## Solution to 2001 Problem 62

 The cyclotron frequency is given in Hz, so we must assume it is not an angular frequency. The formula for the cyclotron frequency (not an angular frequency) is\begin{align*}f = \frac{q B}{2 \pi m}\end{align*}This can be derived as follows. The Lorentz force must equal the force required to maintain circular motion\begin{align*}\frac{v^2}{r} m = q B v \Rightarrow v = \frac{q B r}{m}\end{align*}The frequency is then given by\begin{align*}f = \frac{v}{ 2 \pi r} = \frac{q B r}{2 \pi m r} = \frac{q B}{2 \pi m}\end{align*}So, we solve for $m$ and plug in the given numbers\begin{align*}m = \frac{q B}{2 \pi f} = \frac{2 \cdot 1.60 \cdot 10^{-19} \mbox{ C} \cdot \frac{\pi}{4}}{2 \pi \cdot 1600 \mb...Therefore, answer (A) is correct.