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USAPhO 2026, Полуфинал — B3. The Trojan Cylinder

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Всички материали от USAPhO 2026

2026 USA Physics Olympiad Exam

Внимание: Бележка към темата

The supplied archive folder contains the full question packet and the official interleaved key. The cover titled “Instructions for the Student” contains only a reference sheet and acknowledgements; no administrative instructions, timing, point allocations or blank answer sheets are supplied. These values remain unspecified. Printed source inconsistencies are preserved: A3's key changes $C_P,C_V$ to $c_p,c_v$ , prints an empty equation label (0-5), has a plus sign in radial force balance (0-12) followed by a positive pressure gradient (0-13), and gives the exterior pressure without any radial dependence. B2 refers to “part (2)” despite lettered subparts, and its key uses the angle between the asymptote rays as the deflection angle. B3 refers to “part (1)” despite lettered subparts. The original key is transcribed without mathematical corrections.

AAPT UNITED STATES PHYSICS TEAM — AIP 2026; USA Physics Olympiad Exam — Instructions for the Student — REFERENCE SHEET

Fundamental Constants

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$

Useful Approximations

(1+x)^n\approx1+nx+n(n-1)x^2/2\quad\text{for }|nx|\ll1

e^x\approx1+x+x^2/2+x^3/6\quad\text{for }|x|\ll1

\sin\theta\approx\theta-\theta^3/6\quad\text{for }|\theta|\ll1

\cos\theta\approx1-\theta^2/2\quad\text{for }|\theta|\ll1

You may use this sheet for both parts of the exam.

We acknowledge the US Physics Team coaches and other people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Eddie Chen, Kellan Colburn, Evan Erickson, Harry Kroft, Natalie LeBaron, Rishab Parthasarathy.

Условие

In alternating-current systems, the quality factor $Q$ can be defined as

Q = 2\pi \times \frac{\text{maximum energy stored}}{\text{energy lost per cycle}}.

(a) A circuit consists of an inductor of inductance $L$ and resistance $R$ , carrying a current

I(t) = I_0\cos(\omega t).

Using the energy definition of $Q$ , express $Q$ in terms of $\omega$ , $L$ , and $R$ .

(b) A long solenoid of inductance $L$ is made of superconducting wire and is tightly wound around a thin conducting cylindrical shell of radius $a$ , thickness $t \ll a$ , conductivity $\sigma$ , and length $\ell$ . Ignore end effects and skin effects.

The current in the solenoid is

I(t) = I_0\cos(\omega t).

Using the definition of $Q$ , determine $Q$ in terms of $\omega$ , $\sigma$ , $t$ , $a$ , and $L$ .

(c) Using your result from part (1), determine the effective resistance of the system.

(d) Now suppose the solenoid is connected in series with an external resistor $R$ . Determine the total quality factor of the system in the presence of the conducting cylinder.

(e) Compare the frequency dependence of the quality factor $Q$ for:

an $RL$ circuit consisting of an ideal inductor $L$ in series with a resistor $R$ ,
the same circuit with the conducting cylinder inserted.

Explain physically why the behavior differs.

Решение

(a) The maximum stored energy is

U_{\max} = \frac{1}{2}LI_0^2.

The average power dissipated is

\langle P\rangle = \frac{1}{2}RI_0^2.

The energy lost per cycle is

E_{\mathrm{loss}} = \langle P\rangle\frac{2\pi}{\omega} = \frac{\pi R I_0^2}{\omega}.

Thus,

Q = 2\pi\frac{U_{\max}}{E_{\mathrm{loss}}} = \frac{\omega L}{R}.

(b) The magnetic field inside the solenoid is

B = \mu_0\frac{N}{\ell}I.

The magnetic flux through the cylinder is

\Phi = B\pi a^2 = \mu_0\frac{N}{\ell}\pi a^2 I.

Thus, the induced emf around the cylinder is

\mathcal{E} = -\frac{d\Phi}{dt} = -\mu_0\frac{N}{\ell}\pi a^2\frac{dI}{dt}.

Since

\frac{dI}{dt} = -\omega I_0\sin(\omega t),

the amplitude of the emf is

\mathcal{E}_0 = \mu_0\frac{N}{\ell}\pi a^2\omega I_0.

The resistance of the cylindrical shell is

R_{\mathrm{cyl}} = \frac{2\pi a}{\sigma t\ell}.

Thus, the induced current amplitude is

I_{\mathrm{ind},0} = \frac{\mathcal{E}_0}{R_{\mathrm{cyl}}} = \frac{\mu_0\frac{N}{\ell}\pi a^2\omega I_0}{\frac{2\pi a}{\sigma t\ell}} = \frac{\mu_0 N\sigma t a\omega I_0}{2}.

The average power dissipated is

\langle P\rangle = \frac{1}{2}R_{\mathrm{cyl}}I_{\mathrm{ind},0}^2.

Substituting,

\langle P\rangle = \frac{1}{2}\cdot\frac{2\pi a}{\sigma t\ell}\cdot\left(\frac{\mu_0 N\sigma t a\omega I_0}{2}\right)^2.

Simplifying,

\langle P\rangle = \frac{\pi\mu_0^2 N^2\sigma t a^3\omega^2 I_0^2}{4\ell}.

The energy lost per cycle is

E_{\mathrm{loss}} = \langle P\rangle\frac{2\pi}{\omega} = \frac{\pi^2\mu_0^2 N^2\sigma t a^3\omega I_0^2}{2\ell}.

The maximum stored energy is

U_{\max} = \frac{1}{2}LI_0^2.

Thus,

Q = 2\pi\frac{U_{\max}}{E_{\mathrm{loss}}} = 2\pi\frac{\frac{1}{2}LI_0^2}{\frac{\pi^2\mu_0^2 N^2\sigma t a^3\omega I_0^2}{2\ell}} = \frac{2L\ell}{\pi\mu_0^2 N^2\sigma t a^3\omega} = \frac{2}{\mu_0\sigma t a\omega}.

(c) From part (1),

Q = \frac{\omega L}{R_{\mathrm{eff}}},

R_{\mathrm{eff}} = \frac{\omega L}{Q}.

Substituting the expression for $Q$ ,

R_{\mathrm{eff}} = \omega L\cdot\frac{\mu_0\sigma t a\omega}{2} = \frac{1}{2}\mu_0\sigma t a L\omega^2.

(d)

Q = \frac{\omega L}{R+R_{\mathrm{eff}}}.

(e) For a standard $RL$ circuit,

Q = \frac{\omega L}{R},

so $Q$ increases linearly with frequency.

With the conducting cylinder inserted,

Q = \frac{\omega L}{R+R_{\mathrm{eff}}}, \qquad R_{\mathrm{eff}} = \frac{1}{2}\mu_0\sigma t a L\omega^2.

Therefore,

Q = \frac{\omega L}{R+\frac{1}{2}\mu_0\sigma t a L\omega^2}.

Thus:

at low frequency, $R_{\mathrm{eff}}\ll R$ , so

Q\approx\frac{\omega L}{R}\propto\omega;

at high frequency, $R_{\mathrm{eff}}\gg R$ , so

Q\approx\frac{\omega L}{R_{\mathrm{eff}}}\propto\frac{1}{\omega}.

So the conducting cylinder causes the quality factor to reach a maximum and then decrease. Physically, this happens because rapidly changing magnetic fields induce strong eddy currents in the cylinder, leading to energy dissipation that grows like $\omega^2$ .

Оригинал в Архива: 2026-USAPhO.pdf · официални решения: 2026-USAPhO-Solutions.pdf