题目内容
如图所示,在
轴上方有一匀强电场,场强大小为
,方向竖直向下。在
轴下方有一匀强磁场,磁感应强度为
,方向垂直于纸面向里。在
轴上有一点
,离原点距离为
。现有一带电量为
,质量为
的粒子,不计重力,从
区间某点由静止开始释放后,能经过
点。试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323181828399.png)
(1)释放瞬间粒子的加速度;
(2)释放点的坐标
应满足的关系式?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317668271.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317699322.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317668271.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317792316.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317668271.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317886319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317995287.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318026335.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318058346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318120486.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132317886319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323181828399.png)
(1)释放瞬间粒子的加速度;
(2)释放点的坐标
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318229409.png)
(1)
,方向沿
轴负方向;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318276669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318323313.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
分析:重力忽略不计的带电粒子在匀强电场中在电场力作用下,做匀变速直线运动;而在匀强磁场中在洛伦兹力作用下,做匀速圆周运动.在电场中由动能定理可得速率与位移的关系;在磁场中由洛伦兹提供向心力得半径与速率的关系再根据几何关系可以求出.
解答:解:(1)粒子在电场中有
,得
,因为粒子带正电,所以所受电场力的方向跟场强方向相同,即方向沿
轴负方向
(2)粒子在电场与磁场中的运动轨迹如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323185103660.png)
设粒子在Y轴上的坐标为y,进入磁场时的速度为v,Eq?y=
mv2
∴v=
,由 qvB=
,得R=
=
--① 令a-x=2Rn---②
解①②得![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
故答案为:(1)
,方向沿
轴负方向;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
点评:本题主要考查了带电粒子在混合场中运动的问题,要求同学们能正确分析粒子的受力情况,再通过受力情况分析粒子的运动情况,熟练掌握圆周运动及平抛运动的基本公式,难度适中.
解答:解:(1)粒子在电场中有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318432587.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318276669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318323313.png)
(2)粒子在电场与磁场中的运动轨迹如图所示,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323185103660.png)
设粒子在Y轴上的坐标为y,进入磁场时的速度为v,Eq?y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318541338.png)
∴v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318588763.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318635571.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318682581.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318713904.png)
解①②得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
故答案为:(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318276669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132318323313.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241323183541521.png)
点评:本题主要考查了带电粒子在混合场中运动的问题,要求同学们能正确分析粒子的受力情况,再通过受力情况分析粒子的运动情况,熟练掌握圆周运动及平抛运动的基本公式,难度适中.
![](http://thumb2018.1010pic.com/images/loading.gif)
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