(A) When a charged particle enters a perpendicular magnetic field,the magnetic force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$qvB = \frac{mv^2}{R}$
Rearranging for angular velocity $\omega = \frac{v}{R}$:
$\frac{v}{R} = \frac{qB}{m}$
Since $\omega = \frac{v}{R}$ and $q = e$,we have $\omega = \frac{eB}{m}$.
Now,checking the dimensions:
$[\omega] = \frac{[e][B]}{[m]}$
From $F = qvB$,the dimension of $B$ is $[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{IT \cdot LT^{-1}} = [M I^{-1} T^{-2}]$.
Substituting the dimensions:
$[\omega] = \frac{[I T] \cdot [M I^{-1} T^{-2}]}{[M]} = \frac{M I^0 T^{-1}}{M} = [T^{-1}]$.
Thus,the dimensions of cyclotron frequency are $[T]^{-1}$.