题目内容
(10分)质量为m=0.02 kg的通电细杆ab置于倾角为θ=37°的平行放置的导轨上,导轨的宽度d=0.2 m,杆ab与导轨间的动摩擦因数μ=0.4,磁感应强度B=2 T 的匀强磁场与导轨平面垂直且方向向下,如图所示.现调节滑动变阻器的触头,试求出为使杆ab静止不动,通过ab杆的电流范围为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022376064997.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022376064997.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237621708.png)
试题分析:当电流较大时,导体有向上的运动趋势,所受静摩擦力向下,当静摩擦力达到最大时,磁场力为最大值
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237637333.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237652353.png)
两种情况下的受力图如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250022376689617.png)
由平衡条件得:
对第一种情况有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237684831.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237699816.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237730561.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237746688.png)
解上述方程得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237762691.png)
对第二种情况有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237777893.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237808825.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237824586.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237855694.png)
解上述方程得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237871644.png)
则通过ab杆的电流范围为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002237621708.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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