题目内容
如图3-11所示,单匝线圈在匀强磁场中绕OO′轴从图示的位置开始匀速转动,已知从图示位置转过π/6时,线圈中的电动势大小为6 V,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241207267651855.jpg)
图3-11
(1)交变电动势的最大值、有效值;
(2)设线圈的电阻为R="1" Ω,角速度ω="100" rad/s,线圈由图示位置转过π/2的过程中通过导线截面的电荷量.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241207267651855.jpg)
图3-11
(1)交变电动势的最大值、有效值;
(2)设线圈的电阻为R="1" Ω,角速度ω="100" rad/s,线圈由图示位置转过π/2的过程中通过导线截面的电荷量.
(1)Em="12" V,E=
V (2)q="0.12" C
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120726796428.png)
(1)根据电动势的瞬时值方程e=Emsinωt代入数据得出Em="12" V,则有效值为E=
V.
(2)计算电荷量用平均电动势,转动时为Δt=
s,从最大值里求出磁通量,Em=BSω,从图示位置转过π/2磁通量的变化量为ΔΦ=BS=
,由法拉第电磁感应定律得平均电动势:
,最后使用欧姆定律和电荷量公式得:q=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727124859.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727155852.png)
C="0.12" C.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120726796428.png)
(2)计算电荷量用平均电动势,转动时为Δt=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120726906696.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120726937545.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727046654.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727124859.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727155852.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120727202555.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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