Upload
others
View
7
Download
0
Embed Size (px)
Citation preview
Electromagnetic Field Theory
2nd Year EE Students
Prof. Dr. Magdi El-Saadawiwww.saadawi1.net
2016/201711/8/2016 1Prof. Dr. Magdi El-Saadawi
Chapter 4
Stationary Current Fields4.1. Introduction
4.2. Conduction Current
4.3. Cause of Field through the Impressed Field Intensity
4.4. Boundary Conditions for Steady Electric Currents
4.5. Energy Dissipation and Joule’s Law
4.6. Field Equations for the Electric Field of Stationary Currents
4.7. Electrostatic Simulation
4.8. Equation of Continuity and Kirchhoff’s Current Law
4.9. Resistance Calculations11/8/2016 3Prof. Dr. Magdi El-Saadawi
Stationary Current Fields
4.1. Introduction
In Chapter 3 we dealt with electrostatic problems, field
problems associated with electric charges at rest.
We now consider the charges in motion that
constitute current flow.
11/8/2016 4Prof. Dr. Magdi El-Saadawi
Stationary charges produce electric fields that are
constant in time; (electrostatics fields).
Charges move with constant velocity in solids,
liquids, gasses or in vacuum constitutes a
stationary current flow or stationary current field.
By steady current we mean a flow of charge
which has been going on forever, never
increasing, never decreasing.
Steady currents produce magnetic fields that are
constant in time (magnetostatics fields).
In Summary p. 137
11/8/2016 5Prof. Dr. Magdi El-Saadawi
Two types of electric current are caused by the
motion of electric charges:
(1) Convection current (2) Conduction current
Convection currents: resulting from the motion
of the electron, the ions, or the other charged
particles in vacuum, a liquid or a gas, and these
currents are not governed by Ohms law:
Electron beams in a cathode-ray tube
Violent motions of charged particles in a thunderstorm.
للجسيماث المشحونت في عاصفت رعديت( غير طبيعيت)حركت عنيفت
4.1. Introduction
11/8/2016 6Prof. Dr. Magdi El-Saadawi
Conduction currents: resulting from the motion
of charges in metallic conductor under the action
of an electric field. These currents are caused by
drift motion of conduction electrons or holes and
they obey the Ohm’s law.
Charges move in free space has nothing to impede
but in conductor there is a special vibrating lattice
structure which collide with them
4.1. Introduction
11/8/2016 7Prof. Dr. Magdi El-Saadawi
we will concentrate on conduction currents
When an external electric field is applied on a
conductor, an organized motion of conduction
electrons, which may wander from one atom to
another in a random manner, is produced.
The conduction electrons collide with the atoms
in the course of their motion, dissipating part of
their kinetic energy as heat (thermal radiation).
This phenomenon manifests itself تظهر نفسها as a
damping force قوة تخميد or resistance, to current
flow.
4.2. Conduction Current
11/8/2016 8Prof. Dr. Magdi El-Saadawi
4.2. Conduction Current
Table 4.1 shows the conductivities of several media in
S/m
11/8/2016 12Prof. Dr. Magdi El-Saadawi
A stationary field is a field which reaches to a state of
independence on time (constant state) and is coupled
with an energy transformation.
This energy must come from a non-conservative field.
The source of non-conservative field may be electric
batteries (conservation of chemical energy to electric
energy) or electric generator (conservation of mechanical
energy to electric energy) or other devices.
These electrical energy sources, when connected in an
electric circuit, provide a driving force for the charge
carries, and manifests تتجلى فى itself as equivalent impressed
field intensity E′
4.3. Cause of Field through the Impressed Field Intensity (electromotive force)
11/8/2016 14Prof. Dr. Magdi El-Saadawi
In a conducting medium, the collision of free electrons
with the atomic lattice will generate thermal energy, and
this is an irreversible energy conversion process.
The impressed source has to compensate the energy
dissipation in order to maintain the steady electric
current.
4.5. Energy Dissipation in Steady Electric Current Fields
11/8/2016 17Prof. Dr. Magdi El-Saadawi
In a steady electric current field, we construct a small
cylinder of length and end face area , and assume
the two end faces of the cylinder are equipotential
surfaces.
Under the influence of the electric
field, electric charge dq is moved
to the right end face from the left
end face in dt, with the
Corresponding work done by the electric force as
4.5. Energy Dissipation in Steady Electric Current Fields
11/8/2016 18Prof. Dr. Magdi El-Saadawi
A parallel plate capacitor consists of two imperfect dielectrics in series. Their
permittivities are 1 and 2 , the conductivities are 1 and 2 , and the
thickness are d1 and d2, respectively. If the impressed voltage is U, find the
electric field intensities, the electric energies per unit volume, and the power
dissipations per unit volume in the two dielectrics.
Solution: Since no current exists outside the
capacitor, the electric current lines in the capacitor
can be considered to be perpendicular to the
boundaries. Then we have
J1n= J2n
2211 EE
Solving the two equations we get:
Udd
E1221
21
U
ddE
1221
12
1 1
2 2
d1
d2
U
Example pp. 150
UdEdE 2211
11/8/2016 20Prof. Dr. Magdi El-Saadawi
The electric energies per unit volume in two dielectrics, respectively, are
2
222e
2
111e2
1 ,
2
1EwEw
The power dissipations per unit volume in two dielectrics, respectively, are
2
222
2
111 , EpEp ll
Two special cases are worth noting:
If , then , , , .02 01 E 0e1 w 01 lp2
2d
UE
If , then , , , .01 1
1d
UE 02 E 02e w 02 lp
d1
d2
1= 0
E 2= 0
UE 1= 0
2= 0
U
11/8/2016 21Prof. Dr. Magdi El-Saadawi
The analogy between the electric current field and the
electrostatic field is explained by Table 4.2.
4.7. Electrostatic Simulation pp.153
The electric current density J corresponds to the electric field
intensity E, and the electric current lines to the electric field lines.
11/8/2016 23Prof. Dr. Magdi El-Saadawi
Based on this similarity, the solution of the steady
electric current field can be found directly from
the results of the electrostatic field.
In some cases, since the steady electric current
field is easy to be constructed and measured, the
electrostatic field can be investigated based on the
steady electric current field with the same
boundary conditions, and this method is called
electrostatic simulation.
4.7. Electrostatic Simulation p. 153
11/8/2016 24Prof. Dr. Magdi El-Saadawi
The electrostatic field and the steady electric current field
between two electrodes as follows:
PN
Steady electric current field
PN
Electrostatic field
The calculation of the resistance of conducting media can be
determined based on the results of the corresponding
electrostatic field.
11/8/2016 25Prof. Dr. Magdi El-Saadawi
Based on the equations for two fields, we can find the
resistance and conductance between two electrodes as
If the capacitance between two electrodes is known, from
the above equations the resistance and the conductance
between two electrodes can be found out.
11/8/2016 26Prof. Dr. Magdi El-Saadawi
تصحيح
155ص
d
A
d
AG
The capacitance of a parallel plate capacitor of plate
area A and separation d is .
If the conductivity of the imperfect dielectric is , the
leakage conductance G between two electric plates of
the parallel plate capacitor is
d
AC
11/8/2016 27Prof. Dr. Magdi El-Saadawi
The capacitance of a coaxial line per unit length is
where b is the inner radius of the outer conductor, and a is
the radius of the internal conductor. If the conductivity of
the filled dielectric is , the leakage conductance per unit
length G1 is
)/ln(
π21
abC
)/ln(
π21
abG
11/8/2016 28Prof. Dr. Magdi El-Saadawi