Why does higher momentum give a larger radius of curvature in a magnetic field?

[Matthew35]

I was trying to visualize the path of a charged particle in a uniform magnetic field, and I keep getting stuck on how the curvature relates to its momentum. I can follow the math for the Lorentz force, but picturing why a faster proton traces a wider arc, not a tighter one, makes my intuition falter.

[Layla.L]

I kept picturing a faster proton squeezing into a tighter circle, but the math kept shouting back that the radius grows with speed. The Lorentz force grows with v, but the needed centripetal acceleration grows with v^2, so to balance m v^2 / R = q v B you get R = m v / (q B). In plain terms, faster means the curve loosens up, not tightens.

[Stephen19]

In an old classroom demo I used a bead on a string to mimic the turn. The bead only changes direction because the sideways pull is doing the turning, and when I spun it faster the bead rode a longer circle. It clicked that the force is what keeps turning; speed sets how long the turn has to be to satisfy the balance.

[Mia9]

I did a tiny quick sim, kept B and q fixed and varied the momentum. The arc got wider as momentum went up; curvature dropped. It wasn’t elegant, but the numbers lined up with the circle intuition, even if it still felt odd at first.

[MadisonDJ]

Is your velocity truly perpendicular to the field in the drawing? If there’s any component along B, you start a helix and the circular picture just doesn’t apply as neatly.