......................2分20.(16分)解:(1)从P到Q,x负方向v0t=ry负方向:=r.............................................................................2a=得:E=......................................................................................2(2)在Q点,粒子沿负y方向的速度vy=at粒子与x负方向的夹角α,tanα=..................................................2粒子在第二象限做匀速直线运动,由几何关系tanα=得:xM=﹣r........................................................................................2(3)粒子进入磁场,若粒子轨迹恰好与磁场下边界相切,如图所示,由几何关系:Rcosα+R=r..........................................................................................2R=r此时,Rsinα+R=r<|xM|.................................2粒子能进入第二象限,在磁场中运动的最大半径R=r由qvB=mv=得:B=...................................................................................3若使粒子进入第二象限,应该B≥.................................1