Kinetic energy theorem qu = 12mv2
Alternative data: v=20m/s
(2) Charged particles only do uniform circular motion under the Lorentz force,
Yes: qvB=mv2R
Get r = mvqb
Alternative data: r = 0.50m
And OPC OS 53 = 0.50m m.
So the center of the circle must be on the X axis, and the trajectory is as shown in the figure.
According to the geometric relationship, OQ = r+rsin53.
So OQ = 0.90 meters
(3) Charged particles are not emitted from the X axis (as shown in the figure),
From the geometric relationship: op > r'+r' cos 53 ①.
r′= mvqB′②
163t = 5.33t？
Answer: (1) The velocity V of charged particles when they reach point P is 20m/s;
(2) If the magnetic induction intensity B=2.0T and the particles leave the magnetic field from the Q point on the X axis, the distance of QO is 0.9m；;
(3) If the particles can't enter above the X-axis, the magnetic induction intensity b' satisfies the condition of more than 5.33T t. 。