Answer
(a) $f = 1.2 \times 10^7~Hz$
(b) $f = 2.2 \times 10^{10}~Hz$
Work Step by Step
(a) $F = \frac{mv^2}{r}$
$F = qvB$
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 frequency:
$f = \frac{speed}{distance}$
$f = \frac{v}{2\pi~r}$
$f = \frac{(\frac{q~B~r}{m})}{2\pi~r}$
$f = \frac{q~B}{2\pi~m}$
$f = \frac{(1.6\times 10^{-19}~C)(0.80~T)}{(2\pi)~(1.67\times 10^{-27}~kg)}$
$f = 1.2 \times 10^7~Hz$
(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 frequency:
$f = \frac{speed}{distance}$
$f = \frac{v}{2\pi~r}$
$f = \frac{(\frac{q~B~r}{m})}{2\pi~r}$
$f = \frac{q~B}{2\pi~m}$
$f = \frac{(1.6\times 10^{-19}~C)(0.80~T)}{(2\pi)~(9.1\times 10^{-31}~kg)}$
$f = 2.2 \times 10^{10}~Hz$
