题目内容
如图所示,空间分布着方向平行于纸面且与场区边界垂直的有界匀强电场,电场强度为E、宽度为L。在紧靠电场右侧的圆形区域内,分布着垂直于纸面向外的匀强磁场,圆形磁场区域半径为r。当一带正电的粒子(质量为m,电荷量为q)从A点静止释放后,在M点离开电场,并沿半径方向射入磁场区域,磁感应强度为B,粒子恰好从N点射出,O为圆心,∠MON=120°,粒子重力忽略不计。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342146764928.jpg)
(1)粒子经电场加速后,进入磁场时速度v的大小;
(2)匀强磁场的磁感应强度B的大小和粒子在电场、磁场中运动的总时间t;
(3)若粒子在离开磁场前某时刻,磁感应强度方向不变,大小突然变为B1,此后粒子恰好被束缚在该磁场中,则B1的最小值为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342146764928.jpg)
(1)粒子经电场加速后,进入磁场时速度v的大小;
(2)匀强磁场的磁感应强度B的大小和粒子在电场、磁场中运动的总时间t;
(3)若粒子在离开磁场前某时刻,磁感应强度方向不变,大小突然变为B1,此后粒子恰好被束缚在该磁场中,则B1的最小值为多少?
(1)
;(2)t
;(3)
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134214691821.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342147541200.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342149881227.png)
(1)设粒子经电场加速后的速度为v,根据动能定理有qEL=
mv2 (2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342151445226.jpg)
解得:
(2分)
(2)粒子在磁场中完成了如图所示的部分圆运动,设其半径为 R,因洛伦兹力提供向心力,
所以有qvB=
(1分)
由几何关系得
(2分)
所以
(1分)
设粒子在电场中加速的时间为
,在磁场中偏转的时间为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215643331.png)
粒子在电场中运动的时间t1=
=
(2分)
粒子在磁场中做匀速圆周运动,其周期为
(1分)
由于∠MON=120°,所以∠MO'N=60°故粒子在磁场中运动时间
t2=
(2分)
所以粒子从A点出发到N点离开磁场经历的时间:
t=t1+t2 =
+![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134216142680.png)
(1分)
(3)如图所示,当粒子运动到轨迹与O O'连线交点处改变磁场大小时,粒子运动的半径最大,既B1对应最小值(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342162365808.jpg)
由几何关系得此时最大半径有
(2分)
所以
(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215066321.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342151445226.jpg)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134214691821.png)
(2)粒子在磁场中完成了如图所示的部分圆运动,设其半径为 R,因洛伦兹力提供向心力,
所以有qvB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215284571.png)
由几何关系得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215518703.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215565971.png)
设粒子在电场中加速的时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215596292.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215643331.png)
粒子在电场中运动的时间t1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215705534.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215783752.png)
粒子在磁场中做匀速圆周运动,其周期为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342158301009.png)
由于∠MON=120°,所以∠MO'N=60°故粒子在磁场中运动时间
t2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134216064876.png)
所以粒子从A点出发到N点离开磁场经历的时间:
t=t1+t2 =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134215783752.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134216142680.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342147541200.png)
(3)如图所示,当粒子运动到轨迹与O O'连线交点处改变磁场大小时,粒子运动的半径最大,既B1对应最小值(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342162365808.jpg)
由几何关系得此时最大半径有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134216282645.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241342149881227.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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