题目内容
如题23-1图所示,边长为L、质量为m、总电阻为R的正方形导线框静置于光滑水平面上,处于与水平面垂直的匀强磁场中,匀强磁场磁感应强度B随时间t变化规律如题23-2图所示.求:
(1)在t=0到t=t0时间内,通过导线框的感应电流大小;
(2)在t=
时刻,a、b边所受磁场作用力大小;
(3)在t=0到t=t0时间内,导线框中电流做的功。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241334268006685.png)
(1)在t=0到t=t0时间内,通过导线框的感应电流大小;
(2)在t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426612440.png)
(3)在t=0到t=t0时间内,导线框中电流做的功。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241334268006685.png)
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241334269401020.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426846864.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426893851.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241334269401020.png)
根据法拉第电磁感应定律求出电动势,再利用欧姆定律求电流。根据电流利用安培力公式求作用力,最近根据电功的公式求解。
(1)由法拉第电磁感应定律得,导线框的感应电动势
(4分)
通过导线框的感应电流大小:
(4分)
(2)ab边所受磁场作用力大小:
(4分)
(3)导线框中电流做的功:
(4分)
(1)由法拉第电磁感应定律得,导线框的感应电动势
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426987973.png)
通过导线框的感应电流大小:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426846864.png)
(2)ab边所受磁场作用力大小:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133427065484.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133426893851.png)
(3)导线框中电流做的功:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241334269401020.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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