中文谷如图*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...
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问题详情：
如图*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*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=Imsin
t，则在0~
T时间内斜向上拉力对杆ab做的功W。
【回答】
【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 · 
sin
t ·························································································· 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*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。
知识点：专题四 功和能
题型：综合题
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