电磁场与电磁波第四课课件.ppt

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 const

2、itutes 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 Existence conditi

3、ons 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 Currents Consider a reg

4、ion 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 J drift velocityv4 St

5、eady 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 Steady Electric Current

6、s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 materials(20)73.54 1035 1075.

7、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 medium.The current intensity is The pot

8、ential 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 spherical shells of radii a and b.The cond

9、uctivity 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 potential difference is2()4rqarbrEe2()4rqarbrJE

10、e22200dsin 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 outward flow of charge per second crossing the

11、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 obtain4 Steady Electric Currents The differ

12、ential(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 in a region must be accompanied by a flow of cha

13、rge 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 a continuous or solenoidal field.The lines of stea

14、dy 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：所取积分路径不经过电源 Constitutive relationship0iiU Definition:scalar

15、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 AAAggg4 Steady Electric CurrentsElectric source:提供非静电力将非

16、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 electric source.(V)The total work along a loop done by the force

17、 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 different conductivities and .Normal component of J Construct a cyl

18、indrical 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()0neJJg1122nnEE4 Steady Electric Currents 2 The Tangential Component of E is the

19、 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 CurrentsApplying we have or d0,CElg12d0ttCElElElg12ttEE12()0teEEg(continuous)t 2t

20、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 components are negligibly small.The normal component of E in the good conductor is

21、 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 approaches to zero.Thus,1212nn1122nnEE212121d0Elg12214 Steady Electric Currents4.5

22、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 field force exerted on the charge within is If in time the charges will move a distance s

23、uch 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 volume.Point(or differential)form of Joules law.For a linear conductor,the power densit

24、y 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 conductivity The radii of the inner and outer conductors of the cable are a and b,respectively

25、.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 from inner conductor to outer conductor is I.The magnitude of current density at which the radius

26、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 per unit length isln2baR4 Steady Electric Currents4.6 Analogy Between D and J4 Steady Electric Current

27、sTable The relationship of the two fields0E0Jg0E0DgJEDE20()E20()E12nnJJ12ttEE121212nn12nnDD12ttEE121212nnEquationSteady electric currents(outside electric source)Electrostatics(in a charge-free region)Field equationsConstitutive relationshipLaplaces equationBoundary conditions4 Steady Electric Curre

28、ntsanalogy 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

29、 G is G and C are analogous in pairs.ddddSSbbabaaIGJSESElElgggg蜒ddddSSbbabaaqCDSESElElgggg蜒GC4 Steady Electric CurrentsCalculation of Calculation of ConductanceMethod 1.Definition formula（steady current field）OrIUJ J ElUlE dUIG/E JESISJ d UIG/Method 2.Analogy methodGCCGMethod 3.Laplaces equationUE J

30、ESISJ 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 the plates is U,as shown in Figure.If the conducting medium between the plates is characterized by permittivity

31、 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 conducting medium is The electric flux density in the medium is Using the analogy between J and D for a charge-free medium,

32、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 of the medium.dSAUIUdRJSgdRA4 Steady Electric Currents Example 4.6.2 The region between a very long coaxial cable is

33、 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 the conductors.4 Steady Electric CurrentsMethod 1：SolutionIThe Conductance is2lnIlGbUa图 同轴电缆lIJ22IEl dlU Eld2baIl ln2Ibla

34、Let 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 obtain 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

35、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

36、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）与流经接地体而注入大地土壤的流散电流之比就称为接地电阻。即4 Steady Electric Currents Example 4.6.3 A spherical grounding resistor is embedded in the earth deeply,as shown in Figure.Find the grounding resistance.图2.5.3 深埋球形接地器4 Steady Electri

37、c Currents图 深埋球形接地器Method 1:通过电流场计算电阻24 rIJI24IEr2d44aIIUrra14URIa以r为半径作一球面，由于对称，此面上各点的电流密度矢量大小相同，且均与球面垂直。球面电位4 Steady Electric Currents Method 2:analogy method Solution The capacitance of a isolated spherical conductor of radius a is 4Ca114RGaSubstituting for we obtain the conductanceThe grounding

38、resistance is,4Ga图 深埋球形接地器4 Steady Electric CurrentsGrounding resistance 12Ra22 rIJ Solution Example:浅埋半球形接地器daIu2aElI22IErIwhere is conductivity of the soil.U4 Steady Electric Currents图 跨步电压示意 当跨步电压超过某一安全电压值时，将出现人身伤亡事故。以半球接地体为例研究。跨步电压跨步电压 Step voltage 当电流从接地体流入地中时，特别在发生事故的情况下，经接地体流入地中的电流很大，此电流将沿地面流

39、动而造成地面各点具有较高的电位，此时人若走在电极附近区域，则其两脚将承受一地面电压UBC，由于此电压为人跨步时两脚所承受的电压，故称为跨步电压。4 Steady Electric Currents2d2x bxIUrr人体跨步b约0.8m图 半球形接地器的危险区2()bIx xb22JIEr 距接地器越近（x越小），跨步电压越大，危险性越大。人体的安全电压人体的安全电压U040V跨步电压为22Ibx02IbxU设对人体危险的安全电压设对人体危险的安全电压为为U0，则危险区半径则危险区半径由于x远大于b4 Steady Electric Currents 可根据最大短路电流I，跨步b，安全跨步电压值U0，求得危险区域的半径x，从而采取必要的安全措施。但是决定跨步电压的因素较多，例如赤足者与穿有绝缘鞋者，其能承受的安全跨步电压就有较大差异。