Question

在光滑的直角坐標系xOy水平面的第一象限內(nèi)分布有磁感應(yīng)強度的大小為B、方向垂直紙面向內(nèi)的勻強磁場。在xOy平面內(nèi)放置一單匝矩形導(dǎo)線框abcd，線框邊長ab = L、ad = 2L，電阻為R，質(zhì)量為m。t = 0時，bc邊與Oy軸重合，線框以初速度υ₀沿x軸正方向進入磁場，不計空氣阻力。
（1）求cd邊剛進入磁場時，c、d間的電勢差U；
（2）試討論求線框最終速度大小及對應(yīng)的初速度υ₀的范圍；

（3）求線框進入磁場的過程中通過導(dǎo)線橫截面的電荷量q大��；
【猜題理由】電磁感應(yīng)問題是近年江蘇高考的必考的內(nèi)容，往年高考中沒有同時考查瞬時感應(yīng)電動勢和平均感應(yīng)電動勢，2010年高考很可能以討論運動狀態(tài)、微積分等難度設(shè)置高門檻作為壓軸題，以法拉第電磁感應(yīng)定律、閉合電路歐姆定律、部分電路歐姆定律、動量定理為規(guī)律命題。

Answer 1

【標準解答】⑴線框cd邊剛進入磁場時，切割磁感線的速度為υ₀，線框中電動勢大小  

E = B Lυ₀············································································································· ①（1分）

導(dǎo)線中的電流大小
I =  ············································································································· ②（1分）

c、d間的電勢差
U = I · R = BLυ₀  ······················································································· ③（1分）

（2）線框進入磁場的過程中速度為υ時，受到的安培力
F = BiL = B L = ········································································· ④（1分）
在t→t + Δt時間內(nèi)，由動量定理
     – FΔt＝mΔυ··································································································· ⑤（1分）
求和得  – ∑υ△t = ∑mΔυ
若x < 2L時，線框速度為零，以后保持靜止狀態(tài)，則
 ∑△x = x = mυ₀ ········································································· ⑥（1分）
解得   x =   ························································································ ⑦（1分）
即線框的初速度υ₀滿足 0 < υ₀ < 時，線框最終速度為零。  ················· ⑧（1分）
若x ≥ 2L時，線框速度不為零，而速度υ沿x軸正方向做勻速直線運動，則
∑△x =  · 2L = mυ₀ – mυ································································· ⑨（1分）
得υ = υ₀ – ···························································································· ⑩（1分）
即線框的初速度υ₀滿足υ₀ ≥ 時，線框最終速度大小為υ₀ – 。 （11）（1分）
（3）導(dǎo)線框的平均感應(yīng)電動勢為
=   ··································································································· （12）（1分）

導(dǎo)線框中的電流為
 = ······································································································· （13）（1分）
線框進入磁場的過程中通過導(dǎo)線橫截面的電荷量
q = △t ····································································································· （14）（1分）
得 q =
當 0 < υ₀ < 時，△Ф= B · Lx = ，得q = ···················· （15）（1分）
當υ₀ ≥ 時，△Ф= B · 2L²，得  q = ······································· （16）（1分）

【思維點拔】對于（1）、（2）兩問要搞清瞬時感應(yīng)電動勢和平均感應(yīng)電動勢的區(qū)別，c、d間的電勢差U是指作為電源的cd邊的端電壓，求通過導(dǎo)線橫截面的電荷量時要用平均感應(yīng)電動勢來解。本題難點在于要知道線框最終可能停止運動，也可能勻速運動，所以要進行討論。進入磁場的過程，受到的安培力是變力，無法用動力學(xué)觀點直接求解，可以用動量定理結(jié)合微積分求出位移，然后算出對應(yīng)的電荷量。