B due to current in circular loop dB=L。Idl sin An dlF,.0=90 B=8-6a-台8白0 r2 Rdl 4R2 4元 R2+x22 2R2+x2 3/2 B due to current in circular loop x P x R dBx dBy  r dB  I 2 sin 4 r Idl dB o     2 sin 90 cos 4 cos r Idl B B dB o x           dlr ,   90        3 2 2 2 2 0 3 2 2 2 2 0 4 2 R x IR R x Rdl B R o         