题目内容
如图所示,在倾角为30°的绝缘斜面上,固定两条无限长的平行光滑金属导轨,匀强磁场B垂直于斜面向上,磁感应强度B=0.4T,导轨间距L=0.5m,两根金属棒ab、cd与导轨垂直地放在导轨上,金属棒质量mab=0.1kg,mcd=0.2kg,每根金属棒的电阻均为r=0.2W,导轨电阻不计.当用沿斜面向上的拉力拉动金属棒ab匀速向上运动时.cd金属棒恰在斜面上保持静止。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026504352769.png)
(1) 金属棒cd两端电势差Ucd;
(2) 作用在金属棒ab上拉力的功率。(g取10m/s2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026504352769.png)
(1) 金属棒cd两端电势差Ucd;
(2) 作用在金属棒ab上拉力的功率。(g取10m/s2)
(1)-1V (2)15W
试题分析:(1)cd金属棒恰在斜面上保持静止,cd棒所受重力沿斜面的分力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002650451887.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002650466837.png)
,由右手定则可知电流方向为由d到c,所以
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026504822434.png)
(2)以ab和cd构成的整个系统为研究对象,由于二者均做匀速直线运动,所以整体的合外力为零.则作用在ab上的外力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026504981891.png)
金属棒ab向上匀速运动时,它切割磁感线产生的感应电动势为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002650513634.png)
所以cd棒受的安培力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026505291190.png)
因为cd不动,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002650560829.png)
联立解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026505762281.png)
外力的功率为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002650591910.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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