《电磁场与电磁波》课程教学资源（习题）第一章习题

第一章习题 11()e4=(e2+e,-e3)1+4+92=(e+ey-23)/14 (2)A-B=c+6-44-B=(1+36+16)2=5 (3)A.·B=-8-3=-11 11 (4)0R=COS 14√17 =1800-44.5=1355 (5) ACOSO=V14(-0.713)=-267 (6)AxC=-24-13-210 (7)A·(B×C)=(4×B)oC=-42

第一章习题 1 2 1 2 1 0 1.1 (1) ( 3) /(1 4 9) ( 3) / 14 (2) 6 4 (1 36 16) 53 (3) . 8 3 11 11 (4) 180 44. 14 17 A x y z x y z x y z AB e e e e e e e A B e e e A B A B  COS − = + − + + = + − − = + − − = + + = • = − − = −   − = = −      0 0 5 135.5 (5) 14 ( 0.713) 2.67 (6) 4 13 10 (7) ( ) ( ) 42 AB x y z ACOS A C e e e A B C A B C  = =  − = −  = − − − •  =  • = −

(8)(A×B)xC=B(AoC)-A(B·C)=e12-e40 Ax(B×C)=B(A·C)-C(AB)=e15-e44-2e11 1.2A12=e14+e10-eA3=e2+e1+8 A3=-216-e-e7A12·A3=0 S==A 12 A23=1712 1.3A 5-e3 PP 144=√29B=√77AB=-31COS0= A●B 0.65 AB 6=180-49=1314CO9=√29(0656)=-3533

12 23 (8) ( ) ( ) ( ) 2 40 ( ) ( ) ( ) 55 44 11... 1.2 4 0 2 8 x y x y z x y z x y z A B C B A C A B C e e A B C B A C C A B e e e A e e e A e e e A   = • − • = −   = • − • = − − = + − = + + ' 13 12 23 12 23 6 7 0 1 17.12 2 1.3 5 3 . 1.4 29 77 31 0.656 180 49 131 x y z P P x y z A e e e A A S A A A A e e e e A A B A B A B COS AB   = − − − • = =  = = − − = • = = • = − = = − = − = 0 . 29 ( 0.656) 3.533 ACOS =  − = −

1.12dA·ds=.A·ds+dA·ds+dA·ds 2 =Ae, 5dpz+Aoe rdp.dz-Aoerdpdr 2丌 125do dz+ rdr 8d0+0 25 2丌·125.4+.8·2兀=1200兀 VA=3r+2 V Adr=l(3r+2)rdrdodz=1200T 115φAde=φ(Adx+A,dy+Adz) xdx+ 4dy+ xdx+ Ody=8 0 0 0

1 2 3 1 1 3 2 4 5 2 0 0 0 0 1.12 . . . . . 5 . . . 125 8 0 25 2 125 4 8 2 1200 2 3 s s s s r z z s s s A d s A d s A d s A d s A e d dz A e rd dz A e rd dz d dz rdr d A r           • = • + • + • = • + • − • = + + =   +   =  • =            2 2 0 2 0 0 2 0 2 (3 2) 1200 1.15 ( ). 4 . 0 8. x y z Ad r rdrd dz A de A dx A dy A dz xdx dy xdx dy  +  • = + =    • = + + = + + + =        

V 2 vE +e2x yz ∫V×A·ds=jvxA·eb=]2b=8 123V×V=0VV×F=0 而V×A=0V·A=0,所以A可以用另 矢量旋度表示,也可以用另一标量梯度表示. V×B=0V·B=2rSim0, B以用另一标量梯度表示 V×C=e2(2x-6y)VC=0, C可以用另一矢量旋度表示

.2 2 2 . . 8 1.23 0 0; 0 0, 0 2 , x z z s s A e yz e x A d s A e dxdy x dx dy u F A A A B B rSin B    = +   • =   • = =   =  •  =   =  • =   =  • =    而 所以 可以用另一 矢量旋度表示,也可以用另一标量梯度表示. 可以用另一标量梯度表示 (2 6 ) 0, C e x y C z C   = −  • = 可以用另一矢量旋度表示