题目内容
(15分)如图所示,水平放置的平行金属板A和D间的距离为d,金属板长为
,两板间所加电压为U,D板的右侧边缘恰好是倾斜挡板NM上的一个小孔K,NM与水平挡板NP成
角,且挡板足够长,K与N间的距离为
.现有一质量为m、电荷量为q的带正电的粒子,从A、D的中点O沿平行于金属板方向
以某一速度射入,不计粒子的重力.该粒子穿过金属板后恰好穿过小孔K:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535128103146.jpg)
(1)求该粒子从O点射入时的速度大小v0;
(2)若两档板所夹的整个区域存在一垂直纸面向外的匀强磁场,粒子经过磁场偏转后能垂直打在水平挡板NP上,求该磁场的磁感应强度的大小B0;
(3)若磁场方向变为垂直纸面向里,且只存在于两档板所夹间的某一区域内,同样使该粒子经过磁场偏转后能垂直打在水平挡板NP上(之前与挡板没有碰撞),求满足条件的磁感应强度的最小值Bmin.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512279547.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512373382.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512435542.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512747363.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535128103146.jpg)
(1)求该粒子从O点射入时的速度大小v0;
(2)若两档板所夹的整个区域存在一垂直纸面向外的匀强磁场,粒子经过磁场偏转后能垂直打在水平挡板NP上,求该磁场的磁感应强度的大小B0;
(3)若磁场方向变为垂直纸面向里,且只存在于两档板所夹间的某一区域内,同样使该粒子经过磁场偏转后能垂直打在水平挡板NP上(之前与挡板没有碰撞),求满足条件的磁感应强度的最小值Bmin.
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535129661377.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512825826.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512888941.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535129661377.png)
试题分析:(1)粒子在电场中做类平抛运动:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513028645.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513106945.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513184949.png)
代入
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153512279547.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535133241051.png)
(2)射入的粒子在进入K时竖直方向的分速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513387350.png)
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513496698.png)
竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513558683.png)
可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513668907.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513792468.png)
粒子到达K点时速度为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513902815.png)
由几何关系可得:粒子圆周运动半径为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153513995555.png)
且满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153514073967.png)
可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535141511177.png)
(3)磁场反向,如图,粒子从K点入射后做匀速直线运动从D点开始进入磁场,根据对称性,需偏转300o后从E点射出,做匀速直线运动垂直打到NP挡板上。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535148068184.jpg)
根据:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153514853763.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153514947857.png)
要使B最小,则需使半径r最大,临界情况是轨迹刚好和挡板相切………………………..1分
由几何关系可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153515009538.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824153515087860.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241535151811554.png)
点评:难度较大,涉及到的运动过程较多,首先分析粒子的受力过程,在分析物体的运动过程,求解交叉点的速度大小以及方向,详细确定粒子的运动轨迹,把整段过程分解,逐个过程列公式分析
![](http://thumb2018.1010pic.com/images/loading.gif)
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