# Increase in the strength of a magnetic field in synchrotrons

I have a very simple question

I'm working on a small project, and I worked out sometime that I can't seem to verify online

Basically, I'm trying to find how the magnetic field strength inside a synchrotron has to increase as a function of the velocity of the particles. I understand that there are other things going on in synchrotrons, but I'm really looking at the basics for the sake of this project

My thoughts go as follows:

Since the Centripetal force required to keep the particles in a ring of constant radius is Fc=m.v^2/r

and that the magnetic force applied on a particle by a magnetic field perpendicular to its velocity is FB = V.B

Then we would need the strength of the magnetic field proportionally to the first power of the velocity. This way the product V.B would increase proportionally to the square of the velocity, as we wanted originally

Is all of this correct?

If there's something i'm missing, can you tell me what that is?

Thank you so much

The basic formula you are discussing hold in the non-relativsitic limit. You can combine them to give the so called gyroradius: $$r_g= m v /q B$$ (I have added in the charge $$q$$ which you accidentally omitted). You can re-write this as $$r_g= p/qB$$ where $$p$$ is the particle momentum (tangent to the circular orbit). This suggests the relativistic generalization* $$r_g= \gamma m v/ qB$$ where $$\gamma=\sqrt{1/(1-(v/c)^2)}$$ is the relativistic factor such that $$p=\gamma m c$$ is the relativistic momentum. Indeed this is correct, and in accelerator physics we often use the engineering formula $$B r_g = 3.335\text{[Tm]}p\text{[GeV/c]}/q\text{[elementary charge]}$$ to estimate the particle energy. Thus if an electron or proton $$|q|=1$$ has 1TeV of energy and the dipole (bending magnets) have 1T then the radius will 3.335km. You can see why rings become quite large!
*This correction occurs because $$v$$ will max out at $$c$$ but the particle's energy will continue to increase.