题目内容
(20分)如图所示,磁感应强度大小为B=0.15T,方向垂直纸面向里的匀强磁场分布在半径R=0.20m的圆形区域内,圆的左端跟y轴相切于直角坐标系原点O,右端与边界MN相切于x轴上的A点.MN右侧有平行于x轴负方向的匀强电场.置于坐标原点O的粒子源,可沿x轴正方向射出速度
的带正电的粒子流,比荷为
.不计粒子重力.右侧电场强度大小为
现以过O点并垂直于纸面的直线为轴,将圆形磁场区域按逆时针方向缓慢旋转90
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250040218086712.jpg)
求:
(1)粒子在磁场中运动的半径;
(2)在旋转磁场过程中,粒子经过磁场后,途经MN进入电场,求粒子经过MN时离A点 最远的位置B到A点的距离L1;
(3)通过B点的粒子进人电场后,再次经过MN时距B点的距离L 2为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021746798.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021761975.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021777823.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021777189.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250040218086712.jpg)
求:
(1)粒子在磁场中运动的半径;
(2)在旋转磁场过程中,粒子经过磁场后,途经MN进入电场,求粒子经过MN时离A点 最远的位置B到A点的距离L1;
(3)通过B点的粒子进人电场后,再次经过MN时距B点的距离L 2为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021824468.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004021839543.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022042530.png)
试题分析:⑴ 设粒子做圆周运动的半径为r,由洛伦兹力提供向心力得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022058761.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022073839.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250040221057324.jpg)
⑵ 由图知:sinα=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022120537.png)
设B到A的距离为L1,由几何关系知:
L1=(2R-rtanα)tan2α
得:L1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022151991.png)
⑶ 粒子经过B点时的速度与MN夹角为θ,则θ=90°-2α=30°
粒子经B点时速度的水平分量vx和竖直分量vy分别为:
vx=vsinθ=1.5×104m/s
vy=vcosθ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022167907.png)
进入电场后,粒子水平方向先向右做匀减速运动后反向匀加速,竖直方向为匀速直线运动,设再次经过MN时的位置为D,经历的时间为t,
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250040221831108.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022198934.png)
竖直方向:L2=vyt
解得:L2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004022229900.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目