题目内容
如图所示,真空中有以O′为圆心,r为半径的圆形匀强磁场区域,磁场方向垂直纸面向外,磁感应强度为B。圆的最下端与x轴相切于直角坐标原点O,圆的右端与平行于y轴的虚线MN相切,在虚线MN右侧x轴上方足够大的范围内有方向竖直向下、场强大小为E的匀强电场,在坐标系第四象限存在方向垂直纸面向里、磁感应强度大小也为B的匀强磁场,现从坐标原点O沿y轴正方向发射速率相同的质子,质子在磁场中做半径为r的匀速圆周运动,然后进入电场到达x轴上的C点。已知质子带电量为+q,质量为m,不计质子的重力、质子对电磁场的影响及质子间的相互作用力。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423555353311668.jpg)
(1)质子刚进入电场时的速度方向和大小;
(2)OC间的距离;
(3)若质子到达C点后经过第四限的磁场后恰好被放在x轴上D点处(图上未画出)的一检测装置俘获,此后质子将不能再返回电场,则CD间的距离为多少。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423555353311668.jpg)
(1)质子刚进入电场时的速度方向和大小;
(2)OC间的距离;
(3)若质子到达C点后经过第四限的磁场后恰好被放在x轴上D点处(图上未画出)的一检测装置俘获,此后质子将不能再返回电场,则CD间的距离为多少。
(1)
方向沿x轴正方向 (2)r+
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235556341922.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235554328582.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235555623942.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235556341922.png)
试题分析:(1)质子在磁场中做匀速圆周运动,洛伦兹力提供向心力,
根据牛顿第二定律得qvB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235556637552.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235557932675.png)
质子运动轨迹如下图,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423555924212680.png)
因为圆形匀强磁场区域的半径为r,质子在磁场中做匀速圆周运动的半径也为r,所以四边形
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235600787579.png)
(2)质子沿y轴正方向射入磁场,则以N为圆心转过
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235602362303.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235603642623.png)
由牛顿第二定律得 qE=ma
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235605092831.png)
在x方向上,由题意可知x1=ON=r
电场中x2=NC=v
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235606528267.png)
所以OC间的距离为x=x1+ x2 =r+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235555623942.png)
(3)设质子出电场时在竖直方向的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235609195350.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235610771308.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235612112500.png)
质子合速度与x轴正向夹角
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235612549297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235614047428.png)
质子到达C点后进入第四限的磁场的运动轨迹如下图所示,设质子在第四限磁场中运动的轨道半径为R.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423561540423607.png)
根据圆的性质,由几何知识得:
x3="CD=" 2R sinθ
质子在磁场中做匀速圆周运动,洛伦兹力提供向心力,根据牛顿第二定律得qvB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235616387581.png)
运动半径
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235617572726.png)
以上各式联立解得:x3="CD=" 2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235618867617.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235620318182.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235614047428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235622627741.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824235556341922.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目