Do particle accelerators centrifuge the quarks of a proton? Is it right that circular particle accelerators use magnetic fields to deflect the particle beam?
Using the simple equation of "a charged particle in a magnetic field":


*

*$\vec{f}=q\vec{v}\times\vec{B}$


Say the particle accelerators is currently used for a proton beam. $\vec{f}$ will be directed to the center of the particle accelerators (centripetal force).
The proton itself is constituted of two up quarks and one down quark, which have electric charge:


*

*$\frac{2}{3}e$ for one up quark.

*$-\frac{1}{3}e$ for one down quark.


Now if I look at the effect of the magnetic field at quark level. The force is:


*

*$\frac{2}{3}\vec{f}$ for one up quark (centripetal force).

*$-\frac{1}{3}\vec{f}$ for one down quark (centrifugal force).


I imagine it as a centrifugation of the proton, the down quarks are pushed to the exterior side of the accelerator when the up ones are attracted to the center.
Is this effect real and taken into account in particle accelerators?
What about the mass ? The down quark is at least 2 times heavier than the up quark, does the rotation of the proton "separate" the quarks of the proton?
 A: The centripetal acceleration that the protons feel as they circulate in the LHC is roughly:
$$ a = \gamma^2 \frac{v^2}{r} $$
This is the usual equation for centripetal acceleration but multiplied by a factor of $\gamma^2$ to allow for the time dilation the protons experience. The speed $v$ is approximately $c$. The radius of the LHC is about 4.3km but the bending radius is sharper thna this because the protons are deflected only at certain points around the ring. The bending radius is about 2.8km. Finally, the Lorentz factor is about 7000 (the ratio of the proton enery to its rest mass). Feed all these into my equation and you get:
$$ a \approx 1.6 \times 10^{21} \,\text{m}\,\text{s}^{-2} $$
This seems an awfully large number, but what we're really interested in is the potential energy change across the width of a proton, and this is given by:
$$ U = m\, d\, a $$
where $m$ is the mass of the object, $d$ is the width of the proton and $a$ is the acceleration we've just calculated. The width of a proton is easy as that's around $1.7 \times 10^{-15}$ m. The mass is harder because we have to decide what object we're moving.  Let's take the mass to be 1% the mass of a proton or about $1.7 \times 10^{-29}$kg. This is a bit heavier than the bare mass of the up and down quarks but it's of the same order of magnitude. Anyhow, if we feed these into the equation above, along with our calculated value for $a$, we get:
$$ U \approx 5 \times 10^{-12} \,\text{J} \approx 0.0002 \,\text{eV} $$
And 0.2meV is utterly insignificant compared to the energies inside the proton, so we can be confident that the protons don't get centrifuged by their passage round the ring.
A footnote: honesty compels me to admit I'm not sure if the equation for $a$ should contains a factor of $\gamma^2$ or $\gamma^3$ (though it makes no difference to the conclusion anyway). For linear acceleration you need $\gamma^3$ as you get a factor of $\gamma$ for each power of $t$ due to the time dilation and another factor of $\gamma$ due to the length contraction. For motion in a circle I'm not sure whether length contraction would introduce another factor of $\gamma$. Any contributions on this point are gratefully received.
