题目内容
(12分)如图所示,水平地面上有一辆固定有竖直光滑绝缘管的小车,管的底部有一质量m=0.2g、电荷量q=8×10-5C的小球,小球的直径比管的内径略小.在管口所在水平面MN的下方存在着垂直纸面向里、磁感应强度B1= 15T的匀强磁场,MN面的上方还存在着竖直向上、场强E=25V/m的匀强电场和垂直纸面向外、磁感应强度B2=5T的匀强磁场.现让小车始终保持v=2m/s的速度匀速向右运动,以带电小球刚经过场的边界PQ为计时的起点,测得小球对管侧壁的弹力FN随高度h变化的关系如图所示.g取10m/s2,不计空气阻力.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028245038296.png)
(1)小球刚进入磁场B1时的加速度大小a;
(2)绝缘管的长度L;
(3)小球离开管后再次经过水平面MN时距管口的距离![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824519383.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028245038296.png)
(1)小球刚进入磁场B1时的加速度大小a;
(2)绝缘管的长度L;
(3)小球离开管后再次经过水平面MN时距管口的距离
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824519383.png)
(1)2m/s2(2)1m(3)(
)m
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824535549.png)
试题分析:(1)以小球为研究对象,竖直方向小球受重力和恒定的洛伦兹力f1,故小球在管中竖直方向做匀加速直线运动,加速度设为a,由牛顿第二定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824550635.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824566579.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028245811208.png)
(2)由小球对管侧壁的弹力FN随高度h变化的图象知,在小球运动到管口时,FN=2.4×10-3N,设v1为小球竖直分速度及由v1使小球受到的洛伦兹力
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824613348.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824628626.png)
由水平方向平衡知
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824644544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824659668.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824675981.png)
由竖着方向小球做匀加速直线运动
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824691576.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824706623.png)
(3)小球离开管口进入复合场,其中qE=2×10-3N,mg=2×10-3N.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028247378643.png)
故电场力与重力平衡,小球在复合场中做匀速圆周运动,合速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824753745.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824753745.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824784856.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824800698.png)
小球离开管口开始计时,到再次经过
MN所通过的水平距离由几何关系得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824815700.png)
对应时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250028248311091.png)
小车运动距离为x2,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824800698.png)
故△x= x1- x2=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002824862545.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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