无限长载流直导线 讨 a1=0&2=π 论 2π 半无限长载流直导线 a1=π/2&，=π B= 4π0 •直导线延长线上 a=0 C1=C2=0 0 B= cosa,)= B= 4 0 (cosa-cosa2 4π0•无限长载流直导线 0 1  =  =  2 a I B   2 0 = •半无限长载流直导线 2 1  =   =  2 a I B   4 0 = •直导线延长线上 (cos cos ) 4 1 2 0     = − a I B I B  讨 论 0 0 (cos cos ) 4 1 2 0 =  −  =   a I B a = 0 1 =2 = 0 ?