题目内容
如图为俯视图,abcd是一边长为l的匀质正方形导线框,总电阻为R,放在光滑的水平面上,今使线框在外力作用下以恒定速度v水平向右穿过方向垂直于水平面向里的匀强磁场区域.已知磁感应强度大小为B,磁场宽度为3l,求
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241403333324381.png)
(1)线框在进入磁场区和穿出磁场区的两个过程中的感应电流方向和感应电动势的大小
(2)线框在进入磁场区和穿出磁场区的两个过程中a、b两点间电势差的大小.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241403333324381.png)
(1)线框在进入磁场区和穿出磁场区的两个过程中的感应电流方向和感应电动势的大小
(2)线框在进入磁场区和穿出磁场区的两个过程中a、b两点间电势差的大小.
(1)E=Blv(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333379529.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333379529.png)
(1)感应电流方向 进入时:adcba(或逆时针)穿出时:dabcd(或顺时针)
感应电动势的大小均为 E=Blv
(2)导线框在进入磁区过程中,ab相当于电源,等效电路如下图甲所示.
E=Blv,r=
R,R外=
R,I=
=
,
Uab为端电压;所以Uab=IR外=
.
导线框在穿出磁区过程中,cd相当于电源,等效电路如下图乙所示.
E=Blv,r=
R,R外=
R,I=
=
,
Uab=IRab=
×
R=
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241403343772474.png)
说明:第一、二问各9分
本题考查导线框在磁场中运动的问题,根据右手定则判定感应电动势方向,根据速度计算大小,再由闭合电路欧姆定律计算出电流和电压即可;
感应电动势的大小均为 E=Blv
(2)导线框在进入磁区过程中,ab相当于电源,等效电路如下图甲所示.
E=Blv,r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333410303.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333504385.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333582609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333660550.png)
Uab为端电压;所以Uab=IR外=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333738583.png)
导线框在穿出磁区过程中,cd相当于电源,等效电路如下图乙所示.
E=Blv,r=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333410303.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333504385.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333582609.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333660550.png)
Uab=IRab=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333660550.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333410303.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824140333379529.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241403343772474.png)
说明:第一、二问各9分
本题考查导线框在磁场中运动的问题,根据右手定则判定感应电动势方向,根据速度计算大小,再由闭合电路欧姆定律计算出电流和电压即可;
![](http://thumb2018.1010pic.com/images/loading.gif)
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