题目内容
如图甲所示,一边长L=2.5m、质量m=0.5kg的正方形金属线框,放在光滑绝缘的水平面上,整个装置放在方向竖直向上、磁感应强度B=0.8T的匀强磁场中,它的一边与磁场的边界MN重合。在水平力F作用下由静止开始向左运动,经过5s线框被拉出磁场。测得金属线框中的电流随时间变化图像如乙图所示,在金属线框被拉出过程中。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026510443650.png)
⑴求通过线框导线截面的电量及线框的电阻;
⑵写出水平力F随时间变化的表达式;
⑶已知在这5s内力F做功1.92J,那么在此过程中,线框产生的焦耳热是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026510443650.png)
⑴求通过线框导线截面的电量及线框的电阻;
⑵写出水平力F随时间变化的表达式;
⑶已知在这5s内力F做功1.92J,那么在此过程中,线框产生的焦耳热是多少?
(1)R = 4Ω (2)F=(0.2 t+0.1)N (3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651059560.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651059560.png)
试题分析:⑴根据q =
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651075268.png)
又根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651075268.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651106753.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651137580.png)
⑵由电流图像可知,感应电流随时间变化的规律:I=0.1t
由感应电流
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651153610.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651168789.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002651184613.png)
⑶ t=5s时,线框从磁场中拉出时的速度v5 =" at" =1m/s
线框中产生的焦耳热
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026512151106.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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