作业13-25 如如0l,P2 +d-rsin od8 n d+rsin e 因此穿过圆形回路的磁通量为: ①=如b=为,n,R-4 +d-Rsin g]de n d+Sine 2 d+r d-R [-2√d2-R2(arcg d-R +actg d+R+dz ① =pld-yd2-R2]→M d-vd-Rfl 注:rcg4 actg a2“/+m)d-R_z Vd-R Vd+r 2作业13—25      d R d d R I R d d sin ] sin [ 2 2 0 + − + −  = 因此穿过圆形回路的磁通量为：   − + − + −  =  = 2 2 2 2 0 sin ] sin [        d R d d R I R d d [ 2 ( ) ] 0 2 2    d d R d R arctg d R d R d R arctg I + + − + − + = − − [ ] 2 2 = 0 I d − d − R [ ] 2 2 0 d d R I M = − −   =  2 2 1   = + − + − + + =  d R d R arctg d R d R arctg A 注：arctgA arctg