v0tan450=Eqmt
t=2L2v0
E=2mv02qL
(2) The particle moves in a uniform circle in Zone II, and its orbital radius R is large = L ..
Because qvB=mv2r
v=2v0
B small = =2mv0qL
After the particle enters the ⅲ region, its trajectory NQ is symmetrical with PM, then
Horizontal displacement x = v02t
t=2L2v0x=24L
xQ=.O.Q=O.P=22L+x=22L+24L=324L
(3) The radius r of a circular magnetic field region is equal to the radius r of its trajectory circle, that is, r = r = l.
When all particles come out of the magnetic field, the velocity directions are parallel, and their falling points are between two points GH on the straight line OB, as shown in the figure.
GH=2rsinθ=2Lsinθ
Answer: (1) The magnitude of electric field intensity E of uniform electric field E = 2mv20QL.
(2) The minimum magnetic induction intensity of this magnetic field is 2mv0qL, and the particles reach the abscissa of point Q on the X axis through region III;
(3) The area length of all particles passing through OB is 2Lsinθ. ..