The centripetal force that keeps the proton in circular orbit is provided by the Lorentz force exerted by the magnetic field:
[tex]m \frac{v^2}{r} = qvB[/tex]
where m is the mass of the proton, v its speed, q its charge and B the magnetic field intensity.
By re-arranging the formula, we have:
[tex]r= \frac{mv}{qB} [/tex]
Using B=0.66 T and [tex]v=6.46 \cdot 10^5 m/s[/tex], we find the radius of the proton's orbit:
[tex]r= \frac{(1.67 \cdot 10^{-27}kg)(6.46 \cdot 10^5 m/s)}{(1.6 \cdot 10^{-19}C)(0.66 T)}=9.8 \cdot 10^{-3}m=9.8 mm [/tex]
