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如图所示，通电导体棒静止于水平导轨上，棒的质量为m，长为L，通过的电流大小为I且垂直纸面向里，匀强磁场的磁感强度B的方向与导轨平面成θ角，则导体棒受到的（　　）

A．安培力大小为BIL

B．安培力大小为BILsinθ

C．摩擦力大小为BILsinθ

D．支持力大小为mg-BILcosθ

答案

以导体棒AC为研究对象，分析受力，如图．

其中，安培力大小FA=BIl，根据平衡条件得：

G+FAcosθ=N…①

f=FAsinθ…②

由①得：N=G+FAcosθ=mg+BIlcosθ

由②得：f=BIlsinθ

由牛顿第三定律得，导轨受到AC棒的压力和摩擦力各为：

N=mg+BIlcosθ和f=BIlsinθ．所以选项AC正确，选项BD错误．

故选：AC

单项选择题

胃大部分切除术后可能出现

A．铁和钙不能吸收

B．糖类食物消化和吸收障碍

C．脂肪类食物消化和吸收障碍

D．维生素B12吸收障碍

E．胃蛋白酶缺乏而导致食物中蛋白质不能消化和吸收
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