如圖所示，放在傾角為30°的固定斜面上的正方形導線框abcd與重物之間用足夠長的細線跨過光滑的輕質定滑輪連接，...

問題詳情：

如圖所示，放在傾角為30°的固定斜面上的正方形導線框abcd與重物之間用足夠長的細線跨過光滑的輕質定滑輪連接，線框的邊長為L、質量為m、電阻為R，重物的質量為m。現將線框沿bc方向拋出，穿過寬度為D、磁感應強度為B的勻強磁場，磁場的方向垂直斜面向上．線框向下離開磁場時的速度剛好是進人磁場時速度的（n大於1），線框離開磁場後繼續下滑一段距離，然後上滑並勻速進人磁場．線框與斜面間的動摩擦因數為，不計空氣阻力，整個運動過程中線框始終與斜面平行且不發生轉動，斜細線與bc邊平行，cd邊磁場邊界平行．求：⑴線框在上滑階段勻速進人磁場時的速度υ2大小；⑵線框在下滑階段剛離開磁場時的速度υ1大小；  ⑶線框在下滑階段通過磁場過程中產生的焦耳熱Q．

【回答】

【標準解答】（1）線框勻速上滑進入磁場，則有

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

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

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

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

（2）對系統，由動能定理知，                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            離開磁場後的下滑階段：  mgsin30° · h – mgh–μmgcos30° · h = 0 –(m + m)υ12  ······························· ⑤（2分）

進人磁場前的上滑階段：  –mgsin30° · h + mgh–μmgcos30° · h = (m + m)υ22 – 0 ······························ ⑥（2分）

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

（3）設剛進入磁場時速度為υ0。線框在下滑穿越磁場的過程，由系統能量守恆有： (m + m)υ02 –(m + m)υ12 = mg(L + D) – mgsin30° ·(L + D) + μmgcos30° ·(L + D) + Q  ························································································································ ⑧（4分）                                                                                                                                                      由題設知υ0 = nυ1

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

【思維點拔】本題的關鍵在於理解多過程和多對象設置情景，先找出解題的突破口即先從受力平衡的線框勻速上滑進入磁場的過程來分析，列出平衡方程，由於線框勻速上滑進入磁場，所以對應的重物也是勻速運動，細線的拉力與重物的重力大小相等。而其他過程，系統受力不平衡，所以細線的拉力不再與重物的重力大小相等，由於線框切割磁感線的速度在不斷變化，導致安培力和細線中的拉力大小在不斷變化，無法用動力學觀點來求解，但可以用能量觀點來解決變力做功問題，這時宜以線框和重物整體為研究對象。

知識點：專題八 電磁感應

題型：綜合題