P2-20 Given a vector function F=axy+a,x-y) P1(5,6) evaluate the integral Fdi from R(5,6)to P2(3,3)inFg235 a)along the direct path p P2(3,3 (2)}A b)along path PAP 5 Solution Fd=[ax+a(3x-y2)(a+a小)=对+(3x-y2) he equation of P, P2 is y=(x-1) PD阶Fd=+(3xy1]2(x1)+(2y+3-yb=10 P Pat②2 2=1(15-y xdx=18-24=-6
x y A P1(5,6) P2 (3,3) 0 1 5 (1) (2) (2) 2 (3 ), F a xy a x y = + − x y F dl 1 P(5,6) 2 P (3,3) PP1 2 P AP 1 2 P.2-20 Given a vector function evaluate the integral from to in Fig.2-35 a)along the direct path b) along path 2 2 (3 ) ( ) (3 ) F dl a xy a x y a dx a dy xydx x y dy x y x y = + − + = + − Solution The equation of P1P2 is 3 ( 1) 2 y x = − 2 2 1 1 2 3 2 2 5 6 3 (3 ) ( 1) (2 3 ) 10 2 P P P P Path F dl xydx x y dy x x dx y y dy = + − = − + + − = − ① 2 1 3 3 2 6 5 (15 ) 3 18 24 6. P P Path F dl y dy xdx = − + = − = − ②
Example: Assuming that the electric field intensity is E=a100x(/m) Find the total electric charge contained inside a cubical volume 100 mm on a side centered symmetrically at the origin. E·dS ds E·dS=EdS dS+E·dS+|E E ∫2a1004+a.00x-a45 +m1004s+100-)4S +a100x:(-a,S+a2100 =0+0+[a100x:adS+[a100x:(-a,dS+0+0 100xds-100xds=5 dS+5 ds=0.1 C
Example: Assuming that the electric field intensityis E a ˆ 100x(V / m) = x Find the total electric charge contained inside a cubical volume 100 mm on a side centered symmetricallyat the origin. x y z o ds 0 Q E dS S = xdS xdS dS dS C a x a dS a x a dS a x ( a )dS a x a dS a x a dS a x ( a )dS a x a dS a x（ a dS） E dS E dS E dS E dS E dS E dS E dS 前 前 前 x x x x 左 x y 右 x y 前 x x 后 x x 上 x z 下 x z S 100 100 5 5 0.1 0 0 ˆ 100 ˆ ˆ 100 ( ˆ ) 0 0 ˆ 100 ˆ ˆ 100 ˆ ˆ 100 ˆ ˆ 100 ˆ ˆ 100 ˆ ˆ 100 ˆ = − = + = = + + + − + + + − + + + − = + − + + = + + + 后 后 后 左 右 上 下 前 后
E xample A very long, hollowH23 conductor of inner radius a and outer radius b located along the z axis and carries a current l in the z direction, as depicted in figure. if the current distribution is uniform#je, determine the magnetic flux density磁通密度 at any point in space 71
Example A very long, hollow中空 conductor of inner radius a and outer radius b is located along the z axis and carries a current I in the z direction, as depicted in figure. if the current distribution is uniform均匀, determine the magnetic flux density磁通密度at any point in space. b a y x
解:(1)对称性分析:具有圆柱对称性的问题彡 (2)建立圆柱坐标系:导体沿z轴放置,则磁通密度B将是沿φ向的 B只有q分量),且沿围绕-的任何圆形路径上的B为常量 (3)建立安培环路:选择垂直于导线平面上,以导线为中心半径为的 园为积分回路 (4)安培环路定理求解 y 对整个求解区坷分区域讨论 Region1:rK≤a无源区域 Region2:a≤rsb有源区域 Region3:b≤r无源区域 As the current is uniformly distributed we can express it in terms of the volume current density as a≤r≤b)
解：(1) 对称性分析：具有圆柱对称性的问题 (2) 建立圆柱坐标系：导体沿z轴放置，则磁通密度B 将是沿φ方向 的 （B 只有φ分量），且沿围绕z轴的任何圆形路径上的B 为常量． (3) 建立安培环路：选择垂直于导线平面上，以导线为中心半径为r的 圆为积分回路 (4) 安培环路定理求解 对整个求解区域分区域讨论 Region 1 ：ra 无源区域 Region2： a r b 有源区域 Region 3： b r 无源区域 As the current is uniformly distributed ,we can express it in terms of the volume current density as ( ) 2 2 ˆ ( ) z I J a a r b b a = − b a y x
Following the symmetry arguments, we expect that the field lines must be concentric circles, the magnetic flux density must be in the o direction and B, has a constant magnitude along each circle. there are three regions of interested b=aB dl=a dr +ando+a.dz B·dl=Brdp X 2丌 B·dl B,rdo= b, 2r
Following the symmetry arguments, we expect that the field lines must be concentric circles, the magnetic flux density must be in the direction ,and B has a constant magnitude along each circle. There are three regions of interested. 2 0 ˆ ˆ ˆ ˆ 2 r z C B a B dl a dr a rd a dz B dl B rd B dl B rd B r = = + + = = = b a y x