题目内容
(14分)在xOy平面第Ⅰ、Ⅱ象限中,存在沿y轴正方向的匀强电场,场强为E=
,在第Ⅲ、Ⅳ象限中,存在垂直于xOy平面方向的匀强磁场,如图所示,磁感应强度B1=B,B2=2B.带电粒子a、b分别从第Ⅰ、Ⅱ象限的P、Q两点(图中没有标出)由静止释放,结果两粒子同时分别进入匀强磁场B1、B2中,再经过时间t第一次经过y轴时恰在点M(0,-
l)处发生碰撞,碰撞时两粒子的速度在同一直线上,碰撞前带电粒子b的速度方向与y轴正方向成60°角,不计粒子重力和两粒子间相互作用.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500112058010354.png)
(1)两带电粒子的比荷及在磁场中运动的轨道半径;
(2)带电粒子释放的位置P、Q两点坐标及释放的时间差.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120549609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120564344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500112058010354.png)
(1)两带电粒子的比荷及在磁场中运动的轨道半径;
(2)带电粒子释放的位置P、Q两点坐标及释放的时间差.
2l ![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120611483.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120611483.png)
试题分析: (1)粒子运动轨迹如图.两粒子在磁场中运动时间相等且为t,即t1=t2=t
而t1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120627420.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120642731.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120658461.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120674792.png)
代入B2=2B1=2B得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120798495.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120861526.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120876548.png)
由几何关系知R1=R2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120892722.png)
(2)由qvB=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120908452.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120923600.png)
所以v1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120954677.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120970593.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120970696.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120986581.png)
由Eqy=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121001338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121017700.png)
所以y1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121032756.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121064763.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250011210795923.jpg)
由几何关系知x1=R-Rcos 60°=l,
x2=-(R+Rcos 60°)=-3l
所以P、Q的坐标分别为P(l,2l)、Q(-3l,8l).
粒子在电场中运动的时间为t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121095373.png)
其中加速度a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121110547.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121142612.png)
故两粒子由静止释放的时间差Δt=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001121157601.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001120611483.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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