题目内容
如图,两光滑导体框ABCD与EFGH固定在水平面内,在D点平滑接触,A、C分别处于FE、HG的沿长线上,ABCD是边长为a的正方形;磁感强度为B的匀强磁场竖直向上;导体棒MN置于导体框上与导体框良好接触,以速度v沿BD方向从B点开始匀速运动,已知线框ABCD及棒MN单位长度的电阻为r,线框EFGH电阻不计。求:
⑴导体棒MN在线框ABCD上运动时,通过MN电流的最大值与最小值;
⑵为维持MN在线框ABCD上的匀速运动,必须给MN施加一水平外力,用F(t)函数表示该力;
⑶导体棒达D点时立即撤去外力,则它还能前进多远(设EF、GH足够长)?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344024722629.png)
⑴导体棒MN在线框ABCD上运动时,通过MN电流的最大值与最小值;
⑵为维持MN在线框ABCD上的匀速运动,必须给MN施加一水平外力,用F(t)函数表示该力;
⑶导体棒达D点时立即撤去外力,则它还能前进多远(设EF、GH足够长)?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344024722629.png)
⑴![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025031006.png)
⑵![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025501207.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025031006.png)
⑵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025501207.png)
⑴设某时刻棒MN交线框于P、S点,令PS长为l,则
此时电动势E = Blv
MN左侧电阻
MN右侧电阻![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402659774.png)
则![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344027061324.png)
故:I=
……………………………………………④
因导线框ABCD关于AC对称,所以通MN的电流大小也具有对称性,所以
当l = 0时,电流最小值![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402799946.png)
当l =
时,电流最大值![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025031006.png)
⑵设MN到达B的时间为t0,则t0=
,到达D点用时2t0,
当0≤t≤t0时,由④式得:
(其中vt ="l" )
代入F=BIl得:F =![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344030491119.png)
当t0≤t≤2t0时,将
代入④式得:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403127813.png)
代入F=BIl得:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344031431225.png)
⑶导线框进入矩形磁场后,由牛顿第二定律得:
取任意△t时间有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025501207.png)
此时电动势E = Blv
MN左侧电阻
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402597598.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402659774.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344027061324.png)
故:I=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344027681424.png)
因导线框ABCD关于AC对称,所以通MN的电流大小也具有对称性,所以
当l = 0时,电流最小值
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402799946.png)
当l =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402877409.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025031006.png)
⑵设MN到达B的时间为t0,则t0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134402955510.png)
当0≤t≤t0时,由④式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344029871100.png)
代入F=BIl得:F =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344030491119.png)
当t0≤t≤2t0时,将
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403080724.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403127813.png)
代入F=BIl得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344031431225.png)
⑶导线框进入矩形磁场后,由牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403174587.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403236934.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344032831129.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344033771067.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344034391173.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824134403501959.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241344025501207.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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