题目内容
如图所示,有一垂直于纸面向外的有界匀强磁场,磁场的磁感应强度为B,其边界为一等腰直角三角形(边界上有磁场),ACD为三角形的三个顶点,AC=AD=L.今有一质量为m、电荷量为+q的粒子(不计重力),以速度v=
从CD边上的某点P既垂直于CD边又垂直于磁场的方向射入,然后从AD边上某点Q射出,则有( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250045269452438.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004526914648.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250045269452438.png)
A.DP<![]() | B.DP<![]() | C.DQ≤![]() | D.DQ≤![]() |
试题分析:粒子在磁场中做匀速圆周运动,洛伦兹力提供向心力,临界轨迹是恰好与AC边相切,然后结合几何关系分析即可.
解:A、B、粒子在磁场中做匀速圆周运动,洛伦兹力提供向心力,轨道半径为:
r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527241335.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527257514.png)
临界轨迹圆恰好与AC边相切,如图所示:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500452727213532.png)
结合几何关系,有:
DP=CD﹣CP=CD﹣(CO﹣OP)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527304458.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527319411.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527335642.png)
故DP小于
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527335642.png)
C、D、结合几何关系,根据勾股定理,有:
EO2+QE2=QO2
即(r﹣AQ)2+QE2=r2
(r﹣AQ)2+(L﹣r)2=r2
解得:AQ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527366260.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527382336.png)
故粒子可以从AD边离D的
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004527413337.png)
故选:AC.
点评:本题关键是先结合洛伦兹力提供向心力列式求解出轨道半径,然后画出临界轨迹,最后结合几何关系列式求解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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