题目内容
如图甲所示,一长为L = 2m的金属“U”型框与两平行金属板AB相连,两板之间用一绝缘光滑水平杆相连,一质量为M=0.1kg,电量大小为q=0.1c可看成质点的带电小球套在杆中并靠近A板静止,从t=0时刻开始,在 “U”型框宽为d = 1m内加入垂直纸面向外且大小随时间变化的磁场(如图乙所示)后,发现带电小球可以向右运动.求:
1.小球带何种电荷
2.小球达到B板时的速度大小
3.通过分析计算后在丙图坐标系中作出小球在AB杆上的V-t图象.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140235768235.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140235908209.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241402359715178.png)
1.小球带何种电荷
2.小球达到B板时的速度大小
3.通过分析计算后在丙图坐标系中作出小球在AB杆上的V-t图象.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140235768235.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140235908209.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241402359715178.png)
(1)小球带负电荷(2)1m/s(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236095923.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236095923.png)
(1)小球带负电荷 (3分)
(2)两极板间的电势差U=LdΔB/Δt = 1v (2分)
E="U/L" (1分) F="Eq" (1分) a=F/M(1分)
a=
=
=
m/s2 (1分)
S1=
=1m (2分)
V1=at=1m/s (2分)
以后一直匀速运动,即达B板速度为1m/s (1分)
(3)加速距离S=1m
匀速运动时间T=(2-1)/1=1s (1分)
作图如下 (3分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236095923.png)
本题考查电场力做功,由法拉第电磁感应定律可知电压U=1V,由牛顿第二定律求出加速度,由运动学公式求出末速度
(2)两极板间的电势差U=LdΔB/Δt = 1v (2分)
E="U/L" (1分) F="Eq" (1分) a=F/M(1分)
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236173660.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236220575.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236345346.png)
S1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236407563.png)
V1=at=1m/s (2分)
以后一直匀速运动,即达B板速度为1m/s (1分)
(3)加速距离S=1m
匀速运动时间T=(2-1)/1=1s (1分)
作图如下 (3分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140236095923.png)
本题考查电场力做功,由法拉第电磁感应定律可知电压U=1V,由牛顿第二定律求出加速度,由运动学公式求出末速度
![](http://thumb2018.1010pic.com/images/loading.gif)
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