Question

如圖所示,平行水平面放置的導軌上連有電阻R,處于垂直軌道平面的勻強磁場中.今從靜止起用力拉金屬棒ab,若拉力恒定,經過時間t₁后ab速度為v,加速度為a₁,最終速度可達2v;若改用功率恒定的拉力作用,經過時間t₂后ab的速度也為v,加速度為a₂,最終速度也可達2v.求a₁∶a₂的值.(導軌光滑,摩擦不計)

Answer 1

1∶3

解析：用恒力拉ab棒時,設恒力為F₁,最終平衡時速度為2v,有

F₁=BIl                                                                       ①

I₁=Bl2v/R                                                                    ②

由①②得F₁=2B²l²v/R

當速度為v時,ab棒受安培力為F₂,可得

F₂=                                                                   ③

此時有F₁-F₂=ma₁                                                             ④

聯立解得a₁=

當拉力的功率恒定時,設此功率為P,則有P=P_電==                ⑤

當速度為v時外力為F₁′,P=F₁′v                                             ⑥

此時F₁′-F₂=ma₂                                                            ⑦

由上述等式可得a₂=3B²l²v/mR

故a₁∶a₂=1∶3.