题目内容
如图所示,两根竖直固定的金属导轨ad和bc相距l=0.2m,另外两根水平金属杆MN和EF可沿导轨无摩擦地滑动,MN杆和EF杆的电阻分别为0.2Ω(竖直金属导轨的电阻不计),EF杆放置在水平绝缘平台上,回路NMEF置于匀强磁场内,磁场方向垂直于导轨平面向里,磁感应强度B=1T,试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241539301481844.png)
(1)EF杆不动,MN杆以0.1m/s的速度向上运动时,杆MN两端哪端的电势高?MN两端电势差为多大?
(2)当MN杆和EF杆的质量均为m=10-2kg。MN杆须有多大的速度向上运动时,EF杆将开始向上运动?此时拉力的功率为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241539301481844.png)
(1)EF杆不动,MN杆以0.1m/s的速度向上运动时,杆MN两端哪端的电势高?MN两端电势差为多大?
(2)当MN杆和EF杆的质量均为m=10-2kg。MN杆须有多大的速度向上运动时,EF杆将开始向上运动?此时拉力的功率为多大?
(1)M端 0.01V (2)1m/s 0.2W
试题分析:(1)由右手定则可知M端电势高,N端电势低。
由法拉第电磁感应定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930226577.png)
由闭合电路欧姆定律可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241539303821118.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930460721.png)
(2)由平衡条件知当EF杆开始运动时有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930522607.png)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930662731.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241539307561067.png)
以此时拉力大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930818571.png)
则拉力功率为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241539308811029.png)
点评:在电磁感应现象中,判断电势的高低常常要区分是电源和外电路,根据电源的正极电势比负极电势高,在外电路中,顺着电流方向,电势降低,运用楞次定律判断电流方向,确定电势的高低。本题计算功率的关键是EF杆开始运动时有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153930522607.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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