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Field and Wave ElectromagneticsField and Wave ElectromagneticsLecture 12李融林R.L. Li4.5Magnetic Fields in Materials and MagneticCircuitsBehavior of Magnetic MaterialsNonmagnetic if(Vacuum)Diamagnetic,if(Silver,Lead,Copper,Water)Paramagnetic, if(Air, Aluminum)Ferromagnetic, if(Cobalt, Nickel, Iron, Silicon iron, Mumetal)The orbiting electrons cause circulating currents and form microscopic magnetic dipoles.In the absence of an external magnetic field the magnetic dipoles of the atoms of most materials have random orientations,resulting in no net magnetic moment.The application of an external magnetic field causes both an alignment of the magnetic moments of the spinning electrons and an induced magnetic moment due to a change in the orbital motion of electrons.4.5.1 Magnetization and Equivalent Current DensitiesWe define a magnetization vector,M ,aswhich is the volume density of magnetic dipole moment. The magnetic dipole moment d m = M dv’will produce a vector magnetic potential(Note thatfor a magnetic dipole of moment .)Usingwe can rewrite d A aswhere V’is the volume of the magnetized material. We now use the vector identityto rewriteThe following vector identity enables us to change the volume integral of the curl of a vector into a surface integral:where F is any vector with continuous first derivatives.Proof: Applying divergence theorem to (F x C), where C is a constant vector, we haveSinceandwe haveWe havewhere is the unit outward normal vector from ds’and S’is the surface bounding the volume V’.The effect of the magnetization vector isequivalent to both a volume current densityand a surface current densityThe problem of finding the magnetic flux density B caused by a given volume density of magnetic dipole moment M is then reduced to finding the equivalent magnetization current densities J m and J ms .Figure 4.5.1The cross-sectionof a magneticmaterial.Example4.5.1Determine the magnetic flux density on the axis of a uniformly magnetized circular cylinder of a magnetic material.The cylinder has a radius b, length L,and axial magnetization .Figure4.5.2A uniform magnetized circular cylinder.SolutionSince Mis a constant,The equivalent magnetization surface current density on the side wallisThe magnet is then like a cylindrical sheet with a surface current density of M(A/m).To find B at P(0,0,z),we consider a differential length dz’with a current andobtain4.5.2 Equivalent Magnetization Charge DensitiesIn a current-free region we may define a scalar magnetic potential V m , from which the magnetic flux density B can be found as . In terms of magnetization vector Mwe may write the scalar magnetic potentialIntegrating this equation over a magnetized body carrying no current, we have We know the gradient of 1/Rwith respect to the primed coordinates isHence(Note thatfor a magnetic dipole of moment.)Recalling the vector identityWe obtainwhere is the outward normal to the surface element ds’.We can conclude that, for field calculation,a magnetized body may be replaced by an equivalentmagnetization surface charge densityρms and an equivalent magnetizationvolume charge densityρmsuch that4.5.3 Boundary Conditions for Magnetostatic FieldsBoundary Condition for BApplying to a small pillbox: the normal component of B is continuous across an interface;For linear media, B 1= µ1H 1 and B 2= µ2H 2Boundary Condition for HSimilarly, applying to a small rectangular closed path, we have orWhere is the outward unit normal from medium 2at the interface .Thus the tangential component of the H field is discontinuous across an interface where a free surface current exists.When the conductivities of both media are finite,current are defined by volume current densities and free surface currents do not exist on the interface.Hence J s =0,and the tangential component of H is continuous across the boundary of almost all physical media;it is discontinuous only when an interface with an ideal perfect conductor or a superconductor is assumed.Figure 4.5.3Boundary condition for H .Example 4.5.2Two magnetic media with permeabilities µ1and µ2have a common boundary,as shown in Figure 4.5.4.Determine the magnitude and the direction of the magnetic field intensity at point P 2in medium 2.SolutionFigure 4.5.4Boundary conditions for Hand B.The desired unknown quantities are H 2andα2.Continuity of the normal component of BfieldrequiresThe tangential component of H field is continuous.WehaveDivision of the second equation by the first equation giveswhich describes the refraction property of the magnetic field.If medium 1is nonmagnetic and medium 2is ferromagnetic,then µ2>>µ1and,α2will be nearly 90degrees.If medium 1is ferromagnetic medium and medium 2is air,then α2will be nearly zero.We obtainThe magnitude of H 2isExample4.5.3Find the image currents of a long straight line current I above an interface between two magnetic media with permeabilitiesµ1andµ2SolutionFigure4.5.5A long straight line current I above an interface between two magnetic media(a);Image current for medium1(b)and image current for medium2(c).Since there is no surface current at the interface which leads toUsing yieldsororSolving the equations givesForForFigure4.5.6Magnetic field lines of a long straight line current in two media.Example 4.5.4Sketch the magnetic flux lines both inside and outside a cylindrical bar magnet having a uniform axial magnetization.Figure 4.5.2A uniform magnetizedcircular cylinder.SolutionWe know that the problem of a cylindrical barmagnet could be replaced by that of amagnetization current sheet having a surfacecurrent density J ms =M 0(J m =0).It is obvious from above equations that the magnetic flux density along the axis at the end faces of the magnet is less than that at the center.Fromwe getOn the side of the magnet there is a surface current given byHence according tothe axial component of B changes by anamount equal to µ0M 0.It must be remarked that while H =B/µ0outside the magnet,H and B inside themagnet are far from being proportionalvectors in the same direction.From H =B /µ0–M ,and the fact that B/µ0along theaxis inside is less than M 0,we observe thatH and B are in opposite directions alongthe axis inside.For a long,thin magnet,L>>b,B p0~µ0M 0.From H =B/µ0–M,weobtain H p0~0.Figure 4.5.7Magnetic flux linesaround a cylindrical bar magnet.4.5.4 Magnetic CircuitsThe curl equation for magnetic fields in materials iswhere J(A/m2) is the volume density of free current.The corresponding integral form is obtained asorwhich is the Ampere’s circuital law in materials,where C is the contour bounding the surface S and I is the total free current passing through S. Ampere’s circuital law states that the circulation of the magnetic field intensity around any closed path is equal to the free current flowing through the surface bounded by the path.Ampere’s circuital law is useful on determining the magnetic field in magnetic circuits.The quantity V mmf (=NI )is called a magnetomotive force (mmf).We define reluctance R as the ratio of the magnetic voltage to the flux ;thusAmpere’s Law for an N-turn magnetic circuit becomesSimilar to Kirchhoff’s voltage law,we may write,for any closed path in a magneticcircuit,Around a closed path in a magnetic circuit the algebraic sum of ampere-turns is equal to the algebraic sum of the products of the reluctances andfluxes.Kirchhoff’s current law for a junction is consequence of. Similarly leads to. Thus we havewhich states that the algebraic sum of all the magnetic fluxes flowing out of a junction in a magnetic circuit is zero.Example 4.5.5N turns of wire are wound around a toroidal core of aferromagnetic material with permeabilityµ.Determine Bf ,in the ferromagneticcore;Hf in the core;and Hgin the airgap.Figure4.5.8Coil on ferromagnetictoroid with air gap.Applying Ampere’s circuital lawSolutionIf flux leakage is neglected,the same totalflux will flow in both the ferromagnetic coreand in the air gap.If the fringing effect ofthe flux in the air gap is also neglected,themagnetic flux density B in both the core andthe air gap will also be the same.However,because of the different permeabilities,themagnetic field intensities in both parts willbe different.Wehavewhere the f and g denote ferromagnetic and gap, respectively.In the ferromagnetic core,and, in the air gap,Ampere’s law yieldsWe haveSimilarlyIf the radius of the cross section of the core is much smaller than the mean radius of the toroid,the magnetic flux density B in the core is approximately constant,and the magnetic flux in the circuitiswhere S is the cross-sectional area of the core.where R f and R g are the reluctances of the ferromagnetic core and the air gap, respectively .Figure 4.5.9Equivalent magneticcircuit and analogous electric circuitfor toroidal coil with air gap.Therefore,or,mfmfThe reluctances are:The two loop equations areSolving these simultaneous equations, we haveExample 4.5.6Find the flux linked with coil N 1in a magnetic circuit shown in Figure 4.5.10.Figure 4.5.10A magnetic circuit.Solution.HomeworkProblems P.6-26 and P.6-27References &Acknowledgements1.M.J.Rhee’s Lectures on Electromagnetic Theory,2005.2.W.H.Hayt,J.A.Buck,Engineering Electromagnetics,7th Ed.,McGraw-Hill,2006.3.U.S.Inan,A.S.Inan,Engineering Electromagnetics,AddisonWesley Longman,2000.4. D.Cheng,Field and Wave Electromagnetics,Second Edition,Addison Wesley,1992.。
Course code: 131300112Title: Electromagnetic Field and Electromagnetic waveCredit rating: 3.5Time: Semester SixBrief description:This course makes students master the theorem and the physical meaning of the Maxwell equations and mathematical expressions. It includes the electromagnetic field and electromagnetic wave. Part one is the electromagnetic field. It makes students to learn using the method of vector analysis on the basis of electromagnetism course to describe the essential physical concept of electrostatic field and constant magnetic field, and giving the basic law of electromagnetic field based on summarizing the basic law of experiment, and studying the method to solve problems in the static field. Electromagnetic wave part mainly introduces about the propagation rules of electromagnetic waves in a variety of media and the basic theory of antenna.Syllabus1.Vector analysisVector algebra, three kinds of commonly used orthogonal coordinate system, the gradient of a scalar field, vector field flux and the divergence of the vector field of circulation and curl, irrotational field and solenoidal field, Laplace operation with green's theorem.2.The basic rule of electromagnetic fieldCharge conservation law, the basic rule of electrostatic field in vacuum, the basic law of constant magnetic field in vacuum, electromagnetic properties of medium, the law of electromagnetic induction and the displacement current, Maxwell's equations, boundary conditions of electromagnetic field.3. Static electromagnetic field and its solution of boundary value problemsElectrostatic field analysis, a conductive medium constant electric field analysis, constant magnetic field analysis, the boundary value problem ofa static field and uniqueness theorem of solution, image method, separation variable method, finite difference method.4. Time-varying electromagnetic fieldWave equation, the electromagnetic field of a function, the law of conservation of electromagnetic energy, the uniquenesstheorem ,time-harmonic electromagnetic field5. Uniform plane wave propagation in unbounded spaceIdeal medium uniform plane wave, polarization of electromagnetic wave, Uniform plane wave propagation in conductive medium, Uniform plane wave propagation in anisotropic medium6. Uniform plane wave reflection and transmissionThe uniform plane wave vertical incidence on the plane, the uniform plane wave vertical incidence in multilayer dielectric plane, the uniform plane wave oblique incidence in the ideal dielectric plane, the uniform plane wave oblique incidence in the ideal conductor plane7. Guided electromagnetic waveIntroduction to guide line of electromagnetic wave, rectangular waveguide, cylindrical waveguide, coaxial waveguide, and resonant cavity8. The electromagnetic radiationRetarded potential, Electric dipole radiationRecommended TextbooksXie Chu-fang and Rao Ke-jin, Electromagnetic Field and Electromagnetic Waves, Higher Education Press, 2006。
电磁场与电磁波课程简介（英文版）Course code: 131300112Title: Electromagnetic Field and Electromagnetic waveCredit rating: 3.5Time: Semester SixBrief description:This course makes students master the theorem and the physical meaning of the Maxwell equations and mathematical expressions. It includes the electromagnetic field and electromagnetic wave. Part one is the electromagnetic field. It makes students to learn using the method of vector analysis on the basis of electromagnetism course to describe the essential physical concept of electrostatic field and constant magnetic field, and giving the basic law of electromagnetic field based on summarizing the basic law of experiment, and studying the method to solve problems in the static field. Electromagnetic wave part mainly introduces about the propagation rules of electromagnetic waves in a variety of media and the basic theory of antenna.Syllabus1.Vector analysisVector algebra, three kinds of commonly used orthogonal coordinate system, the gradient of a scalar field, vector field flux and the divergence of the vector field of circulation and curl, irrotational field and solenoidal field, Laplace operation with green's theorem.2.The basic rule of electromagnetic fieldCharge conservation law, the basic rule of electrostatic field in vacuum, the basic law of constant magnetic field in vacuum,electromagnetic properties of medium, the law of electromagnetic induction and the displacement current, Maxwell's equations, boundary conditions of electromagnetic field.3. Static electromagnetic field and its solution of boundary value problemsElectrostatic field analysis, a conductive medium constant electric field analysis, constant magnetic field analysis, the boundary value problem ofa static field and uniqueness theorem of solution, image method, separation variable method, finite difference method.4. Time-varying electromagnetic fieldWave equation, the electromagnetic field of a function, the law of conservation of electromagnetic energy, the uniqueness theorem ,time-harmonic electromagnetic field5. Uniform plane wave propagation in unbounded spaceIdeal medium uniform plane wave, polarization of electromagnetic wave, Uniform plane wave propagation in conductive medium, Uniform plane wave propagation in anisotropic medium6. Uniform plane wave reflection and transmissionThe uniform plane wave vertical incidence on the plane, the uniform plane wave vertical incidence in multilayer dielectric plane, the uniform plane wave oblique incidence in the ideal dielectric plane, the uniform plane wave oblique incidence in the ideal conductor plane7. Guided electromagnetic waveIntroduction to guide line of electromagnetic wave, rectangular waveguide, cylindrical waveguide, coaxial waveguide, and resonant cavity8. The electromagnetic radiationRetarded potential, Electric dipole radiationRecommended TextbooksXie Chu-fang and Rao Ke-jin, Electromagnetic Field and Electromagnetic Waves, Higher Education Press, 2006。
《电磁场与电磁波》讲义一、什么是电磁场与电磁波在我们的日常生活中，电和磁的现象无处不在。
从电动机的转动到手机的通信，从微波炉的加热到卫星的导航，都离不开电磁场与电磁波的作用。
电磁场，简单来说，就是由带电物体产生的一种物理场。
电荷的运动或者静止都会产生电场，而电流的流动则会产生磁场。
当电场和磁场相互作用、相互影响时，就形成了电磁场。
电磁波呢，则是电磁场的一种运动形态。
它是由同相且互相垂直的电场与磁场在空间中以波的形式移动，其传播方向垂直于电场与磁场构成的平面。
二、电磁场的基本原理要理解电磁场，首先得了解库仑定律和安培定律。
库仑定律描述了两个静止点电荷之间的电场力的大小和方向，它表明电场力与两个电荷的电荷量成正比，与它们之间的距离的平方成反比。
安培定律则阐述了电流元之间的磁场相互作用规律。
通过这两个定律，我们可以初步认识到电场和磁场的产生和作用方式。
麦克斯韦方程组是电磁场理论的核心。
这组方程由四个方程组成，分别描述了电场的高斯定律、磁场的高斯定律、法拉第电磁感应定律和安培麦克斯韦定律。
电场的高斯定律表明，通过一个闭合曲面的电通量等于这个闭合曲面所包围的电荷量除以真空介电常数。
磁场的高斯定律指出，通过任何一个闭合曲面的磁通量恒为零，这意味着不存在磁单极子。
法拉第电磁感应定律说明，当穿过一个闭合回路的磁通量发生变化时，会在回路中产生感应电动势。
安培麦克斯韦定律则将安培定律进行了扩展，引入了位移电流的概念，使得在时变电磁场中，磁场的旋度不仅与传导电流有关，还与位移电流有关。
三、电磁波的特性电磁波具有很多独特的特性。
首先是波动性，它以正弦波的形式传播，具有波长、频率和波速等特征。
波长是指相邻两个波峰或波谷之间的距离，频率则是单位时间内电磁波振动的次数，而波速等于波长乘以频率。
电磁波在真空中的传播速度是恒定的，约为 3×10^8 米/秒。
不同频率的电磁波在介质中的传播速度会有所不同。
电磁波还具有偏振性。
Field and Wave ElectromagneticsField and Wave ElectromagneticsField and Wave ElectromagneticsChapter 1Lecture 1李融林 R.L. LiMaxwell’s Theory ofField and Wave Electromagnetics李融林 R.L. Li1.1 Maxwell Equations in Integral FormOutline¾ Maxwell’s Equations in ¾ Maxwell’s Equations in1.1.1 Gauss' Law for the Electric FieldIntegral Form Differential Form Displacement Current Constitutive Relations∫ D ⋅ ds = ∫SvvVρ v dvC / m2 C / m3¾ Continuity Equation and ¾ Material Parameters and ¾ Boundary Conditions ¾ Summaryρ v is the volume charge density, inV is bounded i the th volume l b d d by b S. 1 e=-1.602×10 C In free space,-19v D is the electric flux density, inFigure 1.1.1 Illustration of Gauss’ law for the electric field.v v 1 ε 0 = 8.854 ×10 −12 F / m ≅ × 10 −9 F / m D = ε0E 36π v E is the electric field intensity, in V / m. is the permittivity of free-space.Gauss’ law for the electric field states that electric charges give rise to electric field. Specifically, the electric flux emanating from a closed surface S is equal to the charge enclosed by that surface.1.1.2 Gauss' Law for the Magnetic Field1.1.3 Faraday's LawS v B is the magnetic flux density, inor in a newer SI unit, tesla (T).v v B ∫ ⋅ ds = 0Wb / m2v v v ∂B v ∫CE ⋅ dl = − ∫S ∂t ⋅ dsA older unit for magnetic flux density is the gauss (G). 1 T=1 Wb/m =10,000 G MRI: 0.2-3 T, Earth: 0.3-0.6 GFigure 1.1.2 Illustration of Gauss’ law for the magnetic field.2S is any surface bounded by C.Figure 1.1.3 Illustration of Faraday’s law.Gauss’ law for the magnetic field states that the magnetic flux emanating from a closed surface S is equal to zero.Faraday’s law is a consequence of the experimental finding by Michael Faraday in 1831 that a time-varying magnetic field gives rise to an electric field. Specifically, the electromotive force around a closed path C is equal to the negative of the time rate of increase of the magnetic flux enclosed by that path.11.1.4 Ampere’s Law1.2 Maxwell Equations in differential Form● Gauss' divergence theoremv v v v v ∂D v ∫CH ⋅ dl = ∫SJ ⋅ ds + ∫S ∂t ⋅ dsv H is the magnetic field intensity, in A / m v J is the electric current density, in A / m 2 v v In free space, B = μ0 H . μ0 = 4π ×10−7 H / mis the permeability of free-space.v v v G ⋅ d s = ∇ ⋅ G ∫ ∫ dvS VFigure 1.1.4 Illustration of Ampere’s circuital law.● Stokes’ theoremε 0 μ0 Ampere’s circuital law is a combination of an experimental finding of Oersted that electric currents generate magnetic fields and a mathematical contribution of Maxwell that time-varying electric fields give rise to magnetic fields. Specifically, the magnetomotive force (mmf) around a closed path C is equal to the sum of the current enclosed by that path due to actual flow of charges and the displacement current due to the time rate of increase of the electric flux (or displacement flux) enclosed by that path.The speed of light: c =1≅ 3.0 ×108 m / sv v v v G ∫ ⋅ dl = ∫ ∇ × G ⋅ dsC SThe del operator ∇ is defined as a vector operator,∇=∂ ∂ ∂ ˆ+ ˆ+ z ˆ x y ∂x ∂y ∂z1.2.1 Gauss’ Law for the Electric Field1.2.2 Gauss' Law for the Magnetic Fieldv v ∫ D ⋅ ds = ∫ ρv dvS VUsing Gauss' divergence theorem,v v B ∫ ⋅ ds = 0SUsing Gauss' divergence theorem,d =∫ ∫ ∇ ⋅ DdvVvVρ v dv dd =0 ∫ ∇ ⋅ BdvVvv ∇ ⋅ D = ρvv ∇⋅B = 01.2.3 Faraday's Lawv v v ∂B v ∫CE ⋅ dl = − ∫S ∂t ⋅ ds1.2.4 Ampere’s Lawv v v v v ∂D v ∫CH ⋅ dl = ∫SJ ⋅ ds + ∫S ∂t ⋅ dsUsing Stokes’ theorem,Using Stokes’ theorem,v v v ∂B v ∫S∇ × E ⋅ ds = −∫S ∂t ⋅ ds v v ∂B ∇×E = − ∂tv v v v v ∂D v ∫S∇ × H ⋅ ds = ∫SJ ⋅ ds + ∫S ∂t ⋅ dsv v v ∂D ∇× H = J + ∂t21.3 Continuity Equation and Displacement Current● Conservation of charge and continuity equation● Ampere’s law and continuity equationS Using Gauss' divergence theorem, we have∫ J ⋅ ds = − dt ∫vvvdVρ v dvdvv Using vector identity ∇ ⋅ ∇ × G ≡ 0v v v ∂D ∇× H = J + ∂t∫ ∇ ⋅ Jdv = −∫V∂ρ v V ∂tr v ∇ ⋅∇× H = ∇ ⋅ J = 0Continuity equation:v ρv ∇ ⋅ J = − ∂∂ tThe law of conservation of charge states that the current due to flow of charges emanating from a closed surface S is equal to the time rate of decrease of the charge inside the volume V bounded by that surface.v ρv ∇ ⋅ J = − ∂∂ tThe principle of conservation of charge requires● Generation of interdependent electric and magnetic fields Maxwell equations:v v v ∇× H = J + JDv Using vector identity, ∇ ⋅ ∇ × G ≡ 0r v v ∇ ⋅ ∇ × H = ∇ ⋅ (J + J D ) = 0r v v ∂ρ ∂ ∇ ⋅ J = −∇ ⋅ J D = − v = − ∇ ⋅ D ∂t ∂tDisplacement current:v v ∂B ∇×E = − ∂t v v v ∂D ∇× H = J + ∂t v ∇ ⋅ D = ρv(1) (2) (3) (4)Continuity equation:v ρv ∇ ⋅ J = − ∂∂ t(5)v ∇⋅B = 0v J(5)Interdependence:(2)v v ∂D JD = ∂tv v H, B(1)ρv(3)v v D, EReferences & Acknowledgements1. N. Narayana Rao’s Book Chapter “Fundamentals of Engineering Electromagnetics Revisited”, Fundamentals of Engineering g g Electromagnetics g , Taylor y & Francis Group, p, LLC, 2006. 2. U. S. Inan, A. S. Inan, Engineering Electromagnetics, Addison Wesley Longman, 2000.3。
