题目内容
(12分)如图所示,坐标平面第Ⅰ象限内存在大小为E=4×105 N/C、方向水平向左的匀强电场,在第Ⅱ象限内存在方向垂直纸面向里的匀强磁场.质荷比为
=4×10-10 kg/C的带正电粒子从x轴上的A点以初速度v0=2×107 m/s垂直x轴射入电场,OA=0.2 m,不计重力.求:
(1)粒子经过y轴时的位置到原点O的距离;
(2)若要求粒子不能进入第三象限,求磁感应强度B的取值范围(不考虑粒子第二次进入电场后的运动情况.)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250033143834671.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314367454.png)
(1)粒子经过y轴时的位置到原点O的距离;
(2)若要求粒子不能进入第三象限,求磁感应强度B的取值范围(不考虑粒子第二次进入电场后的运动情况.)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250033143834671.jpg)
(1)0.4 m (2)B≥(2
+2)×10-2 T
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314414344.png)
试题分析:(1)设粒子在电场中运动的时间为t,粒子经过y轴时的位置与原点O的距离为y,
则:sOA=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314430338.png)
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314445453.png)
E=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314477431.png)
y=v0t (2分)
联立解得a=1.0×1015 m/s2 t=2.0×10-8 s y=0.4 m (1分)
(2)粒子经过y轴时在电场方向的分速度为:vx=at=2×107 m/s
粒子经过y轴时的速度大小为:v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314492574.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314414344.png)
与y轴正方向的夹角为θ,θ=arctan
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314539399.png)
要使粒子不进入第三象限,如图所示,此时粒子做匀速圆周运动的轨道半径为R,则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250033145555591.jpg)
R+
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314570413.png)
qvB=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314601452.png)
联立解得B≥(2
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003314414344.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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