As shown in the figure, the square area abcd is filled with a uniform magnetic field with a magnetic induction intensity of B, which is perpendicular to the paper surface and inward. At t=0, one is located. Solution: Solution: (1) The trajectory of the particle in the initial velocity direction along the od direction is shown in Figure 1, and the center of the trajectory is O 1.

According to Newton's second law: qvB=mv2L,

Period: T=2πLv,

sin∠o 1Pc = o 1co 1P = 12，

So: ∠ OO 1p = ∠ O 1pc = 30,

t0=30 360 T，

Solution: QM = π 6bt0; ?

(2) The trajectory of the particles leaving the magnetic field at last is shown in Figure 2, with the center of the circle being O2 and the velocity deflection angle being α.

According to geometry knowledge, ob=L2+(L2)2,

From Cosine Theorem: Ob2=L2+L2-2L2cosα

Solution: cosα= 38；;

(3) Because ∠ OO 1p = 30, the magnetic field of particles whose velocity deflection angle is less than 30 has been ejected, the trajectory of particles just leaving the magnetic field at t0 is shown in Figure 3, ∠ OO3m = 30, and the geometric relationship shows that the angle between the particle velocity direction and Oa is 30, so the number of particles still in the magnetic field at this time and the total number of particles emitted by the particle source.

n= 180？ 30 180 =56;

Answer: (1) Specific charge of particles: QM = π 6bt0;

(2) The deflection angle of the particle finally leaving the magnetic field is α, and COS α = 38;

(3) Assuming that the particles emitted by the particle source are uniformly distributed in the range of 0- 180, the ratio of the number of particles still in the magnetic field at t0 to the total number of particles emitted by the particle source is 5: 6.