= 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 Lagrangian>
$$
L=-mc^2\sqrt{1-\mathbf v^2/c^2}+q\mathbf A\cdot\mathbf v,
$$
with 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 $B_z=\rho^{-1}\partial_\rho(\rho A_\theta)=B$. Hence
$$
\boxed{L=-mc^2\sqrt{1-\frac{\dot z^2+\dot\rho^2+\rho^2\dot\theta^2}{c^2}}+\frac{qB\rho^2\dot\theta}{2}}.
$$
In 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 $B_\rho=-\rho B'(z)/2$ 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.
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