## College Physics (4th Edition)

$T = \frac{2\pi~m}{q~B}$
$F = \frac{mv^2}{r}$ $F = qvB$ 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.