题目内容
通电导线CD,悬挂在磁感应强度为B(向上)的匀强磁场中,如图15-3-11所示.导线通以电流I后,向纸外移动α角后平衡,若导线长为________L,则所受安培力大小为________,电流大小是________,方向为________.(已知导线CD的质量为m)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209259813774.jpg)
图15-3-11
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209259813774.jpg)
图15-3-11
mgtanα
C→D
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120926043554.gif)
首先画出导线CD侧面受力图,如图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209260592644.jpg)
由于磁场竖直向上且导体静止,故安培力应水平向外,再由左手定则不难判定,电流方向沿C→D方向.
由平衡条件知F=mgtanα
又安培力F=BIL,故I=
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241209260592644.jpg)
由于磁场竖直向上且导体静止,故安培力应水平向外,再由左手定则不难判定,电流方向沿C→D方向.
由平衡条件知F=mgtanα
又安培力F=BIL,故I=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120926043554.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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