题目内容

如图所示,匀强电场方向竖直向上,匀强磁场方向水平且垂直纸面向里,有两个带电小球a和b,a恰能在垂直于磁场方向的竖直平面内做半径r=0.8 m的匀速圆周运动,b恰能以v=2 m/s的水平速度在垂直于磁场方向的竖直平面内向右做匀速直线运动.小球a、b质量ma=10 g,mb=40 g,电荷量qa=1×10-2C,qb= 2×10-2C,g=10m/s2。求:

(1)小球a和b分别带什么电?电场强度E与磁感应强度B的大小?

(2)小球a做匀速周周运动绕行方向是顺时针还是逆时针?速度大小va是多大?

(3)设小球b的运动轨迹与小球a的运动轨迹的最低点相切,当小球a运动到最低点即切点时小球b也同时运动到切点,a、b相碰后合为一体,设为c,在相碰结束的瞬间,c的加速度ac=?

 

【答案】

(1)小球a在互相垂直的匀强电场和匀强磁场中做匀速圆周运动,电场力和重力的合力为零,电场力方向向上,所以小球a带正电,且有

mag=qaE··································································①

E=10 N / C································································②

小球b做匀速直线运动,合力为零,带正电,且有

mbg=qbvB+qbE·····························································③

B=5T·····································································④

(2)小球a做匀速圆周运动绕行方向是逆时针方向。

······························································⑤

va=4 m / s·································································⑥

(3)ab相碰前速度方向相同,设碰后的共同速度为vc,则

ma va+mb v=(ma+mb) vc······················································⑦

vc=2.4 m / s································································⑧

由牛顿第二定律得

mcac=qcE+qc vcBme·····················································⑨

mc=ma+mb

qc=(qa+qb)

解得ac=3.2 m / s2··························································⑩

评分标准:本题满分18分。其中判断ab带电性质各1分,a绕行方向1分,①式2分,②式1分,③式2分,④式1分,⑤式2分,⑥式1分,⑦式2分,⑧式1分,⑨式2分,⑩式1分。

【解析】略

 

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