(2013? Henan Simulation) As shown in the figure, in the area of ??x≥O, there is a uniform magnetic field perpendicular to the xOy plane. The magnitude of the magnetic induction intensity is B and the

Solution: (1) Assume the accelerating voltage is U, the initial velocity of the positive ion is zero, and it is accelerated by the accelerating electric field. According to the kinetic energy theorem:

qU=12mv2

< p>Positive ions make uniform circular motion in the magnetic field, and Lorentz force provides centripetal force:

qvB=mv2R

R=2mqUqB

When the acceleration voltage is relatively high When the ions are small, the radius of uniform circular motion of ions in the magnetic field is small. When the ions happen to hit the lower end point N of the metal plate, the minimum radius of circular motion is Rmin, as shown in Figure 1

According to geometric knowledge, it can be Determine Rmin=d

So: Umin=qB2d22m

When the accelerating voltage is large, the radius of the ions making uniform circular motion in the magnetic field is larger. When the ions happen to hit the metal plate When the upper endpoint is point M, the maximum radius of circular motion is Rmax, as shown in Figure 2

Based on geometric knowledge: Rmax2=d2+(2d-Rmin)2

The solution is Rmax=54d

So Umax=25qB2d232m

So the ions can all hit the metal plate, and the value range of the acceleration voltage is: qB2d22m≤U≤25qB2d232m

(2 ) Assuming that the period of uniform circular motion of ions in a magnetic field is T, according to the law of circular motion:

T=2πRv? ①

And qvB=mv2R?②

The joint solution of ①② shows that T=2πmqB

The period of uniform circular motion of ions in a magnetic field has nothing to do with the accelerating voltage.

The shortest movement time of ions in the trajectory shown in Figure 1 is

tmin=14T

The shortest movement time of ions in the trajectory shown in Figure 2 The length is: