题目内容

【题目】如图所示为一速度选择器，也称为滤速器的原理图。K为电子枪，由枪中沿KA方向射出的电子，速率大小不一。当电子通过方向互相垂直的匀强电场和磁场后，只有一定速率的电子能沿直线前进，并通过小孔S。设产生匀强电场的平行板间的电压为300 V，间距为5 cm，垂直于纸面的匀强磁场的磁感应强度为0．06 T，问：

（1）磁场的指向应该向里还是向外？

（2）速度为多大的电子才能通过小孔S?

【答案】（1）磁场方向垂直于纸面向里 （2）105 m/s

【解析】（1）由题图可知，平行板产生的电场强度E方向向下，带负电的电子受到的电场力FE＝eE，方向向上。

若没有磁场，电子束将向上偏转，为了使电子能够穿过小孔S，所加的磁场对电子束的洛伦兹力必须是向下的。根据左手定则分析得出，B的方向垂直于纸面向里。

（2）电子受到的洛伦兹力为：FB＝evB，它的大小与电子速率v有关。只有那些速率的大小刚好满足洛伦兹力与电场力相平衡的电子，才可沿直线KA通过小孔S。据题意，能够通过小孔的电子，其速率满足下式：evB＝eE

解得：v＝

又因为E＝

所以v＝

将U＝300 V，B＝0．06 T，d＝0．05 m代入上式，得v＝105 m/s

即只有速率为105 m/s的电子才可以通过小孔S

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