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

  • $\begingroup$ 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. $\endgroup$ May 9 at 11:05
  • $\begingroup$ 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 $\endgroup$
    – Crucio
    May 9 at 12:01
  • $\begingroup$ your conversion is in the wrong direction. $\endgroup$ May 9 at 12:40

1 Answer 1


The formula that you obtain is correct, it accomplish with the dimensional test. Please, be sure and check the formula of the referred book, maybe the magnetic induction appears in another Unit System.


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