Question

兩個(gè)沿豎直方向的磁感強(qiáng)度大小相等、方向相反的勻強(qiáng)磁場穿過光滑的水平桌面，它們的寬度均為L.質(zhì)量為m、邊長為L的平放在桌面上的正方形線圈的ab邊與磁場邊界ee′的距離為L，如圖所示.線圈在恒力作用下由靜止開始沿桌面加速運(yùn)動(dòng)，ab邊進(jìn)入右邊的磁場時(shí)恰好做速度為v的勻速直線運(yùn)動(dòng).求：
（1）當(dāng)ab邊剛越過ff′時(shí)線圈的加速度.
(2)當(dāng)ab邊運(yùn)動(dòng)到ff′與gg′之間的正中間位置時(shí)，線圈又恰好做勻速直線運(yùn)動(dòng)，從ab邊剛越過ee′到達(dá)右邊磁場正中間位置的過程中，線圈共產(chǎn)生多少熱量？
               

Answer 1

（1）a==
方向與運(yùn)動(dòng)方向相反.
(2) Q=mv²

（1）設(shè)恒力為F，由線圈開始運(yùn)動(dòng)到ab邊進(jìn)入磁場前瞬間，根據(jù)動(dòng)能定理
W=ΔE_k得F=                                                                                       ①
ab邊剛越過ee′未到達(dá)ff′時(shí)做勻速運(yùn)動(dòng)，處于平衡狀態(tài)，磁場力大小F₁=F，設(shè)磁感強(qiáng)度為B，線圈電阻為R，則感應(yīng)電動(dòng)勢(shì)E₁=BLv
磁場力F₁=BI₁L
因?yàn)?i>I₁=得F=                                                                               ②
當(dāng)ab邊剛越過ff′地，ab、cd邊感應(yīng)電動(dòng)勢(shì)大小、方向相同，總電動(dòng)勢(shì)ε₂=2ε₁,電流I₂=2I₁                                                                                                                                                                                ③
故磁場力的合力F₂=4F₁                                                                                                                                                                        ④
由牛頓第二定律和①、④式得
a==
方向與運(yùn)動(dòng)方向相反.
(2)設(shè)ab邊運(yùn)動(dòng)到ff′與gg′正中間時(shí)速度為v₃,則感應(yīng)電動(dòng)勢(shì)ε₃=2BLv₃，磁場力大小F₃=                                                                                                          ⑤
此時(shí)線圈又做勻速運(yùn)動(dòng)，磁場力的大小F₃=F                                             ⑥
由②、⑤、⑥式得v₃=                                                                               ⑦
根據(jù)動(dòng)能定理W_F-W_B=ΔE_k
根據(jù)能量守恒定律，從ab邊剛越過ee′到達(dá)右側(cè)磁場正中間過程中，線圈產(chǎn)生的熱量與磁場力做的功大小相等，得Q=FL+mv²-mv²₃                                                                                 ⑧
由①、⑦、⑧式得Q=mv²