题目内容
如图11-3-8,一电子以与磁场垂直的速度v从P处沿PQ方向进入长为d、宽为h的匀强磁场区域,从N处离开磁场.若电子质量为m,电荷量为e,磁感应强度为B,则…( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209396312467.jpg)
图11-3-8
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209396312467.jpg)
图11-3-8
A.电子在磁场中运动时间t="d/v" | B.电子在磁场中运动时间t=h/v |
C.洛伦兹力对电子做功为Bevh | D.电子在N处的速度大小也是v |
D
如下图所示,过P点作出速度v的垂线OMP,在MP上取点O,使OP=ON,则O为质子圆周运动的圆心.由图中几何关系知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209396463381.jpg)
R2=d2+(R-h)2,所以R=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120939662476.gif)
圆心角θ的正弦值sinθ=
t=
T
显然A、B错.洛伦兹力不做功,所以电子的线速度大小不变,故C错,D对.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209396463381.jpg)
R2=d2+(R-h)2,所以R=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120939662476.gif)
圆心角θ的正弦值sinθ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120939678492.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120939693382.gif)
显然A、B错.洛伦兹力不做功,所以电子的线速度大小不变,故C错,D对.
![](http://thumb2018.1010pic.com/images/loading.gif)
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