問題詳情：
如圖*所示，間距爲L、足夠長的固定光滑平行金屬導軌MN、PQ與水平面成θ角，左端M、P之間連接有電流傳感器和阻值爲R的定值電阻。導軌上垂直停放一質量爲m、電阻爲r的金屬桿ab，且與導軌接觸良好，整個裝置處於磁感應強度方向垂直導軌平面向下、大小爲B的勻強磁場中。在t= 0時刻，用一沿MN方向的力斜向上拉金屬桿ab，使之從磁場的左邊界由靜止開始斜向上做直線運動，電流傳感器將透過R的電流i即時採集並輸入電腦，可獲得電流i隨時間t變化的關係圖線，電流傳感器和導軌的電阻及空氣阻力均忽略不計，重力加速度大小爲g。
（1）若電流i隨時間t變化的關係如圖乙所示，求t時刻杆ab的速度υ大小；
（2）在（1）問的情況下，請判斷杆ab的運動*質，並求t時刻斜向上拉力的功率P；
（3）若電流i隨時間t變化規律爲i=Imsin
t，則在0~
T時間內斜向上拉力對杆ab做的功W。
【回答】
【標準解答】（1）由乙圖可知，t=t1時刻電路中的感應電流爲I1，則t時刻，電流爲
i=
t ··········································································································· ①（1分）
杆ab切割磁感線產生的感應電動勢爲
e = BLυ··············································································································· ②（1分）
根據閉合電路歐姆定律有
e = i( R + r ) ······································································································ ③（1分）
由以上三式解得
υ = 
t ································································································ ④（1分）
（2）由於
是常量，所以杆ab是做勻加速直線運動，其加速度大小爲
a = 
= ························································································ ⑤（1分）
設t時刻水平拉力大小爲F，根據牛頓第二定律有
F–BiL–mgsinθ=ma ······················································································· ⑥（1分）
又 P=Fυ········································································································· ⑦（1分）
得 P=
t2 + [gsinθ + 
] 
t ···························· ⑧（2分）
（3）設位移x=–Acos 
t，由導數的物理意義可知，有
υ=A · 
sin
t ·························································································· ⑨（1分）
由BLυ=i( R + r )及i=Imsint 得
υ=
sint ···················································································· ⑩（1分）
可見，杆ab做簡諧運動。
所以振幅A=
········································································ （11）（1分）
在0 ~ 
T時間內，重力做功爲W1 =–mg · 2Asinθ········································· （12）（1分）
t=
T時刻，杆ab的速度大小
υ= 0 ·········································································································· （13）（1分）
在0~
T時間內，整個迴路產生的焦耳熱爲
Q= (
)2( R + r ) · T ·············································································· （14）（1分）
安培力對杆ab做的功爲
W2 =–Q ······································································································ （15）（1分）
根據動能定理有
W + W1 + W2=
– 0 ········································································· （16）（1分）
聯立以上四式解得
W=
+ 
Im2( R + r )T ················································· （17）（1分）
【思維點拔】本題的關鍵在於對電流傳感器得到的電流i隨時間t變化的關係圖線的理解，獲取資訊，從加速度定義來確定運動*質，利用數學知識位移的導函數是速度函數來確定簡諧運動的振幅，從而來求重力做功，整個迴路產生的焦耳熱要用電流的有效值來算，然後藉助動能定理加以求解。
知識點：專題四 功和能
題型：綜合題
