题目内容
(16分)如图所示,MN、PQ是两条水平平行放置的光滑金属导轨,导轨的右端接理想变压器的原线圈,变压器的副线圈与电阻R=20Ω组成闭合回路,变压器的原副线圈匝数之比n1 : n2=1 : 10,导轨宽L=5m。质量m=2kg、电阻r=1Ω的导体棒ab垂直MN、PQ放在导轨上,在水平外力F作用下从t=0时刻开始在图示的两虚线范围内往复运动,其速度随时间变化的规律是v=2sin20πt m/s。垂直轨道平面的匀强磁场的磁感应强度B=4T。导轨、导线和线圈电阻不计。求:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214601398.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214617401.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214617401.gif)
(1)从t=0到t1=10 s的时间内,电阻R上产生的热量Q=?
(2)从t=0到t2=0.025 s的时间内,外力F所做的功W=?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241212146485996.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214601398.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214617401.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214617401.gif)
(1)从t=0到t1=10 s的时间内,电阻R上产生的热量Q=?
(2)从t=0到t2=0.025 s的时间内,外力F所做的功W=?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241212146485996.jpg)
(1)4×104J·
(2)124 J
(2)124 J
(1) a b棒产生的是交流,其电动势为E=BLv=40 sin20πt V (1分)
设原线圈上电压的有效值为U1,副线圈上电压的有效值为U2,则
U1=20
V·····(1分)
, U2=200
V··········(2分)
则Q=
·····(2分) Q=4×104J······················(2分)
(2)从t=0到t2=0.025 s,经历了四分之一个周期,设在这段时间内电阻r和R上产生的热量分别是Q1和Q2,在t2=0.025 s时刻,ab棒的速度为v2,则
Q1=
··········(1分) Q2=
···················(1分)
v2=2sin20πt2·········(1分) Q1=20J,Q2=100J,v2=2m/s
W=Q1+Q2+
··(3分) W=124 J····················(2分)
设原线圈上电压的有效值为U1,副线圈上电压的有效值为U2,则
U1=20
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214679215.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214695462.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214679215.gif)
则Q=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214726408.gif)
(2)从t=0到t2=0.025 s,经历了四分之一个周期,设在这段时间内电阻r和R上产生的热量分别是Q1和Q2,在t2=0.025 s时刻,ab棒的速度为v2,则
Q1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214742405.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214757422.gif)
v2=2sin20πt2·········(1分) Q1=20J,Q2=100J,v2=2m/s
W=Q1+Q2+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121214773416.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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