《电磁场与电磁波第四课.ppt》由会员分享，可在线阅读，更多相关《电磁场与电磁波第四课.ppt（56页珍藏版）》请在三一办公上搜索。
1、4 Steady Electric Currents,4.1 Current Density,Current,Conduction 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 constitutes a current.,Current Definition:The charge
2、 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 exist a steady electric field inside
3、the conductor.,Current density J(volume current density)(A/m2)Note The total current passing through a surface S is,图 电流的计算,q,电流面密度,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 c
4、urrent through the surface is Thus,the current density is,Note:Conduction current J drift velocity,Surface current density(A/m)where the line element is perpendicular to the current direction.where is an average velocity of the moving charges.,电流线密度,current element,电流元是电荷元dq以速度 v 运动形成的电流,图 J 与 E 之关系
5、,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.,恒定电流场与恒定电场相互依存。电流J与电场E方向一致。,电路理论中的(积分形式)欧姆定律,The conductivities of common materials(20),Consider a conducting medium.The current intensity is The po
6、tential difference along the length l is Substituting,we obtain,where is the resistance.(unit:)Resistivity:,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 spherica
7、l capacitor.,(s),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,The conductance of the spherical capacitor is,4.2 Continuity of Current,Consider any conducting region V bounded by a close
8、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,The dif
9、ferential(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 continuity,The principle of conservation of charge,Any change of charge in a region must be accompanied by a flow of charge across
10、the surface bounding the region.,4.3 Electric Field for the Conducting Medium,恒定电场（电源外）的基本方程,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.电流线是连续的。,Kirchhoffs current law 基尔霍夫电流定
11、律,The steady electric field must be irrotational or conservative.,Kirchhoffs voltage law 基尔霍夫电压定律,Note：所取积分路径不经过电源,Constitutive relationship,Definition:scalar potential,Substituting into,we have For a uniform medium thus,Substitute,we have Thus,The potential distribution within a conducting medium s
12、atisfies Lapalaces equation.,Electric source:提供非静电力将非电能转为电能的装置。(non-electrostatic force),Non-electrostatic field intensity,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
13、 a loop done by the force exerted on the unit charge is where E is the coulomb electric field.,4.4 Boundary Conditions for Current Density,Boundary(interface)of two conducting media of different conductivities and.Normal component of J Construct a cylindrical pillbox.The height h shrinks to zero.Eac
14、h flat surface is very small.is the unit vector normal to the interface pointing from medium 2 to medium 1.,Applying,we get(continuous),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
15、 of the interface.The height of the closed path h approaches to zero.,Applying we have or,(continuous),et,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 t
16、he good conductor is very small.,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,4.5 Joules Law,Consider a conducting medium in which the charges are moving with an average velocity under the influe
17、nce 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 such that the work done by the electric field force is The power supplied by the electric field is,Definition:Power density p is the power p
18、er unit 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,(W/m3),(W),W,焦耳定律积分形式,Example 4.5.1 The medium between the conductors of a coaxial cable has conductivity The radii of the inner and outer conductors of t
19、he 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.,Solution Assume that the current per unit length from inner conductor to outer conductor is I.The magnitude of current density at which the
20、radius is The electric field is The potential difference is Thus,the current density is,The power dissipation per unit length of the coaxial cable is where the resistance per unit length is,4.6 Analogy Between D and J,Table The relationship of the two fields,analogy method比拟方法,静电场,恒定电场（电源外）,恒定电场,E,静
21、电场,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。,The capacitance C is The conductance G is G and C are analogous in pairs.,Calculation of Conductance,Method 1.Definition formula（steady current field）,Or,Method 2.Analogy m
22、ethod,Method 3.Laplaces equation,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 and c
23、onductivity determine the current through the medium using the analogy between the J and D fields.,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,we can obtain the volume current
24、density in the medium by substituting for as,Hence,the current through the medium iswhere is the resistance of the medium.,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,
25、respectively,determine the conductance per unit length between the conductors.,Method 1：,Solution,The Conductance is,图 同轴电缆,Let,The conductance per unit length,Solution Method 2:Analogy methodThe capacitance per unit length of a coaxial cable isSubstituting for we obtain the conductance per unit len
26、gth as,接地电阻 grounding resistance.,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。,接地电阻的计算,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
27、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,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,图 深埋球形接地器,Method 1:通过电流场计算电阻,以r为半径作一球面
28、，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,Method 2:analogy method Solution The capacitance of a isolated spherical conductor of radius a is,Substituting for we obtain the conductance,The grounding resistance is,图 深埋球形接地器,Grounding resistance,Solution,Example:浅埋半球形接地器,I,where is conductivity of the soil.,
29、U,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 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。,人体跨步b约0.8m,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U040V,跨步电压为,设对人体危险的安全电压为U0，则危险区半径,由于x远大于b,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。,
