Question

（分）平面直角坐標系xOy中，第1象限存在沿y軸負方向的勻強電場，第Ⅳ象限存在垂直于坐標平面向外的勻強磁場，磁感應強度大小為B。一質量為m、電荷量為q的帶正電的粒子從y軸正半軸上的M點以速度v₀垂直于y軸射入電場，經(jīng)x軸上的N點與x軸正方向成60°角射入磁場，最后從y軸負半軸上的P點與y軸正方向成60⁰角射出磁場，如圖所示。不計粒子重力，求

（1）粒子在磁場中運動的軌道半徑R；

（2）粒子從M點運動到P點的總時間t；

（3）勻強電場的場強大小E。

Answer 1

（1）    

（2）  

（3）  

解析:

（1）設粒子過N點時的速度為v，根據(jù)平拋運動的速度關系 

v＝··························································· ①

分別過N、P點作速度方向的垂線，相交于Q點，則Q是粒子在磁場中做勻速圓周運動的圓心，根據(jù)牛頓第二定律  qvB＝··································································· ···· ②

聯(lián)立①②解得軌道半徑  R＝ ····························· ③

（2）設粒子在電場中運動的時間為t₁，有  ON＝v₀t₁ ····················································· ④

由幾何關系得  ON＝Rsin30°＋Rcos30°········································································ ⑤

聯(lián)立③④⑤解得  t₁＝··················································································· ⑥

粒子在磁場中做勻速圓周運動的周期  T＝ ····························································· ⑦

由幾何關系知∠NQP＝150°，設粒子在磁場中運動的時間為t₂   t₂＝T····························· ⑧

聯(lián)立⑦⑧解得  t₂＝···························································································· ⑨

故粒子從M點運動到P點的總時間  t＝t₁＋t₂＝(1＋＋)··········································· ⑩

（3）粒子在電場中做類平拋運動，設加速度為a  qE=ma ··············································· ⑾

設沿電場方向的分速度為v_y    v_y=at₁ ········································································ ⑿

又  v_y=v₀tan60⁰ ··································································································· ⒀

聯(lián)立⑥⑾⑿⒀解得  E＝··············································································· ⒁