v 20 R 1 ; 解得: R 1 = m v 0 eB 质子离开磁场时距O的距离: d 1 =2 R 1 = 2m v 0 eB 同理,α粒子在磁场中的运动半径: R 2 = 4m v 0 2eB = 2m v 0 eB α粒子离开磁场时距O的距离: d 2 =2 R 2 = 4m v 0 eB 由此可以看出质子由P点离开磁场,α粒子由Q点离开磁场P、Q两点间的距离: d= d 2 - d 1 = 2m v 0 eB (2)α粒子在磁场中的运动的周期为T,则: T= 2πR v = 2π•4m 2e•B = 4πm eB 粒子运动的时间: t= T 2 = 2πm eB 答:(1)P、Q两点间的距离 2m v 0 eB ; (2)α粒子在磁场中的运动时间 2πm eB