(20 13? Meizhou second mode) As shown in the figure, there is a uniform electric field between parallel metal plates M and N with a distance of D, and the magnetic induction intensity is Bo when it is

(20 13? Meizhou second mode) As shown in the figure, there is a uniform electric field between parallel metal plates M and N with a distance of D, and the magnetic induction intensity is Bo when it is perpendicular to the paper. (1) Let the field strength of uniform electric field between parallel metal plates M and N be E0,

Then it is: U=Eod

Because ions are injected between the two plates along the direction of the metal plate and move in a straight line at a uniform speed.

Then it is: qEo=qvoBo

Solution: the voltage between the metal plates m and n is U=Bovod.

(2) In the electric field of the first quadrant, the ions do quasi-planar motion,

Then it is: cos 45 o = vov

Therefore, the speed at which ions move to point A: v = 2vo.

Newton's second law: qE=ma

Again? vy=at

And tan45o=vyvo.

It is found that the time required for ions to move to point A in electric field E is t = mvoqe.

(3) Lorentz force provides centripetal force in the magnetic field,

Then it is: qvB=mv2R.

Solution: r = mvqb = 2mvoqb

According to geometry knowledge, AC = 2 rcos405o = 2r = 2mvoqb.

Here we go again. OA = VOT = MVO2QE

Therefore, the distance between the position c where the ions leave the fourth quadrant magnetic field region for the first time and the coordinate origin.

There are:. oc =。 OA+。 AC = mvo2qe+2mvoqb。

Answer: (1) The voltage between the metal plates M and N is Bovod；;

(2) When ions move to point A, the magnitude of velocity V and time mv0qE need to move from point P to point A;

(3) The distance from the position C (not shown in the figure) where ions leave the fourth quadrant magnetic field region for the first time to the coordinate origin, mv20qE+2mv0qB. ..