current and current density

President University Erwin Sitompul EEM 7/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 7

Engineering Electromagnetics

http://zitompul.wordpress.com


Current and Current DensityElectric charges in motion constitute a current.The unit of current is the ampere (A), defined as a rate of

movement of charge passing a given reference point (or crossing a given reference plane).

dQI

dt

Chapter 5 Current and Conductors

Current is defined as the motion of positive charges, although conduction in metals takes place through the motion of electrons.

Current density J is defined, measured in amperes per square meter (A/m2).


Current and Current DensityChapter 5 Current and Conductors

The increment of current ΔI crossing an incremental surface ΔS normal to the current density is:

NI J S

If the current density is not perpendicular to the surface,

I J S

Through integration, the total current is obtained:

SI d J S



Current density may be related to the velocity of volume charge density at a point.

• An element of charge ΔQ = ρvΔSΔL moves along the x axis

• In the time interval Δt, the element of charge has moved a distance Δx

• The charge moving through a reference plane perpendicular to the direction of motion is ΔQ = ρvΔSΔx

QI

t

v

xSt



v xI Sv The limit of the moving charge with respect to time is:

x v xJ v In terms of current density, we find:

vJ v

This last result shows clearly that charge in motion constitutes a current. We name it here convection current.

J = ρvv is then called convection current density.


Continuity of CurrentChapter 5 Current and Conductors

The principle of conservation of charge: “Charges can be neither created nor destroyed.”

But, equal amounts of positive and negative charge (pair of charges) may be simultaneously created, obtained by separation, destroyed, or lost by recombination.

SI d J S

Any outward flow of positive charge must be balanced by a decrease of positive charge (or perhaps an increase of negative charge) within the closed surface.

If the charge inside the closed surface is denoted by Qi, then the rate of decrease is –dQi/dt and the principle of conservation of charge requires:

i

S

dQI d

dt J S

• The Continuity Equation in Closed Surface

• The Integral Form of the Continuity Equation


v

t

J


Continuity of CurrentThe differential form (or point form) of the continuity equation is

obtained by using the divergence theorem:

vol( )

Sd dv J S J

vol vol( ) v

ddv dv

dt J

We next represent Qi by the volume integral of ρv:

If we keep the surface constant, the derivative becomes a partial derivative. Writing it within the integral,

vol vol( ) vdv dv

t

J

( ) vv vt

J

• The Differential Form (Point Form) of the Continuity Equation



Continuity of CurrentExample

The current density is given by .21A mt

rer

J a

• Total outward current at time instant t = 1 s and r = 5 m.r rI J S 1 1 2

5( )(4 5 )r re a a 23.11 A

• Total outward current at time instant t = 1 s and r = 6 m.

r rI J S 1 1 2

6( )(4 6 )r re a a 27.74 A

• Finding volume charge density:

v

t

J 2

2

1 1( )tr e

r r r

2

1 ter

2

1 tv e dt

r 2

1( )te K r

r

, 0vt ( ) 0K r 32

1C mt

v er

rr

v

Jv

m sr



Metallic ConductorsThe energy-band structure of three types of materials at 0 K is

shown as follows:

Energy in the form of heat, light, or an electric field may raise the energy of the electrons of the valence band, and in sufficient amount they will be excited and jump the energy gap into the conduction band.



Metallic ConductorsFirst let us consider the conductor.Here, the valence electrons (or free conductive electrons) move

under the influence of an electric field E.An electron having a charge Q = –e will experiences a force:

eF E

d ev E

e e J E

J E

In the crystalline material, the progress of the electron is impeded by collisions with the lattice structure, and a constant average velocity is soon attained.

This velocity vd is termed the drift velocity. It is linearly related to the electric field intensity by the mobility of the electron μe:

e e • The Point Form of Ohm’s Law



Metallic ConductorsThe application of Ohm’s law in point form to a macroscopic

region leads to a more familiar form.Assuming J and E to be uniform, in a cylindrical region shown

below, we can write:

V IR

LR

S

SI d JS J S

a

ab bV d E L

a

bd E L

ba E L ab E L

V ELI

J ES

V

L

LV I

S

abVRI

a

b

S

d

d

E L

E S


Conductor Properties and Boundary ConditionsChapter 5 Current and Conductors

Property 1:The charge density within a conductor is zero (ρv = 0) and the surface charge density resides on the exterior surface.

Property 2:In static conditions, no current may flow, thus the electric field intensity within the conductor is zero (E = 0).

Now our next concern is the fields external to the conductor.The external electric field intensity and electric flux density are

decomposed into the tangential components and the normal components.


Conductor Properties and Boundary ConditionsChapter 5 Current and Conductors

The tangential component of the electric field intensity is seen to be zero Et = 0 Dt = 0. If not, then a force will be applied to the surface charges,

resulting in their motion and no static conditions.

The normal component of the electric flux density leaving the surface is equal to the surface charge density in coulombs per square meter (DN = ρS). According to Gauss’s law, the electric flux leaving an

incremental surface is equal to the charge residing on that incremental surface.

The flux cannot penetrate into the conductor since the total field there is zero.

It must leave the surface normally.



Conductor Properties and Boundary Conditions

0d E L0

b c d a

a b c d

1 1,at b ,at a2 2

0t N NE w E h E h

0, 0th w E w

0tE

Sd Q D S

top bottom sidesQ

N SD S Q S

N SD

0t tD E 0N N SD E



Conductor Properties and Boundary ConditionsExample

Given the potential V = 100(x2–y2) and a point P(2,–1,3) that is predefined to lie on a conductor-to-free-space boundary, find V, E, D, and ρS at P, and also the equation of the conductor surface.

2 2100((2) ( 1) )PV 300 V

E V 200 200x yx y a a400 200 V mP x y E a a

0P PD = E23.542 1.771 nC mx y a a

2 2Conductor surface is equipotential

The surface equation is 300 100( )x y 2 2 3x y

23.96 nC mN PD = D =2

, 3.96 nC mS P ND =

• Carefully examine the surface direction


Homework 6D5.2. D5.3. All homework problems from Hayt and Buck,

6th or 7th Edition.

Deadline: 10.03.11, at 07:30 am.


current and current density

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