NBPhO 2025

Overview

A uniformly charged rod in a uniform magnetic field is, despite appearances, an almost trivial system once one notices that every infinitesimal slice has the same charge-to-mass ratio $q/m$. Several remarkable simplifications follow from that single fact:

1. Every slice has the same cyclotron frequency $\omega_c = qB/m$.
2. The total Lorentz force on the rod is $q\,\vec v_C\times\vec B$ — the centre of mass orbits like a single point charge.
3. The total Lorentz torque about the centre of mass vanishes, so the spin rate $\Omega$ about the CoM is conserved.
4. The Lorentz force does no work, so kinetic energy is conserved.

Once these are in hand, the motion of the rod is the superposition of two independent rigid rotations: the CoM goes around a cyclotron circle at $\omega_B = qB/m$, while the rod spins about the CoM at a constant $\Omega$ set by the initial conditions. Part (i) asks for the special case $\Omega = \omega_c$ in which these two motions become “the same motion”, and the rod glides through the field with no internal stress. Part (ii) is a geometry problem: when, in the superposed motion, can a specified end of the rod return to a specified point?

The relevant dimensionless parameter is

$$\beta \;=\; \frac{\Omega}{\omega_B} \;=\; \frac{m v}{q B l},$$

the ratio of the spin rate to the cyclotron rate. Three special values matter: $\beta = 0$ (rod just orbits without spinning), $|\beta|=1/2$ (the answer to part (ii)), and $|\beta|=1$ (the pathological “blue-end-fixed” case).

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Part (i) — Tension-free rotation rate

Why every slice has the same cyclotron frequency

The charge-to-mass ratio is uniform along the rod: a slice of length $dr$ carries mass $dm = (m/l)\,dr$ and charge $dq = (q/l)\,dr$, so $dq/dm = q/m$. A free particle with this charge-to-mass ratio would orbit in $\vec B = B\hat z$ at the cyclotron frequency

$$\omega_c \;=\; \frac{q B}{m},$$

with the gyration clockwise when viewed from $+\hat z$ for $q>0$ (since $\vec a = (q/m)\vec v\times\vec B$ pulls the particle to the right of its motion).

Setting up the force balance on a slice

Place the rod momentarily along $\hat x$ and let it rotate about its CoM with angular velocity $\vec\omega = \omega\hat z$ (we keep $\omega$ signed so that the sense of rotation will fall out automatically). A slice at signed position $r'\hat x$ (with $r'\in[-l/2, l/2]$) has velocity

$$\vec v \;=\; \vec\omega \times r'\hat x \;=\; \omega r'\,\hat y.$$

Its Lorentz force per unit length is

$$\frac{d\vec F_{\mathrm{L}}}{dr'} \;=\; \frac{q}{l}\,(\omega r'\hat y)\times(B\hat z) \;=\; \frac{q\,\omega\,B}{l}\,r'\,\hat x.$$

For $\omega>0$ (counterclockwise spin, $q>0$) this force points outward, away from the rod’s centre, on every slice. For $\omega<0$ it points inward.

Each slice undergoes uniform circular motion at radius $|r'|$, so its required centripetal acceleration is $-\omega^2 r'\,\hat x$ — i.e. inward for any sign of $\omega$. Newton’s law on the slice reads

$$\frac{d m}{dr'}\,(-\omega^{2} r')\,\hat x \;=\;\frac{d\vec F_{\mathrm{L}}}{dr'} \;+\;\frac{d\vec T}{dr'},$$

where $d\vec T/dr'$ is the gradient of the longitudinal tension along the rod. Eliminating tension everywhere is asking too much; what the problem requests is that the midpoint be tension-free.

Translating “zero tension at the midpoint” into a force balance

Cut the rod at its centre. The right half is a rigid body of mass $m/2$ and charge $q/2$ rotating about the CoM at angular speed $\omega$. If the cut is to carry no force, the net Lorentz force on that half must by itself supply the centripetal force needed by the half.

Net Lorentz on the right half:

$$\vec F_{\mathrm{L}}^{\,(\text{right})} \;=\;\int_{0}^{l/2}\frac{q\,\omega\,B}{l}\,r'\,dr'\,\hat x \;=\;\frac{q\,\omega\,B\,l}{8}\,\hat x.$$

The right half’s CoM is at $r' = l/4$, moving in a circle of radius $l/4$ at angular speed $\omega$. Its centripetal force is

$$\vec F_{c}^{\,(\text{right})} \;=\; -\,(m/2)\,\omega^{2}\,(l/4)\,\hat x \;=\; -\,\frac{m\,\omega^{2}\,l}{8}\,\hat x.$$

Setting Lorentz equal to centripetal,

$$\frac{q\,\omega\,B\,l}{8} \;=\; -\,\frac{m\,\omega^{2}\,l}{8} \quad\Longrightarrow\quad \omega \;=\; -\,\frac{q B}{m}.$$

The negative sign just records that the rod must spin in the same sense as the cyclotron motion of a free positive charge — clockwise viewed from $+\hat z$.

$$\boxed{\;|\omega| \;=\; \frac{q B}{m}\;}$$

Why the same $\omega$ kills the tension everywhere, not just at the centre

The slice-level equation derived above can be reorganised to give the tension gradient explicitly:

$$\frac{d\vec T}{dr'} \;=\;\frac{m}{l}\bigl(-\omega^{2} - \tfrac{q}{m}\,\omega\,B\bigr)\,r'\,\hat x.$$

The bracketed factor is exactly $-(\omega + qB/m)\,\omega$, which vanishes identically at $\omega = -qB/m$. So at the cyclotron rate the longitudinal tension gradient is zero on every slice, and since the tension also vanishes at the free ends, it vanishes throughout the rod. The midpoint is not special — the question could equally have asked about any point.

Physical interpretation

At $\omega = qB/m$, every slice of the rod is doing its own cyclotron orbit of radius $|r'|$ about the CoM. The rigid-body picture and the “swarm of independent free charges” picture give exactly the same trajectory. No internal force is required to hold the rod together because the elastic, particle-by-particle motion already keeps neighbouring slices at fixed relative positions.

This is a special case of a more general theorem: a rigid body with uniform $q/m$ in a uniform $\vec B$ admits “torque-free, tension-free” rotations at $\omega_c$, in addition to the trivial CoM cyclotron orbit.

Consistency check — dimensions and limits

$\omega = qB/m$ has units of $[\mathrm{C}\cdot\mathrm{T}/\mathrm{kg}] = [\mathrm{C}\cdot \mathrm{kg}\,\mathrm{s}^{-1}\,\mathrm{C}^{-1}/\mathrm{kg}] = [\mathrm{s}^{-1}]$, correct. Limits: $B\to 0$ gives $\omega\to 0$ (no field, the only tension-free rotation is “no rotation”, as the rod would otherwise need a centripetal pull); $q\to 0$ gives $\omega\to 0$ for the same reason. Both pass.

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Part (ii) — Smallest return time

Geometry of the problem

The rod is initially the segment from $(0,0)$ to $(l,0)$, with the blue end at the origin (at rest) and the red end at $(l,0)$ moving with velocity $v\hat y$. We are asked for the smallest later time at which the red end again coincides with the origin.

Three results from the structural facts in the Overview will do all the heavy lifting.

The CoM orbits like a point charge

For a body with uniform $q/m$, the total Lorentz force is

$$\vec F_{\text{tot}} \;=\;\int dq\,\vec v\times\vec B \;=\;\frac{q}{m}\int dm\,\vec v\times\vec B \;=\;\frac{q}{m}(M\vec v_C)\times\vec B \;=\;q\,\vec v_C\times\vec B.$$

So the CoM equation of motion is identical to that of a single particle of mass $m$ and charge $q$. The CoM traces a circle of radius

$$R \;=\; \frac{m\,v_C}{q\,B}$$

at angular frequency $\omega_B = qB/m$.

Initial CoM data. The rod is rigid, so the velocity field is $\vec v(\vec r) = \vec v_B + \vec\Omega\times(\vec r - \vec r_B)$. From $\vec v_R - \vec v_B = v\hat y = \vec\Omega\times l\hat x$ we read

$$\vec\Omega \;=\;\frac{v}{l}\,\hat z, \qquad \Omega \;=\;\frac{v}{l}.$$

The CoM sits at $\vec r_C = (l/2)\hat x$, with velocity

$$\vec v_C \;=\; \vec v_B + \vec\Omega\times(l/2)\hat x \;=\; \tfrac{v}{2}\,\hat y.$$

Hence

$$R \;=\; \frac{m v}{2 q B}, \qquad v_C \;=\; \frac{v}{2}.$$

The cyclotron sense is clockwise (for $q>0$), so the orbit centre lies in the direction of $q\vec v_C\times\vec B = (qvB/2m)\hat x$, i.e. the centre is at

$$\vec C \;=\; \bigl(l/2 + R,\; 0\bigr).$$

The spin rate is conserved

Two independent arguments make the spin angular speed about the CoM constant.

Argument 1 — energy conservation. The Lorentz force is everywhere perpendicular to velocity, so it does no work. By König’s theorem,

$$T \;=\; \tfrac{1}{2}m v_C^{2} + \tfrac{1}{2}I\,\Omega^{2}$$

is conserved. The first term is conserved separately (the CoM is on a cyclotron orbit, so $|v_C|$ is constant). Therefore $\Omega^{2}$ — and, by continuity, $\Omega$ itself — is constant.

Argument 2 — torque about the CoM vanishes. Write $\vec r' = \vec r - \vec r_C$ and $\vec v = \vec v_C + \vec\Omega\times\vec r'$. The total torque about the CoM is

$$\vec\tau \;=\; \int \vec r' \times d\vec F \;=\; \int dq\,\vec r'\times\!\bigl[(\vec v_C + \vec\Omega\times\vec r')\times\vec B\bigr].$$

The $\vec v_C$ part gives $\bigl(\!\int dq\,\vec r'\bigr)\times(\vec v_C\times\vec B) = 0$ because the centre of charge coincides with the centre of mass for uniform $q/m$. The $\vec\Omega\times\vec r'$ part: with $\vec\Omega \parallel \vec B \parallel \hat z$ and $\vec r'$ in the plane, one easily checks $(\vec\Omega\times\vec r')\times\vec B = \Omega B\,\vec r'$, so $\vec r'\times[\,\Omega B\,\vec r'\,] = 0$. The torque vanishes identically and $\Omega$ is constant.

(Note: the angular momentum about the origin is not conserved, because the magnetic field exerts a non-central force on the rod. It is the angular momentum about the CoM, and the spin alone, that are conserved here.)

Trajectory of the red end

Combine the two motions. Take the spin angle about the CoM, measured from $+\hat x$, to be $\theta(t) = \Omega t$ (with $\Omega = v/l$, so positive $v$ gives counterclockwise spin). The CoM, on its clockwise cyclotron orbit, is parameterised by

$$\vec r_C(t) \;=\; \bigl(l/2 + R - R\cos\omega_B t,\; R\sin\omega_B t\bigr).$$

(Check: at $t=0$, $\vec r_C = (l/2, 0)$, $\dot{\vec r}_C = (0, R\omega_B) = (0, v/2)$, ✓.)

Since the rod has length $l$ and the red end is initially at $+\hat x$ from the CoM, at time $t$ the red end is at

$$\vec r_R(t) \;=\; \vec r_C(t) \;+\;\frac{l}{2}\bigl(\cos\Omega t,\; \sin\Omega t\bigr).$$

Red end at the origin — a geometry problem in disguise

We want $\vec r_R(t) = 0$ with $t>0$. Expand:

$$\begin{aligned} 0 &\;=\; l/2 + R - R\cos\omega_B t + (l/2)\cos\Omega t, \\ 0 &\;=\; R\sin\omega_B t + (l/2)\sin\Omega t. \end{aligned}$$

The first equation rearranges to

$$R\,(1 - \cos\omega_B t) \;+\; (l/2)\,(1 + \cos\Omega t) \;=\; 0. \tag{$\ast$}$$

For $R \geq 0$ both summands are non-negative, so each must vanish individually:

$$1 - \cos\omega_B t \;=\; 0 \quad\text{and}\quad 1 + \cos\Omega t \;=\; 0.$$

These force

$$\omega_B t \;=\; 2\pi n, \qquad \Omega t \;=\; (2k+1)\pi, \qquad n,k\in\mathbb Z_{\geq 0}.$$

The $y$-equation is automatically satisfied because both $\sin\omega_B t$ and $\sin\Omega t$ vanish under those conditions.

Why the $y$-equation kills the “tangent” geometry trap

A purely geometric way to see the same thing: the red end at the origin requires the CoM to lie at distance $l/2$ from the origin (since it is the rod’s midpoint and the rod has length $l$). The CoM’s cyclotron orbit is a circle of radius $R$ around $\vec C = (l/2 + R, 0)$, and $\vec C$ is at distance $l/2 + R$ from the origin, exactly the sum of the two radii. The two circles are therefore externally tangent, touching only at the single point $(l/2, 0)$ — the rod’s starting point. So the CoM is at distance $l/2$ from the origin only at integer multiples of the cyclotron period, $t = n T_B$.

The geometry-only argument is what proves that no $t < T_B$ is allowed; the algebraic argument above is what fixes the rod’s orientation at those allowed times.

Reading off the smallest solution

Smallest $n$ is $n=1$, smallest $k$ is $k=0$. Then

$$t \;=\;\frac{2\pi}{\omega_B}\;=\;T_B, \qquad \Omega T_B \;=\;\pi \;\Longrightarrow\; \Omega \;=\;\frac{\omega_B}{2}.$$

Translating $\Omega = v/l$ and $\omega_B = qB/m$:

$$\frac{v}{l} \;=\; \frac{q B}{2 m} \quad\Longrightarrow\quad v \;=\; \frac{q B l}{2 m},$$

and

$$t_{\min} \;=\; \frac{2\pi m}{q B}.$$ $$\boxed{\;t_{\min} \;=\; \frac{2\pi m}{q B},\qquad v \;=\; \frac{q B l}{2\,m}\;}$$

Picture of the motion

With these values, $R = mv/(2qB) = l/4$, so the CoM orbits a circle of radius $l/4$ centred at $(3l/4,\,0)$:

                   ↑ y
                   |
                   |     ___...___
                   |  ,-'         `-,
              l/4  |.'               `.
                  /·                   ·\
           ──────●─────────●─────────●──────→ x
                 |  l/2    3l/4      l
                  \·                   ·/
              -l/4 `.                .'
                   |  `-,_________,-'
                   |

The CoM (dot at the left of the orbit) starts at $(l/2, 0)$, moves up and to the right (clockwise), passes through $(l, 0)$ at $t = T_B/2$, then back through $(l/2, 0)$ at $t = T_B$.

Meanwhile, the rod spins counterclockwise about the CoM at $\Omega = \omega_B/2$. In one cyclotron period $T_B$, the spin angle advances by $\pi$: the rod has flipped end-over-end. So at $t = T_B$ the CoM is back at $(l/2,0)$ but the rod is reversed, with the red end now at $(0,0)$ and the blue end at $(l,0)$. The rod has interchanged its two ends.

The pathological "$|\beta|=1$" case

The dimensionless ratio $\beta = \Omega/\omega_B = mv/(qBl)$ encodes everything. Could a smaller $t$ be obtained by taking $\beta$ negative (i.e. the red end’s initial speed in $-\hat y$, so that the spin is clockwise like the cyclotron orbit)? In equation $(\ast)$, allowing $R<0$ formally (let $\rho = mv/(2qB)$ be a signed orbit radius) replaces the equation by

$$\rho\,(1 - \cos\omega_B t) \;+\; (l/2)\,(1 + \cos\Omega t) \;=\; 0,$$

and for $\rho<0$ the two terms can cancel against each other.

A short geometric inspection shows the CoM orbit is internally tangent to the $|\vec r|=l/2$ circle for $-l/2 < \rho < 0$ (and again touches only at the start), and internally coincident for $\rho = -l/2$. The only new structure appears at $\rho = -l/2$, i.e. $\beta = -1$ — that is, $|v| = qBl/m$.

In that case the CoM is permanently on the circle $|\vec r| = l/2$, and a calculation along the same lines as above gives

$$\vec r_C(t) \;=\; (l/2)\bigl(\cos\omega_B t,\,-\sin\omega_B t\bigr), \qquad \vec r_R(t) \;=\; 2\,\vec r_C(t),$$ $$\vec r_B(t) \;\equiv\; 0.$$

The blue end stays nailed to the origin (which is consistent with its initial velocity $\vec v_B(0) = 0$), and the red end traces a circle of radius $l$ around the origin. The red end never reaches the origin in this case.

This case is therefore a near miss — at first sight $\beta=-1$ allows the CoM to be at distance $l/2$ from the origin at any time, suggesting some $t < T_B$ might work, but the spin direction is wrong: rod and CoM rotate at the same rate in the same sense, locking the blue end at the origin instead of bringing the red one there.

Sanity checks

• Dimensions. $t_{\min} = 2\pi m/(qB)$ has units $[\mathrm{kg}/(\mathrm{C}\cdot\mathrm{T})] = [\mathrm{s}]$ ✓; $v = qBl/(2m)$ has units $[\mathrm{C}\cdot\mathrm{T}\cdot\mathrm{m}/\mathrm{kg}] = [\mathrm{m/s}]$ ✓.
• Energy conservation. Initial kinetic energy: blue end at rest, red end at $v\hat y$, intermediate slices linearly interpolated; $T_0 = \tfrac{1}{2}m v_C^{2} + \tfrac{1}{2}I\Omega^{2} = \tfrac{1}{2}m(v/2)^{2} + \tfrac{1}{2}(ml^{2}/12)(v/l)^{2} = \tfrac{m v^{2}}{8} + \tfrac{m v^{2}}{24} = \tfrac{m v^{2}}{6}$. At $t = T_B$ the rod has flipped, so red and blue have swapped roles — by symmetry the KE is unchanged. Consistent with Lorentz doing no work.
• Limits. Strong field $B\to\infty$: $t_{\min}\to 0$ and $v\to\infty$. Both are sensible — a tiny cyclotron orbit completes almost instantly, but the spin rate must be huge to flip the rod in that short time. Weak field $B\to 0$: $t_{\min}\to\infty$ (no force) and $v\to 0$; the only way to “return” in the no-field case is infinitely slowly. Long rod $l\to\infty$: $v\to\infty$ — to flip a long rod in one cyclotron period one needs a correspondingly larger angular speed, hence a faster red-end speed.
• A second solution at the same $t$. $\beta = +1/2$ and $\beta = -1/2$ give the same $t_{\min}$ by symmetry (mirror reflection across $\hat x$). The problem fixes the sign by specifying the red end’s velocity direction, so a single answer.

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Summary of results

Part	Quantity	Result
(i)	Tension-free angular speed	$\;\displaystyle \lvert\omega\rvert \;=\; \dfrac{qB}{m}\;$ (cyclotron sense)
(ii)	Smallest return time	$\;\displaystyle t_{\min} \;=\; \dfrac{2\pi m}{qB}\;$
(ii)	Required initial red-end speed	$\;\displaystyle v \;=\; \dfrac{q B l}{2\,m}\;$

One sentence to take away. A uniformly-charged rigid body in a uniform magnetic field decouples cleanly into a CoM cyclotron orbit and a free spin about the CoM; both rotate at constant rates, and any “return” question is the geometric question of when those two rotations re-align.