wavelength, except when r = 0. Everything depends on the ratio r, which is the ratio of λ/2π to the skin depth. If r is small, then λ is the actual wavelength, and k" = (2π/λ)(r/2). This is the case of weak damping.
On the other hand, if r is large, then k' = k" = 1/δ. This means the amplitude absorption is e-z/δ
, so the wave will be absorbed almost completely in the distance of a few skin depths. In this case, the conduction current completely dominates the total current, and the fields diffuse into the material, like heat. This is the case with metals at low and moderate frequencies. Multiply the conductivity in S/m by 9 x 109 to find the conductivity in esu. The conductivity of copper is 5.80 x 107 S/m, or 5.22 x 1017 esu. This makes δ = 6.61ν-1/2 cm. At 60 Hz, the skin depth is 0.85 cm. At 1 MHz, it is 66 μm.
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 22 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
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
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 xk
k
k
h
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 z0
at
0
0
at
0
j zy x
z
A
A
k
x
k
y
e
Substituting
2 2 2 2
2
