题目

如图甲所示，间距为L、足够长的固定光滑平行金属导轨MN、PQ与水平面成θ角，左端M、P之间连接有电流传感器和阻值为R的定值电阻。导轨上垂直停放一质量为m、电阻为r的金属杆ab，且与导轨接触良好，整个装置处于磁感应强度方向垂直导轨平面向下、大小为B的匀强磁场中。在t = 0时刻，用一沿MN方向的力斜向上拉金属杆ab，使之从磁场的左边界由静止开始斜向上做直线运动，电流传感器将通过R的电流i即时采集并输入电脑，可获得电流i随时间t变化的关系图线，电流传感器和导轨的电阻及空气阻力均忽略不计，重力加速度大小为g。 （1）若电流i随时间t变化的关系如图乙所示，求t时刻杆ab的速度υ大小； （2）在（1）问的情况下，请判断杆ab的运动性质，并求t时刻斜向上拉力的功率P； （3）若电流i随时间t变化规律为i = Imsint，则在0~T时间内斜向上拉力对杆ab做的功W。 答案：【标准解答】（1）由乙图可知，t=t1时刻电路中的感应电流为I1，则t时刻，电流为 i = t  ··········································································································· ①（1分） 杆ab切割磁感线产生的感应电动势为 e = BLυ··············································································································· ②（1分） 根据闭合电路欧姆定律有 e = i( R + r ) ······································································································ ③（1分） 由以上三式解得 υ = t  ································································································ ④（1分） （2）由于是常量，所以杆ab是做匀加速直线运动，其加速度大小为 a =  =  ························································································ ⑤（1分） 设t时刻水平拉力大小为F，根据牛顿第二定律有 F – BiL – mgsinθ= ma  ······················································································· ⑥（1分） 又  P = Fυ········································································································· ⑦（1分） 得  P = t2 + [gsinθ +  ] t  ···························· ⑧（2分） （3）设位移x = – Acos t，由导数的物理意义可知，有 υ = A  · sint  ··························································································  ⑨（1分） 由BLυ = i( R + r )及i = Imsint 得 υ = sint  ···················································································· ⑩（1分） 可见，杆ab做简谐运动。 所以振幅A =   ········································································ （11）（1分） 在0 ~ T时间内，重力做功为W1 = – mg · 2Asinθ········································· （12）（1分） t = T时刻，杆ab的速度大小 υ = 0  ·········································································································· （13）（1分） 在0~T时间内，整个回路产生的焦耳热为 Q = ()2( R + r ) · T  ·············································································· （14）（1分） 安培力对杆ab做的功为 W2 = – Q ······································································································ （15）（1分） 根据动能定理有 W + W1 + W2=  – 0  ········································································· （16）（1分） 联立以上四式解得 W =  + Im2( R + r )T  ················································· （17）（1分） 【思维点拔】本题的关键在于对电流传感器得到的电流i随时间t变化的关系图线的理解，获取信息，从加速度定义来确定运动性质，利用数学知识位移的导函数是速度函数来确定简谐运动的振幅，从而来求重力做功，整个回路产生的焦耳热要用电流的有效值来算，然后借助动能定理加以求解。

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