题目内容
(9分)如图所示,重为3N的导体棒,放在间距为d=1m的水平放置的导轨上,其中电源电动势E=6V,内阻r=0.5
,定值电阻R0=11.5
,其它电阻不计。试求:
(1)若磁场方向垂直导轨平面向上,大小为B=2T(图未画出),要使导体棒静止不动,导轨与导体棒间的摩擦力至少为多大?
(2)若磁场大小不变,方向与导轨平面成
角。如图所示,此时导体棒所受的摩擦力多大?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515022012626.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151501826337.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151501826337.png)
(1)若磁场方向垂直导轨平面向上,大小为B=2T(图未画出),要使导体棒静止不动,导轨与导体棒间的摩擦力至少为多大?
(2)若磁场大小不变,方向与导轨平面成
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151502138471.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515022012626.jpg)
(1)1N(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515022321102.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515022321102.png)
试题分析:(1)以金属棒为研究对象,受力分析如图甲
∵棒静止不动 ∴水平方向Ff=F安………………1分
根据闭合电路欧姆定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151502388671.png)
∴Ff=F安=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824151502513975.png)
(2)棒受力分析如图
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515026377187.png)
∵安培力的大小没变而此时棒对轨道的正压力数值是比原来增大。
所以棒与轨道间的最大静摩擦力增大,棒仍静止。……2分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241515022321102.png)
点评:做此类型的题目,需要对导体棒分析受力,然后运动欧姆定律,列出电学和力学的等式关系
![](http://thumb2018.1010pic.com/images/loading.gif)
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