✔ 最佳答案
As there is no diagram given, I presume the magnetic field is pointing OUT of the paper perpendicualr to the direction of motion of the ball down the inclined plan.
The magnetic force (Lorentz force) is opposite to the downward weight component normal to the inclined plane. When the two forces are equal, the ball begins to lift off from the plan. This happens when
q.v.B = mg.cos(x)
where v is the velocity of the ball when lifting occurs and g is the acceleration due to gravity.
thus, v = mg.cos(x)/q.B
Thus, the velocity of the ball increases from zero up to mg.cos(x)/q.B
Since acceleration a = g.sin(x)
using equation of motion, v^2 = u^2 + 2.a.s
[mg.cos(x)/q.B]^2 = 2.(g.sin(x)).s
thus s = [mg.cos(x)/q.B]^2/2.g.sin(x)