题目内容
轻质细线吊着一质量为m=0.32kg,边长为L=0.8m、匝数n=10的正方形线图总电阻为r=1
,边长为
的正方形磁场区域对称分布在线图下边的两侧,如图甲所示,磁场方向垂直纸面向里,大小随时间变化如图乙所示,从t=0开始经时间t0细线开始松驰,g=10m/s2。求:
(1)在前t0时间内线图中产生的电动势;
(2)在前t0时间内线图的电功率;
(3)求t0的值。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241233112306026.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311199215.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311215234.gif)
(1)在前t0时间内线图中产生的电动势;
(2)在前t0时间内线图的电功率;
(3)求t0的值。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241233112306026.jpg)
(1)由法拉第电磁感应定律得:
…………5分![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412331135573.gif)
(2)
…………………………5分
(3)分析线圈受力可知,当细线松弛时有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311464270.gif)
…………………………4分
由图像知:
解得:
……………………4分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241233113081676.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412331135573.gif)
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311371500.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311418597.gif)
(3)分析线圈受力可知,当细线松弛时有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311449633.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311464270.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311480699.gif)
由图像知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311527490.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123311542274.gif)
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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