题目内容
如图所示,质量为m、边长为L的正方形线框,从有界的匀强磁场上方由静止自由下落,线框电阻为R,匀强磁场的高度为H,(L<H),磁感应强度为B,线框下落过程中ab边与磁场边界平行且保持水平。已知ab边刚进入磁场时和ab边刚穿出磁场时线框都做减速运动,加速度大小都为
.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241617102952446.png)
(1)ab边刚进入磁场时和ab边刚出磁场时的速度大小;
(2)cd边刚进入磁场时,线框的速度大小;
(3)线框进入磁场的过程中,产生的热量.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161710014400.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241617102952446.png)
(1)ab边刚进入磁场时和ab边刚出磁场时的速度大小;
(2)cd边刚进入磁场时,线框的速度大小;
(3)线框进入磁场的过程中,产生的热量.
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161711122927.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161710607890.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241617108411543.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161711122927.png)
试题分析:(1)设进入时速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161711434265.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161711715295.png)
则此刻感应电动势
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161712042500.png)
感应电流为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161712276609.png)
ab边刚出去时则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161712479934.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161710607890.png)
(2)根据动能定理设cd刚进入时速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161712729378.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161712854853.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241617108411543.png)
(3)在整个过程中,根据能量守恒定律产生的热量为机械能的损失:即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241617131341133.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824161713259617.png)
点评:本题考察了通过电磁感应定律求安培力的方法,结合牛顿第二定律列示,通过能量观点解决能量转化问题
![](http://thumb2018.1010pic.com/images/loading.gif)
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