Question

如圖14所示,兩根足夠長(zhǎng)的平行金屬導(dǎo)軌,固定在同一水平面上,導(dǎo)軌的電阻很小,可忽略不計(jì).導(dǎo)軌間的距離L=0.2 m.磁感應(yīng)強(qiáng)度B=0.50 T的勻強(qiáng)磁場(chǎng)與導(dǎo)軌所在平面垂直.兩根質(zhì)量均為m=0.10 kg的平行金屬桿甲、乙可在導(dǎo)軌上無(wú)摩擦地滑動(dòng),滑動(dòng)過(guò)程中與導(dǎo)軌保持垂直,每根金屬桿的電阻為R=0.50 Ω.在t=0時(shí)刻,兩根金屬桿并排靠在一起,且都處于靜止?fàn)顟B(tài).現(xiàn)有一與導(dǎo)軌平行、大小為0.20 N 的恒力F作用于金屬桿甲上,使金屬桿在導(dǎo)軌上滑動(dòng).經(jīng)過(guò)t=5.0 s,金屬桿甲的加速度為1.37 m/s²,問(wèn)此時(shí)甲、乙兩金屬桿速度v₁、v₂及它們之間的距離是多少?

圖14

Answer 1

解析：金屬桿甲受的安培力F_安=                                   ①

由牛頓第二定律F-F_安=ma                                                      ②

Ft=mv₁+mv₂                                                                  ③

由①②③三式解得v₁=8.15 m/s,v₂=1.85                                           ④

對(duì)乙:BIl·t=mv₂                                                              ⑤

得QIB=mv₂,Q=1.85 C                                                         ⑥

又Q=,得                                                       ⑦

s_相對(duì)=18.5 m.                                                                 ⑧

答案：18.5 m