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1、Chapter 7 The steady Magnetic Field恒定磁场,Magnetic Flux and Gauss Law Magnetic Force on a Current-Carrying Wire Magnetic Field Due to Current Amperes Law Magnetic Materials,7-1.The steady current恒定电流,7-2.The power source , the electromotive force 电源 非静电力,compass;microphone; computer disk; ,Magnetic ph
2、enomena:,7-3.The magnetic field磁场,A electric charge set up an electric field that can affect other electric charges. A magnetic charge set up a magnetic field that can affect other magnetic charges? magnetic monopoles磁单极子?,The interacting force between moving charges is magnetic force which is trans
3、ferred by magnetic field (运动电荷之间的相互作用力-磁力,由磁场传递).,The magnitude of :,The direction of is along the zero-force.,we define by the acting force of magnetic filed on the moving charges as:,Dimension: I-1MT2.,地磁场Bearth 510-5 T,The principle of superposition of :,SI unit: tesla (特斯拉) (T), gauss or N/(Am).
4、,2011年6月22日，德国亥姆霍兹-德累斯顿-罗森多夫研究中心（HZDR）的研究人员制造出了强度为91.4特斯拉的磁场，打破数年前由美国洛斯阿拉莫斯国家实验室所创造的89特斯拉的纪录，成为目前世界上最强的人造磁场。,7-4. The Biot-Savart Law毕奥-萨伐尔定理,1. The Biot-Savart law,Electric currents create magnetic fields. Oersted 1820,Shortly after 1820 J. B. Biot and F. Savart developed what is known as the law o
5、f Biot and Savart gives a quantitative description of Oersteds magnetic field in terms of the electric current.,A question arises: Is there a general relation between a current in a wire of any shape and the magnetic field around it?,In SI unit:,The current element Idl generates a magnetic field dB
6、given by:,Magnetic permeability constant in a vacuum (真空磁导率):,The principle of superposition of magnetic field, can be applied for a complete circuit:,(Biot-Savart law),(1) A long straight wire (载流长直导线的磁场 ),The magnitude of the differential magnetic field produced at P by the current-length element
7、Idl located a distance r from P is given by:,2. The application of Biot-Savart law,r =R /sin , s = -R cot,Discussions:,The magnetic field from all current element of straight wire are in the same direction and into the screen.同向,For straight and infinite wire (无限长载流直导线), 1=0,2=,then, The magnitude o
8、f magnetic field of infinite straight wire at a point with distance R is in inverse proportion to the distance.,(2) a circular current (载流圆线圈轴线上的磁场),As the figure, given I, R, x,The direction of is perpendicular to the plane made of Idl and r and .,Discussions:,At center O, x=0,For a circular arc of
9、 current (一段圆弧电流圆心处的磁感应强度):,The direction of .,Solution:,(into the screen),Example :,One quarter of a circular ring of wire carries a current I, and two straight sections whose extensions intersect the center C of the arc. Find at point C?,Exercise: Magnetic Field at point P Due to the current in th
10、e square loop.,Example : A thin disk of dielectric material with radius a has a total charge +Q distributed uniformly over its surface. It rotates f times per second about an axis perpendicular to the surface of the disk and passing through its center. Find the magnetic field at the center of the di
11、sk.,Solution:,The charge on the ring with radius r is:,When the ring rotates with f, its corresponding current is:,The magnetic field of the current ring at the center:,The magnetic field of whole charge disk:,The number of turns per unit length of the solenoid(单位长匝数) n. (i, n, R, L, 1, 2 ),(3) A st
12、retched-out solenoid (长直螺线管),choose dI=nidx,x=Rcot , dx=Rcsc2 d , R2+x2= R2 csc2 ,Same direction!,Explanations:,For a “infinitely long” solenoid, The at the axial line:,Direction inside the solenoid is along the axis and right-hand rule.,B-x curve inside the solenoid (L=10R):,7-5. Magnetic Field Lin
13、es and Flux, Gauss law, No origin, no termination and no cross point(无头无尾不相交,闭合曲线);, Wrap around with current与电流套连;, Right-hand rule with current (与电流成右手螺旋关系),1. The features of magnetic field lines,2. Magnetic flux & Gauss law,Let number of magnetic field line threading area dS,Magnetic Flux,1) Def
14、inition: The number of magnetic field lines crossing a given area. SI unit: Wb(韦伯). 1Wb=1T m2=1N m/A,(1) In uniform field,2) Calculations of Electric Flux,(2) Non uniform field,Question:,Divide the surface, and choose any small element dS.,The electric flux through the entire surface is,(Gauss law o
15、f magnetic field),The flux of magnetic field intersects a closed surface:,7-3. Amperes Loop Law (Ampere circuital theorem),Andre Marie Ampere (17751836) French mathematician and physicist.,安培分子环流假说 : The source of magnetism of all natural-matter come from current（一切磁性的根源是电流）.,Ampere: All magnetic ph
16、enomena come from current (一切磁现象都起源于电流).,Find Electric Field : highly symmetric charge distribution: Gausss law Find Magnetic Field: highly symmetric current distribution: Amperes law,1. Amperes loop law安培环路定理,Any line integral of around a closed path was proportional to the current encircled by the
17、 path.,在恒定磁场中，磁感强度沿任一闭合环路的线积分等于穿过该环路所围曲面的所有电流代数和的 倍,Contributed by all current inspace.,Freely choose a closed loop and assign a positive direction.,Explanation:,The net current encircled by the loop (与L套连的电流,回路所围面积截得)accord with right-hand rule (与L绕行方向成右螺电流取正).,Any differential elements on the L, a
18、long the tangent direction to the loop.,In the above case,1. How to Prove the Law?,1) A long, straight wire (I), and the integration path is a circle centered on the conductor.,2) the integration path is any closed line in the plane perpendicular to the conductor.,3) An integration path that does no
19、t enclose the conductor,Discussion:,“B” is the magnetic field on the line;（回路上的磁感应强度由回路内外所有电流共同提供） “Iencl”, the sum of the currents enclosed or linked by the path.（对环流有贡献的只有穿过回路所围面积的电流） Positive direction of current I is decided by right-hand rule.（电流正方向与环流积分方向成右手螺旋定则）,Discussion:,The circular flowi
20、ng (环流量) does not equal zero, this means the magnetic field is not a conservative field保守场, we can not define potential in the field.,Amperes law is valid for any stable magnetic field, but we can use it to calculate B only if there is a highly symmetric current distribution.,One can also use Ampere
21、s law to calculate with symmetrical distribution of current .,2. The application of Amperes law安培环路定理应用,Main steps:,(1) Symmetry analyses (由I的分布，分析分布的对称性).,(2) Choose a loop (使回路上各处B相等，方向特殊，从而可从回路积分中提出B).,(3) Put into the law and calculate B.,Find the inside and outside a long straight wire of radiu
22、s R and uniformly distributed current I over a cross section of the wire 长直圆柱形载流圆柱体(电流I 均匀分布在圆柱的横截面内)内外的磁场.,Solution:,Cylindrical symmetry,Choose a circular loop L1 with r R,Example:,Choose a circular loop L2 with r R,R,Can you draw the curve of Distribution of in whole space,?,In the similar way on
23、e may find (长直圆柱面载流导线内外的磁场) as figure:,Compare with produced by a current i in a long straight wire,= 0,Solution:,Plane symmetry,It is a uniform field, independent of position r,在无限大均匀平面电流两侧的磁场均为均匀磁场，并且大小相等,方向相反。,Example:,Magnetic field of a toroid (载流螺绕环内的磁场), given i, N, R1 , R2 .,Solution:,Exampl
24、e:,From symmetry, the lines of form concentric circles inside the toroid, directed as figure,Choose concentric (同心) circle loop L (R1 r R2 ),(Right-hand rule direction),Discussion:,(1) In contrast to the situation for a solenoid, B is not constant over cross section of a toroid. But if its cross are
25、a is relatively small, then,It is as same as the expression of the one of long solenoid.,(2) so the magnetic field of toroid is limited in its interior.,Find inside a “infinitely long” solenoid (载流长直螺线管内部的磁场)(The number of turns per unit length of the solenoid n and total number of coils N).,Example
26、:,Solution:,B=0,Based on the feature of the distribution of , we choose a rectangular loop abcd,Example : A very large flat conducting sheet carries a uniform current density j throughout. Find the magnetic field (magnitude and direction) at a distance y above the plane. (Assume the plane is infinit
27、ely long and wide),Solution:,7-7. The motions of charged particles in electric and Magnetic fields,1. Lorentzs force洛仑兹力,F=qvBsin (when = 0,F=0),The charged particles moving in magnetic field acted by magnetic force:,2. The motion of charged particle in uniform :, particle straightly move in constan
28、t velocity.,Radius of track:,(2) Particle moves perpendicular to,(1) Particle moves parallel to,F=qvB, Circular motion with constant speed in the plane of perpendicular to,Period:,Radius:,Period:,Pitch(螺距):,Particle follows a spiral path (螺旋线运动).,(3)Magnetic focusing磁聚焦 Particle moves in an angle wi
29、th :,电镜用电子束代替了可见光，用电磁透镜代替了光学透镜并使用荧光屏将肉眼不可见电子束成像。 1931年卢斯卡研制了第一台透视电子显微镜。 1986年卢斯卡为此获得Nobel物理学奖。,电子显微镜 electron microscope,From it one can determine the motion of particles.,Lorentz equation one of the basic equation in physics.,If a particle of charge q moves with velocity in the presence of both mag
30、netic field and the electric field , it will feel a force,(1) Feature of electron,Both and can produce a force on a charged particle:,3. Samples of practical application实际应用,It can be used to measure the ratio m/q of the particles moving through Thomsons apparatus.,电子比荷,(2) The Hall effect (霍尔效应),载流
31、导体放在磁场中，磁场垂直于电流方向，则在磁场与电流垂直方向出现横向电势差。,Hall Hall voltage.,What relation between U & I ?,Hall effect can be used to find B, I, positive or negative and drift speed of charged carriers,半导体类型判别,Magnitude:,Direction: right-hand rule,Directs the current,If I is not perpendicular to,If the magnetic field i
32、s not uniform, or if the wire does not everywhere make the same angle with , then,7-8. The magnetic force acting on a wire with current,1 The Ampere force,Lorentz force on a moving particle:,I,Current element in the wire includes motion point charges of electric quantity q and velocity .,Current Ele
33、ment,？,Example:,A straight, horizontal length of copper wire has a current I=28A through it. What are magnitude and direction of the minimum magnetic field needed to suspend 悬浮the wire (to balance ) on it? The linear density of wire is 46.6g/m.,Solution:,This is about 160 times the strength of Earth
34、s magnetic field.,A half-circular wire of radius R with current I is put in an uniform magnetic field as the figure shown, =30, find the magnetic force acted on the arc (此段圆弧电流受的磁力).,Solution:,Example:,No magnetic force exerted on a current loop in a uniform magnetic field.,(1) The sum of magnetic f
35、orce acted on a whole curved wire = the one of straight wire with same current connected from start to end points (均匀磁场中任意弯曲导线所受的磁场力=相同电流的直导线所受的磁场力).,(2) If points a and b overlapping, then l=0, F=0.,(3) Magnetic force is always perpendicular to the plane defined by length vector and magnetic field.
36、,(均匀磁场中任意闭合导线所受的磁场力=0),F=,F=,F=,EXERCISES,The current in the long, straight wire CD is .The wire AB carries and its length is L. The wire AB lies in the plane perpendicular to CD. Their distance is d.,Find the net force exerted on the AB by the magnetic field created by the wire CD.,Example:,Solutio
37、n,An other exercise: inclined current line,Two long parallel wires carrying currents exert forces (平行载流导线)on each other as the figure,2. Force between two parallel wires,Think about how about the force between two antiparallel wire 平行反向载流导线之间的力怎样(方向、大小)?, repel each other!,The definition of the Ampe
38、re (电流单位“安培”),两无限长平行载流导线相距1m,通有相同电流,单位长互作用力为210 -7 N时,导线内电流为1A.,Proof: when Ia =Ib=1A,Hence,One Ampere is defined as that current flowing in each of two long parallel conductors 1m apart, which results in a force of exactly of length of each conductor.,Proving :the torque on a plane current-carrying
39、 coil of arbitrary given shape in uniform magnetic field is,where ,N is number of turns of the coil,is the area vector of the coil.,3. Torque on a Current Loop作用在线圈上的磁力矩,l1,l2,a,b,c,d,I,Edge View边缘视图 of the Loop,A rectangular current loop in a uniform magnetic field,In uniform magnetic field:,The tw
40、o forces create a torque that trends to rotate the loop clockwise.,M=ISB,M = 0,Discussion,Although the torque expression was derived for a rectangular loop, the result is valid for a loop of any shape.,The equilibrium position of the coil .,the steady equilibrium position.,the unsteady equilibrium p
41、osition.,Under the action of the torque, the magnetic moment 磁矩 always tries to turn to the direction of external magnetic field.,Example,Z,o,X,Y,I,A coil of half circle is set in the uniform magnetic field parallel to the coil plane in the beginning instant, the radius of coil R=0.5m, I=10A, B=0.5T
42、.,Find (1) the torque of magnetic force on the coil in the beginning instant. (2) The work done by the torque of magnetic force when the coil rotated by angle /2 from the initial position.,Example: Show that the magnetic dipole moment p of an electron orbiting a hydrogen atom is related to the orbit
43、al momentum L of the electron by p =eL/(2m).,Solution:,Another way to measure magnetic field:,The one of magnetic dipole moment when a coil is in stable-equilibrium.,Direction of :,Magnitude of :,1. Classification of magnetic materials,When the interior of a solenoid is filled with some material of
44、relative permeability r (相对磁导率),Ferromagnetism (铁磁质),Paramagnetism (顺磁质),Diamagnetism (抗磁质),7-9 The Magnetic Media in a Magnetic Field,2. Magnetization(磁化)of magnetic materials,Cross area of solenoid,When the interior of a solenoid is fully filled with some isotropic material r , it will be magnetiz
45、ed (长直螺线管内部充满均匀的各向同性介质，将被均匀磁化).,Taking the paramagnetic material as an example,each molecular in it is equivalent to a loop current:,To express the extent to which a given medium is magnetized, a vector quantity, magnetization 磁化强度, is introduced:,The unit of is A/m.,It can be proved,The net bound c
46、urrent enclosed in a closed loop L equals to the circulation环流 of magnetization along the closed loop.,and,Take notes!,3. Circulation theorem with magnetic materials,Let,then,The circulation of along any closed loop equals to the algebraic sum of free current encircled by the closed loop L Theorem o
47、f circulation .,(1) If the magnetic material were not present, then the above equation reduce to the Ampere-loop law in vacuum.,Explanations:,(2) For isotropic medium,where is the permeability of magnetic materials.,(3) One may compare with , which is another supplementary physical quantity.,(4) The
48、 unit of in SI system is A/m.,Example:,Solution:,A toroid is fully filled with magnetic medium, =510-4 Tm/A, n=103/m, I=2A. Find: interior the medium.,Is the medium paramagnetism or diamag or ferromag?,So it is ferromagnetism.,As same as the one of long solenoid.,Ferromagnetism:,Iron (and a few other materials), , can be made into strong magnets ferromagnetism.,Magnetic domains (磁畴):,A ferromagnetic material will, in its normal state, be made up of a number of magnetic domains which are tiny regions (about 10-4m) .,Under ,the domains are gradually aligned in the