Question

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

（3）求線(xiàn)框進(jìn)入磁場(chǎng)的過(guò)程中通過(guò)導(dǎo)線(xiàn)橫截面的電荷量q大��；
【猜題理由】電磁感應(yīng)問(wèn)題是近年江蘇高考的必考的內(nèi)容，往年高考中沒(méi)有同時(shí)考查瞬時(shí)感應(yīng)電動(dòng)勢(shì)和平均感應(yīng)電動(dòng)勢(shì)，2010年高考很可能以討論運(yùn)動(dòng)狀態(tài)、微積分等難度設(shè)置高門(mén)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(dòng)量定理為規(guī)律命題。

Answer 1

【標(biāo)準(zhǔn)解答】⑴線(xiàn)框cd邊剛進(jìn)入磁場(chǎng)時(shí)，切割磁感線(xiàn)的速度為υ₀，線(xiàn)框中電動(dòng)勢(shì)大小  

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

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

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

（2）線(xiàn)框進(jìn)入磁場(chǎng)的過(guò)程中速度為υ時(shí)，受到的安培力
F = BiL = B L = ········································································· ④（1分）
在t→t + Δt時(shí)間內(nèi)，由動(dòng)量定理
     – FΔt＝mΔυ··································································································· ⑤（1分）
求和得  – ∑υ△t = ∑mΔυ
若x < 2L時(shí)，線(xiàn)框速度為零，以后保持靜止?fàn)顟B(tài)，則
 ∑△x = x = mυ₀ ········································································· ⑥（1分）
解得   x =   ························································································ ⑦（1分）
即線(xiàn)框的初速度υ₀滿(mǎn)足 0 < υ₀ < 時(shí)，線(xiàn)框最終速度為零。  ················· ⑧（1分）
若x ≥ 2L時(shí)，線(xiàn)框速度不為零，而速度υ沿x軸正方向做勻速直線(xiàn)運(yùn)動(dòng)，則
∑△x =  · 2L = mυ₀ – mυ································································· ⑨（1分）
得υ = υ₀ – ···························································································· ⑩（1分）
即線(xiàn)框的初速度υ₀滿(mǎn)足υ₀ ≥ 時(shí)，線(xiàn)框最終速度大小為υ₀ – 。 （11）（1分）
（3）導(dǎo)線(xiàn)框的平均感應(yīng)電動(dòng)勢(shì)為
=   ··································································································· （12）（1分）

導(dǎo)線(xiàn)框中的電流為
 = ······································································································· （13）（1分）
線(xiàn)框進(jìn)入磁場(chǎng)的過(guò)程中通過(guò)導(dǎo)線(xiàn)橫截面的電荷量
q = △t ····································································································· （14）（1分）
得 q =
當(dāng) 0 < υ₀ < 時(shí)，△Ф= B · Lx = ，得q = ···················· （15）（1分）
當(dāng)υ₀ ≥ 時(shí)，△Ф= B · 2L²，得  q = ······································· （16）（1分）

【思維點(diǎn)拔】對(duì)于（1）、（2）兩問(wèn)要搞清瞬時(shí)感應(yīng)電動(dòng)勢(shì)和平均感應(yīng)電動(dòng)勢(shì)的區(qū)別，c、d間的電勢(shì)差U是指作為電源的cd邊的端電壓，求通過(guò)導(dǎo)線(xiàn)橫截面的電荷量時(shí)要用平均感應(yīng)電動(dòng)勢(shì)來(lái)解。本題難點(diǎn)在于要知道線(xiàn)框最終可能停止運(yùn)動(dòng)，也可能勻速運(yùn)動(dòng)，所以要進(jìn)行討論。進(jìn)入磁場(chǎng)的過(guò)程，受到的安培力是變力，無(wú)法用動(dòng)力學(xué)觀點(diǎn)直接求解，可以用動(dòng)量定理結(jié)合微積分求出位移，然后算出對(duì)應(yīng)的電荷量。