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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: Th
2、e 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 exist a steady electric fi
3、eld inside the conductor.,Current density J ( volume current density)(A/m2)Note The total current passing througha 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 throug
4、h would be The current 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
5、运动形成的电流,图 J 与 E 之关系,Ohms LawFor conduction current in a conducting medium. In a linear mediumwhere 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 curre
6、nt intensity isThe potential difference along the length l isSubstituting , we obtain,where is the resistance. (unit: )Resistivity:,Conductance G:Example 4.1.1A spherical capacitor is formed by two concentric spherical shells of radii a and b. The conductivity between two shells is Determine the con
7、ductance of the spherical capacitor.,(s),SolutionThe electric field intensity between two shells isUsing Ohms law, the current density between two shells isThe current isThe potential difference is,The conductance of the spherical capacitor is,4.2 Continuity of Current,Consider any conducting region
8、 V bounded by a closed 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 re
9、gion is,we obtain,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 continuity,The principle of conservation of charge,Any change of charge in a region must be accompanied
10、by a flow of charge across 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。,
11、Kirchhoffs current law 基尔霍夫电流定律,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 haveFor a uniform medium thus, Substitute , we haveThus,The potential distrib
12、ution within a conducting medium satisfies 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 el
13、ectric 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 .,4.4 Boundary Conditions for Current Density,Boundary (interface) of two conducting media of different conductivities and . Normal component of JConstruct a cylindrica
14、l 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),2 The Tangential Component of Eis the unit vector tangent to the interface. Consider a small closed path. The two li
15、ne segmentsare parallel to and on opposite sides of the interface. The height of the closed path h approaches to zero.,Applying we haveor,(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 the 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,
