电磁场与电磁波第五课

1、5 Magnetostatics5.1 Biot-Savart Law and Amperes Force Law地球北极1820年4月，丹麦物理学家奥斯特发现电流的磁效应。Amperes Force LawThe force exerted by current-carrying circuit 1 upon 2 iswhere R is the distance vector from to . The magnetic force exerted on the moving charge element dq is The magnetic force exerted on the mo

2、ving charge q isLorentz force in electromagnetic fields is载流（ I ）回路 C 在磁场B中所受的力 The magnetic flux density (magnetic induction intensity) produced at point P from a current element is where is the permeability in vacuum and R is the distance vector from source point to field point .Biot-Savart Law In

3、tegrating, we obtain the magnetic flux density at point P due to a wire carrying steady current I. (T)（Wb/m2） The current element may be expressed as The magnetic flux density in terms of J is The magnetic flux density in terms of JS is Example 5.1.1 An infinitely long filamentary wire located along

4、 z axis carries a current in the z direction, as shown in Figure 5.1.3. Find an expression of magnetic flux density at any point in space. Solution Since the wire is infinitely long , the magnetic flux density is not a function of z. We may choose a field point P in the plane. From Figure, we have a

5、nd Thus, Substituting these expressions into The magnetic flux lines are circles surrounding the wire.5.2 Guasss Law and Amperes Circuital LawGuasss Law for Magnetic FieldTaking the divergence, we get Since we have Using we obtain Note that the curl operation represents partial differentiation with

6、respect to x, y, and z and the differential element is a function only of and We therefore have Hence, The magnetic flux density is solenoidal or continuous.Gausss law for magnetic fields Amperes Circuital Law Amperes circuital law states that the line integral of the magnetic flux density around a

7、closed path is proportional to the current enclosed. Integral form of Amperes circuital law where I is the net current intercepted by the area enclosed by the path. I1I2I3I4CNote：Positive current and directed closed path form right-handed screw. An infinitely long wire located along z axis carries a

8、 current I. The magnetic flux density is Consider a circle of radius Using Stokess theorem, we get where S is bounded by the closed contour C. Since the current can be expressed in terms of volume current density J We get Since S must be any arbitrary open surface bounded by the closed loop, we obta

9、in The steady magnetic field is rotational and solenoidal. The volume current density (current) is the source of the magnetic field.恒定磁场的性质是有旋无源,电流是激发磁场的涡旋源。Basic equations for magnetostatic fieldF2不能表示恒定磁场。F1可以表示恒定磁场。解： 例 试判断 能否表示为一个恒定磁场？ Example 5.2.1 A coaxial cable consists of a long cylindrical

10、 conductor of radius a surrounded by a cylindrical shell of inner radius b and outer radius c. The inner conductor and the outer shell each carry equal and opposite currents I uniformly distributed through the conductor, as indicated in Figure 5.2.2. Calculate the magnetic flux density for all regio

11、ns. Solution The lines of magnetic flux must be concentric circle because of the symmetry. So, the magnetic flux density must be in the direction and has a constant magnitude along each circle. We can use Amperes circuital law to determine the magnetic flux density. Consider a circle of radius as sh

12、own in Figure. Region 1, Since the net current enclosed by circle of radius is We have and The magnetic flux density in this region is Region 2, Since the field point is between the inner and outer conductors, the current enclosed is I. Thus, the magnetic flux density in this region is Region 3, The

13、 magnetic flux density is Region 4, For any closed path outside the coaxial cable the current enclosed is zero. 5.3 Magnetic Vector Potential Because , the vector B can be expressed in terms of the curl of another vector. Definition: Magnetic vector potential A (unit: Wb/m) Note: The directions of B

14、 and A will always be perpendicular to each other. From Biot-Savart law where Usingwe haveThus, Thus, The direction of is the same as The current flows through a volume The current flows over a surface Change the order of the operationFrom and the curl of B isApplying we get Choose . Coulomb gaugeWe

15、 obtain Poissons equation Laplaces equationIn Cartesian coordinates, where, and . The flux of B passing through a surface S is Applying Stokestheorem, we getwhere C is the contour bounding the open surface S. Example 5.3.1 A straight wire carries a current I extend in the z direction from to as show

16、n in Figure 5.3.1. Obtain (a) an expression for magnetic vector potential at a point P in free space, and (b) an expression for magnetic flux density at a point P in free space. Solution (cylindrical coordinate system) The current element is . The distance vector R is The magnetic vector potential a

17、t point is 取零参考点 The magnetic flux density at field point P is For an infinitely long current-carrying straight wire, 解: 由上例计算结果, 两导线在 P 点的磁矢位例 应用磁矢位计算两线输电线的磁场。总的磁矢位磁感应强度图 圆截面双线输电线原点处x=0的平面上图 圆截面双线输电线 图 一对反向电流传输线xy 图 一对同向电流传输线 Example 5.3.2 A circular loop of radius a is in the xy plane and carries

18、a current I, as shown in Figure. Obtain an expression for the magnetic flux density at any point.a Solution In spherical coordinate system, the magnetic vector potential can only have an component which is function of r and from the symmetry. Choose a field point P on the plane. Choose two symmetric

19、al current elements and located at and respectively. The magnitude of magnetic vector potential produced by the current element at field point is The magnitude of A then produced by and at point is Integrating, we obtain the magnitude of A produced by the current loop is ?where Since we haveThus,The

20、 vector form is Finally, the magnetic flux density produced by the current loop is Definition: magnetic dipole moment where S is the area of the loop. We have The expression of A can be rewritten asI5.4 Amperes Circuital Law in Magnetic Materialsmolecular magnetic moment 分子磁矩orbital magnetic moment

21、轨道磁矩spin magnetic moment 自旋磁矩 molecule current 分子电流molecule magnetic moment 分子磁矩 m+-分子磁矩I磁偶极子受磁场力而转动Magnetic materials 磁性物质：(磁介质 magnetic medium）（2）Paramagnet 顺磁体：介质磁化后呈弱磁性。附加磁场B与外场B0同向。合成磁场B B0 , r 1（1）Diamagnet 抗磁体：介质磁化后呈弱磁性。附加磁场B与外场B0反向。 合成磁场B B0 , r B0 , r 1（3）Ferromagnet 铁磁体：介质磁化后呈强磁性。附加磁场B与外场B

22、0同向。 合成磁场B B0 , r 1 介质磁化后，在介质横截面边缘出现环形电流，称为“磁化电流” magnetizing current （Im ）。ImParamagnet:存在分子固有磁矩。无外磁场：外磁场中：Ferromagnet：铁磁质内部存在“自发磁化区”，称为磁畴。 magnetic domain 磁畴大小：每个磁畴所含分子数： Hysteresis loop 磁滞回线 aO HBBr-HsHsbcdefHcOa: 起始磁化曲线Bs : 饱和磁场强度Br : 剩余磁感应强度Bs铁磁材料的分类：Soft magnetic material软磁材料：磁滞回线细而窄容易磁化，容易退磁，

23、适用于交变磁场。如制造电机，变压器等的铁心。Hard magnetic material 硬磁材料磁滞回线较宽，剩余磁感应强度比较大。适合于制造永磁体。Permanent magnetic material 永磁材料HB矩磁材料：磁滞回线接近于矩形，剩余磁感应强度Br接近于饱和磁感应强度Bs。适合于制作记录磁带及计算机的记忆元件。磁介质种类种 类温度相对磁导率r 1铋汞铜氢（气）293K293K293K116.610-512.910-511.010-513.8910-5r 1氧（液）氧（气）铝铂90K293K293K293K1+769.910-51+334.910-51+1.6510-51+2

24、6.010-5r 1铸钢铸铁硅钢坡莫合金2.2103（最大值）4102（最大值）7102（最大值）1105（最大值）where is the magnetic moment of the i th molecule. Macroscopic magnetization phenomena. Definition: Magnetization M is the magnetic dipole moment per unit volume. unit：Thus, the bound volume current density is The bound surface current densit

25、y is where en is the unit normal vector of a surface S.Thus, the bound volume current density is The bound surface current density is where en is the unit normal vector of a surface S.where is the magnetic moment of the i th molecule. Definition: Magnetization M is the magnetic dipole moment per uni

26、t volume. unit：The magnetic moment for an elemental volume is The dA is produced by Using we obtainApplying the vector identitywhere volume is bounded by the surfaceIf the volume A of the magnetized material is We obtainThus, the bound volume current density is The bound surface current density is w

27、here en is the unit normal vector of a surface S.Thus, the bound volume current density is The bound surface current density is where en is the unit normal vector of a surface S. The B field is determined by Definition : magnetic field intensity H Amperes circuital law For a linear homogeneous and i

28、sotropic medium, is called the magnetic susceptibility. is the permeability of the medium. is called the relative permeability of the medium. 铁磁材料B 和H 呈非线性关系， 不是一个恒量。Constitutive relationship Magnetic Scalar Potential In source-free region, Definition: magnetic scalar potential (magnetostatic potent

29、ial) (unit: A) Substituting into , we obtain Laplaces equation Example 5.4.1 A magnetic circuit with a square cross section has a tightly coil with N turns, as shown in Figure. The inner and outer radii of the core are a and b, respectively. If the current in the coil is I and the relative permeabil

30、ity of the magnetic material isbacalculate (a) the magnetic field intensity, (b) the magnetic flux density, (c) the magnetization vector, (d) the bound volume current density and (e) the bound surface current density. Solution (Ignore magnetic leakage) The magnetic field intensity has component only

31、. Consider a circle of radius within the ring. The total current enclosed is NI.baz From Amperes law, the magnetic field intensity is The magnetic flux density is The polarization vector is The bound volume current density is The bound surface current density on the top surface isbaz The bound surfa

32、ce current density on the bottom surface is The bound surface current density on the surface at is The bound surface current density on the surface at is5.5 Boundary Conditions The boundary (interface) between two media having different permeabilities. The Normal Component of B Construct a Gaussian

33、surface in the form of a pillbox. The height h shrinks to zero. Each flat surface S is very small. The unit vector is normal to the interface from medium 2 to medium 1. Applying Gausss law, we get Thus, (continuous) or Since we can also write (5.5.2a) in terms of the normal components of the H field

34、. That is, (discontinuous) The Tangential Component of H Consider a rectangle path shown in Figure. The two segments are parallel to and on opposite sides of the interface. The height h approaches to zero. If , and are three mutually perpendicular unit vectors.电流方向与矩形回路形成右手螺旋关系 Applying Amperes circ

35、uital law, we obtain Assuming that the boundary may carry a surface current density so we can write or Thus, (discontinuous) If the surface current density is zero, continuous orRefraction law is Medium 1 is air and medium 2 is steel with a relative permeability of 2400. The direction of B in air is

36、 almost normal to the boundary.Boundary conditions of magnetic vector potential A Boundary conditions of magnetic scalar potential Magnetic circuit 磁路We call a closed path followed by the magnetic flux in a magnetic material a magnetic circuit. Magnetic materials Many devices such as transformers, e

37、lectric motors, generators, relays. Leakage fluxWe assume thatThe magnetic flux is restricted to flow through the magnetic material with no leakage.The magnetic flux density is uniform within the magnetic material.There is no spreading or fringing of the magnetic flux in the air gap region.The magne

38、tic flux in the magnetic material is equal to the magnetic flux in the air gap.magnetomotive force, (mmf)磁通势或磁动势单位:A or At 为磁路的磁阻reluctance ， 单位： ampere-turns per weber (At/Wb) or (A/Wb)。 Ohms law for magnetic circuitHl mmf drop 磁位降或磁压降。 Consider a magnetic circuit with a cross-sectional area S and

39、the mean length l. The permeability of the magnetic material is . The coil has N turns and carries a current IReluctance method对于任意复杂磁路，磁路的每一个分支上所连接各支路的磁通代数和等于0。在磁路的任意闭合回路中，各段磁路上磁压的代数和等于闭合回路中磁动势。图 磁通计算 Example 5.5.1 A toroidal core of a ferromagnetic material with the relative permeability are wound

40、 N turns of wire. The inner and outer radii of the magnetic core are a and b, respectively. The length of the narrow air gap is l0, as shown in Figure 5.5.5. If a steady current I flows in the wire, calculate (a) the magnetic flux, (b) the magnetic flux density, and (c) the magnetic field intensity

41、in the core and the air gap. Solution Since the permeability of the magnetic material is a constant and applied mmf is known, we can use the reluctance method to determine the magnetic flux density in the core. The total reluctance in the magnetic circuit is The flux in the magnetic circuit isThe ma

42、gnetic flux density of each region isThe magnetic field intensity of each region is法向方向的B连续Bi=B0Two types of problems to the magnetic circuits (1)calculate mmf(2)calculate flux or magnetic flux densityNote:Linear magnetic circuit: is a constant.Nonlinear magnetic circuit: is a variable.We need know

43、the magnetization characteristic or the B-H curve.Calculation of Linear Magnetic CircuitThe magnetomotive force isThe current isThe reluctance of each region issolution:求电流I？,若在磁路中产生 , 例 已知磁路 l=20cm ，截面积 ， ，A/WbA/Wb侧柱对称性解: 中间柱例 有一对称磁路，中间柱截面积为 ,求侧柱的磁通。 两侧柱截面积侧柱磁通Wb图 磁通计算Calculation of the nonlinear m

44、agnetic circuit解：查磁化曲线，H=300 A/m磁势例 一圆环形磁路及基本磁化曲线如图所示，平均磁路长度 l = 100 cm ，截面积 A= 5 cm2，若要求产生 210-4 Wb 的磁通，求磁势为多少？图 磁路计算B/T图 非线性磁路计算B/T3000.4已知线圈匝数N=1000，电流 I = 1A，试求磁通 为多少？查磁化曲线，Wb解 B=1.05T图 非线性磁路计算5.6 Inductance and Energy in a Magnetic Field Inductance 电感 The magnetic flux is directly proportional

45、to the current. If a current I flows in an N-turn coil, the flux linkage is defined as (1) Definition: The inductance (or self-inductance) is the ratio of the total flux (or flux linkage) to the current which they link. (H) (2)磁链 For a circuit having a single turn, we assume that the current centers

46、 upon the axis loop of the circuit. We take the loop of the circuit to determine the flux. External inductance 外自感 电流 I 产生的磁矢位A为与 C2 交链的磁通为The self-inductance is Consider two circuits carrying the currents Definition: mutual inductance where denotes the flux (produced by current I1) through the path

47、 C2. Similarly, where denotes the flux （produced by I2) through the path C1.回路C1对C2间的互感系数回路C2对C1间的互感系数Consider two linear circuits each having a single turn. Thus, The mutual inductance is设回路 1 通以电流 I1，回路 1在空间任意点的磁矢位为A1。Neumann formula穿过回路 2 的磁链为 Example 5.6.1 Two long straight, parallel conductors

48、of radius a carry equal and opposite currents of I, as shown in Figure. The separation between the conductors is D . Calculate the self inductance per unit length of the transmission-line.D Solution The magnetic flux density on the plane containing two lines is The magnetic flux density and are in t

49、he same direction between the lines and normal to the plane containing two lines. The magnitude of the total magnetic flux density is The magnetic flux per unit length is The self inductance per unit length of the transmission-line isD Example 5.6.2 Two coaxial circular coils of radii a and b are shown in Figure 5.6.4. The separation between them is d. Determine the mutual inductance of the system. rzIC2 Solution The magnetic vector potential produced