如图所示，放在倾角为30°的固定斜面上的正方形导线框abcd与重物之间用足够长的细线跨过光滑的轻质定滑轮连接，...

问题详情：

如图所示，放在倾角为30°的固定斜面上的正方形导线框abcd与重物之间用足够长的细线跨过光滑的轻质定滑轮连接，线框的边长为L、质量为m、电阻为R，重物的质量为m。现将线框沿bc方向抛出，穿过宽度为D、磁感应强度为B的匀强磁场，磁场的方向垂直斜面向上．线框向下离开磁场时的速度刚好是进人磁场时速度的（n大于1），线框离开磁场后继续下滑一段距离，然后上滑并匀速进人磁场．线框与斜面间的动摩擦因数为，不计空气阻力，整个运动过程中线框始终与斜面平行且不发生转动，斜细线与bc边平行，cd边磁场边界平行．求：⑴线框在上滑阶段匀速进人磁场时的速度υ2大小；⑵线框在下滑阶段刚离开磁场时的速度υ1大小；  ⑶线框在下滑阶段通过磁场过程中产生的焦耳热Q．

【回答】

【标准解答】（1）线框匀速上滑进入磁场，则有

mgsin30° + μmgcos30° + BIL =Mg······································································ ①（2分）

而I=··········································································································· ②（1分）

E=BLυ2 ··········································································································· ③（1分）

解得υ2 =  ···························································································· ④（2分）

（2）对系统，由动能定理知，                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            离开磁场后的下滑阶段：  mgsin30° · h – mgh–μmgcos30° · h = 0 –(m + m)υ12  ······························· ⑤（2分）

进人磁场前的上滑阶段：  –mgsin30° · h + mgh–μmgcos30° · h = (m + m)υ22 – 0 ······························ ⑥（2分）

由以上三式得：υ1 =  ·········································································· ⑦（2分）

（3）设刚进入磁场时速度为υ0。线框在下滑穿越磁场的过程，由系统能量守恒有： (m + m)υ02 –(m + m)υ12 = mg(L + D) – mgsin30° ·(L + D) + μmgcos30° ·(L + D) + Q  ························································································································ ⑧（4分）                                                                                                                                                      由题设知υ0 = nυ1

解得：Q =  –mg(L + D) ·················································· ⑨（2分）

【思维点拔】本题的关键在于理解多过程和多对象设置情景，先找出解题的突破口即先从受力平衡的线框匀速上滑进入磁场的过程来分析，列出平衡方程，由于线框匀速上滑进入磁场，所以对应的重物也是匀速运动，细线的拉力与重物的重力大小相等。而其他过程，系统受力不平衡，所以细线的拉力不再与重物的重力大小相等，由于线框切割磁感线的速度在不断变化，导致安培力和细线中的拉力大小在不断变化，无法用动力学观点来求解，但可以用能量观点来解决变力做功问题，这时宜以线框和重物整体为研究对象。

知识点：专题八 电磁感应

题型：综合题