Using relativity to calculate the radius of a cyclotron when accelerating protons to 0.99c

If I need to find the minimum radius necessary to have a fixed target collision of $$10^{10}eV$$ with two protons, and the traveling proton is going 0.99c in a cyclotron of 8 Tesla. Do I need to consider relativity when calculating this radius?

I've tried to calculate the radius by equating the force on the proton from the magnetic field with the centriptal force and solving for $$r$$.

$$qvB={mv^2\over r}$$ $$r={mv\over qB}={(1.67\times10^-27kg)(0.99c)\over (q_e8T)}={\big[(1.67\times10^-27kg)(0.99(3\times10^8m/s))\big]\over \big[(1.602\times10^{-19}C) \times 8T\big]}=3.87\times10^{-47}m$$

I know this is extremely incorrect because a proton traveling at 0.99c in a field of 8 Tesla is approximately what the LHC does, and it is no where near this tiny of a radius. So I thought that maybe I would use a relativistic centripetal force, $${\gamma m_0v^2\over r}$$. This didn't change significantly. Instead of $$3.87\times10^{-47}m$$ I got $$2.74\times10^-8m$$, which makes sense because $$\gamma\approx7$$ and wouldn't affect my result much.

I'm assuming that I need to take relativity into account because the proton is traveling at a relativistic speed, but I'm not sure how. This stack exchange post: Relativistic centripetal force seemed promising, but it didn't answer my question.

Can anyone provide some guidance as to how I can find the radius for this cyclotron?

• The operation you show in that equation does not give that result. Check the orders of magnitude. – secavara Dec 3 '18 at 0:00