NBPhO 2025

Part (i) — 2.0 / 2.0 pts

Criterion	Points	Result
Considers forces on an infinitesimal part of the rod	0.4	✓ slice $dr'$ with $dm=(m/l)\,dr'$, $dq=(q/l)\,dr'$ and explicit $d\vec F_\mathrm{L}/dr'$
Equates, with justification, Lorentz and centrifugal forces $dq\,vB=dm\,\omega^{2}r$	0.4	✓ Newton’s law on the slice with Lorentz, centripetal, and tension-gradient terms, then the half-rod balance $qBl/8\cdot\omega = m\omega^{2}l/8$
Uses $\omega = v/r$	0.4	✓ implicit in $\vec v = \vec\omega\times r'\hat x = \omega r'\hat y$ for the slice velocity
Uses $dq/dm = q/m$	0.4	✓ explicitly stated on the slice
Expresses $\omega = qB/m$	0.4	✓ $\lvert\omega\rvert = qB/m$, with the cyclotron sense identified

Part (ii) — 4.0 / 4.0 pts

Criterion	Points	Result
Deduces, with justification, that the net force on the rod $\vec F = q\vec v_C\times\vec B$	0.5	✓ $\int dq\,\vec v\times\vec B = (q/m)\int dm\,\vec v\times\vec B = q\,\vec v_C\times\vec B$
Uses $v_C = v/2$	0.2	✓ $\vec v_C = \vec v_B + \vec\Omega\times(l/2)\hat x = (v/2)\hat y$
Justifies that the CoM moves on a circular path	0.3	✓ “the CoM equation of motion is identical to that of a single particle of mass $m$ and charge $q$“
Expresses the radius of the path traced by the CoM $R = mv/(2qB)$	0.3	✓ stated alongside $v_C=v/2$
Concludes that the angular velocity of the CoM is $\omega = qB/m$	0.2	✓ $\omega_B = qB/m$
Expresses the angular velocity of the rotation around the CoM $\Omega = v/l$	0.2	✓ from $\vec v_R - \vec v_B = v\hat y = \vec\Omega\times l\hat x$
Justifies that $\Omega$ is conserved	0.3	✓ two independent arguments: König’s theorem with $T$ and $v_C$ both constant; torque about the CoM vanishes
Argues that $t < 2\pi/\omega$ is possible only if $R = l/2$	0.5	✓ algebraic: equation $(\ast)$ forces $R\,(1-\cos\omega_B t)+(l/2)\,(1+\cos\Omega t)=0$, both summands non-negative for $R\ge 0$ so $t\ge T_B$; the only signed-radius branch reaching the origin earlier is $\rho=-l/2$, i.e. $R=l/2$
Justifies that in this case, the red end will never end up at the origin	0.5	✓ “the blue end stays nailed to the origin… the red end traces a circle of radius $l$ around the origin” — locked by the rod and CoM rotating in the same sense at the same rate
Justifies that $\Omega t = \pi + 2\pi k$ with $k\in\mathbb Z_{\ge 0}$	0.4	✓ from $1+\cos\Omega t = 0$, written as $\Omega t = (2k+1)\pi$
Expresses $v = (qBl/m)(1/2 + k)$	0.6	✓ general spin condition $\Omega t = (2k+1)\pi$ combined with $\Omega = v/l$ and $t = nT_B$; smallest case $n=1,k=0$ written out as $v = qBl/(2m)$, the value the problem asks for

Overall score: 6.0 / 6.0 pts — full marks

The two requested numerical answers match the official key exactly: $\lvert\omega\rvert = qB/m$ in Part (i), and $t_{\min} = 2\pi m/(qB)$ with $v = qBl/(2m)$ in Part (ii).

Commentary

Where this solution goes beyond the grading scheme. Part (i) carries a tension-gradient calculation showing that the same $\omega = qB/m$ kills the tension on every slice, not just at the midpoint — with the bracketed factor $-(\omega + qB/m)\,\omega$ identifying $\omega = -qB/m$ as the unique non-trivial root — and then notes the same fact from a “swarm of independent cyclotron orbits” picture. Part (ii) introduces the dimensionless ratio $\beta = \Omega/\omega_B = mv/(qBl)$ as the organising parameter, and the analysis is structured around its three special values $0$, $\pm 1/2$, $\pm 1$. The “two circles externally tangent at $(l/2, 0)$, with centre-to-centre distance $l/2 + R$ equal to the sum of radii” geometric reason why the CoM revisits distance $l/2$ from the origin only at $t = nT_B$ is sharper than just the algebraic count of the cosine zeros, and the ASCII picture with $R = l/4$ and the $T_B/2$ midway-point makes the flip-end-over-end physical motion concrete. Two independent proofs of $\Omega$-conservation are given (König’s theorem and direct torque computation), and the explicit warning that angular momentum about the origin is not conserved (because the magnetic force is not central) heads off a common misstep. Multiple sanity checks close out: dimensions, an energy-conservation cross-check that computes $T_0 = mv^{2}/6$ from the rigid-body decomposition, the symmetric $\beta = \pm 1/2$ pair noted as a mirror reflection that the problem statement fixes, and limits $B\to\infty$, $B\to 0$, $l\to\infty$.

Where the official solution is sharper. Two places. (1) Part (i): the official avoids the half-rod force balance entirely. It opens with “all particles with the same charge-to-mass ratio orbit in a homogeneous magnetic field with the same frequency $\omega_B$”, from which $\omega = qB/m$ is the immediate corollary — every slice does its own orbit and no tension is needed. The official then closes the uniqueness loop with one line (“if $\omega$ were smaller or larger, $Bqv \lessgtr mv\omega$ for all pieces, giving stretching or compressive tension”). Claude reaches both points but only after a half-rod integration, with the cyclotron picture appearing as a “physical interpretation” after the algebraic answer. The official’s ordering — observation first, computation second — is the cleaner pedagogical route. (2) Part (ii): the official gives the general family $v = (qBl/m)(1/2 + k)$ in closed form and only then specialises to the minimum $k=0$. Claude derives the general spin condition $\Omega t = (2k+1)\pi$ but only writes out $v$ for the smallest case. Reporting the family in the final answer keeps the $k$ structure visible to the reader, which is the natural close to the cyclotron-period-counting argument.