Question

如圖所示,足夠長(zhǎng)的平行金屬導(dǎo)軌MN?PQ平行放置,間距為L(zhǎng),與水平面成θ角,導(dǎo)軌與固定電阻R₁和R₂相連,且R₁=R₂=R.R₁支路串聯(lián)開(kāi)關(guān)S,原來(lái)S閉合,勻強(qiáng)磁場(chǎng)垂直導(dǎo)軌平面斜向上.有一質(zhì)量為m的導(dǎo)體棒ab與導(dǎo)軌垂直放置,接觸面粗糙且始終接觸良好,導(dǎo)體棒的有效電阻也為R.現(xiàn)讓導(dǎo)體棒從靜止釋放沿導(dǎo)軌下滑,當(dāng)導(dǎo)體棒運(yùn)動(dòng)達(dá)到穩(wěn)定狀態(tài)時(shí)速率為v,此時(shí)整個(gè)電路消耗的電功率為重力功率的3/4.重力加速度為g,導(dǎo)軌電阻不計(jì).試求:

(1)在上述穩(wěn)定狀態(tài)時(shí),導(dǎo)體棒中的電流I,以及磁感應(yīng)強(qiáng)度B的大小;
(2)當(dāng)斷開(kāi)開(kāi)關(guān)S后,導(dǎo)體棒ab所能達(dá)到的最大速率v′是v的多少倍?

Answer 1

(1)          

(1)當(dāng)導(dǎo)體棒以速率v勻速下降時(shí),電路中的總電阻R_總=     ①
感應(yīng)電動(dòng)勢(shì)  E=BLv                                               ②
導(dǎo)體棒中的電流I=                                            ③
總電功率  P_電=I²R_總                                                                                 ④
重力功率  P_重=mgv·sinθ                                         ⑤
根據(jù)題意有  P_電=P_重                                                                            ⑥
由①②③④⑤⑥解得:                              ⑦
                                             ⑧
(2)由S斷開(kāi)前的情況可知:
·sinθ=μmgv·cosθ                                      ⑨
S斷開(kāi)后,當(dāng)導(dǎo)體棒以速度v′勻速下滑時(shí)
總電阻  R′_總=2R                                                 ⑩
導(dǎo)體棒中的電流  I′=?                                     (11)?
由導(dǎo)體棒受力平衡有  mgsinθ=μmgcosθ+BI′L?                   (12)?
由⑧⑨⑩(11)(12)求得: