Transcript current

Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
Chapter 5. Steady Electric Currents
•Types of electric currents
– Convection currents result from motion of electrons and/or holes in a vacuum or rarefied gas. (electron beams in
a CRT, violent motion of charged particles in a thunder storm). Convection current , the result of hydrodynamic
motion involving a mass transport, are not governed by Ohm’s law.
– Electrolytic current are the result of migration of positive and negative ions.
Electrolysis
chemical decomposition
+ion
Electrolyte
Usually a diluted
salt solution
-ion
current
- Conduction currents result from drift motion of electrons and/or holes in conductors and semiconductors.
 Atoms of the conducting medium occupy regular positions in a crystalline structure and do not move.
 Electrons in the inner shells are tightly bound to the nuclei are not free to move away.
 Electrons in the outermost shells do not completely fill the shells; they are valence or conduction electrons and are
only very loosely bound to the nuclei.
 When an external E field is applied, an organized motion of the conduction electrons will result (e.g. electron current
in a metal wire).
 The average drift velocity of the electrons is very low (10 -4~10-3 m/s) because of collision with the atoms, dissipating
part of their kinetic energy as a heat.
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring
•
Prof. C.W. Baek & H. Kim
Steady current density
– Electric currents : motion of free charges
– Current density : current per unit area
I  dQ / dt [A]=[C/s]
J  dI / ds [A/m2 ]
 Consider a tube of charge with volume charge density v
moving with a mean velocity u along the axis of the tube.
 Over a period t, the charges move a distance l = u t.
 The amount of charge that crosses the tube's cross-sectional
surface s' in time t is therefore
q '  v v  v l s '  vus ' t
 If the charges are flowing through a surface s whose surface
normal n̂ is not necessarily parallel to u,
q  v u   st
I 
q
 v u   s  J   s
t
J   u [C/m3 m/s=A/m 2 ]
The total current flowing through an arbitrary surface S is I 
Chapter 5 : Steady Electric Currents
: (volume) current density

S
J  d s [A]
Static Electromagnetics, 2008 Spring
In vacuum,
F  qE  me a
u   adt
Prof. C.W. Baek & H. Kim
Example 5-1.
– In conductors and semiconductors, electrons and/or holes can not be accelerated due to the collision.
– The drift velocity is proportional to the applied E field.
u  e E (e : electron mobility)
J   u [C/m3 m/s=A/m 2 ]
J  e e E   E
: point form of Ohm's law σ: conductivity (S/m)
For semiconductors :    e e  h h ( e  h )
cf) resistivity :   1 (  m)
For metal,

 Ohmic media : material following Ohm's law
V12  E
 E  V12 /
I   J  d s  JS
S
 J  I /S


V12  
 I  RI

S


The resistance of a material having a straight length , uniform
cross section area S, and conductivity  : R 
()
S
J  E 
Chapter 5 : Steady Electric Currents
V12

I
S

Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
•Electromotive force (emf)
Static (conservative) electric field :
For an ohmic material :

1
C


E
C
E d  0
J d  0
This equation tells us that a steady current cannot be maintained in the same direction in a closed
circuit by an electrostatic field (Charge carriers collide with atoms and therefore dissipate energy
in the circuit).
 This energy must come from a nonconservative field source for continuous current flow (e.g.
battery, generator, thermocouples, photovoltaic cells, fuel cells, etc.).
 These energy sources, when connected in an electric circuit, provide a driving force to push a
current in a circuit : impressed electric field intensity Ei .
– EMF of a battery : the line integral of the impressed field intensity Ei from the negative to the
positive electrode inside the battery.
V =
ò
1
ur
r
E i ×d l = -
2
ò
1
ur r
E ×d l
2
ur r
òÑE ×d l =
C
ò
2
ur r
E ×d l +
1
ur r
E ×d l = 0
2
Outside
the source
Inside
the source
ò
1
ò
1
2
ur r
E ×d l -
ò
1
ur
r
E i ×d l = 0
2
Inside
the source
1
1
1
2
2
 V   E i  d   E  d  V12  V1  V2
V   E  d
2
1
V   Ei  d
2
Chapter 5 : Steady Electric Currents
(inside source, for emf )
Current flows from (-) to
(+) inside source!
Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
• Kirchhoff's voltage law
When a resistor is connected between terminal 1 and 2 of the battery to complete the
circuit : the total electric field intensity (E + Ei) must be used in the point form of
Ohm's law.

J   E  Ei
1
 J d  
C


C

E  Ei 
E  E d  
i
C
R
J

E  d   Ei  d
C
1
  Ei  d V
2
If the resistor has a conductivity , length
V 

1
C
J d 

, and uniform cross section S, J = I / S.
1 I
   RI
 S
– Kirchhoff's voltage law : Around a closed path in an electric circuit, the algebraic
sum of the emf’s (voltage rises) is equal to the algebraic sum of the voltage drops
across the resistances.
 V  R I
j
j
Chapter 5 : Steady Electric Currents
k k
k
(V)
Static Electromagnetics, 2008 Spring
•
Prof. C.W. Baek & H. Kim
Equation of continuity and Kirchhoff's current law
– Principle of conservation of charge : If a net current I flows across the surface out of (into)
the region, the charge in the volume must decrease (increase) at a rate that equals the current.
dQ
d

    dv     J dv   
dv
S
V
V
V
dt
dt
t

  J  
(A/m 3 ) : Equation of continuity
t
I   J ds  
For steady currents,

=0
t

and
 Jtherefore
0
S
J ds  0
I
j
 0 (A)
j
– Kirchhoff's current law : Algebraic sum of all the dc currents flowing out of (into) a junction
in an electric circuit is zero.
For ac currents,

 0  J  0 and therefore
t
Charge relaxation
  J    ( E )    E  

t
 
E   / 
  0
t 
Chapter 5 : Steady Electric Currents

S
J ds  0
I
j
 0 (A)
j
Really?
Quasi-static case
(at low frequency =0)
  0e



 ( /  ) t
3
(C/m )
 For a good conductor
(e.g. copper),
 = 1.5210-19 [s]
: relaxation time
decay to 1/e (36.8% value)
Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
• Power dissipation and Joule's law
– Power dissipated in a conducting medium in the presence of an electrostatic field E
 Microscopically, electrons in the conducting medium moving under the influence of an
electric field collide with atoms or lattice sites  Energy is thus transmitted from the
electric field to the atoms in thermal vibration.
The work W done by an electric field E in moving a charge q a distance   is
 
w  qE  
w
 qE  u
t 0 t


dP   pi  E    N i qi u i dv  E  J dv
i
 i

 p  lim
dP
 EJ
dv
(W/m 3 )
: Power density under steady-current conditions
For a given volume V, the total electric power converted into heat is
P   E  J dv
V
(W)
: Joule’s law
In a conductor of a constant cross section, dv  dsd, with d
P   Ed
L

S
Jds  VI
Since V = RI,
P  I 2R
Chapter 5 : Steady Electric Currents
(W)
measured in the direction J.
Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
[HW] Solve Example 5-3.
• Boundary conditions for current density
– For steady current density J in the absence of nonconservative energy sources
Differential form
 J  0
J




0


Integral form
 J ds  0
S

C
1

J d  0
(1) Normal component : the normal component of a divergenceless vector field is
continuous.
J1n  J 2 n (A/m 2 )
(2) Tangential component : the tangential component of a curl-free vector field is
continuous across an interface.
J1t  1

J 2t  2
The ratio of the tangential components of J at two sides
of an interface is equal to the ratio of the conductivities.
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
• Resistance calculations
We have calculated the resistance of a conducting medium having a straight length
, uniform cross section area S, and conductivity .
R
S
()
This equation can not be used if the S of the conductor is not uniform  How can
we calculate the resistance?
V   E  d   LE  d 
R  L

I
 J  d s  E  d s
S
S
Q S  E  d s
cf) C  
V  E  d
L
 RC 


– Procedures for resistance calculation
(1) Choose an appropriate coordinate system for the given geometry.
(2) Assume a potential difference V0 between the conductor material.
(3) Find E within the conductor (by solving Laplace's equation and taking E  V ).
(4) Find the total current I from I 
 J  d s   E  d s
S
(5) Find resistance R by taking the ratio V0 / I.
Chapter 5 : Steady Electric Currents
S
Static Electromagnetics, 2008 Spring
Prof. C.W. Baek & H. Kim
• Example 5-6 : Resistance of a conducting flat circular washer
Sol.
(1) Choose a coordinate system : CCS

(2) Assume a potential difference V0.
(3) Find E .
d 2V
0
d 2

V  c1  c2
V  0 at   0
V  V0 at    / 2
2V
V
E  V  ˆ
 ˆ 0
r
r
Boundary conditions are :
V
2V0

,
+ V0
-
(4) Find the total current I.
2 V0
J   E  ˆ
r
(5) Find R.
I   J ds 

S
R 
Chapter 5 : Steady Electric Currents
2 V0

V0


I 2h ln b
a
h
b
a
dr 2 hV0 b

ln (at  = /2 surface)
r

a