题目内容
如图所示,在第二象限和第四象限的正方形区域内分别存在着匀强磁场,磁感应强度均为B,方向相反,且都垂直于xOy平面.一电子由P(-d,d)点,沿x轴正方向射入磁场区域Ⅰ.(电子质量为m,电荷量为e,sin 53°=
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036016313360.jpg)
(1)求电子能从第三象限射出的入射速度的范围.
(2)若电子从
位置射出,求电子在磁场 Ⅰ 中运动的时间t.
(3)求第(2)问中电子离开磁场Ⅱ时的位置坐标.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601615346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036016313360.jpg)
(1)求电子能从第三象限射出的入射速度的范围.
(2)若电子从
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601647713.png)
(3)求第(2)问中电子离开磁场Ⅱ时的位置坐标.
(1)
<v<
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036017091131.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601662662.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601678606.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601693794.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036017091131.png)
(1)电子能从第三象限射出的临界轨迹如图甲所示.电子偏转半径范围为
<r<d
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036017402724.jpg)
由evB=m
得v=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601771557.png)
故电子入射速度的范围为
<v<
.
(2)电子从
位置射出的运动轨迹如图乙所示.设电子在磁场中运动的轨道半径为R,则R2=
2+d2
解得R=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601849472.png)
则∠PHM=53°
由evB=mR
2解得T=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601881644.png)
电子在磁场Ⅰ中运动的时间t=
T=
.
(3)如图乙所示,根据几何知识,带电粒子在射出磁场区域Ⅰ时与水平方向的夹角为53°,在磁场区域Ⅱ位置N点的横坐标为
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036019437075.jpg)
由△NBH′可解得NB的长度等于d,则QA=d-![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601959504.png)
由勾股定理得H′A=
d,H′B=Rcos 53°=![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003602005506.png)
所以电子离开磁场Ⅱ的位置坐标为
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601725433.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036017402724.jpg)
由evB=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601756424.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601771557.png)
故电子入射速度的范围为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601662662.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601678606.png)
(2)电子从
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601647713.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601834748.png)
解得R=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601849472.png)
则∠PHM=53°
由evB=mR
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601865703.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601881644.png)
电子在磁场Ⅰ中运动的时间t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601896541.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601693794.png)
(3)如图乙所示,根据几何知识,带电粒子在射出磁场区域Ⅰ时与水平方向的夹角为53°,在磁场区域Ⅱ位置N点的横坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601927527.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036019437075.jpg)
由△NBH′可解得NB的长度等于d,则QA=d-
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601959504.png)
由勾股定理得H′A=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003601990507.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003602005506.png)
所以电子离开磁场Ⅱ的位置坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250036017091131.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目