Suppose that a positively charged particle with charge q moves in a circular path of radius r in a constant magnetic field of strength B. If the magnetic field strength is doubled to 2B, what effect does this have on the radius of the circular path that this charge takes?
In this case, the magnetic force is acting as the centrifugal force which helps to move in a circular path.
So that,
We can apply F = ma to the particle, where F = Bqv and a=v^2 / r
Then,
by applying F =ma,
F = ma
Bqv = m * V^2 / r
r = (mv / Bq)
So, we can see that if m,v,q are constants, then r is inversely proportional to the B.
If B is doubled, then r should be half (1/2) of it.