Question

如圖所示，相距為d、板間電壓為U0的平行金屬板間有方向垂直紙面向里、磁感應(yīng)強(qiáng)度大小為B0的勻強(qiáng)磁場(chǎng)；OP和x軸的夾角α＝45°，在POy區(qū)域內(nèi)有垂直紙面向外的勻強(qiáng)磁場(chǎng)，POx區(qū)域內(nèi)有沿x軸正方向的勻強(qiáng)電場(chǎng)，場(chǎng)強(qiáng)大小為E；一質(zhì)量為m、電荷量為q的正離子沿平行于金屬板、垂直磁場(chǎng)的方向射入板間并做勻速直線運(yùn)動(dòng)，從坐標(biāo)為(0，L)的a點(diǎn)垂直y軸進(jìn)入磁場(chǎng)區(qū)域，從OP上某點(diǎn)沿y軸負(fù)方向離開磁場(chǎng)進(jìn)入電場(chǎng)，不計(jì)離子的重力．

(1)離子在平行金屬板間的運(yùn)動(dòng)速度v0；

(2)POy區(qū)域內(nèi)勻強(qiáng)磁場(chǎng)的磁感應(yīng)強(qiáng)度B；

(3)離子打在x軸上對(duì)應(yīng)點(diǎn)的坐標(biāo)．

 

 

 

 

Answer 1

 

解析　(1)正離子在平行金屬板間勻速運(yùn)動(dòng)，根據(jù)平衡條件有Eq＝B0qv0                ①

根據(jù)平行金屬板間的場(chǎng)強(qiáng)和電勢(shì)差的關(guān)系有

E＝                                                                                                                 ②

由①②式解得

v0＝                                                                                                                    ③

(2)在磁場(chǎng)中，由幾何關(guān)系有

L＝R＋Rtan α                                                                                                              ④

洛倫茲力充當(dāng)向心力，根據(jù)牛頓第二定律有

Bqv0＝m                                                                                                             ⑤

由③④⑤式解得

B＝                                                                                                          ⑥

(3)在電場(chǎng)中正離子做類平拋運(yùn)動(dòng)，則有

沿y軸負(fù)方向：Rtan α＝v0t                                                                                        ⑦

沿x軸正方向：x0＝at2                                                                                            ⑧

離子在電場(chǎng)中運(yùn)動(dòng)的加速度a＝                                                                            ⑨

離子打在x軸上對(duì)應(yīng)點(diǎn)的橫坐標(biāo)

x＝x0＋R                                                                                                                     ⑩

由③④⑦⑧⑨⑩式解得

離子打在x軸上對(duì)應(yīng)點(diǎn)的坐標(biāo)為(＋，0)                                                  ⑪