Question

如圖所示,一半徑為r的圓形導(dǎo)線框內(nèi)有一勻強磁場,磁場方向垂直導(dǎo)線框所在平面,導(dǎo)線框的左端通過導(dǎo)線接一對水平放置的平行金屬板,兩板間距離為d,板長為l.t=0時,磁場的磁感應(yīng)強度B從B₀開始均勻增大,同時,在板2的左端且非�？拷�板2的位置有一質(zhì)量為m、帶電荷量為-q的液滴以初速度v₀水平向右射入兩板間,該液滴可視為質(zhì)點.

(1)要使該液滴能從兩板間射出,磁感應(yīng)強度隨時間的變化率k應(yīng)滿足什么條件?

(2)要使該液滴能從兩板間右端的中點射出,磁感應(yīng)強度B與時間t應(yīng)滿足什么關(guān)系?

Answer 1

解析：(1)由題意可知：板1為正極，板2為負(fù)極                                ①

兩板間的電壓

U==S=Sk                                                          ②

而S=π·r²                                                                   ③

帶電液滴受的電場力F=qE=                                               ④

故F-mg=-mg=ma

a=-g.                                                                   ⑤

（a）若a＞0，液滴向上偏轉(zhuǎn),做類平拋運動

y=at²=(-g)t²                                                         ⑥

當(dāng)液滴剛好能射出時

有l(wèi)=v₀t，t=，y=d

故d=(-g)·()²                                                        ⑦

由②③⑦知k₁=(g+)                                                ⑧

要使液滴能射出,必須滿足y＜d,故k＜k₁.

（b）若a=0，液滴不發(fā)生偏轉(zhuǎn),做勻速直線運動.此時，

a=-g=0                                                                  ⑨

由②③⑨知k₂=                                                         ⑩

液滴能射出,必須滿足k=k₂.

（c）若a＜0,液滴將被吸附在板2上

綜上所述，液滴能射出

k應(yīng)滿足≤k＜(g+).                                          

(2)B=B₀+kt

當(dāng)液滴從兩板間中點射出時,滿足條件（a）的情況,則

用替換⑧式中的d，

k=(g+)                                                         

即B=B₀+(g+)t.                                                  

答案：(1)≤k＜(g+)

(2)B=B₀+(g+)t