题目内容
如图所示,MN是一荧光屏,当带电粒子打到荧光屏上时,荧光屏能够发光.MN的上方有磁感应强度为B的匀强磁场,磁场方向垂直纸面向里,P为屏上的一小孔,PQ与MN垂直.一群质量为m、带电荷量q的粒子(不计重力),以相同的速率v,从P点沿垂直于磁场方向射入磁场区域,其入射方向分布在以PQ为中心,夹角为2θ的范围内,不计粒子间的相互作用,以下说法正确的是( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412395789218817.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412395789218817.png)
A.荧光屏上将出现一圆形亮斑,其半径为![]() |
B.荧光屏上将出现一条亮线,其长度为![]() |
C.荧光屏上将出现一条亮线,其长度为![]() |
D.荧光屏上将出现一条亮线,其最右端距P点为![]() |
B
粒子仅受洛伦兹力,做匀速圆周运动,分析找出粒子的一般轨迹后得到荧光屏上亮纹的范围.
解:粒子做匀速圆周运动,洛伦兹力提供向心力,得到
qvB=m
解得
r=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123958048495.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239580796968.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239580957772.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239581266889.png)
正粒子沿着右侧边界射入,轨迹如上面左图,此时出射点最近,与边界交点与P间距为:2rcosθ;
正粒子沿着左侧边界射入,轨迹如上面中间图,此时出射点最近,与边界交点与P间距为:2rcosθ;
正粒子垂直边界射入,轨迹如上面右图,此时出射点最远,与边界交点与P间距为:2r;
故范围为在荧光屏上P点两侧,将出现一条亮线,
其长度为:2r-2rcosθ=2r(1-cosθ)=
故选B .
解:粒子做匀速圆周运动,洛伦兹力提供向心力,得到
qvB=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123958001396.png)
r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123958048495.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239580796968.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239580957772.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239581266889.png)
正粒子沿着右侧边界射入,轨迹如上面左图,此时出射点最近,与边界交点与P间距为:2rcosθ;
正粒子沿着左侧边界射入,轨迹如上面中间图,此时出射点最近,与边界交点与P间距为:2rcosθ;
正粒子垂直边界射入,轨迹如上面右图,此时出射点最远,与边界交点与P间距为:2r;
故范围为在荧光屏上P点两侧,将出现一条亮线,
其长度为:2r-2rcosθ=2r(1-cosθ)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123958141756.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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