1、5 Magnetostatics5.1 Biot-Savart Law and Amperes Force Law地球北極1820年4月，丹麥物理學家奧斯特發現電流的磁效應。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載流（ I ）回路 C 在磁場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.恒定磁場的性質是有旋無源,電流是激發磁場的渦旋源。Basic equations for magnetostatic fieldF2不能表示恒定磁場。F1可以表示恒定磁場。解： 例 試判斷 能否表示為一個恒定磁場？ 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 取零參考點 The magnetic flux density at field point P is For an infinitely long current-carrying straight wire, 解: 由上例計算結果, 兩導線在 P 點的磁矢位例 應用磁矢位計算兩線輸電線的磁場。總的磁矢位磁感應強度圖 圓截面雙線輸電線原點處x=0的平面上圖 圓截面雙線輸電線 圖 一對反向電流傳輸線xy 圖 一對同向電流傳輸線 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、軌道磁矩spin magnetic moment 自旋磁矩 molecule current 分子電流molecule magnetic moment 分子磁矩 m+-分子磁矩I磁偶極子受磁場力而轉動Magnetic materials 磁性物質：(磁介質 magnetic medium）（2）Paramagnet 順磁體：介質磁化后呈弱磁性。附加磁場B與外場B0同向。合成磁場B B0 , r 1（1）Diamagnet 抗磁體：介質磁化后呈弱磁性。附加磁場B與外場B0反向。 合成磁場B B0 , r B0 , r 1（3）Ferromagnet 鐵磁體：介質磁化后呈強磁性。附加磁場B與外場B

22、0同向。 合成磁場B B0 , r 1 介質磁化后，在介質橫截面邊緣出現環形電流，稱為“磁化電流” magnetizing current （Im ）。ImParamagnet:存在分子固有磁矩。無外磁場：外磁場中：Ferromagnet：鐵磁質內部存在“自發磁化區”，稱為磁疇。 magnetic domain 磁疇大小：每個磁疇所含分子數： Hysteresis loop 磁滯回線 aO HBBr-HsHsbcdefHcOa: 起始磁化曲線Bs : 飽和磁場強度Br : 剩余磁感應強度Bs鐵磁材料的分類：Soft magnetic material軟磁材料：磁滯回線細而窄容易磁化，容易退磁，

23、適用于交變磁場。如制造電機，變壓器等的鐵心。Hard magnetic material 硬磁材料磁滯回線較寬，剩余磁感應強度比較大。適合于制造永磁體。Permanent magnetic material 永磁材料HB矩磁材料：磁滯回線接近于矩形，剩余磁感應強度Br接近于飽和磁感應強度Bs。適合于制作記錄磁帶及計算機的記憶元件。磁介質種類種 類溫度相對磁導率r 1鉍汞銅氫（氣）293K293K293K116.610-512.910-511.010-513.8910-5r 1氧（液）氧（氣）鋁鉑90K293K293K293K1+769.910-51+334.910-51+1.6510-51+2

24、6.010-5r 1鑄鋼鑄鐵硅鋼坡莫合金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. 鐵磁材料B 和H 呈非線性關系， 不是一個恒量。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.電流方向與矩形回路形成右手螺旋關系 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)磁通勢或磁動勢單位:A or At 為磁路的磁阻reluctance ， 單位： ampere-turns per weber (At/Wb) or (A/Wb)。 Ohms law for magnetic circuitHl mmf drop 磁位降或磁壓降。 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對于任意復雜磁路，磁路的每一個分支上所連接各支路的磁通代數和等于0。在磁路的任意閉合回路中，各段磁路上磁壓的代數和等于閉合回路中磁動勢。圖 磁通計算 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連續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:求電流I？,若在磁路中產生 , 例 已知磁路 l=20cm ，截面積 ， ，A/WbA/Wb側柱對稱性解: 中間柱例 有一對稱磁路，中間柱截面積為 ,求側柱的磁通。 兩側柱截面積側柱磁通Wb圖 磁通計算Calculation of the nonlinear m

44、agnetic circuit解：查磁化曲線，H=300 A/m磁勢例 一圓環形磁路及基本磁化曲線如圖所示，平均磁路長度 l = 100 cm ，截面積 A= 5 cm2，若要求產生 210-4 Wb 的磁通，求磁勢為多少？圖 磁路計算B/T圖 非線性磁路計算B/T3000.4已知線圈匝數N=1000，電流 I = 1A，試求磁通 為多少？查磁化曲線，Wb解 B=1.05T圖 非線性磁路計算5.6 Inductance and Energy in a Magnetic Field Inductance 電感 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)磁鏈 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 電流 I 產生的磁矢位A為與 C2 交鏈的磁通為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對C2間的互感系數回路C2對C1間的互感系數Consider two linear circuits each having a single turn. Thus, The mutual inductance is設回路 1 通以電流 I1，回路 1在空間任意點的磁矢位為A1。Neumann formula穿過回路 2 的磁鏈為 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