The induced electromotive force generated by rectangular coil E=200V②.
Because the voltage between AB plates is equal to the electromotive force generated by the coil,
So what? UAB = 200 volts ③
(2) Charged particles do quasi-flat throwing motion between AB plates, and the vertical velocity when they leave the edge of the lower plate is vy.
Then: a = quabmd4 horizontal direction: L=v0t⑤.
Vertical direction: d =12at26
Substitute data from ④ ⑤ ⑤: v0 = 2.0× 104m/s ⑦.
(3) The moment when the particle enters the magnetic field: vy=at=2.0× 104m/s⑧
Charged particles do uniform circular motion in a circular magnetic field,
Lorentz force as centripetal force: qvb = mv2r ⑨ v = v02+vy2 ⑩.
Substituting data from ⑧ ⑨ ⑩: r = 0.22m
As shown in the figure, the length of the chord NQ is. From the geometric relationship, NQ = r = 0.22.
Among the circles passing through NQ two points, the circle with the diameter of chord NQ is the smallest.
The minimum radius of circular magnetic field area is: r = r/2 = 0.12m = 0.14m.
Answer: (1) The voltage between AB plates is 200 V;
(2) 2) the size of v0 is 2.0×104m/s;
(3) The minimum radius of the circular magnetic field region is 0. 14m.