Question

（分）平面直角坐標(biāo)系xOy中，第1象限存在沿y軸負(fù)方向的勻強(qiáng)電場，第Ⅳ象限存在垂直于坐標(biāo)平面向外的勻強(qiáng)磁場，磁感應(yīng)強(qiáng)度大小為B。一質(zhì)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(diǎn)以速度v₀垂直于y軸射入電場，經(jīng)x軸上的N點(diǎn)與x軸正方向成60°角射入磁場，最后從y軸負(fù)半軸上的P點(diǎn)與y軸正方向成60⁰角射出磁場，如圖所示。不計(jì)粒子重力，求

（1）粒子在磁場中運(yùn)動的軌道半徑R；

（2）粒子從M點(diǎn)運(yùn)動到P點(diǎn)的總時(shí)間t；

（3）勻強(qiáng)電場的場強(qiáng)大小E。

Answer 1

（1）    

（2）  

（3）  

解析:

（1）設(shè)粒子過N點(diǎn)時(shí)的速度為v，根據(jù)平拋運(yùn)動的速度關(guān)系 

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

分別過N、P點(diǎn)作速度方向的垂線，相交于Q點(diǎn)，則Q是粒子在磁場中做勻速圓周運(yùn)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(shè)粒子在電場中運(yùn)動的時(shí)間為t₁，有  ON＝v₀t₁ ····················································· ④

由幾何關(guān)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(yùn)動的周期  T＝ ····························································· ⑦

由幾何關(guān)系知∠NQP＝150°，設(shè)粒子在磁場中運(yùn)動的時(shí)間為t₂   t₂＝T····························· ⑧

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

故粒子從M點(diǎn)運(yùn)動到P點(diǎn)的總時(shí)間  t＝t₁＋t₂＝(1＋＋)··········································· ⑩

（3）粒子在電場中做類平拋運(yùn)動，設(shè)加速度為a  qE=ma ··············································· ⑾

設(shè)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