[題目] 103年新竹實驗中學教師甄試
[領域] 電磁感應
[來源] 教師甄試
[題目] http://ppt.cc/vD9r Ans: 條件不足，送分。
[瓶頸]
一、假設一開始距離d
二、對m做力分析
由ΣF = ma
可以得知[(LB)^2(V1)]/R = m[d(V1)/dt]
解得V1 = Ae^[-(LB)^2t/R]
代t=0時V1 = Vo
所以V1 = Voe^[-(LB)^2t/R]
積分後再代X1(0) = 0
X1 = R/[(LB)^2]Vo{1-e^[-(LB)^2t/R]}
三、對2m做力分析
由ΣF = 2ma
即ILB = 2m(dV/dt)
其中I = LBV1/R
可以得知[(LB)^2(V1)]/2R = d(V2)/dt
代V1 = Voe^[-(LB)^2t/2R]，以及V2(0) =0
解得V2 = Vo{1- e^[-(LB)^2t/2R]}
積分後代X2(0) =d
解得X2 = Vot - R/[(LB)^2]Vo{1-e^[-(LB)^2t/2R]} + d
四、由於時間無限大為終端速度
但是代入以後X2會變成無限大
不知道哪邊出問題
謝謝!!
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