電磁場(chǎng)與電磁波第四課

1、4 Steady Electric Currents共五十七頁4.1 Current Density共五十七頁CurrentConduction current (傳導(dǎo)電流(chun do din li)The motion of charges in a conducting medium ( or metal conductor) Convention current (運(yùn)流電流)The motion of charged particles in vacuum (or free space).The motion of charges constitutes a current. 共五十

2、七頁 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共五十七頁 Current density J ( volume current density) (A/m2) Note The total current passing through a surface S is 圖 電流的計(jì)算Sq電流(dinli)面密度共五十七頁 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共五十七頁 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. 電流線密度(md)共五十七頁current element電流元是電荷元dq以速度(sd) v 運(yùn)動(dòng)形成的電流共五十七頁圖 J 與 E 之關(guān)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)電流場(chǎng)與恒定(hngdng)電場(chǎng)相互依存。電流J與電場(chǎng)E方向一致。 電路理論中的(積分(jfn

6、)形式)歐姆定律共五十七頁 The conductivities of common materials (20)MaterialConductivity (S/m)MaterialConductivity (S/m)AluminumClayCopperWater (fresh)GoldWater (Sea) 5SilverSoil (sandy)NickelRubberZincQuartzIronMarble共五十七頁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:共五十七頁 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、七頁 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共五十七頁The conductance of the spherical capacitor is共五十七頁4.2 Continuity of Current共五十七頁 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、頁 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.共五十七頁4.3 Electric Field for the Conducting Medium恒定電場(chǎng)(din chng)（電源外）的基本方程共五十七頁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.電流線是連續(xù)(li

12、nx)的。Kirchhoffs current law 基爾霍夫電流(dinli)定律共五十七頁 The steady electric field must be irrotational or conservative. Kirchhoffs voltage law 基爾霍夫電壓(diny)定律 Note：所取積分(jfn)路徑不經(jīng)過電源 Constitutive relationshipDefinition: scalar potential共五十七頁 Substituting into , we have For a uniform medium thus, Substitute ,

13、we have Thus, The potential distribution within a conducting medium satisfies Lapalaces equation.共五十七頁Electric source:提供(tgng)非靜電力將非電能轉(zhuǎn)為電能的裝置。 (non-electrostatic force)Non-electrostatic field intensity共五十七頁 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 . 共五十七頁4.4 Boundary Conditions for Current Density共五十七頁 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) 共五十七頁 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. 共五十七頁Applying we have or (continuous)et共五十七頁 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. 共五十七頁 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, 共五十七頁4.5 Joules Law共五十七頁 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 共五十七頁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焦耳定律積分(jfn)形式共五十七頁 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. 共五十七頁 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共五十七頁 The power dissipation per

22、unit length of the coaxial cable is where the resistance per unit length is共五十七頁4.6 Analogy Between D and J共五十七頁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共五十七頁analogy method比擬(bn)方法靜電場(chǎng)恒定(hngdng)電場(chǎng)（電源外）恒定電場(chǎng)E靜電場(chǎng)ED共五十七頁 在均勻媒質(zhì)情況下，當(dāng)兩種場(chǎng)的邊界條件（邊界形狀及邊界賦值）完全相同時(shí)，它們的 場(chǎng)與 場(chǎng)是完全(wnqun)相同的，而 場(chǎng)與 場(chǎng)則是彼此相似的，這是從唯一性定理所得到的結(jié)論。 運(yùn)用它們彼此間的相似關(guān)系，將一種場(chǎng)的求解方法過渡到另一種場(chǎng)中來，這種方法稱之為場(chǎng)的比擬法。共五十七頁 The capacitance C is The conductance G is G and C are analogous in pairs.

24、共五十七頁Calculation of ConductanceMethod 1. Definition formula（steady current field）OrMethod 2. Analogy methodMethod 3. Laplaces equation共五十七頁 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.共五十七頁 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共五十七頁Hence, the current through the medium iswhere is the resistance of the medium. 共五十七頁 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.共五十七頁Method 1：SolutionThe Conductance is圖 同軸電纜(tn zhu din ln)Let The conductance per unit l

28、ength共五十七頁Solution Method 2: Analogy methodThe capacitance per unit length of a coaxial cable isSubstituting for we obtain the conductance per unit length as 共五十七頁接地(jid)電阻 grounding resistance. 接地(jid) 在電力設(shè)備的實(shí)際運(yùn)行中，為了設(shè)備及人身的安全和電力系統(tǒng)需要，電氣設(shè)備的接地是必不可少的。這種保護(hù)人身及設(shè)備安全的接地措施，稱之為保護(hù)接地。 “接地”就是電氣設(shè)備和地之間的導(dǎo)體連接。 如當(dāng)變壓器的

29、絕緣損壞時(shí)，變壓器的外殼將可能具有對(duì)地的高電位，此時(shí)當(dāng)工作人員觸及變壓器外殼時(shí)，將承受這一對(duì)地高壓而發(fā)生人身傷亡事故。如果將外殼接地，則外殼與地電位相等，就不會(huì)出現(xiàn)這種危險(xiǎn)。共五十七頁接地電阻(dinz)的計(jì)算 接地裝置由連接導(dǎo)線和埋入地中的接地體（或稱接地電極）組成。 在通常情況下，連接導(dǎo)體本身的電阻，連接導(dǎo)線與接地體間的接觸電阻，以及接地體本身所具有的電阻值都是非常小的，因?yàn)樗麄兌际橇紝?dǎo)體。 因而接地電阻，主要是電流從接地體流入地中時(shí)，所具有的電阻值，亦即從接地體流入地中的流散電流所遇到的電阻。 因此(ync)接地體的電位（無限遠(yuǎn)處電位為零）與流經(jīng)接地體而注入大地土壤的流散電流之比就稱為接地電阻。即共五十七頁 Example 4.6.3 A spherical grounding resistor is embedded in the earth deeply, as shown in Figure. Find the grounding resistance