题目内容
(10分)如图所示,一质量为m、电荷量为q的带负电的粒子,从A点射入宽度为d、磁感应强度为B的匀强磁场,MN、PQ为该磁场的边界线,磁场方向垂直于纸面向里,带电粒子射入时的初速度方向与下边界成θ=45°,且粒子恰好没有从MN边界射出,不计粒子所受重力,求该带电粒子初速度v0。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230280814445.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230280814445.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230287051029.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230300471035.png)
试题分析:由题意知,带电粒子恰好不能从MN边界射出时有两种可能的情况均满足题意。若带电粒子初速度向左上方和带电粒子初速度向右上方时粒子轨迹与磁场上边界相切,作出两种情况的粒子运动轨迹,如图所示。 ①
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230308749295.png)
设带电粒子初速度向右上方时轨迹半径为R1,则由牛顿第二定律和几何关系有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824223031732863.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824223032668828.png)
联解②③得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230287051029.png)
设粒子初速度向左上方时轨迹半径为R2,则由牛顿第二定律和几何关系有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824223034930906.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824223036084853.png)
联解⑤⑥得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242230300471035.png)
评分参考意见:本题共10分,其中①②⑤式2分,③④⑥⑦式各1分;若有其他合理解法且答案正确,可同样给分。
![](http://thumb2018.1010pic.com/images/loading.gif)
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