Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S19 / 1 / a

Use transverse coordinates $(x,x^{\prime })$ and define a positive signed focal length as focusing in that plane. A thin quadrupole magnet of length $\ell$ and signed gradient $G$ has integrated strength $1/f=|e|G\ell /p_0$; the electron charge and chosen field orientation are absorbed into the sign of $G$. Its beam transfer matrix isFor total electron energy $E_0$, $p_0=\sqrt{E_0^2-m_e^2{c}^4}/c$, so $f\simeq E_0/(|e|cG\ell )$ in the ultrarelativistic limit. The other transverse plane has the opposite signed focal length. A quadrupole magnet therefore focuses one plane while defocusing the other; the longitudinal magnetic field component is negligible in this paraxial idealization.

At fixed gradient and magnet length, $f\propto p$. If $E$ is total energy,Here ${\beta }_0=v_0/c$, not one of the Courant-Snyder parameters. For ultrarelativistic electrons this becomes $f\simeq f_0(1+\delta )$, or $1/f\simeq (1-\delta )/f_0$: higher-energy particles are bent and focused less strongly. If the symbol $E$ instead denotes kinetic energy $K$, the coefficient is $(K+m{c}^2)/(K+2m{c}^2)$, approaching $1/2$ in the nonrelativistic limit. Stating the energy convention avoids applying the ultrarelativistic result to a slow beam.