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Eleg 840 Lecture 2

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To derive the magnetic field boundary condition, we us Ampere's Law: ... Now cast ME into frequency domain, in which case. Then we have: ... – PowerPoint PPT presentation

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Title: Eleg 840 Lecture 2


1
Eleg 840Lecture 2
  • Dr. Dennis Prather

2
  • Last Time
  • Discussed course objective and syllabus.
  • Derived Maxwells equations from empirical
    laws Coulomb, Bio-Savart, Ampere, Faraday, and
    Gauss.
  • Today
  • Introduce constitutive parameters
  • Derive boundary conditions
  • Discuss Poyntings Theorem
  • Classification of EM problems according to
    their differential operator

3
In the lecture, we derived ME in terms of
currents and charges. Now we want to further
define the type of sources.
Displacement current
Impressed current
Conductive current
Magnetic displacement current
Impressed magnetic current
4
We have introduced migmv to balance MEs.
Although they are not physical, they do allow
for the simplification of some physical problems.
Electron charge density
Constitutive Parameters Materials inherently
contain charges, thus when they are subjected to
EM fields, they interact in such a way as to
modify the fields.
5
To account for these interactions on a
macroscopic level, we introduce constitutive
relations.
Electric flux density
Electric field
Electric permittivity
Displacement current
Magnetic flux density
Magnetic field
Magnetic permeabilitiy
6
Electric field
Electric conductivity of the medium
Electric current density
NOTE These equations are only valid for
ISOTROPIC (spatially independent) materials.
7
For anisotropic media
Where
and
are now tensors, 3x3 matrices in Cartesian
co-ordinates.
8
(No Transcript)
9
Time Dependence In general, for time dependent
phenomena, such as wave propagation, we need only
focus on two of MEs
10
One of the most important ramifications of MEs
is the ability to describe EM wave propagation
11
Wave Equation
12
Vector identity
13
Assume
14
Frequency Domain The frequency domain can
represent time dependent fields as time harmonic
signals using phasor notation.
The real part is understood!!
In this case
15
We can replace
Helmholtz Equation
16
Boundary Conditions ME represent a set of
self-consistent partial differential equations.
As such, they require the specification of
boundary conditions in order to provide a UNIQUE
solution to a given problem. To derive the
magnetic field boundary condition, we us Amperes
Law
17
Now, apply this to a material interface
If we allow the surface, S, to shrink to zero
18
This results because D is everywhere finite with
S.
However,
because we can represent a current
localization in space.
interface
19
Then we have
S
As
we get
20
So, we have the following scenarios
1) Perfect Conductor
2) Perfect Dielectric
3) Partially Conducting Surface
For the electric field, we use Faradays law
21
1) PEC ?
2) Dielectric
PEC
22
For normal field components, we use Gausss law
1
2
In the limit
23
Noting that
we get
24
Similarly, for the magnetic field
  • Power flow in EM fields
  • EM Waves carry energy as the propagate
  • To quantify that, we use Poyntings theorem

Faradays
Amperes
25
Vector Identity
26
Now, integrate over a volume, v
, from the divergence theorem
Note
from the product rule of derivatives
27
Then we have
is the energy stored per unit volume in the
electric field, similar to
is the energy stored per unit volume in the
magnetic field, similar to
is ohmic power loss, as in
28
As a result
is the power flow into volume, v.
is the power flow Out of volume v.
Likewise,
Now, we define
as the Poynting vector.
This says that EM power only flows in the a
direction perpendicular to both E and H.
Time Harmonics Power Flow Now cast ME into
frequency domain, in which case
29
Then we have
Where denotes complex conjugate!
Integrate over volume
30
Now, because E and H are sinusoidal, we define
power flow in terms of a time average.
Power Flow Example
V?E
Ohmic Loss

-
H
R
L
31
Assume only loss is across the resistor. The
electric field across the resistor is
The magnetic field is obtained from Amperes law
Diameter of the resistor
Power flow through the resistor is
32
Now let S be a cylindrical surface enclosing the
resistor
Because
is in the radial direction. Then we have
E
H
33
Therefore, we can think of power flow,
, as flowing into the resistor and
being converted into heat.
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