## College Physics (4th Edition)

$F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find the minimum required value for $r$: $\frac{mv^2}{r} = qvB$ $r = \frac{mv}{qB}$ $r = \frac{(1.67\times 10^{-27}~kg)(1.0\times 10^7~m/s)}{(1.6\times 10^{-19}~C)(0.50~T)}$ $r = 0.21~m$ The minimum radius is 21 cm