NBPhO 2025

Difficulty hard

Prerequisites

• Lorentz force on a moving charge, $\vec F = q\vec v\times\vec B$
• Cyclotron motion of a point charge: $\omega_B = qB/m$, $R = mv/(qB)$
• Splitting rigid-body motion into CoM translation plus rotation about the CoM
• König's theorem, $T = \tfrac{1}{2}M v_C^2 + \tfrac{1}{2}I\Omega^2$
• Uniform circular motion and centripetal force

Learning objectives

• Prove that for a body with uniform $q/m$, the net Lorentz force is $q\,\vec v_C\times\vec B$ and the CoM moves as a point charge
• Derive the zero-tension rotation rate $\omega = qB/m$ from local centripetal balance on a slice
• Use energy conservation (Lorentz does no work) plus König's theorem to argue that $\Omega$ about the CoM is constant
• Superpose two simultaneous rotations — CoM orbit at $\omega_B$ and spin at $\Omega$ — and solve the geometric return condition
• Recognise the tangency of the CoM orbit to the circle $|\vec r| = l/2$ and the pathological blue-end-stays-at-origin case

Watch out for

• The rod's spin rate $\Omega = v/l$ about the CoM is not in general equal to the cyclotron frequency $\omega_B = qB/m$. Part (i)'s special value applies only to part (i); do not carry it over to part (ii).
• The smallest return time is a full cyclotron period $T_B$, not $T_B/2$. The CoM orbit is tangent (not secant) to the $|\vec r| = l/2$ circle, so the CoM revisits distance $l/2$ from the origin only at $t = n T_B$.
• The special case $R = l/2$ looks like it would allow $t < T_B$, but it corresponds to the orbit-direction branch where the rod and the CoM rotate in the same sense at equal rates — the blue end then sits fixed at the origin and the red end never arrives. Missing this case (or miscounting it as an extra solution) is the main 0.5-pt trap in the grading scheme.
• Angular momentum about the origin is not conserved — the magnetic field exerts a non-central force. Use energy conservation, not $L$-conservation, to pin down that $\Omega$ is constant.