2)、严格证明 B k xr=P0-kx 2Ir 2Tr 2Tr k×(x+y)=k1 (xj-yi 2Tr 2Tr Bdl 2m2(-y).d(x+y)= (xdy-ydx) 2π(x2+y Ly=rsing dx=cosedr-rsinede x=rcog ldy=sinOr+ cos0 rde xdy=rose(sinedr+coserde) ydx=sine(cosedr-rsinedo) xdy-ydx=rde=(x+y)de →∮B.d= 2π(x2+y b-y)=20 X,y de 图8.12、任意路径的安培环路定律22)、严格证明 ( ) 2 ( ) 2 2 2 2 2 0 2 0 0 0 0 xj yi r I k xi yj r I r r k r I k r r I r I B            = ´ + = - = = ´ = ´ p m p m p m p m a p m d I I xdy ydx x y I B dl xdy ydx r d x y d ydx r dr r d xdy r dr rd dy dr rd dx dr r d y r x r xdy ydx x y I xj yi d xi yj r I B dl 0 2 0 0 2 2 0 2 2 2 2 2 0 2 0 2 ( ) 2 ( ) ( ) sin (cos sin ) cos (sin cos ) sin cos cos sin sin cos ( ) 2 ( ) ( ) ( ) 2 q m p m p m q q q q q q q q q q q q q q q q q J p m p m p - = = + Þ = Þ - = = + î í ì = - = + î í ì = + = - Þ î í ì = = - + = - + = ò ò ò ò ò ò · · ·         • • x q dq dl y 图 8.12、任意路径的安培环路定律 2 P(x,y)