Electromagnetic Field Theory - saadawi1 · This energy must come from a non-conservative field. The source of non-conservative field may be electric batteries (conservation of chemical

Electromagnetic Field Theory

2nd Year EE Students

Prof. Dr. Magdi El-Saadawiwww.saadawi1.net

[email protected]

2016/201711/8/2016 1Prof. Dr. Magdi El-Saadawi

http://www.saadawi1.net/



mailto:[email protected]



Chapter 4

Stationary Current Fields

11/8/2016 2Prof. 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.


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


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


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


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.






10

11/8/2016 Prof. Dr. Magdi El-Saadawi 11


Table 4.1 shows the conductivities of several media in

S/m



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)






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


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



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


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


4.6. Equations for the Electric Field of Stationary Currents


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.


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


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.


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.


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


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




4.9. Resistance Calculations p.159


Ex. 1

p.161

solution








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Electromagnetic Field Theory - saadawi1 · This energy must come from a non-conservative field. The source of non-conservative field may be electric batteries (conservation of chemical