题目
12 )载有电流I的导线如图放置,在圆心O处的磁-|||-感应强度B为:4172几-|||-(A) .dfrac ({mu )_(0)I}(4R)+dfrac ({mu )_(0)I}(4pi R) . (B) .dfrac ({mu )_(0)I}(2pi R)+dfrac (3{mu )_(0)I}(8R)-|||-(c) dfrac ({mu )_(0)I}(4pi R)-dfrac (3{mu )_(0)I}(8R) (D) dfrac ({mu )_(0)I}(4R)+dfrac ({mu )_(0)I}(2pi R)-|||-<100-|||-21-|||-2-|||-R-|||-(005)-|||-X K-|||-21.18= 小
题目解答
答案
:本题考查安培环路定理.由安培环路定理,圆心O处的磁感应强度为
B=B1+B2=B1+B3+B4=B1+B3+B5+B6=B1+B3+B5+B7+B8
由于圆弧形导线在O点的磁感应强度为零,即B1=0,所以
B=B3+B5+B7+B8
又因为B3=B5=B7=B8=,所以
B=B3+B5+B7+B8=4B3=4×=
答案 B
B=B1+B2=B1+B3+B4=B1+B3+B5+B6=B1+B3+B5+B7+B8
由于圆弧形导线在O点的磁感应强度为零,即B1=0,所以
B=B3+B5+B7+B8
又因为B3=B5=B7=B8=,所以
B=B3+B5+B7+B8=4B3=4×=
答案 B