OPhO 2024, Теория — Задача 1. Penned Particles

a.

The electrodes are equipotentials. The potential of the endcaps is $V=V_0\frac{z_0^2}{2d^2}$, while the potential of the ring is $V_0\frac{-r_0^2}{2d^2}$. Thus, $d^2=\frac12(z_0^2+r_0^2)$. The equation for the endcaps is $z^2=z_0^2+r^2$, and for the ring is $z^2=r^2-r_0^2$.

Grading Scheme: 5.0 pts total.

• 1.5 points for realizing the electrodes are equipotentials, or equivalent.
• 1.5 points for correct equations for the cross sections: $z^2=z_0^2+r^2$ and $z^2=r^2-r_0^2$, OR $z^2=z_0^2+\frac12r^2$ and $z^2=\frac12r^2-\frac12r_0^2$.
• 2.0 points for calculation of potential of endcaps and and ring, and using this to find an expression for $d$: $d=\frac12(z_0^2+r_0^2)$, OR $d=\frac12(2z_0^2+r_0^2)$.

b.

Because $E=-V_0\frac z{d^2}$, we find $\omega_z=\sqrt{\frac{qV_0}{md^2}}$. Then, the equations of motion are:

Some contestants wrote their equations in terms of $x$ and $y$ coordinates; we will also accept those as valid.

We have $\ddot{\mathbf r}=q\dot{\mathbf r}\times\mathbf B/m$. It’s evident the particle performs uniform circular motion at frequency $\omega_c=\frac{qB_0}m$. Note that $\omega_c$ points opposite $\mathbf B$.

Grading Scheme: 7.0 pts total.

• 2 points for correct axial angular frequency: $\omega_z=\sqrt{\frac{qV_0}{md^2}}$ OR $\omega_z=\sqrt{\frac{2qV_0}{md^2}}$.
• 2 points for correct differential equations.
• 3 points for the correct cyclotron frequency.

c.

To start, we have $\ddot{\mathbf r}=\omega_z^2\mathbf r+\dot{\mathbf r}\times\boldsymbol\omega_c$. Suppose we have $\mathbf r=(r,\omega t)$ in polar coordinates. Then $\dot{\mathbf r}=\omega r\hat\theta$, and $\ddot{\mathbf r}=-\omega^2\mathbf r$. We plug these into the differential equation, to get

Using the quadratic equation, this gives us

Recall that when $\omega_z=0$, the cyclotron motion has frequency $\omega_c$, so we take

This is less than $\omega_c$ because of the radial potential pushing the ion outwards. Then, the magnetron frequency is

We thus have $\omega_c\gg\omega_z\gg\omega_m$.

Grading Scheme: 8.0 pts total.

• 3 points for guessing a circular/periodic/complex exponential solution to the differential equation.
• 4 points for solving the differential equation and correctly deriving the two possible frequencies.
• 1 point for correctly identifying which frequency is the magnetron frequency and which is the cyclotron frequency.

d.

The original formula for power provided was missing a factor of $\pi$; we will not penalize solutions that are missing this factor. Note that $E=\frac12m\dot r^2=\frac12m\omega_c^2r^2$. The acceleration is $\omega_c^2r$, so

Thus,

The energy is $qV_0\frac{z_m^2}{2d^2}=\frac12m\omega_z^2z_m^2$, where $z_m$ is the amplitude of the oscillation. We have $z=z_m\sin\omega_zt$, and $a=-\omega_z^2z$. Averaged over one oscillation, $\langle a^2\rangle=\frac12\omega_z^4z_m^2$. We then have

We then have

Grading Scheme: 5.0 pts total.

• 0.5 points for correct expression for the energy of the cyclotron motion.
• 1 point for correct expression for the power.
• 0.5 points for correct damping constant $\gamma_c$.
• 1 point for correct expression for the energy of the axial motion.
• 1.5 points for the correct expression for the time-averaged power.
• 0.5 points for correct damping constant $\gamma_z$.

e.

Supposing that the endcaps were infinite flat planes, the electric field will be $\frac{IR}{2z_0}$. So we have $E=\frac{\kappa IR}{2z_0}$.

The power on the ion is $F\dot z=qE\dot z$, which we equate to $RI^2$:

Then

So we have

Grading Scheme: 5.0 pts total.

• 1.5 points for the correct electric field at the center of the trap.
• 1.5 points for correct power on the ion, and equating it to the power lost through the resistor.
• 1.5 points for solving for current in terms of $\dot z$, or equivalent derivation of the force in terms of $\dot z$.
• 0.5 point for correct expression for $\zeta$.

f.

Note that:

Since $\omega_z\gg\omega_m$ this is obviously negative.

Since the system is in equilibrium, over the long run the change in entropy must be zero. When photons interact with the ion, we can view it as adding heat $Q_z=\hbar\omega_z$ and $Q_m=\hbar\omega_m$ to reservoirs at temperatures $T_z$ and $T_m$ respectively. The change in entropy is

. So, we have

Many teams reported $T_m$ to be positive. Here is a subtle point about negative temperature. Earlier in the problem, we derived that the total energy of the magnetron motion was negative. If we were to add energy, the radius would shrink. Very informally, temperature is defined such that $T\,dS=dE$, and intuitively, an increase in the radius of the orbit is an increase in entropy. (This can all be formalized with quantum mechanics). Adding energy decreases the entropy, which is why $T_m$ is negative.

Grading Scheme: 5.0 pts total.

• 1 point for correct expression for the energy (up to first order).
• 3 points for conserving entropy.
• 1 point for correct expression for $T_m$. $-0.5$ points if positive.