Question

如圖所示，在矩形ABCD內(nèi)對角線BD以上的區(qū)域存在有平行于AD向下的勻強(qiáng)電場，對角線BD以下的區(qū)域存在有垂直于紙面的勻強(qiáng)磁場（圖中未標(biāo)出），矩形AD邊長L，AB邊長為L。一個(gè)質(zhì)量為m、電荷+q的帶電粒子（不計(jì)重力）以初速度v₀從A點(diǎn)沿AB方向進(jìn)入電場，在對角線BD的中點(diǎn)P處進(jìn)入磁場，并從DC邊上的Q點(diǎn)垂直于DC離開磁場，試求：

（1）       電場強(qiáng)度的大小

（2）       帶電粒子經(jīng)過P點(diǎn)時(shí)速度的大小和方向

（3）       磁場的磁感應(yīng)強(qiáng)度的大小和方向

Answer 1

【標(biāo)準(zhǔn)解答】

（1）       帶電粒子受電場力作用做類平拋運(yùn)動(dòng)，則

L=at²··································································································· ①（1分）

L=v₀t···································································································· ②（1分）

Eq=ma········································································································ ③（1分）

得 a=，場強(qiáng)為································································· ④（2分）

（2）在豎直方向上做勻變速運(yùn)動(dòng)，Y方向分速度為v_y，則有

2 a= v_y²   得v_y==v₀································································ ⑤（2分）

到P點(diǎn)時(shí)速度為V==v₀························································· ⑥（1分）

速度與水平方向的夾角θ滿足 =·········································· ⑦（2分）

得此時(shí)速度與水平方向的夾角為θ=arctan············································· ⑧（1分）

（3）BD與水平方向的夾角滿足 ··························································· ⑨（1分）

則  有v⊥BD············································································ ⑩（1分）

粒子在磁場中運(yùn)動(dòng)軌跡的圓心就在D點(diǎn)，則R=BD=L··················· （1分）

由V=v₀，qvB=m········································································· （2分）

得·························································································· （1分）

方向垂直紙面向外·················································································· （1分）

【思維點(diǎn)拔】

    若題中邊長未知，求BD與CD間的夾角又變成新題