Electromagnetics P10-1

Lesson 10 Steady Electric Currents

10.1

■ Definition

Consider a group of charged particles (each has charge q ) of N (m-3),

v 2 v moving across an elemental surface an Δs (m ) with 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 Δv = (uΔt)⋅(anΔs) (Fig. 10-1): v v ΔQ = NqΔv = ρΔt(u ⋅ anΔs), where ρ (C/m3) denotes the volume . The corresponding is:

ΔQ I = = ρ()uv ⋅ av Δs . Δt n v The current I can be regarded as the “” of a volume current density J , i.e.,

v v I = J ⋅()anΔs . By comparing the above two relations, we have:

v 2 J = ρuv (A/m ) (10.1)

Fig. 10-1. Schematic of derivation of current density.

■ Convection currents

Convection currents result from motion of charged particles (e.g. , ) in

“vacuum” (e.g. cathode ray tube), involving with mass transport but without collision.

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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 , resulting in a v convection current flow. Find the relation between the steady-state current density J and the bias 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 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 = ayu(y) , respectively. Under the space-charge limited condition: (i) u(0) = 0 ,

v v (ii) the net electric field E = −ay Ey (y) at the cathode ( y = 0 ) is zero: Ey (0) = 0 , ⇒

V ′(0) = 0 .

v J (1) In steady state, J = −av J is constant. By eq. (10.1), J = −ρ(y)u(y) , ⇒ ρ(y) = − . y u(y) 1 (2) By the energy conservation: eV (y) = mu 2 (y) , where m is the mass of an . ⇒ 2

2eV (y) m u(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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d 2V ρ(y) J m 2 = − = . dy ε 0 ε 0 2eV (y)

(4) It is unnecessary to solve this “nonlinear” ordinary differential equation to get V (y ) if dV only the relation of J(V ) is of interest. Instead, we multiply both sides by 2: 0 dy

dV d 2V J dV m 2 2 = 2 , dy dy ε 0 dy 2eV (y) then integrate with respect to y :

2 2J m −1/ 2 ⎛ dV ⎞ 4J mV (y) ∫ 2V ′(y)⋅V ′′(y)dy = (∫V dV ), ⇒ ⎜ ⎟ = + c . ε 0 2e ⎝ dy ⎠ ε 0 2e

1/ 4 dV J ⎡mV (y)⎤ By boundary conditions: V (0) = 0 , V ′(0) = 0 , ⇒ c = 0 , ⇒ = 2 ⎢ ⎥ , dy ε 0 ⎣ 2e ⎦

1/ 4 J ⎛ m ⎞ V −1/ 4dV = 2 ⎜ ⎟ dy . ε 0 ⎝ 2e ⎠

1/ 4 V0 J ⎛ m ⎞ d (5) V −1/ 4dV = 2 ⎜ ⎟ ⎜⎛ dy⎟⎞ , ⇒ ∫ 0 ∫ 0 ε 0 ⎝ 2e ⎠ ⎝ ⎠

4ε 2e V 3 / 2 J = 0 0 (10.2) 9 m d 2

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 diodes exhibits a stronger

nonlinearity: I ∝ eαV .

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■ 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 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 m uv − eE (e > 0) to accelerate the electrons, (2) frictional force − n d due to collisions with τ v immobile ions, where ud 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 (), ⇒

v eτ v v ud = − E = −μe E , mn

eτ 2 where the μe = (m /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, ud 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)

-1 where σ = −ρeμe [(Ωm) ] (ρe < 0 , σ > 0) denotes the electric conductivity, ρe means the charge density of the electrons. For , both electrons and holes contribute to conduction currents, ⇒

σ = −ρeμe + ρhμh .

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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 ⋅dl V12 ∫ L R = = v , (10.4) I σ E ⋅ dsv ∫ 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 S (Fig. 10-3b), eq. EL (10.4) gives R = , ⇒ σES L R = (10.5) σS

Fig. 10-3. A conductor of (a) arbitrary shape (after Inans’ book), and (b) uniform cross-sectional area S and length L (after DKC).

■ 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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v v v v v v charges) alone, ⇒ J = σE , E ⋅ dl = (J σ )⋅dl = 0 [eq. (6.4)], ⇒ no steady “loop” ∫ C ∫ C 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 of

v Ei from electrode 2 to electrode 1:

1 v v V ≡ Ei ⋅ dl , ∫ 2 v v describes the strength of the non-conservative source. By Einside = −Ei and eq. (6.4), we

v v 2 v v 1 v v 2 v v 1 v v have: E ⋅ dl = Eoutside ⋅dl + Einside ⋅dl = Eoutside ⋅ dl − Ei ⋅dl = 0 , ⇒ ∫ C ∫ 1 ∫ 2 ∫ 1 ∫ 2

2 v v V = Eoutside ⋅ dl = V1 −V2 () (10.6) ∫ 1

L If the two terminals are connected by a uniform conducting wire of resistance R = , the σS total field:

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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 2 v v ⎛ J ⎞ v IL J = σEoutside , V = Eoutside ⋅dl = ⎜ ⎟⋅dl = , ⇒ V = RI . For a closed path with ∫ 1 ∫ C ⎜ ⎟ ⎝σ ⎠ σS multiple sources and resistors, we get the Kirchhoff’s voltage law:

∑V j = ∑ Rk I k (10.7) j 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:

dQ v d I = − , ⇒ J ⋅dsv = − ρdv . dt ∫ S dt ∫ V By the [eq. (5.24)] and assuming that the volume V is stationary (does

v ∂ρ not moving with time), ⇒ ()∇ ⋅ J dv = − dv , leading to the equation of continuity: ∫ V ∫ V ∂t v ∂ρ ∇ ⋅ J = − (10.8) ∂t

∂ρ For “steady currents”, = 0 , eq. (10.8) is reduced to: ∂t v ∇⋅ J = 0 (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 I j = J ⋅ ds = (∇ ⋅ J )dv = 0 , ⇒ ∑ ∫ S ∫ V j

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∑ 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 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 e−t τ , where ∂t ε 0 ε τ = (10.11) σ represents the life time of free charges inside the conductor (for the initial charge density ρ0 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 ud . Collisions among free electrons and immobile 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

Δw v v to a dissipation of: p = lim= QE ⋅ud . For a conductor of free charge density ρ Δt→0 Δt v where an electric field E exists, the power dissipation in a differential volume dv

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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 with inhomogeneous E(rv ) and σ (rv ) 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 , we can derive the boundary conditions for (conduction) current density v J across an interface between two media with different conductivities σ1 , 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 J v 2) Transform the equations into their integral forms: (i) J ⋅ dsv = 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 2n (10.13) J J 1t = 2t (10.14) σ 1 σ 2

Example 10-3: Represent the density ρs on the interface between two lossy

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media with ε1 , ε 2 and conductivities σ1 , 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), ⇒

ρs = D1n − D2n = ε1E1n − ε 2 E2n ,

v v v where Din , Ein (i = 1, 2) denote the projections of Di , Ei along an2 , respectively. (2) By

σ 2 σ1 eq’s (10.3) and (10.13), ⇒ J1n = J 2n , σ1E1n = σ 2 E2n , ⇒ E1n = E2n , or E2n = E1n . ⇒ σ1 σ 2

⎛σ ε ⎞ ⎛ σ ε ⎞ ⎜ 2 1 ⎟ ⎜ 1 2 ⎟ ρs = ⎜ −1⎟D2n = ⎜1− ⎟D1n (10.15) ⎝σ1ε 2 ⎠ ⎝ σ 2ε1 ⎠

σ1 σ 2 Eq. (10.15) means that ρs = 0 only if (1) = , which is rare in reality; (2) σ1 = 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 → ∞ , ρs = 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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Fig. 10-6. Parallel-plate 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 v J between the plates. By eq. (10.3), ⇒ Ei = = −ay Ei , where Ei = . Since V0 = ∑ Eidi σ i σ i i=1,2 d V = J ∑ i , ⇒ J = 0 . i=1,2σ i ()d1 σ1 + ()d2 σ 2

σ 2V0 σ 1V0 (2) E1 = , E2 = . σ 2d1 + σ1d2 σ 2 d1 + σ 1d 2

ε1σ 2V0 ε 2σ1V0 (3) By eq. (7.15), ρs1 = D1 = ε1E1 = ; ρs2 = −D2 = −ε 2 E2 = − . σ 2d1 + σ1d2 σ 2d1 + σ1d2 (4) By eq. (10.15), the surface charge density between the two lossy media is:

⎛ σ ε ⎞ (ε σ − ε σ )V ⎜ 1 2 ⎟ 2 1 1 2 0 ρsi = ⎜1− ⎟ε1(−E1) = . ⎝ σ 2ε1 ⎠ σ 2d1 + σ1d2

1) ρs1 ≠ ρs2 , but ρs1 + ρs2 + ρsi = 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 ) coexist,

causing “electromagnetostatic” field. However, the magnetic field is a consequence and

does not enter into the calculation of the electric field.

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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 Find the potential distribution V (rv ) by solving boundary-value problem. (3) Find E by

v v v V E = −∇V . (4) Find the total current by I = J ⋅ dsv = σE ⋅ dsv . (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,

2V0 v v 2V0 1 ⎛ 1 ⎞ v v v 2V0σ 1 v v ⇒ V (φ) = φ , ⇒ E = −∇V = −aφ ⎜∝ ⎟ , J = σE = −aφ , I = J ⋅ ds π π r ⎝ r ⎠ π r ∫ S b 2V σ 2V σh ⎛ b ⎞ π = 0 ⋅ hdr = 0 ln⎜ ⎟ , R = . ∫ a πr π ⎝ a ⎠ 2σhln()b a

■ Resistance between two perfect conductors

The resistance R between two perfect conductors immersed in some homogeneous lossy

Edited by: Shang-Da Yang Electromagnetics P10-13 medium with permittivity ε and conductivity σ can be derived by evaluating the corresponding C and applying the relation: ε RC = (10.16) σ

Fig. 10-8. Evaluation of resistance by RC constant (after DKC).

v D ⋅ dsv V Q Q Q ∫ S ε Proof: By definitions, RC = ⋅ = . By eq’s (7.10), (10.3), (7.12), ⇒ = v = , I V I I J ⋅ dsv 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 medium of conductivity σ is filled between the conductors. 2πεL ε ln(b a) Ans: By eq. (9.4), C = . By eq. (10.16), R = = . ln()b a σC 2πσL

Since the range of dielectric constant of available materials is very limited (1-100), 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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