题目内容
如图所示,在匀强磁场中匀速转动的矩形线圈的周期为T,转轴O1O2垂直于磁场方向,线圈电阻为2
。从线圈平面与磁场方向平行时开始计时,线圈转过60°时的感应电流为1A。那么
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421832311375.png)
A.线圈消耗的电功率为4W
B.线圈中感应电流的有效值为2A
C.任意时刻线圈中的感应电动势为e = 4cos![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
D. 任意时刻穿过线圈的磁通量为
=
sin![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218277335.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421832311375.png)
A.线圈消耗的电功率为4W
B.线圈中感应电流的有效值为2A
C.任意时刻线圈中的感应电动势为e = 4cos
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
D. 任意时刻穿过线圈的磁通量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218370273.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218386444.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
AC
分析:绕垂直于磁场方向的转轴在匀强磁场中匀速转动的矩形线圈中产生正弦或余弦式交流电,
由于从垂直中性面开始其瞬时表达式为i=Imcosθ,由已知可求Im=
,
根据正弦式交变电流有效值和峰值关系可求电流有效值
根据P=I2R可求电功率
根据Em=Imr可求感应电动势的最大值
任意时刻穿过线圈的磁通量为Φ=BSsin
根据Em=NBSω可求Φm=BS=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218542526.png)
解答:解:从垂直中性面开始其瞬时表达式为i=Imcosθ,
则Im=
=
=2A
感应电动势的最大值为Em=Imr=2×2=4V
电功率为P=I2r=
2 r=
2×2W=4W
任意时刻穿过线圈的磁通量为Φ=BSsin
Φm=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218901877.png)
Φ=
sin
t
故选AC
由于从垂直中性面开始其瞬时表达式为i=Imcosθ,由已知可求Im=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218511481.png)
根据正弦式交变电流有效值和峰值关系可求电流有效值
根据P=I2R可求电功率
根据Em=Imr可求感应电动势的最大值
任意时刻穿过线圈的磁通量为Φ=BSsin
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218542526.png)
解答:解:从垂直中性面开始其瞬时表达式为i=Imcosθ,
则Im=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218511481.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218589578.png)
感应电动势的最大值为Em=Imr=2×2=4V
电功率为P=I2r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218682399.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218854405.png)
任意时刻穿过线圈的磁通量为Φ=BSsin
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218339592.png)
Φm=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218901877.png)
Φ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218932472.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124218947468.png)
故选AC
![](http://thumb2018.1010pic.com/images/loading.gif)
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