题目内容
(7分)某同学自制一电流表,其原理如图所示。质量为m的均匀细金属杆MN与一竖直悬挂的绝缘轻弹簧相连,弹簧的劲度系数为k,在矩形区域abcd内有匀强磁场,磁感应强度大小为B,方向垂直纸面向外。MN的右端连接一绝缘轻指针,可指示出标尺上的刻度。MN的长度大于ab,当MN中没有电流通过且处于静止时,MN与矩形区域的ab边重合,且指针指在标尺的零刻度;当MN中有电流时,指针示数可表示电流强度。MN始终在纸面内且保持水平,重力加速度为g。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250021576235293.png)
(1)当电流表的示数为零时,求弹簧的伸长量;
(2)为使电流表正常工作,判断金属杆MN中电流的方向;
(3)若磁场边界ab的长度为L1,bc的长度为L2,此电流表的量程是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250021576235293.png)
(1)当电流表的示数为零时,求弹簧的伸长量;
(2)为使电流表正常工作,判断金属杆MN中电流的方向;
(3)若磁场边界ab的长度为L1,bc的长度为L2,此电流表的量程是多少?
(1)
(2) M
N (3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157669715.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157638682.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157654219.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157669715.png)
试题分析:(1)设当电流表示数为零时,弹簧的伸长量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157685365.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157716904.png)
(2)根据左手定则,金属杆中的电流方向为:M
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002157654219.png)
(3)设电流表满偏时通过MN的电流为Im,则有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250021577321408.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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