如何描述线1周围的用来决定对线2作用力的力场？Note that in the third case (perpendicular currents), I2 is in the same direction as H, so that their cross product (and the resulting force) is zero. The actual force computation involves a different field quantity, B, which is related to H through B = μ0H in free space. This will be taken up in a later lecture. Our immediate concern is how to find H from any given current distribution.第三种情况，磁场与电流平行，叉乘=0特别注意与距离的平方成反比, 而且叉矢量指向纸内（右手螺旋法则决定）Note the similarity to Coulomb’s Lawa point charge of magnitude dQ1at Point 1 would generate electric field at Point 2 given by:The units of H are [A/m]To determine the total field arising from the closed circuit path,we sum the contributions from the current elements that make upthe entire loop, orThe contribution to the field at P from any portion of the current will be just the above integral evalated over just that portion...and so the differential current quantity thatappears in the Biot-Savart law becomes:The magnetic field arising from a currentsheet is thus found from the two-dimensionalform of the Biot-Savart law:of three-dimensional current elements, and so the Biot-Savartand so..so that:Integrate this over the entire wire:..after carrying out the cross productExample: concludedfinally:Current is into the page.Magnetic field streamlinesare concentric circles, whose magnitudesdecrease as the inverse distance from the 线是如何画？（力线疏密反应强如何画？定值/ρ 2，所以，磁力线的间隔怎么画？..after a few additional steps (see Problem 7.8), we find:carry out the cross products to find:but we must include the angle dependence in the radialunit vector注意:rho的单位矢量不是常量！！！with this substitution, the radial component will integrate to zero, meaning that all radial components will cancel on the z axis. rho分量消了，仅仅z分量Note the form of the numerator: the product ofthe current and the loop area. We define this asthe magnetic moment:环可定义磁矩似偶极矩的定义横向涡旋磁场路径a和b的磁场环流= 总电流I路径c的磁场环流= 总电流I的一部分ρso that:as before.solid conductors that carry equal and oppositecurrents, I.The line is assumed to be infinitely long, and thecircular symmetry suggests thatφ-directed, and will vary only with radiusOur objective is to find the magnetic fieldfor all values of ρ无限长内外导体等量异向电流仅沿rho变化的phi向磁场求全径向磁场分布导体间磁场内导体可看作无限长线电流束考虑1、2处的电流丝产生的磁场叠加仅phi向存在ab间场直接为But now, the current enclosed isor finally:体外的场As the current is uniformly distributed, and since wehave circular symmetry, the field would have tobe constant over the circular integration path, and so itmust be true that:+外导体圈内反向电流百分比..and so finally:同轴线磁场强度的径向分布s circuital law to the path we find:In other words, the magnetic field is discontinuous across the current sheet by the magnitude of the surface current density.生的磁场（续）电场的均匀分布If instead, the upper path is elevated to the line between and , the same current is enclosed and we would havefrom which we conclude thatconstant in each region(above and below the current plane)a N is the unit vector that is normal to thecurrent sheet, and that points into the region inwhich the magnetic field is to be evaluated.引申为等效电流aN X H = aN X (0.5KXaN)=0.5KKHHKWe will now use this result as a building blockto construct the magnetic field on the axis ofa solenoid --formed by a stack of identical currentloops, centered on the z axis.contribution to the total field from a stack of N closely-spacedloops, each of which carries current I . The length of the stack so therefore the density of turns will be N/d.Now the current in the turns within a differential length, dz , will bez -d/2d/2so that the previous result for H from a single loop:now becomes:in which z is measured from the center of the coil,where we wish to evaluate the field.We consider this as our differential “loop current ”z-d/2d/2长螺线管的场近似z-d/2d/2We now have the on-axis field at the solenoid midpoint (z = 0):Note that for long solenoids, for which , theresult simplifies to:( )This result is valid at all on-axis positions deep within long coils --at distances from each end of several radii.d/2Therefore:In other words, the on-axis field magnitude near the center of a cylindricalcurrent sheet, where current circulates around the z axis, and whose lengthis much greater than its radius, is just the surface current density.Solenoid Field --Off-Axis螺线管非轴线上的场用安培环路定律来求解:The illustration below shows the solenoid cross-section, from a lengthwise cut through thethe windings flows in and out of the screen in the circular current path. Each turn carries currentfield along the z axis is NI/d as we found earlier.Where allowance is made for the existence of a radial H component,The radial integrals will now cancel, because they are oppositely-directed, and because in the long coil,Conclusion: The magnetic field within a long solenoid is approximately constant throughout the coil cross-section, and is H z = NI/d.绕组被建N个流回路，每个承载电流I.镯形环形绕组，横截面可是圆形或其他任意形状individual windings. Under this condition, we would assume:This approximation improves as the density of turns gets higher(using more turns with finer wire).Ampere’s Law now takes the form:so that….Performing the same integrals over contours drawnin the regions or willlead to zero magnetic field there, because no currentis enclosed in either case. 环外场= 0Consider a sheet current molded into a doughnut shape, as shown.The current density at radius crosses the plane in the zdirection and is given in magnitude byleading to…inside the toroid…. and the field is zero outside as before.where:基于Taylor 级数展开的近似And therefore:向路径段的贡献The contribution from the opposite side is:Note the path directions as specified in the figure, andhow these determine the signs used .This leaves the left and right sides…..…and the contribution from the left side (path 4-1) is:The next step is to add the contributions of all four sides to find the closed path integral:合路径积分and using our previous results, the becomes:Dividing by the loop area, we now have:The expression becomes exact as the loop areaapproaches zero:其他方向的结果类似：The same exercise can be carried with the rectangular loop in the other two orthogonal orientations.下的结果.方向的确定遵循右手螺旋法则：四指指向积分路径方向，大拇指指向旋度方向The curl component in the direction N, normal to the plane of the integration loop is:is taken using the right-hand convention: With fingers of the right hand orientedin the direction of the path integral, the thumb points in the direction of the normal (or curl).An easy way to calculate this is to evaluate the following determinant:which we see is equivalent to the cross product of the del operator with the field:Look these up as needed….…..静磁场的麦克斯韦方程（思考总磁场的形式）:同时也是安培环路定律的点形式（微分形式）.总电场的形式）:(applies to a static electric field) Recall the condition for a conservative field: that is, its closed path integral is zero everywhere.Therefore, a field is conservative if it has zero curlis paritioned into sub-regions, each of small areaThe curl component that is normal to a surface element canWe now apply this to every partition on the surface, and add the results…..cancellation here:Theorem.We now take our previous result, and take the limit asIn the limit, this side becomes the path integral of H over the outer perimeter because all interior paths cancel 等效于外围闭合路径积分In the limit, this sidebecomes the integralof the curl of H over surface S等效于在整个S 上积分两个定律均可直接从纯高等数学的角度直接展开证明）This is a valuable tool to have at our disposal, because it gives us two ways to evaluate the same thing!积分联系起来了Integrate both sides over surface..in which the far right hand side is found from the left hand sideusing Stokes’Theorem. The closed path integral is taken around theperimeter of S. Again, note that we use the right-hand convention inchoosing the direction of the path integral.so we are left with the integral form of Ampere应强度）in which the electric flux density in free space is:and where the free space permittivity ismagnetic flux in units of Webers (Wb):磁通密度（磁感应强度）in free space is:and where the free space permeability is有关。