题目内容
如图所示,一束电子(电量为e)以速度V0垂直射入磁感应强度为B,宽为d的匀强磁场中,电子穿出磁场的速度方向与电子原来的入射方向的夹角为30°,(电子重力忽略不计)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241523416972844.jpg)
求:(1)电子的质量是多少?
(2)穿过磁场的时间是多少?
(3)若改变初速度大小,使电子刚好不能从A边射出,则此时速度v是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241523416972844.jpg)
求:(1)电子的质量是多少?
(2)穿过磁场的时间是多少?
(3)若改变初速度大小,使电子刚好不能从A边射出,则此时速度v是多少?
(1)
(3分) (2)
(3分) (3)
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152341775791.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152341884609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152341947446.png)
试题分析:(1)电子垂直射入匀强磁场中,只受洛伦兹力作用做匀速圆周运动,画出轨迹,由几何知识得到,轨迹的半径为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152342009853.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152342087764.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152341775791.png)
(2)由几何知识得到,轨迹的圆心角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152342212551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152342274950.png)
(3)使电子刚好不能从A边射出,则运动半径为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152342352937.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824152341947446.png)
点评:关键要画出轨迹,根据圆心角求时间,由几何知识求半径是常用方法.
![](http://thumb2018.1010pic.com/images/loading.gif)
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