电磁场与电磁波第四课

1、4 Steady Electric Currents共五十七页4.1 Current Density共五十七页CurrentConduction current (传导电流(chun do din li)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 constitutes a current. 共五十

2、七页 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.Existence conditions of steady current in a conductor:There must e

3、xist a steady electric field inside the conductor.Sq共五十七页 Current density J ( volume current density) (A/m2) Note The total current passing through a surface S is 图 电流的计算Sq电流(dinli)面密度共五十七页 Consider a region with . The charge are moving with an average velocity . Choose a surface element is normal t

4、o the velocity. The total charge moveing through would be The current through the surface is Thus, the current density is Note: Conduction current J drift velocity共五十七页 Surface current density (A/m) where the line element is perpendicular to the current direction. where is an average velocity of the

5、 moving charges. 电流线密度(md)共五十七页current element电流元是电荷元dq以速度(sd) v 运动形成的电流共五十七页图 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. 恒定(hngdng)电流场与恒定(hngdng)电场相互依存。电流J与电场E方向一致。 电路理论中的(积分(jfn

6、)形式)欧姆定律共五十七页 The conductivities of common materials (20)MaterialConductivity (S/m)MaterialConductivity (S/m)AluminumClayCopperWater (fresh)GoldWater (Sea) 5SilverSoil (sandy)NickelRubberZincQuartzIronMarble共五十七页Consider a conducting medium.The current intensity is The potential difference along the

7、 length l is Substituting , we obtainwhere is the resistance. (unit: )Resistivity:共五十七页 Conductance G: Example 4.1.1 A spherical capacitor is formed by two concentric spherical shells of radii a and b. The conductivity between two shells is Determine the conductance of the spherical capacitor.(s)共五十

8、七页 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 is共五十七页The conductance of the spherical capacitor is共五十七页4.2 Continuity of Current共五十七页 Consider any conducting region V bounded by a close

9、d surface S. An outward 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 we obtain共五十七

10、页 The differential (or point) form of the equation of continuity. The points of changing charge density are sources of volume current density. The intergral form of the equation of continuityThe principle of conservation of chargeAny change of charge in a region must be accompanied by a flow of char

11、ge across the surface bounding the region.共五十七页4.3 Electric Field for the Conducting Medium恒定电场(din chng)（电源外）的基本方程共五十七页For a conducting medium to sustain a steady current, Thus, The steady current field is a continuous or solenoidal field. The lines of steady current are always continuous.电流线是连续(li

12、nx)的。Kirchhoffs current law 基尔霍夫电流(dinli)定律共五十七页 The steady electric field must be irrotational or conservative. Kirchhoffs voltage law 基尔霍夫电压(diny)定律 Note：所取积分(jfn)路径不经过电源 Constitutive relationshipDefinition: scalar potential共五十七页 Substituting into , we have For a uniform medium thus, Substitute ,

13、we have Thus, The potential distribution within a conducting medium satisfies Lapalaces equation.共五十七页Electric source:提供(tgng)非静电力将非电能转为电能的装置。 (non-electrostatic force)Non-electrostatic field intensity共五十七页 Electromotive force (emf): It is the work done by non-electrostatic force on unit positive ch

14、arge from negative to positive pole within the electric 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 . 共五十七页4.4 Boundary Conditions for Current Density共五十七页 Boundary (interface) of two conducting media of different cond

15、uctivities 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.Applying , we get (continuous) 共五十七页 2 The Tangential Component of E is the unit vecto

16、r 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. 共五十七页Applying we have or (continuous)et共五十七页 Medium 1 is a poor conductor and medium 2 is a good conductor. J an

17、d 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 very small. 共五十七页 Boundary conditions in terms of the potential Since the height h approaches to zero, the line integral from point 1 to point 2 appr

18、oaches to zero. Thus, 共五十七页4.5 Joules Law共五十七页 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

19、 distance such that the work done by the electric field force is The power supplied by the electric field is 共五十七页Definition: Power density p is the power per unit volume. Point (or differential) form of Joules law. For a linear conductor, the power density is Thus, the power dissipation with a volu

20、me V is (W/m3)(W) W焦耳定律积分(jfn)形式共五十七页 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. If the potential difference between the conductors is U. Determine the power dissipation per u

21、nit length of the coaxial cable. 共五十七页 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 is The electric field is The potential difference is Thus, the current density is共五十七页 The power dissipation per

22、unit length of the coaxial cable is where the resistance per unit length is共五十七页4.6 Analogy Between D and J共五十七页Table The relationship of the two fieldsEquationSteady electric currents (outside electric source)Electrostatics(in a charge-free region)Field equationsConstitutive relationshipLaplaces eq

23、uationBoundary conditions共五十七页analogy method比拟(bn)方法静电场恒定(hngdng)电场（电源外）恒定电场E静电场ED共五十七页 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(wnqun)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。共五十七页 The capacitance C is The conductance G is G and C are analogous in pairs.

24、共五十七页Calculation of ConductanceMethod 1. Definition formula（steady current field）OrMethod 2. Analogy methodMethod 3. Laplaces equation共五十七页 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 pla

25、tes 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.共五十七页 Solution The electric field intensity in the conducting medium is The electric flux

26、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 as共五十七页Hence, the current through the medium iswhere is the resistance of the medium. 共五十七页 Example 4.6.2 The region between a very long coa

27、xial 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 the conductors.共五十七页Method 1：SolutionThe Conductance is图 同轴电缆(tn zhu din ln)Let The conductance per unit l

28、ength共五十七页Solution Method 2: Analogy methodThe capacitance per unit length of a coaxial cable isSubstituting for we obtain the conductance per unit length as 共五十七页接地(jid)电阻 grounding resistance. 接地(jid) 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

29、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。共五十七页接地电阻(dinz)的计算 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(ync)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。即共五十七页 Example 4.6.3 A spherical grounding resistor is embedded in the earth deeply, as shown in Figure. Find the grounding resistance