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【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006 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 2105 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=006 T,d=005 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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