Question

如圖所示,在xOy平面內(nèi),第Ⅰ象限中有勻強(qiáng)電場,場強(qiáng)大小為E,方向沿x軸負(fù)方向,在第Ⅱ象限和第Ⅲ象限有勻強(qiáng)磁場,方向垂直于紙面向里.今有一個(gè)質(zhì)量為m、,電荷量為e的質(zhì)子(不計(jì)重力),從x軸上的P點(diǎn)以初速度v₀垂直于電場方向進(jìn)入電場.經(jīng)電場偏轉(zhuǎn)后,沿著與y軸正方向成45°角的方向進(jìn)入磁場,從磁場中飛出時(shí)恰好能返回到原出發(fā)點(diǎn)P.試求:

(1)P點(diǎn)離坐標(biāo)原點(diǎn)的距離l;

(2)磁場的磁感應(yīng)強(qiáng)度B;

(3)質(zhì)子第二次在電場中運(yùn)動的時(shí)間t.

Answer 1

解:(1)質(zhì)子運(yùn)動軌跡如圖所示.質(zhì)子進(jìn)入電場從P到A做類平拋運(yùn)動,進(jìn)入磁場從A到C再到D做勻速圓周運(yùn)動,離開磁場從D到P做勻速直線運(yùn)動.

質(zhì)子經(jīng)過A點(diǎn)的速度v_A=v₀/cos45°=v₀                                        ①

質(zhì)子從P到A,由動能定理:qEl=mv_A²-mv₀²                                     ②

解得l=                                                                  ③

(2)設(shè)OA的距離為s₁,OD的距離為s₂,則s₁=v₀t                                      ④

l=(v₀tan45°)t/2                                                                ⑤

解得s₁=2l                                                                     ⑥

質(zhì)子從磁場中飛出時(shí)的方向與y軸亦成45°,所以s₂=l                               ⑦

所以,質(zhì)子在磁場中的軌跡半徑為R=                                ⑧

由于R=                                                                  ⑨

解得B=                                                               ⑩

(3)質(zhì)子第二次在P點(diǎn)時(shí),速度大小為v₀,速度方向與x軸正方向成45°.質(zhì)子第二次在電場中運(yùn)動,在x方向做初速度為v₀、加速度大小為a=的勻減速直線運(yùn)動.             

則有-l=v₀t-·t²                                                         

解得t=(1+)