Question

如圖所示,坐標(biāo)系xOy位于豎直平面內(nèi),所在空間有沿水平方向垂直于紙面向里的勻強(qiáng)磁場(chǎng),磁感應(yīng)強(qiáng)度大小為B,在x<0的空間內(nèi)還有沿x軸負(fù)方向的勻強(qiáng)電場(chǎng),場(chǎng)強(qiáng)大小為E.一個(gè)帶電油滴經(jīng)圖中x軸上的M點(diǎn),沿著直線MP做勻速運(yùn)動(dòng),圖中α=30°,經(jīng)過(guò)P點(diǎn)后油滴進(jìn)入x>0的區(qū)域,要使油滴在x>0的區(qū)域內(nèi)做勻速圓周運(yùn)動(dòng),需在該區(qū)域內(nèi)加一個(gè)勻強(qiáng)電場(chǎng).若帶電油滴沿PN做勻速圓周運(yùn)動(dòng),并垂直于x軸通過(guò)軸上的N點(diǎn).已知重力加速度為g.

(1)判斷油滴的帶電性質(zhì);

(2)求油滴運(yùn)動(dòng)的速率;

(3)求在x>0的區(qū)域內(nèi)所加電場(chǎng)的場(chǎng)強(qiáng);

(4)求油滴從M點(diǎn)出發(fā)運(yùn)動(dòng)到N點(diǎn)所用的時(shí)間.

Answer 1

解析:(1)油滴帶正電.

(2)油滴受三力作用(見(jiàn)圖)沿直線勻速運(yùn)動(dòng)

由平衡條件有qvBsin30°=qE                                                     ①

mgtan30°=qE                                                                  ②

由①式解得v=.                                                            ③

(3)在x>0的區(qū)域,油滴要做勻速圓周運(yùn)動(dòng),其所受的電場(chǎng)力必與重力平衡,由于油滴帶正電,所以場(chǎng)強(qiáng)方向豎直向上.

設(shè)該電場(chǎng)的場(chǎng)強(qiáng)為E′,則有qE′=mg                                               ④

②④式聯(lián)立解得E′=E.                                                      ⑤

(4)見(jiàn)下圖,?為油滴做圓周運(yùn)動(dòng)在x>0、y<0 區(qū)域內(nèi)形成的圓弧軌道所對(duì)應(yīng)的弦,PO′是過(guò)P點(diǎn)所作的垂直于MP的直線,由于MO=NO,由幾何關(guān)系容易知道O′點(diǎn)一定是圓心,且∠PO′N=120°                                                                 ⑥

設(shè)油滴從M點(diǎn)到P點(diǎn)和從P點(diǎn)到N點(diǎn)經(jīng)歷的時(shí)間分別為t₁和t₂

做勻速圓周運(yùn)動(dòng)時(shí)有qvB=                                                 ⑦

由②③⑦式解得R=                                                    ⑧

t₁=                                                     ⑨

t₂=                                              ⑩

全過(guò)程經(jīng)歷的時(shí)間為t=t₁+t₂=.                                    11