# Converting units in this formula for charged particle oribiting in a magnetic field

A particle with charge $$q$$ in a magnetic field $$B$$ perpendicular to its momentum $$p$$ will have a circular trajectory with radius $$R$$, and this relation should hold $$p = qRB$$

and in the relativistic limit $$p \simeq E/c$$ thus $$\frac{E}{c} = q R B \implies B = \frac{E}{qRc}$$ which I assume will hold for SI units so

$$B(T) = \frac{E(J)}{q(C)R(m)c(m/s)}$$

now my textbook reference and lecture slides say that if we want to use this formula with practical and useful dimensions, it turns into this $$B(T) = \frac{E(GeV)}{0.3 R(m)}$$ assuming that $$q=|e|$$.

I was not able to reproduce this conversion, it seems rather simple and I do know how to convert from $$eV$$ to $$J$$ and such, but somehow it does not work in the end. If someone could show the steps explicitly it would be very useful

• The electron charge in q and the electron charge in eV cancel, leaving you with Giga on top and speed of light on the bottom. One is 10^9. the other is 3E8, leaving you with 0.3 below. May 9 at 11:05
• I see your reasoning but this looks wrong to me. Because if I start from eV and want to convert to GeV, I have to multiply by 10^-9 right? So it does not cancel out May 9 at 12:01
• your conversion is in the wrong direction. May 9 at 12:40