北京交通大学电磁波教案(3)

6.解：法(一)：对A?B= A?C两边取A的叉乘： A?(A?B)= A ?(A?C) (A·B)A-(A·A)B= (A·C) A-( A·A)C 将A·B= A·C代入即得B=C 法(二):：A·B= A·C ? A·(B-C)=0 ? A? (B-C) A?B= A?C ? A?(B-C)=0 ? A?? (B-C) ?只有B-C=0即B=C 法(三)： A·B= A·C ? A·(B-C)=0 ?A? B-C? cos?=0 A?B= A?C ? A?(B-C)=0 ? A? B-C? sin?=0 两式平方相加: A2? B-C?2=0： ? B=C 法(四)：A·B= A·C ? ABcos?1= ACcos?2 A?B= A?C ? ABsin?1 = ACsin?2 两式相比，得：tan?1=tan?2 ∵?1、?2∈[0,π] ∴?1=?2 ∴B= C 又由A?B= A?C可知B、 C在A的同侧 ∴B= C 返回 7.解：P=A?X ? A?P= A?(A?X) = (A·X)A-( A·A)X 代入p= A·X得 A?P= p A—A2X ? X= (p A- A?P)?A2 返回 8.解：(1)在直角坐标中： x=rcos?=2 y=rsin?=2 z=3 ?(-2,2 ,3) (2)在球坐标中： rs=(x2+y2+z2)1?2=5 ?=tg-1(r?z) ?53.1? ?=tg-1(y?x)=120? ?(5,53.1?,120?) 返回 9.解：(1) r2=x2+y2+z2=(-3)2+42+(-5)2=50 E=25?r2=0.5 Ex=E x?r?-0.2121 (2) E=Ear=(axx+ayy+azz)?r=(-ax3+ay4-az5)? cos?=E·B?EB?-0.8955 ??98.2? 返回 10.证：球坐标与直角坐标的关系为 x=rsin?cos? y=rsin?sin? z=rcos? 因此，在直角坐标中， R1=ax r1sin?1cos?1+ay r1sin?1sin?1 +az r1cos?1 R2=ax r2sin?2cos?2+ay r2sin?2sin?2 +az r2cos?2 cos?= R·R2?R1R2 =( r1sin?1cos?1 r2sin?2cos?2+ r1sin?1sin?1 r2sin?2sin?2+ r1cos?1 r2cos?2)?r1r2 = sin?1sin?2(cos?1cos?2+ sin?1sin?2)+ cos?1cos?2 = sin?1sin? cos(?1-?2)+ cos?1cos?2 返回 11.解：∮s (ar3 sin?)·dS=∮s 3sin?dSr = 3 sin?r2 sin?d?d?=75?2 返回 12.解：?·A=1?r ?(rAr)??r+1?r ?A????+?Az??z=3r+2 ???·Ad?=(3r+2)rdrdzd?=1200? ∮s A·dS=∮s r2dSr+2zdSz= r2 rdzd?+ 2zrdrd?=1200? ????·Ad?=∮s A·dS 从而验证了散度定理。 返回 13.解：(1) ?·A=?(Ax)??x+ ?Ay??y+?Az??z=2x+2x2y+72x2y2z2 (2) ???·Ad?= (2x+2x2y+72x2y2z2)dxdydz=1?24 (3)∮s A·dS=?(Ax?x=0.5-Ax?x=-0.5)dydz+(Ay?y=0.5-Ay?y=-0.5)dxdy+(Az?z=0.5-Az?z=-0.5)dxdy =1?24 ? ???·Ad?=∮s A·dS 返回 14.解：∮s r·dS= r r2 sin?d?d?=4?a3 ???·rd?=??1?r2 ?(r2 r)??r r2 sin?drd?d?=4?a3 返回 15.解：∮c A·dl=∮c Ax dx+ Aydy+ Azdz = Ax?y=0dx+ Ay?x=2dy+ Ax?y=2dx+ Ay?x=0dy=8 ??A=ax2yz+az2x ?s??A·dS=( ax2yz+az2x)·azdxdy=8 ?∮c A·dl= ?s??A·dS 从而验证了斯托克斯定理。 返回 16.解：x=rcos? y=rsin? dl=ad?a? a?=-axsin?+aycos? ∮c A·dl=(axx2＋ay xy2)·a(-axsin?+aycos?)d? = (-a2cos2?sin?+a3cos?sin2?cos?)ad?=?a4?4 ??A=azy2=azr2sin2? ?s??A·dS=azr2sin2?·azrdrd?=?a4?4 ?∮c A·dl= ?s??A·dS 返回 17.证：(1)? ·R=?Rx??x+?Ry??y+?Rz??z=3 (2)??R=0 (3) ?(A·R)= ?(xox+yoy+z0z)=axx0+ayy0+azz0=A 返回 18.解：?·F=1?r2 ?(r2Fr)??r=0 ??r2 f (r)???r=0 r2 f (r)=c f (r)=c? r2 返回 19.解：(1) ?c1E·dl=?c1( axy＋ayx) ·( axdx＋aydy)= ?c1ydx＋xdy = (y4ydy+2y2dy)=14 (2)直线方程为x-6y+4=0 并有dx=6dy ?c2E·dl=?c2( axy＋ayx) ·( axdx＋aydy)= ?c2ydx＋xdy = ?y 6dy+(6y-4)dy?=14 ?·E=?y??x+ay?x?y=0 ?这个E是保守场。 返回 20.解：??=ax????x+ay????y+az????z= ax2xyz+ayx2z+azx2y 方向导数d??dl=??·al= (6xyz+4x2z+5x2y) ? d??dl?(2,3,1)=112? 返回 21.解：球坐标中的?算符为 ?=ar???r+a??r ????+a??rsin? ???? ?ar ??r=0 ?ar???=a? ?ar???=a?sin? ?a???r=0 ?a????= -a? ?a????=a?cos? ?a? ??r=0 ?a????=0 ?a????= -arsin? -a?cos? ?·A=( ar???r+a??r ????+a??rsin? ????)·( arAr+ ar???r+a?A?+a?A?) = ar·( ar?Ar??r +A?? a???r +a??A???r +A??a? ??r+a??A???r) + a??r·( Ar?ar???+ ar?Ar???+ A??a????+a??A????+A??a????) + a??rsin? ·(Ar?ar???+ ar?Ar ???+A??a????+a??A????+A??a????+a??A????) =(?Ar??r+2?r Ar)+(1?r ?A????+ cos??rsin? A?)+1?rsin? ?A???? =1?r2 ?(r2Ar)??r+1?rsin? ?(sin?A?)???+1?rsin? ?A???? 返回 22.解：?u=ax2x? a2+ay2y? b2+az2z? c2 -? n=?u???u ?= (ax2x? a2+ay2y? b2+az2z? c2)(x2? a4+y2? b4+z2? c4) 返回 23.解：(1)??A=0 ?·A=0 ??B=0 ?·B=2rsin?