题目内容
如图在第一象限存在匀强磁场,第四象限存在正交电场和磁场,磁感应强度均为B,一个电子从y轴上的c点平行x轴射入磁场,经x轴的P点沿PC直线射出第四象限,已知AC的长度为L;∠CAP=30°;电子质量为m,电量为q。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250048230852385.jpg)
(1)电子射入磁场时的速度v;
(2)电子在第一象限运动时间;
(3)电场强度E的大小和方向;
(4)电子在第四象限运动时间.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250048230852385.jpg)
(1)电子射入磁场时的速度v;
(2)电子在第一象限运动时间;
(3)电场强度E的大小和方向;
(4)电子在第四象限运动时间.
(1)
(2)
(3)
,30°(4)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823147591.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823100648.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823116715.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823131699.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823147591.png)
试题分析:分析如图
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250048231634375.jpg)
(1)设电子在第一象限做圆周运动的半径为R,由几何知识得∠ACP=30° OC=2OP
又OP=R ,OC=L-R;(2分)所以L-R=2R;R=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823178327.png)
由qvB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823194570.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823100648.png)
(2) 由几何知识得,电子在第一象限做圆周运动转过的圆心角α=120°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823225382.png)
电子在第一象限运动时间t1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250048232561153.png)
(3)由qE=qvB得E=vB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004823131699.png)
由二力平衡知,电场强度的方向在纸面内斜向下与x轴成30°(2分)
(4)电子在第四象限运动时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250048232871300.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目