B=Holsin 0 2πR 4元2 Idl P R 因r2=R2+x2,sin0= R (R2+x2)% 所以B= HoIR2 Ho IS 2(R2+x2)% 2元 (R2+x2)为 S=元R2 上觉子觉道司退欢 上页 下页 返回 退出 R r I B 2π 4π sin 2 0  = 2 1 ( ) ; sin 2 2 2 2 2 R x R r R r R x + = +  = = 2 3 2 3 2( ) 2π ( ) 2 2 0 2 2 2 0 R x IS R x IR B + = + =   2 S = πR I O R x P  I dl r B  ⊥ d B  d  d B//  因 所以