The ratio of cyclotron frequency f to the radius of the path of a...

The radius of the circular path of a charged particle in a uniform magnetic field is given by $R = \dfrac{mv}{qB}$.

The cyclotron frequency is given by $f = \dfrac{qB}{2\pi m}$.

Taking the ratio of the cyclotron frequency to the radius, we get:

$\dfrac{f}{R} = \dfrac{\dfrac{qB}{2\pi m}}{\dfrac{mv}{qB}} = \dfrac{q^2 B^2}{2\pi m^2 v}$

It is given that $\dfrac{f}{R} = \dfrac{q^2 B^2}{2\pi m^2}x$.

Comparing the two expressions, we get $x = \dfrac{1}{v} = v^{-1}$.

Answer: $v^{-1}$