A particle of mass m and charge q moves at speed v perpendicular to a uniform field B. What is the radius of its circular path?

1. For motion perpendicular to B the magnetic force has magnitude qvB, since sin(theta) is 1. 2. That force supplies the centripetal force, which is mv^2/r. 3. Setting qvB = mv^2/r and cancelling one factor of v gives qB = mv/r. 4. Rearranging yields r = mv/(qB). 5. A heavier or faster particle bends less, while a larger charge or stronger field bends it more, which matches this form. _Source: OpenStax College Physics (CC BY 4.0), Ch 22 "Magnetism", section 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications_