Past exam of the physics 2 paper of the University of Oxford / 2014 resit / 7

The Lorentz force law isIts rate of work is $dT/dt=\mathbf{F}\cdot \mathbf{v}=q\mathbf{E}\cdot \mathbf{v}$, because $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})=0$. Thus, with $\mathbf{E}=0$, the kinetic energy cannot increase: a static magnetic field changes the direction of motion but does no work.

Take the initial velocity perpendicular to a uniform magnetic field. The magnetic Lorentz force is perpendicular to the motion, has constant magnitude $|q|vB$, and supplies the centripetal force:The rotation sense is fixed by $q\mathbf{v}\times \mathbf{B}$. A component of velocity parallel to the field remains constant and produces a helical motion in a uniform magnetic field; a circle requires that component to vanish. The speed-independent cyclotron frequency here assumes nonrelativistic motion.

For a proton, the fundamental cyclotron resonance condition isA half-orbit takes $\pi m_p/(eB)$, during which the applied voltage reverses sign once. The proton then crosses the gap in the opposite direction and receives another accelerating impulse. The electric force in the gap, rather than the magnetic bending inside a dee, provides the kinetic energy. The injection phase must be chosen so the first crossing accelerates the proton.

At the largest admitted orbit, $r=R$ and $m_p{v}_{\max }=eBR$. ConsequentlyUsing $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_p\simeq 1.67262\times 10^{-27}\mathrm{kg}$ with the specified $B$ and $R$ yieldsThis is about $0.160c$; the displayed results use the nonrelativistic approximation explicitly requested by the paper. The drive amplitude controls the energy gain per gap crossing and hence the number of turns, but the ideal orbit-radius limit is set by $B$ and $R$.