题目

在光滑的直角坐标系xOy水平面的第一象限内分布有磁感应强度的大小为B、方向垂直纸面向内的匀强磁场。在xOy平面内放置一单匝矩形导线框abcd，线框边长ab = L、ad = 2L，电阻为R，质量为m。t = 0时，bc边与Oy轴重合，线框以初速度υ0沿x轴正方向进入磁场，不计空气阻力。 （1）求cd边刚进入磁场时，c、d间的电势差U； （2）试讨论求线框最终速度大小及对应的初速度υ0的范围； （3）求线框进入磁场的过程中通过导线横截面的电荷量q大小； 【猜题理由】电磁感应问题是近年江苏高考的必考的内容，往年高考中没有同时考查瞬时感应电动势和平均感应电动势，2010年高考很可能以讨论运动状态、微积分等难度设置高门槛作为压轴题，以法拉第电磁感应定律、闭合电路欧姆定律、部分电路欧姆定律、动量定理为规律命题。 答案：【标准解答】⑴线框cd边刚进入磁场时，切割磁感线的速度为υ0，线框中电动势大小   E = B Lυ0············································································································· ①（1分） 导线中的电流大小 I =  ············································································································· ②（1分） c、d间的电势差 U = I · R = BLυ0  ······················································································· ③（1分） （2）线框进入磁场的过程中速度为υ时，受到的安培力 F = BiL = B L = ········································································· ④（1分） 在t→t + Δt时间内，由动量定理      – FΔt＝mΔυ··································································································· ⑤（1分） 求和得  – ∑υ△t = ∑mΔυ 若x < 2L时，线框速度为零，以后保持静止状态，则  ∑△x = x = mυ0 ········································································· ⑥（1分） 解得   x =   ························································································ ⑦（1分） 即线框的初速度υ0满足 0 < υ0 < 时，线框最终速度为零。  ················· ⑧（1分） 若x ≥ 2L时，线框速度不为零，而速度υ沿x轴正方向做匀速直线运动，则 ∑△x =  · 2L = mυ0 – mυ································································· ⑨（1分） 得υ = υ0 – ···························································································· ⑩（1分） 即线框的初速度υ0满足υ0 ≥ 时，线框最终速度大小为υ0 – 。 （11）（1分） （3）导线框的平均感应电动势为 =   ··································································································· （12）（1分） 导线框中的电流为  = ······································································································· （13）（1分） 线框进入磁场的过程中通过导线横截面的电荷量 q = △t ····································································································· （14）（1分） 得 q = 当 0 < υ0 < 时，△Ф= B · Lx = ，得q = ···················· （15）（1分） 当υ0 ≥ 时，△Ф= B · 2L2，得  q = ······································· （16）（1分） 【思维点拔】对于（1）、（2）两问要搞清瞬时感应电动势和平均感应电动势的区别，c、d间的电势差U是指作为电源的cd边的端电压，求通过导线横截面的电荷量时要用平均感应电动势来解。本题难点在于要知道线框最终可能停止运动，也可能匀速运动，所以要进行讨论。进入磁场的过程，受到的安培力是变力，无法用动力学观点直接求解，可以用动量定理结合微积分求出位移，然后算出对应的电荷量。

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