Magnetic Fields in an accelerator

A revised calculation for the Fermilab Tevatron B field.

The magnetic force provides the centripetal acceleration in large acclerators, i.e.
QvB/=mv^2/r ;
so for a proton held in circular motion by a magnetic field we have:
r =mv/eB and
B= mv/(r e) ( where "e" is the charge on a proton). At relativistic energy mv = E/c; where c = speed of light = 3E8 m/s. For a proton in the Tevatron E = 1E12 eV (where 1 ev =1.6E-19 J) and the radius is about 1000m
Thus we have:

B= [1E12x1.6E-19/3E8]/[1.6E-19x1000m]= 3.3Tesla or about 66000 times the earths magnetic field and about half of what the big coils on a fusion reactor would create.