题目内容
如图所示,导体杆ab的质量为m,电阻为R,放置在水平夹角为θ的倾斜金属导轨上。导轨间距为d,电阻不计,系统处于竖直向上的匀强磁场中,磁感应强度为B,电源内阻不计,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857048891776.png)
(1)若导轨光滑,电源电动势E为多大时能使导体杆静止在导轨上?
(2)若杆与导轨之间的动摩擦因数为μ,且不通电时导体杆不能静止在导轨上,现要使导体杆静止在导轨上,电源电动势取值范围为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857048891776.png)
(1)若导轨光滑,电源电动势E为多大时能使导体杆静止在导轨上?
(2)若杆与导轨之间的动摩擦因数为μ,且不通电时导体杆不能静止在导轨上,现要使导体杆静止在导轨上,电源电动势取值范围为多少?
(1)E=
(2)
、![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857057471355.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185705170734.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857054661355.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857057471355.png)
试题分析:导体杆静止在导轨上,受到重力、支持力和安培力三个力作用,安培力大小为F=BIL,根据平衡条件和闭合电路欧姆定律结合求解电源的电动势.
(1)对导体棒受力分析如图,由平衡条件得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857060752662.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185706090598.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185706356696.png)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185706683701.png)
由以上三式解得 E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185705170734.png)
(2)由两种可能:一种是E偏大,I偏大,F偏大,2杆有上滑趋势,摩擦力沿斜面向下,由平衡条件
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857071821075.png)
根据安培力公式有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824185707448623.png)
以上两式联立解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857054661355.png)
另一种可能是E偏小,摩擦力沿斜面向上,同理可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241857057471355.png)
点评:本题是通电导体在磁场中平衡问题,分析受力情况,特别是安培力的情况是关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目