Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 2 / a / Solution

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.