## College Physics (4th Edition)

We can find an expression for the radius of the path: $F = \frac{mv^2}{r}$ $qvB = \frac{mv^2}{r}$ $r = \frac{mv^2}{qvB}$ $r = \frac{mv}{qB}$ We can see that the doubling the charge $q$ will result in a radius that is half the original radius. Therefore, the singly charged ion has a path of larger radius by a factor of 2.