题目内容
在如图所示的直角坐标系中,x轴的上方存在与x轴正方向成45°角斜向右下方的匀强电场,场强的大小为E=×104 V/m.x轴的下方有垂直于xOy面向外的匀强磁场,磁感应强度的大小为B=2×10-2 T.把一个比荷为
=2×108 C/kg的正电荷从坐标为(0,1)的A点处由静止释放.电荷所受的重力忽略不计.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035568894840.jpg)
(1)求电荷从释放到第一次进入磁场时所用的时间;
(2)求电荷在磁场中做圆周运动的半径;(保留两位有效数字)
(3)当电荷第二次到达x轴时,电场立即反向,而场强大小不变,试确定电荷到达y轴时的位置坐标.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556842451.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035568894840.jpg)
(1)求电荷从释放到第一次进入磁场时所用的时间;
(2)求电荷在磁场中做圆周运动的半径;(保留两位有效数字)
(3)当电荷第二次到达x轴时,电场立即反向,而场强大小不变,试确定电荷到达y轴时的位置坐标.
(1)10-6 s (2)0.71 m (3)(0,8)
(1)如图,电荷从A点匀加速运动到x轴上C点的过程:
位移s=AC=
m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035569204646.jpg)
加速度a=
=2
×1012 m/s2
时间t=
=10-6 s.
(2)电荷到达C点的速度为
v=at=2
×106 m/s
速度方向与x轴正方向成45°角,在磁场中运动时
由qvB=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556982570.png)
得R=
=
m
即电荷在磁场中的偏转半径为0.71 m.
(3)轨迹圆与x轴相交的弦长为Δx=
R=1 m,所以电荷从坐标原点O再次进入电场中,且速度方向与电场方向垂直,电荷在电场中做类平抛运动.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035570455443.jpg)
设电荷到达y轴的时间为t′,则:
tan 45°=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003557060676.png)
解得t′=2×10-6 s
则类平抛运动中垂直于电场方向的位移
L=vt′=4
m
y=
=8 m
即电荷到达y轴时位置坐标为(0,8).
位移s=AC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556904344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035569204646.jpg)
加速度a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556920547.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556904344.png)
时间t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556951537.png)
(2)电荷到达C点的速度为
v=at=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556904344.png)
速度方向与x轴正方向成45°角,在磁场中运动时
由qvB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556982570.png)
得R=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556998583.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003557013413.png)
即电荷在磁场中的偏转半径为0.71 m.
(3)轨迹圆与x轴相交的弦长为Δx=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556904344.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250035570455443.jpg)
设电荷到达y轴的时间为t′,则:
tan 45°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003557060676.png)
解得t′=2×10-6 s
则类平抛运动中垂直于电场方向的位移
L=vt′=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003556904344.png)
y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003557091614.png)
即电荷到达y轴时位置坐标为(0,8).
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目