Chap7-MagFie(zq).ppt
《Chap7-MagFie(zq).ppt》由会员分享,可在线阅读,更多相关《Chap7-MagFie(zq).ppt(76页珍藏版)》请在三一文库上搜索。
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
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- Chap7 MagFie zq
链接地址:https://www.31doc.com/p-3478745.html