Answer
$T = \frac{2\pi~m}{q~B}$
Work Step by Step
$F = \frac{mv^2}{r}$
$F = qvB$
We can equate the two expressions for $F$ to find an expression for the speed $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
We can find an expression for the period:
$T = \frac{distance}{speed}$
$T = \frac{2\pi~r}{v}$
$T = \frac{2\pi~r}{(\frac{q~B~r}{m})}$
$T = \frac{2\pi~m}{q~B}$
We can see that the period is independent of the particle's speed.
