题目内容
![](http://thumb.1010pic.com/pic3/upload/images/201201/21/c3b93ffd.png)
分析: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.
解答: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/pic3/upload/images/201301/17/e890b02d.png)
![](http://thumb.1010pic.com/pic3/upload/images/201301/17/f44a5807.png)
![](http://thumb.1010pic.com/pic3/upload/images/201301/17/d3ffb066.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;
负离子受力与正离子相反,轨迹与正粒子轨迹关于PQ对称;
故范围为在荧光屏上P点两侧,将出现两个相等长度条形亮线,其长度为:2r-2rcosθ=2r(1-cosθ)=2
(1-cosθ).
故选A.
qvB=m
v2 |
r |
解得
r=
mv |
qB |
![](http://thumb.1010pic.com/pic3/upload/images/201301/17/e890b02d.png)
![](http://thumb.1010pic.com/pic3/upload/images/201301/17/f44a5807.png)
![](http://thumb.1010pic.com/pic3/upload/images/201301/17/d3ffb066.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;
负离子受力与正离子相反,轨迹与正粒子轨迹关于PQ对称;
故范围为在荧光屏上P点两侧,将出现两个相等长度条形亮线,其长度为:2r-2rcosθ=2r(1-cosθ)=2
mv |
qB |
故选A.
点评:本题关键通过作图分析粒子可能出现的运动轨迹,然后综合考虑在荧光屏上的落点,得到亮纹的范围.
![](http://thumb2018.1010pic.com/images/loading.gif)
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