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Solution: Solution: (1) The particle moves in a uniform circular motion in a uniform magnetic field, and the orbital radius is r. According to the geometric relationship:

r2-(r-d2)2=L2

And qv0B=mv20r.

Solution: B=4dmv0q(d2+4L2)

(2) When a particle moves along a straight line OO', there are:

qv0B=qE

Solution: E=Ud

Solution: U=4d2mv20q(d2+4L2)

(3) When the magnetic field is removed, the particles do similar flat throwing motion under the action of electric field force, assuming that the particles can fly out of the parallel plate, then:

L=v0t

y= 12gt2

q=qEm

Solution: y = 2dl2d2+4l2 < D2.

Therefore, the hypothesis holds, that is, the time for the particles to move between the two poles t=Lv0, and the displacement along the electric field force direction y = 2dl2d 2+4 L2；;

Answer: (1) If the particles just fly out from the right edge of the N plate, the magnetic induction intensity B of the uniform magnetic field is 4 dmv0q (D2+4L2);

(2) If a voltage U is applied to the M and N plates, the particles can pass through the parallel plate capacitor along a straight line OO', and the size of U is 4d2mv20q (D2+4L2);

(3) If the magnetic field is removed, the voltage between the two plates and the incident velocity of the particles in (2) remain unchanged, the time t for the particles to move between the two plates is Lv0, and the displacement y along the electric field force direction is 2dL2d2+4L2. ..