题目内容
如图所示,在光滑水平面上,有竖直向下的匀强磁场,分布在宽度为L的区域内,两个边长均为a(a<L)的单匝闭合正方形线圈甲和乙,分别用相同材料不同粗细的导线绕制而成(甲为细导线),将线圈置于光滑水平面上且位于磁场的左边界,并使两线圈获得大小相等、方向水平向右的初速度,若甲线圈刚好能滑离磁场,则( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239441791222.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239441791222.png)
A.乙线圈也刚好能滑离磁场 |
B.两线圈进入磁场过程中通过导线横截面积电量相同 |
C.两线圈进入磁场过程中产生热量相同 |
D.甲线圈进入磁场过程中产生热量Q1与离开磁场过程中产生热量Q2之比为![]() |
AD
对甲线圈进入的过程和离开磁场的过程,将两过程一并处理,就有
(动量定理),以
代入就得到
,所以
,而
与线圈横截面积无关(D为密度),由此可见由于乙线圈唯一与甲线圈的区别就是粗细不同,而
与线圈横截面积无关,所以线圈乙也刚好能离开磁场,A正确;
因为乙的电阻小于甲的电阻,进人时
相同
由公式q=
可知q不相同
假设初速度为v0完全进入磁场时v1
根据F安=
=ma=m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944647379.png)
进入过程:
=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944679443.png)
∑V△t=m∑
∑V△t="a" (边长a) 求和公式
a=m
——(1)
出去过程:同理:
a=m(v1-0) ——(2)
所以:v0=2v1
Q1=
m(
-
)=
m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944866360.png)
Q2=
m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944866360.png)
最终:Q1:Q2=3:1
因为甲乙质量不同根据动能定理可知进入时热量不同
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944351925.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944367662.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239443981119.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944429906.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239445071248.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241239445071248.png)
因为乙的电阻小于甲的电阻,进人时
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944585371.png)
由公式q=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944616395.png)
假设初速度为v0完全进入磁场时v1
根据F安=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944632421.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944647379.png)
进入过程:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944632421.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944679443.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944710359.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944725343.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944710359.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944788462.png)
出去过程:同理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944710359.png)
所以:v0=2v1
Q1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944819284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944835366.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944866360.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944959279.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944866360.png)
Q2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944819284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123944866360.png)
最终:Q1:Q2=3:1
因为甲乙质量不同根据动能定理可知进入时热量不同
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目