题目内容
(18分)一圆筒的横截面如图所示,其圆心为O。筒内有垂直于纸面向里的匀强磁场,磁感应强度为B。圆筒下面有相距为d的平行金属板M、N,其中M板带正电荷,N板带等量负电荷。质量为m、电荷量为q的带正电粒子自M板边缘的P处由静止释放,经N板的小孔S以速度v沿半径SO方向射入磁场中,粒子与圈筒发生两次碰撞后仍从S孔射出,设粒子与圆筒碰撞过程中没有动能损失,且电荷量保持不变,在不计重力的情况下,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250041513994685.jpg)
(1)M、N间电场强度E的大小;
(2)圆筒的半径R;
(3)保持M、N间电场强度E不变,仅将M板向上平移
,粒子仍从M板边缘的P处由静止释放粒子自进入圆筒至从S孔射出期间,与圆筒的碰撞次数n。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250041513994685.jpg)
(1)M、N间电场强度E的大小;
(2)圆筒的半径R;
(3)保持M、N间电场强度E不变,仅将M板向上平移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151415507.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151446825.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151462846.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151477535.png)
试题分析: (1)设两极板间的电压为U,由动能定理得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151508795.png)
由匀强电场中电势差与电场强度的关系得 U="Ed"
联立上式可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151446825.png)
(2)粒子进入磁场后做匀速圆周运动,运用几何关系做出圆心O’, 圆半径为r,设第一次碰撞点为A,由于粒子与圆筒发生两次碰撞又从S孔射出,因此SA弧所对圆心角
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151540765.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250041515555533.jpg)
由几何关系得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151571723.png)
粒子运动过程中洛伦兹力充当向心力,由牛顿第二定律,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151586760.png)
联立得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151462846.png)
(3)保持M、N间电场强度E不变,M板向上平移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151415507.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151633344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151649767.png)
设粒子进入S孔时的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151664287.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151680614.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151696581.png)
设粒子做圆周运动的半径为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151711285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151727837.png)
设粒子从S到第一次与圆筒碰撞期间的轨道所对圆心角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151742297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151758414.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004151477535.png)
粒子须经过这样的圆弧才能从S孔射出,故 n=3
![](http://thumb2018.1010pic.com/images/loading.gif)
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