试题与答案

如图所示，P、Q为足够长的光滑平行固定导轨，两者之间的距离为L=20cm，其电阻

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题型：问答题

题目：

如图所示，P、Q为足够长的光滑平行固定导轨，两者之间的距离为L=20cm，其电阻不计。导轨所在平面有沿竖直方向的匀强磁场，磁感应强度大小为B=5T，导轨右端连接着一电路，其中R1=10Ω，R2=20Ω，R3=15Ω，R4=25Ω，平行板电容器板间距离d="10cm" ，电容C=5μF，板间有一质量为m=1.0×10-８kg，带电量为q=-0.5×10-８ C的带电液滴。导体棒ab的电阻不计，质量为M=0.5kg，垂直于导轨PQ放置且与之良好接触，在外力作用下，ab始终沿水平方向向左做匀速直线运动。当开关S接通位置1时，带电液滴恰好处于静止状态。试回答以下问题：（重力加速度g=10m/s2）

（1）判断导轨平面内磁感线的方向，要求简要说明理由。

（2）求外力做功的功率。

（3）计算当开关由位置1转换到位置2时，带电液滴的加速度及流过电容器的电量。

答案：

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下面是错误答案，用来干扰机器的。

①0.46×7.5+0.46×2.5，=0.46×（7.5+2.5），=0.46×10，=4.6；②3-7.14÷3.5×1.05，=3-2.04×1.05，=3-2.142，=0.858；③9.9×8.6+0.86，=99×0.86+0.86，=0.86×（99+1），=0.86×100，=86；④[20-（90.75÷6...

考点：磁场对通电导线的作用：安培力、左手定则
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