题目内容
磁悬浮列车是一种高速低耗的新型交通工具。它的驱动系统简化为如下模型,固定在列车下端的动力绕组可视为一个矩形纯电阻金属框,电阻为
,金属框置于xOy平面内,长边MN长为
平行于y轴,宽为
的NP边平行于x轴,如图5-1所示。列车轨道沿Ox方向,轨道区域内存在垂直于金属框平面的磁场,磁感应强度
沿Ox方向按正弦规律分布,其空间周期为
,最大值为
,如图5-2所示,金属框同一长边上各处的磁感应强度相同,整个磁场以速度
沿Ox方向匀速平移。设在短暂时间内,MN、PQ边所在位置的磁感应强度随时间的变化可以忽略,并忽略一切阻力。列车在驱动系统作用下沿Ox方向加速行驶,某时刻速度为
(
)。
(1)简要叙述列车运行中获得驱动力的原理;
(2)为使列车获得最大驱动力,写出MN、PQ边应处于磁场中的什么位置及
与d之间应满足的关系式;
(3)计算在满足第(2)问的条件下列车速度为
时驱动力的大小。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082411461015324257.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114609903199.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114609950191.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114609965192.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114609981203.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610012179.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610059207.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610075193.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610090181.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610106224.gif)
(1)简要叙述列车运行中获得驱动力的原理;
(2)为使列车获得最大驱动力,写出MN、PQ边应处于磁场中的什么位置及
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610012179.gif)
(3)计算在满足第(2)问的条件下列车速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610090181.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082411461015324257.jpg)
(1)由于列车速度与磁场平移速度不同,导致穿过金属框的磁通量发生变化,由于电磁感应,金属框中会产生感应电流,该电流受到的安培力即为驱动力。
(2)
或![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610184689.gif)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610199676.gif)
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610168582.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610184689.gif)
(3)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610199676.gif)
(1)由于列车速度与磁场平移速度不同,导致穿过金属框的磁通量发生变化,由于电磁感应,金属框中会产生感应电流,该电流受到的安培力即为驱动力。
(2)为使列车获得最大驱动力,MN、PQ应位于磁场中磁感应强度同为最大值且反向的地方,这会使得金属框所围面积的磁通量变化率最大,导致框中电流最强,也会使得金属框长边中电流受到的安培力最大,因此,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610215198.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610231231.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610168582.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610184689.gif)
(3)由于满足第(2)问条件,则MN、PQ边所在处的磁感应强度大小均为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610059207.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610309214.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610480357.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610496340.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610511243.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610309214.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610543521.gif)
在此
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610309214.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610652682.gif)
同理,在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610309214.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610699679.gif)
故在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610309214.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610730628.gif)
根据法拉第电磁感应定律,金属框中的感应电动势大小:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610761476.gif)
根据闭合电路欧姆定律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610777302.gif)
根据安培力公式,MN边所受的安培力:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610886480.gif)
PQ边所受的安培力:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610901476.gif)
根据左手定则,MN、PQ边所受的安培力方向相同,此时列车驱动力的大小:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610917657.gif)
联立解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114610199676.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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