电磁场与电磁波第四课

1、4 Steady Electric Currents4 Steady Electric Currents4.1 Current Density4 Steady Electric CurrentsCurrentConduction current (传导电流)The motion of charges in a conducting medium ( or metal conductor) Convention current (运流电流)The motion of charged particles in vacuum (or free space).The motion of charges

2、 constitutes a current. 4 Steady Electric Currents Current Definition: The charge quantity passing through a given cross section per unit time (A) Note It is in the direction of the motion of the positive charges. Steady current (direct current DC): The current is constant in time.0dlimdtqqItt Exist

3、ence conditions of steady current in a conductor:There must exist a steady electric field inside the conductor.Sq4 Steady Electric Currents Current density J ( volume current density) (A/m2) Note The total current passing through a surface S is 0limsIJSdSI JSg图 电流的计算SSq电流面密度4 Steady Electric Current

4、s Consider a region with . The charge are moving with an average velocity . Choose a surface element is normal to the velocity. The total charge moveing through would be The current through the surface is Thus, the current density is vSVVVqVt SlSvggddVVVqIttSlSSv=vgggVJvSIVSNote: Conduction current

5、J drift velocityv4 Steady Electric Currents Surface current density (A/m) where the line element is perpendicular to the current direction. where is an average velocity of the moving charges. 0limslIJlSSJvlv电流线密度l4 Steady Electric Currentscurrent element电流元是电荷元dq以速度 v 运动形成的电流dddddddVSSlVVqSSIJ Jl l4

6、 Steady Electric Currents图 J 与 E 之关系 Ohms Law For conduction current in a conducting medium. In a linear medium where is the conductivity of the medium. (S/m). differential form of Ohms law. JE恒定电流场与恒定电场相互依存。电流J J与电场E E方向一致。 电路理论中的(积分形式)欧姆定律URI4 Steady Electric Currents The conductivities of common

7、materials (20)73.54 1035 1075.8 1031071.10 1076.20 1051071.45 10151 1071.67 10171 1071.00 1081 10MaterialConductivity (S/m)MaterialConductivity (S/m)AluminumClayCopperWater (fresh)GoldWater (Sea) 5SilverSoil (sandy)NickelRubberZincQuartzIronMarble4 Steady Electric CurrentsConsider a conducting mediu

8、m.The current intensity is The potential difference along the length l is Substituting , we obtainIlUIRSdSIJSJSgdlUElElgJEwhere is the resistance. (unit: )Resistivity:lRS1(m)gJI SEU lIUSlURI4 Steady Electric Currents Conductance G: Example 4.1.1 A spherical capacitor is formed by two concentric sphe

9、rical shells of radii a and b. The conductivity between two shells is Determine the conductance of the spherical capacitor.IGU(s)4 Steady Electric Currents Solution The electric field intensity between two shells is Using Ohms law, the current density between two shells is The current is The potenti

10、al difference is2()4rqarbrEe2()4rqarbrJEe22200dsin d d4SqqIrr JSg4 Steady Electric CurrentsThe conductance of the spherical capacitor is4411IabGUbaab4 Steady Electric Currents4.2 Continuity of Current4 Steady Electric Currents Consider any conducting region V bounded by a closed surface S. An outwar

11、d flow of charge per second crossing the closed surface S must be equal to the rate at which the charge is diminishing in the bounded region V. where q is the total charge enclosed by the surface at any time . Assume that the volume charge density in the region is dddSqt JSgdSJSg.VdVVqVddVVVVVt Jgwe

12、 obtain4 Steady Electric Currents The differential (or point) form of the equation of continuity. The points of changing charge density are sources of volume current density. Vt JgddVSVVt JSgThe intergral form of the equation of continuityThe principle of conservation of chargeVAny change of charge

13、in a region must be accompanied by a flow of charge across the surface bounding the region.4 Steady Electric Currents4.3 Electric Field for the Conducting Medium恒定电场（电源外）的基本方程4 Steady Electric CurrentsFor a conducting medium to sustain a steady current, Thus, d0SJSg0Jg0VtThe steady current field is

14、a continuous or solenoidal field. The lines of steady current are always continuous.电流线是连续的。0iiI Kirchhoffs current law 基尔霍夫电流定律S I1 I2 I3 I4 节点电流定律 Vt Jg4 Steady Electric Currents The steady electric field must be irrotational or conservative. d0CElgJE0EKirchhoffs voltage law 基尔霍夫电压定律 Note：所取积分路径不经

15、过电源 Constitutive relationship0iiU Definition: scalar potential E4 Steady Electric Currents Substituting into , we have For a uniform medium thus, Substitute , we have Thus, The potential distribution within a conducting medium satisfies Lapalaces equation.JE(0),0Jg0Eg2()0 Egg20()0 EEEggg EEgg E()uuu

16、 AAAggg4 Steady Electric CurrentsElectric source:提供非静电力将非电能转为电能的装置。 (non-electrostatic force)FEqNon-electrostatic field intensity4 Steady Electric Currents Electromotive force (emf): It is the work done by non-electrostatic force on unit positive charge from negative to positive pole within the elec

17、tric source. (V) The total work along a loop done by the force exerted on the unit charge is where E is the coulomb electric field . () dCeEElgdABeElgdCeElg4 Steady Electric Currents4.4 Boundary Conditions for Current Density4 Steady Electric Currents Boundary (interface) of two conducting media of

18、different conductivities and . Normal component of J Construct a cylindrical pillbox. The height h shrinks to zero. Each flat surface is very small . is the unit vector normal to the interface pointing from medium 2 to medium 1.1(2)neApplying , we get (continuous) d0SJSg12d0nnSJSJSJSg12nnJJ12()0neJJ

19、g1122nnEE4 Steady Electric Currents 2 The Tangential Component of E is the unit vector tangent to the interface. Consider a small closed path. The two line segments are parallel to and on opposite sides of the interface. The height of the closed path h approaches to zero. lte124 Steady Electric Curr

20、entsApplying we have or d0,CElg12d0ttCElElElg12ttEE12()0teEEg(continuous)t 2t 1EE 1 11222coscosEE1122sinsinEEet1122ttJJ12nnJJ1122tantan124 Steady Electric Currents Medium 1 is a poor conductor and medium 2 is a good conductor. J and E in medium 1 are almost normal to the interface. The tangential co

21、mponents are negligibly small. The normal component of E in the good conductor is very small. 1212nnEE12=12=725 10 s/m2110s/m1122tantan1212nnEE12=4 Steady Electric Currents Boundary conditions in terms of the potential Since the height h approaches to zero, the line integral from point 1 to point 2

22、approaches to zero. Thus, 1212nn1122nnEE212121d0Elg12214 Steady Electric Currents4.5 Joules Law4 Steady Electric Currents Consider a conducting medium in which the charges are moving with an average velocity under the influence of an electric field E. If the volume charge density is the electric fie

23、ld force exerted on the charge within is If in time the charges will move a distance such that the work done by the electric field force is The power supplied by the electric field is VVFEVWV tV t FlEJEvgggWPVtJEgv,VVtl, t lv4 Steady Electric CurrentsDefinition: Power density p is the power per unit

24、 volume. Point (or differential) form of Joules law. For a linear conductor, the power density is Thus, the power dissipation with a volume V is p JEg,JE2pEJEEEgg2dddVVVPp VVEVJEg(W/m3)(W)PU I W焦耳定律积分形式4 Steady Electric Currents Example 4.5.1 The medium between the conductors of a coaxial cable has

25、conductivity The radii of the inner and outer conductors of the cable are a and b, respectively. If the potential difference between the conductors is U. Determine the power dissipation per unit length of the coaxial cable. .4 Steady Electric Currents Solution Assume that the current per unit length

26、 from inner conductor to outer conductor is I. The magnitude of current density at which the radius is The electric field is The potential difference is Thus, the current density is2IJab2IE4 Steady Electric Currents The power dissipation per unit length of the coaxial cable is where the resistance p

27、er unit length isln2baR4 Steady Electric Currents4.6 Analogy Between D and J4 Steady Electric CurrentsTable The relationship of the two fields0E0Jg0E0DgJEDE20() E20() E12nnJJ12ttEE121212nn12nnDD12ttEE121212nnEquationSteady electric currents (outside electric source)Electrostatics(in a charge-free re

28、gion)Field equationsConstitutive relationshipLaplaces equationBoundary conditions4 Steady Electric Currentsanalogy method比拟方法(0)V静电场0 DEDdSq DS020E恒定电场（电源外）JESISJ d0 J0E02恒定电场JIE静电场EDq4 Steady Electric Currents 在均匀媒质情况下，当两种场的边界条件（边界形状及边界赋值）完全相同时，它们的 场与 场是完全相同的，而 场与 场则是彼此相似的，这是从唯一性定理所得到的结论。 运用它们彼此间的相

29、似关系，将一种场的求解方法过渡到另一种场中来，这种方法称之为场的比拟法比拟法。EDJ4 Steady Electric Currents The capacitance C is The conductance G is G and C are analogous in pairs. ddddSSbbabaaIGJSESElElgggg蜒ddddSSbbabaaqCDSESElElgggg蜒GC4 Steady Electric CurrentsCalculation of Calculation of ConductanceMethod 1. Definition formula（stead

30、y current field）OrIUJ J ElUlE dUIG/E JESISJ d UIG/Method 2. Analogy methodGCCGMethod 3. Laplaces equationUE JESISJ d UIG/4 Steady Electric Currents Example 4.6.1 Two infinitely conducting parallel plates, each of cross-sectional area S, are separated by a distance d. The potential difference between

31、 the plates is U, as shown in Figure. If the conducting medium between the plates is characterized by permittivity and conductivity determine the current through the medium using the analogy between the J and D fields.,4 Steady Electric Currents Solution The electric field intensity in the conductin

32、g medium is The electric flux density in the medium is Using the analogy between J and D for a charge-free medium, we can obtain the volume current density in the medium by substituting for aszUd EezUd DezUd Je4 Steady Electric CurrentsHence, the current through the medium iswhere is the resistance

33、of the medium. dSAUIUdRJSgdRA4 Steady Electric Currents Example 4.6.2 The region between a very long coaxial cable is filled with a material of conductivity and permittivity If the radii of the inner and outer conductors are a and b, respectively, determine the conductance per unit length between th

34、e conductors.4 Steady Electric CurrentsMethod 1：SolutionIThe Conductance is2lnIlGbUa图 同轴电缆lIJ22IEl dlU Eld2baIl ln2IblaLet The conductance per unit length2ln( / )Gb a4 Steady Electric CurrentsSolution Method 2: Analogy methodThe capacitance per unit length of a coaxial cable isSubstituting for we ob

35、tain the conductance per unit length as 2ln( / )Cb a2ln( / )Gb a,4 Steady Electric Currents接地电阻 grounding resistance. 接地接地 在电力设备的实际运行中，为了设备及人身的安全和电力系统需要，电气设备的接地是必不可少的。这种保护人身及设备安全的接地措施，称之为保护保护接地接地。 “接地”就是电气设备和地之间的导体连接。 如当变压器的绝缘损坏时，变压器的外壳将可能具有对地的高电位，此时当工作人员触及变压器外壳时，将承受这一对地高压而发生人身伤亡事故。如果将外壳接地，则外壳与地电位相等，就不会出现这种危险。4 Steady Electric Currents接地电阻的计算接地电阻的计算URI 接地装置由连接导线和埋入地中的接地体（或称接地电极）组成。 在通常情况下，连接导体本身的电阻，连接导线与接地体间的接触电阻，以及接地体本身所具有的电阻值都是非常小的，因为他们都是良导体。 因而接地电阻，主要是电流从接地体流入地中时，所具有的电阻值，亦即从接地体流入地中的流散电流所遇到的电阻。 因此接地体的电位（无限远处电位为零）与