Electromagnetics
P10-1
Lesson 10 Steady Electric Currents
10.1 Current Density
■
Definition
Consider a group of charged particles (each has charge q ) of number density N (m-3),
v
v
moving across an elemental surface a n Δs (m2) with velocity u (m/sec). Within a time
interval Δt , the amount of charge ΔQ passing through the surface is equal to the total
v
v
charge within a differential parallelepiped of volume Δv = (u Δt ) ⋅ (an Δs ) (Fig. 10-1):
v v
ΔQ = NqΔv = ρΔt (u ⋅ an Δs ) ,
where ρ (C/m3) denotes the volume charge density. The corresponding electric current is:
I=
ΔQ
v v
= ρ (u ⋅ an Δs ) .
Δt
v
The current I can be regarded as the “flux” of a volume current density J , i.e.,
v v
I = J ⋅ (an Δs ) . By comparing the above two relations, we have:
v
v
J = ρu (A/m2)
(10.1)
Fig. 10-1. Schematic of derivation of current density.
■
Convection currents
Convection currents result from motion of charged particles (e.g. electrons, ions) in
“vacuum” (e.g. cathode ray tube), involving with mass transport but without collision.
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Electromagnetics
P10-2
Example 10-1 (Optional): In vacuum-tube diodes, some of the electrons boiled away from the
incandescent cathode are attracted to the anode due to the external electric field, resulting in a
v
convection current flow. Find the relation between the steady-state current density J and
the bias voltage V0 . Assume the electrons leaving the cathode have zero initial velocity. This
is the “space-charge limited condition”, arising from the fact that a cloud of electrons (space
charges) is formed near the hot cathode, repulsing most of the newly emitted electrons.
Fig. 10-2. Schematic of vacuum-tube diode.
Ans: Assume the linear dimension of cathode and anode is much larger than the length of
tube d , planar symmetry leads to: (i) potential distribution is V ( y ) (with boundary
conditions: V (0) = 0 , V (d ) = V0 ), (ii) volume charge density is ρ ( y ) (<0), (iii) charge
v v
velocity is u = a y u ( y ) , respectively. Under the space-charge limited condition: (i) u (0) = 0 ,
v
v
(ii) the net electric field E = − a y E y ( y ) at the cathode ( y = 0 ) is zero: E y (0) = 0 , ⇒
V ′(0) = 0 .
v
J
v
(1) In steady state, J = − a y J is constant. By eq. (10.1), J = − ρ ( y )u ( y ) , ⇒ ρ ( y ) = −
.
u( y)
(2) By the energy conservation: eV ( y ) =
u( y) =
1
mu 2 ( y ) , where m is the mass of an electron. ⇒
2
m
2eV ( y )
, ρ ( y) = − J
.
m
2eV ( y )
(3) Since there is free charge density ρ ( y ) inside the tube, the potential V ( y ) has to
satisfy with Poisson’s equation [eq. (8.1)]:
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Electromagnetics
P10-3
d 2V
ρ ( y) J
=−
=
2
dy
ε0
ε0
m
.
2eV ( y )
(4) It is unnecessary to solve this “nonlinear” ordinary differential equation to get V ( y ) if
only the relation of J (V0 ) is of interest. Instead, we multiply both sides by 2
2
dV d 2V
J dV
=2
2
dy dy
ε 0 dy
dV
:
dy
m
,
2eV ( y )
then integrate with respect to y :
2J
∫ 2V ′( y) ⋅V ′′( y)dy = ε
0
(
)
⎛ dV
⎜⎜
⎝ dy
m
V −1 / 2 dV , ⇒
2e ∫
2
⎞
4J
⎟⎟ =
ε0
⎠
mV ( y )
+c .
2e
dV
J ⎡ mV ( y ) ⎤
By boundary conditions: V (0) = 0 , V ′(0) = 0 , ⇒ c = 0 , ⇒
=2
ε 0 ⎢⎣ 2e ⎥⎦
dy
1/ 4
,
1/ 4
V
(5)
∫
V0
0
1/ 4
V
−1 / 4
J ⎛m⎞
dV = 2
⎜ ⎟
ε 0 ⎝ 2e ⎠
−1 / 4
J ⎛m⎞
dV = 2
⎜ ⎟
ε 0 ⎝ 2e ⎠
dy .
⎛⎜ d dy ⎞⎟ , ⇒
⎝ ∫0 ⎠
4ε
J= 0
9
2e V03 / 2
m d2
(10.2)
<Comment>
1) The nonlinear I − V relation of vacuum-tube diode differs from the Ohm’s law ( I ∝ V ),
which describes the “conduction” current in conductors.
2) The I − V
relation of forward-biased semiconductor diodes exhibits a stronger
nonlinearity: I ∝ eαV .
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Electromagnetics
■
P10-4
Conduction (drift) currents
As discussed in Lesson 7, the electrons of conductors only partially fill the conduction band
(Fig. 7-1)
and can be easily released from parent nuclei as free electrons by thermal
excitation at room temperatures. The velocities of individual free electrons are high in
magnitude (~105 m/s at 300K) but random in direction, resulting in no net “drift” motion nor
net current.
v
In the presence of static electric field E , the free electrons experience: (1) electric force
v
v
mn ud
− eE ( e > 0 ) to accelerate the electrons, (2) frictional force −
due to collisions with
τ
v
immobile ions, where u d is the drift (average) velocity of electrons, mn and τ represent
the effective mass of conduction electrons and mean scattering time between collisions
(considering the influence of crystal lattice), respectively. In steady state, these two forces
balance with each other (Drude model), ⇒
v
eτ v
v
ud = −
E = − μe E ,
mn
where the electron mobility μe =
eτ
(m2/V/sec) describes how easy an external electric
mn
field can influence the motion of electrons in the conductor. For typical conductors and
v
strength of electric fields, u d
is much slower (~mm/sec) than the speed of individual
electrons. By eq. (10.1), the conduction current density is:
v
v
J = σE (A/m2)
(10.3)
where σ = − ρ e μe [(Ωm)-1] ( ρ e < 0 , σ > 0 ) denotes the electric conductivity, ρe means
the charge density of the electrons. For semiconductors, both electrons and holes contribute
to conduction currents, ⇒
σ = − ρe μe + ρ h μ h .
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Electromagnetics
P10-5
10.2 Microscopic and Macroscopic Current Laws
■
Ohm’s law
Eq. (10.3) is the microscopic form of Ohm’s law. Consider a piece of (imperfect) conductor
of arbitrary shape and homogeneous (finite) conductivity σ (Fig. 10-3a). The potential
v v
difference between the two equipotential end faces A1 , A2 is: V12 = V1 − V2 = ∫ E ⋅ dl ,
L
where L is some path starting from A1 and ending at A2 . The total current flowing through
v v
v v
some surface A between A1 and A2 is: I = ∫ J ⋅ ds = ∫ σE ⋅ ds . The resistance R of
A
A
the conductor is defined as:
v v
E
V
∫ ⋅ dl
R = 12 = L v v ,
I
σ ∫ E ⋅ ds
(10.4)
A
which is a constant independent of V12 and I (but depending on the geometry and material
of the conductor). For a conductor of “uniform” cross-sectional area S (Fig. 10-3b), eq.
(10.4) gives R =
EL
,⇒
σES
R=
L
σS
(10.5)
Fig. 10-3. A conductor of (a) arbitrary shape (after Inans’ book), and (b) uniform
cross-sectional area S and length L (after DKC).
Electromotive force and Kirchhoff’s voltage law
v
v
If conduction current density J is driven by a conservative electric field E (created by
■
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Electromagnetics
P10-6
v
v
charges) alone, ⇒ J = σE ,
∫
C
v v
E ⋅ dl = ∫
v
v
C
(J σ )⋅ dl
= 0 [eq. (6.4)], ⇒ no steady “loop”
current exists. Therefore, a non-conservative field produced by batteries, generators …etc. is
required to drive charge carriers in a closed loop.
Fig. 10-4. Electric fields inside an electric battery (after DKC).
Consider an open-circuited battery, where some positive and negative charges are
accumulated in electrodes 1 and 2 due to chemical reaction (Fig. 10-4). Inside the battery, an
v
impressed field Ei (not an electric field, but a “force”) produced by chemical reaction
v
balances the electrostatic field Einside arising from the accumulated charges, preventing
charges from further movement. The electromotive force (emf), defined as the line integral of
v
Ei
from electrode 2 to electrode 1:
v
1 v
V ≡ ∫ Ei ⋅ dl ,
2
v
v
describes the strength of the non-conservative source. By Einside = − Ei and eq. (6.4), we
have:
∫
C
v 1v
v
v 1v v
v v
2 v
2 v
E ⋅ dl = ∫ Eoutside ⋅ dl + ∫ Einside ⋅ dl = ∫ Eoutside ⋅ dl − ∫ Ei ⋅ dl = 0 , ⇒
1
2
1
2
v
2 v
V = ∫ Eoutside ⋅ dl = V1 − V2 (Volt)
(10.6)
1
If the two terminals are connected by a uniform conducting wire of resistance R =
L
, the
σS
total field:
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Electromagnetics
P10-7
v
v
v v ⎧⎪ Einside + Ei = 0, inside the battery
E + Ei = ⎨ v
⎪⎩ Eoutside , outside the battery
v
drives a loop current I of volume density J (where J = I S ). In the conducting wire,
v
v
v
v
2 v
⎛ J ⎞ v IL
J = σEoutside , V = ∫ Eoutside ⋅ dl = ∫ ⎜⎜ ⎟⎟ ⋅ dl =
, ⇒ V = RI . For a closed path with
1
C
σS
⎝σ ⎠
multiple sources and resistors, we get the Kirchhoff’s voltage law:
∑V
j
j
■
= ∑ Rk I k
(10.7)
k
Equation of continuity and Kirchhoff’s current law
Consider a net charge Q confined in a volume V bounded by a closed surface S . Based
on the principle of conservation of charge (a fundamental postulate of physics), a net current
I flowing out of V must result in decrease of the enclosed charge:
I =−
v
dQ
,⇒
dt
v
d
∫ J ⋅ ds = − dt ∫
S
V
ρdv .
By the divergence theorem [eq. (5.24)] and assuming that the volume V is stationary (does
not moving with time), ⇒
∂ρ
dv , leading to the equation of continuity:
V ∂t
v
∫ (∇ ⋅ J )dv = − ∫
V
v
∂ρ
∇⋅ J = −
∂t
For “steady currents”,
∂ρ
= 0 , eq. (10.8) is reduced to:
∂t
v
∇⋅ J = 0
(10.8)
(10.9)
v
This means there is no steady current source/sink, and the field lines of J always close
upon themselves. The total current flowing out of a circuit junction enclosed by surface S
becomes:
v v
v
(
)dv = 0 , ⇒
I
=
J
⋅
d
s
=
∇
⋅
J
∑ j ∫
∫
j
S
V
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Electromagnetics
P10-8
∑I
j
=0
(10.10)
j
The Kirchhoff’s current law eq. (10.10) is the macroscopic form of eq. (10.8) in steady state.
Example 10-2: Show the dynamics (time dependence) of free charge density ρ inside a
homogeneous conductor with constant electric conductivity σ and permittivity ε .
Ans: By eq’s (10.3), (10.8),
v
v
v
∂ρ
∇ ⋅ J = ∇ ⋅ (σE ) = σ (∇ ⋅ E ) = −
.
∂t
By eq’s (7.8), (7.12),
r
v
v
∇ ⋅ D = ∇ ⋅ (εE ) = ε (∇ ⋅ E ) = ρ .
⇒
∂ρ σ
+ ρ = 0 , ρ (t ) = ρ 0e −t τ , where
∂t ε
τ=
ε
σ
(10.11)
represents the life time of free charges inside the conductor (for the initial charge density ρ 0
will decay to its 1/e within a time interval of τ . For a good conductor like copper, τ ≈ 10-19
sec. Therefore, the dynamics is too fast to be considered.
■
Joule’s law
v
In the presence of an electric field E , free electrons in a conductor have a drift (average)
v
velocity u d . Collisions among free electrons and immobile atoms transfer energy from the
v
electric field to thermal vibration. Quantitatively, the work done by E in moving an amount
v
v v
of charge Q for a differential “drift” displacement Δld is: Δw = QE ⋅ Δld , corresponding
v v
Δw
= QE ⋅ ud . For a conductor of free charge density ρ
Δt →0 Δt
to a power dissipation of: p = lim
v
where an electric field E exists, the power dissipation in a differential volume dv
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Electromagnetics
P10-9
v v
v v
v v
v v
( Q = ρdv ) becomes: Δp = ρdv(E ⋅ ud ) = (E ⋅ ρud )dv = (E ⋅ J )dv . This means that E ⋅ J
(W/m3) represents the (ohmic) power density, and the total power dissipated in a volume V
v v
v
with inhomogeneous E (r ) and σ (r ) is described by the Joule’s law:
v v
P = ∫ (E ⋅ J )dv (W)
(10.12)
V
If we apply a voltage difference V12 across a homogeneous conductor of uniform
cross-sectional area S and length L (Fig. 10-3b), ⇒
v
v v
v v
v
P = ∫ (E ⋅ J )dv = ∫ (E ⋅ J )Sdl = ∫ E ⋅ Idl = V12 I .
V
L
L
10.3 Boundary Conditions
■
Derivation
As in electrostatics, we can derive the boundary conditions for (conduction) current density
v
J across an interface between two media with different conductivities σ 1 , σ 2 by:
v
v
1) Deduce the divergence and curl relations of J : (i) For steady currents, ∇ ⋅ J = 0 [eq.
v
v
(10.9)]. (ii) By eq’s (10.3) and (6.2), ∇ × E = ∇ × (J σ ) = 0 .
v
v v
J v
2) Transform the equations into their integral forms: (i) ∫ J ⋅ ds = 0 . (ii) ∫
⋅ dl = 0 .
S
C
σ
3) Apply the integral relations to a differential (i) pillbox, and (ii) rectangular loop across the
interface, respectively. We can arrive at (i) normal, and (ii) tangential components of
boundary conditions:
J1n = J 2 n
J 1t
σ1
=
J 2t
σ2
(10.13)
(10.14)
Example 10-3: Represent the surface charge density ρ s on the interface between two lossy
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Electromagnetics
P10-10
media with permittivities ε 1 , ε 2 and conductivities σ 1 , σ 2 by either of the two normal
v
components of D .
Fig. 10-5. Interface between two lossy media.
v
v
Ans: Use the normal boundary conditions of E and J : (1) By eq’s (7.12), (7.15), ⇒
ρ s = D1n − D2 n = ε 1 E1n − ε 2 E2 n ,
v
v
v
where Din , Ein ( i = 1, 2) denote the projections of Di , Ei along a n 2 , respectively. (2) By
eq’s (10.3) and (10.13), ⇒ J1n = J 2 n , σ 1 E1n = σ 2 E2 n , ⇒ E1n =
σ2
σ
E2 n , or E2 n = 1 E1n . ⇒
σ1
σ2
⎞
⎛ σε ⎞
⎛σ ε
ρ s = ⎜⎜ 2 1 − 1⎟⎟ D2 n = ⎜⎜1 − 1 2 ⎟⎟ D1n
⎝ σ 2ε 1 ⎠
⎝ σ 1ε 2 ⎠
Eq. (10.15) means that ρ s = 0 only if (1)
(10.15)
σ1 σ 2
=
, which is rare in reality; (2) σ 1 = σ 2 = 0
ε1 ε 2
(both media are lossless, no free charge exists), where eq. (10.15) fails. For air-conductor
interface (Fig. 7-1), σ 2 → ∞ , ρ s = D1n , consistent with eq. (7.4).
v v
Example 10-4: Find J , E in the two lossy media between two parallel conducting plates
biased by a dc voltage V0 (Fig. 10-6). Also find the surface charge densities on the two
conducting plates and on the interface between the two lossy media, respectively.
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Electromagnetics
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Fig. 10-6. Parallel-plate capacitor filled with two lossy media stacked in series.
v
v
Ans: (1) By planar symmetry and eq. (10.13), there is a constant current density J = − a y J
v
v
J
J
v
= −a y Ei , where Ei =
. Since V0 =
between the plates. By eq. (10.3), ⇒ Ei =
σi
=J∑
i =1, 2
di
σi
,⇒ J =
σi
∑E d
i =1, 2
i
i
V0
.
(d1 σ 1 ) + (d 2 σ 2 )
σ 1V0
σ 2V0
, E2 =
.
σ 2 d1 + σ 1d 2
σ 2 d1 + σ 1d 2
ε 1σ 2V0
ε 2σ 1V0
; ρ s 2 = − D2 = −ε 2 E2 = −
.
(3) By eq. (7.15), ρ s1 = D1 = ε 1 E1 =
σ 2 d1 + σ 1d 2
σ 2 d1 + σ 1d 2
(2) E1 =
(4) By eq. (10.15), the surface charge density between the two lossy media is:
⎛ σε ⎞
(ε σ − ε σ )V
ρ si = ⎜⎜1 − 1 2 ⎟⎟ε 1 (− E1 ) = 2 1 1 2 0 .
σ ε
σ d +σ d
⎝
2 1
⎠
2 1
1 2
<Comment>
1) ρ s1 ≠ ρ s 2 , but ρ s1 + ρ s 2 + ρ si = 0 . By Gauss’s law, there is no electric field outside the
parallel-plate capacitor.
2) In this example, static charge and steady current (i.e., static magnetic field) coexist,
causing “electromagnetostatic” field. However, the magnetic field is a consequence and
does not enter into the calculation of the electric field.
Edited by: Shang-Da Yang
Electromagnetics
P10-12
10.4 Evaluation of Resistance
■
Resistance of single imperfect conductor
The resistance R of a piece of homogeneous lossy medium of finite conductivity σ can
be evaluated by: (1) Assume a potential difference V0 for the two “selected” end faces. (2)
v
v
Find the potential distribution V (r ) by solving boundary-value problem. (3) Find E by
v
v v
v v
V
E = −∇V . (4) Find the total current by I = ∫ J ⋅ ds = ∫ σE ⋅ ds . (5) By eq. (10.4), R = 0 .
S
S
I
Example 10-5: Consider a quarter-circular washer of rectangular cross section (Fig. 10-7) and
finite conductivity σ . Find the resistance if the two electrodes are located at φ = 0 and
φ =π 2.
Fig. 10-7. A quarter-circular conducting washer of rectangular cross section (after DKC).
Ans: Assume: (1) V (φ = 0) = 0 , V (φ = π 2) = V0 , (2) the current flow and electric field are
v
only in aφ -direction, ⇒ V (r , φ , z ) = V (φ ) . Laplace’s equation [eq. (8.2)] in cylindrical
coordinates is simplified as: V ′′(φ ) = 0 , ⇒ V (φ ) = c1φ + c2 . By the two boundary conditions,
⇒ V (φ ) =
2V0
π
v
v
v
v v
v 2V0 1 ⎛ 1 ⎞
v 2V σ 1
, I = ∫ J ⋅ ds
⎜ ∝ ⎟ , J = σE = − aφ 0
S
π r ⎝ r⎠
π r
φ , ⇒ E = −∇V = −aφ
=∫
2V0σ
2V σh ⎛ b ⎞
π
⋅ hdr = 0 ln⎜ ⎟ , R =
.
a πr
π
2σh ln (b a )
⎝a⎠
■
Resistance between two perfect conductors
b
The resistance R between two perfect conductors immersed in some homogeneous lossy
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Electromagnetics
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medium with permittivity ε and conductivity σ can be derived by evaluating the
corresponding capacitance C and applying the relation:
RC =
ε
σ
(10.16)
Fig. 10-8. Evaluation of resistance by RC constant (after DKC).
Q
V Q Q
=
Proof: By definitions, RC = ⋅ = . By eq’s (7.10), (10.3), (7.12), ⇒
I
I V I
v v
D
∫S v ⋅ ds = ε ,
v
∫ J ⋅ ds σ
S
as long as ε and σ have the same spatial dependence.
Example 10-6: Find the leakage resistance between the inner conductor (of radius a ) and
outer conductor (of inner radius b ) of a coaxial cable of length L (Fig. 9-4), where a lossy
dielectric medium of conductivity σ is filled between the conductors.
Ans: By eq. (9.4), C =
ln (b a )
2πεL
ε
=
.
. By eq. (10.16), R =
ln (b a )
σC 2πσL
<Comment>
Since the range of dielectric constant of available materials is very limited (1-100), electric
flux usually cannot be well confined within a dielectric slab. The fringing flux around the
edges of conductors makes the computation of capacitance less accurate. In contrast,
conductivity of available materials differ a lot (10-17−107 S/m), and fringing effect is normally
negligible.
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