Question

如图所示，相距为d、板间电压为U0的平行金属板间有方向垂直纸面向里、磁感应强度大小为B0的匀强磁场；OP和x轴的夹角α＝45°，在POy区域内有垂直纸面向外的匀强磁场，POx区域内有沿x轴正方向的匀强电场，场强大小为E；一质量为m、电荷量为q的正离子沿平行于金属板、垂直磁场的方向射入板间并做匀速直线运动，从坐标为(0，L)的a点垂直y轴进入磁场区域，从OP上某点沿y轴负方向离开磁场进入电场，不计离子的重力．

(1)离子在平行金属板间的运动速度v0；

(2)POy区域内匀强磁场的磁感应强度B；

(3)离子打在x轴上对应点的坐标．

 

 

 

 

Answer 1

 

解析　(1)正离子在平行金属板间匀速运动，根据平衡条件有Eq＝B0qv0                ①

根据平行金属板间的场强和电势差的关系有

E＝                                                                                                                 ②

由①②式解得

v0＝                                                                                                                    ③

(2)在磁场中，由几何关系有

L＝R＋Rtan α                                                                                                              ④

洛伦兹力充当向心力，根据牛顿第二定律有

Bqv0＝m                                                                                                             ⑤

由③④⑤式解得

B＝                                                                                                          ⑥

(3)在电场中正离子做类平抛运动，则有

沿y轴负方向：Rtan α＝v0t                                                                                        ⑦

沿x轴正方向：x0＝at2                                                                                            ⑧

离子在电场中运动的加速度a＝                                                                            ⑨

离子打在x轴上对应点的横坐标

x＝x0＋R                                                                                                                     ⑩

由③④⑦⑧⑨⑩式解得

离子打在x轴上对应点的坐标为(＋，0)                                                  ⑪