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USAPhO 2021, Финален кръг · подбор на отбора — Задача 3. Great Hall

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Всички материали от USAPhO 2021

2021 USAPhO+ — Traveling Team Selection Exam · 8 май 2021 г.

Information About The 2021 USAPhO+

The 2021 USAPhO+ is a 5-hour exam taking place on Saturday, May 8 from noon to 5 PM, Eastern time.
The exam is hosted by AAPT on the platform provided by Art of Problem Solving. It will be proctored by the US Physics Team coaches via Zoom.
Before you start the exam, make sure you are provided with blank paper, both for your answers and scratch work, writing utensils, graph paper and a ruler, a hand-held scientific calculator with memory and programs erased, and a computer for you to log into the USAPhO+ testing page.
At the end of the exam, you have 20 minutes to upload solutions to all of the problems for that part. For each problem, scan or photograph each page of your solution, combine them into a single PDF file, and upload them on the testing platform.
USAPhO+ graders are not responsible for missing pages or illegible handwriting. No late submissions will be accepted.

Congratulations again on your qualification for the USAPhO+. We wish you the best of luck on the challenging problems to follow.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

JiaJia Dong, Mark Eichenlaub, Abijith Krishnan, Kye W. Shi, Brian Skinner, Mike Winer, and Kevin Zhou.

Условие

The classical Hall effect was first measured by Edwin Hall in 1879, shortly after the publication of Maxwell’s equations. In all parts of this problem, materials contain $n_V$ electrons per unit volume, and each electron has charge $q_e < 0$ and mass $m_e$ . You may use these quantities in all of your answers. We will begin by investigating the implications of the classical Hall effect.

1

An infinite plate in the $xy$ plane, with thickness $d$ in the $z$ direction, is placed in a uniform magnetic field $\mathbf{B} = B\hat{z}$ as shown. An electric field $\mathbf{E} = E\hat{x}$ is applied in the plane of the plate and the system is allowed to reach a steady state.

Grey square plate seen from above; a dotted-circle symbol labelled B (field out of the page) on the left, an arrow labelled E pointing right in the middle of the plate, and axes y (up) and x (right) at the right edge.

Hall effect sensor circuit: a square plate of side L and thickness d with upward magnetic field arrows B, minus signs along the far edge and plus signs along the near edge; an ammeter A (current I) and battery with EMF ℰ (+ and − terminals) in series with the plate, and a voltmeter V measuring V_H across the plate.

(a) If the electrons have velocity $\mathbf{v}$ at steady state, what is the current density $\mathbf{J}$ ? Recall that $\mathbf{J}$ is defined as the total flow of charge through a unit cross-section area per unit time.

(b) In the Drude model, electrons are subject to both the Lorentz force and a damping force $-\gamma\mathbf{v}$ , where $\gamma$ is a constant that depends on the material. In the above system, what is the current density in the steady state? Give both the magnitude and direction of $\mathbf{J}$ , e.g. in polar coordinates.

(c) Compute the electrical resistivity,

\rho_0 = \lim_{B\to0}\frac{E}{|J_x|}

and the transverse Hall resistivity

\rho_H = \lim_{\gamma\to0}\frac{E}{|J_y|}.

(d) A Hall effect sensor detects the strength of magnetic fields. Consider the following circuit consisting of a square plate of side length $L$ and thickness $d$ in a perpendicular uniform magnetic field $B$ .

A longitudinal emf $\mathcal{E}$ is applied to the plate. At steady state, a Hall voltage $V_H$ is measured across the plate due to the buildup of charge on either side of the plate. If the electrical resistivity of the plate at zero magnetic field is $\rho_0$ , what is the Hall voltage $V_H$ and the current $I$ through the plate? Express your answer in terms of $\rho_0$ , $\mathcal{E}$ , $B$ , and the dimensions of the plate.

2

Experiments in the 20th century revealed that in many materials, the Hall resistivity could only take certain discrete values. We will now show how this follows from Bohr quantization. (These next parts are independent of the first part of the problem.)

A zero-resistance loop of wire of radius $R$ and cross-sectional area $A_w$ carries a counterclockwise current $I$ . A solenoid through the middle of the loops carries magnetic flux $\Phi$ out of the page, which we define to be the positive $\hat{z}$ direction.

A circular wire loop with an arrowhead marking the direction of current, surrounding a smaller grey disk (the solenoid) labelled Φ at the center.

(a) If the electrons all have the same speed, what is the angular momentum of each electron?

(b) If we allow the flux in the solenoid to change, the usual, “mechanical” angular momentum $L$ of each electron is not conserved. Instead, a quantity called the canonical angular momentum, $L_{\mathrm{can}} = L + Cq_e\Phi$ , for some constant $C$ , is conserved. Find $C$ .

(c) The Bohr quantization condition says that for a closed circular orbit, an integer number of de Broglie wavelengths must fit in its circumference. The de Broglie wavelength is

\lambda = \frac{h}{p_{\mathrm{can}}},

where $h$ is Planck’s constant, and $p_{\mathrm{can}} = L_{\mathrm{can}}/R$ is the canonical momentum. For a given solenoid flux $\Phi$ , what is the set of allowed mechanical angular momenta $L$ ?

(d) What is the minimum possible change in the magnetic flux for which the same set of mechanical angular momenta is allowed? This is known as the flux quantum.

3

Now, consider an annulus held perpendicular to a fixed, uniform external magnetic field $B$ , and suppose an additional, tunable magnetic flux $\Phi$ threads the center of the annulus, with both pointing out of the page. The annulus has a transverse Hall resistance $R_H$ (i.e., an EMF of $\mathcal{E}$ around the annulus generates a perpendicular current $\mathcal{E}/R_H$ via the Hall effect) and you may neglect its self-inductance.

A grey annulus with a white central hole labelled Φ; to its left a dotted-circle symbol labelled B (field out of the page).

(a) Suppose $\Phi$ begins to increase slowly and steadily in time. After a short time, the electrons will begin flowing steadily from one side of the annulus to the other. Do the electrons move inward or outward? Justify your answer.

(b) If the threaded flux increases by $\Delta\Phi$ , how many electrons pass from one edge of the annulus to the other? You may use $R_H$ , among other variables, in your answer.

(c) As we showed in 2(d), if the magnetic flux changes by the flux quantum $\Phi_q$ , the allowed orbits from Bohr quantization are unchanged. Quantum mechanics thus tells us that in conventional materials, if the magnetic flux changes by $\Phi_q$ , an integer number $k$ of electrons must pass from one edge to another. What constraint does this place on the Hall resistance?

Отговори

Решение

1

(a) The current density is given by $n_Vq_e\mathbf{v}$ .

(b) The equation of motion for an electron is

q_eE\hat{\mathbf{x}} - q_ev_xB\hat{\mathbf{y}} + q_ev_yB\hat{\mathbf{x}} - \gamma v_x\hat{\mathbf{x}} - \gamma v_y\hat{\mathbf{y}} = 0.

Solving the system of equations gives us

v_x = \frac{\gamma q_eE}{\gamma^2 + q_e^2B^2},\quad v_y = \frac{-q_e^2EB}{\gamma^2 + q_e^2B^2}.

Then,

J_x = \frac{\gamma n_Vq_e^2E}{\gamma^2 + q_e^2B^2},\quad J_y = \frac{-n_Vq_e^3EB}{\gamma^2 + q_e^2B^2}.

As requested, we compute the magnitude and direction:

|\mathbf{J}| = \frac{q_e^2n_VE}{\sqrt{\gamma^2 + B^2q_e^2}},\quad \theta = -\arctan\left(\frac{Bq_e}{\gamma}\right).

(c) The two resistivities are

\rho_0 = \frac{\gamma}{n_Vq_e^2},\quad \rho_H = \frac{B}{n_Vq_e}.

(d) The Hall voltage is such that the electric field from the external EMF and from the charge buildup causes the current to flow in the $x$ direction. Our equation of motion is thus

q_e\mathcal{E}/L\hat{\mathbf{x}} + q_eV_H/L\hat{\mathbf{y}} - q_ev_xB\hat{\mathbf{y}} - \gamma v_x\hat{\mathbf{x}} = 0.

Then, we have the two equations

V_H = Lv_xB.\quad Lv_x = \frac{q_e\mathcal{E}}{\gamma}.

Substituting gives us

\begin{aligned} V_H &= \frac{q_e\mathcal{E}B}{\gamma} = \frac{\mathcal{E}B}{n_Vq_e\rho_0}, \\ I &= n_Vq_ev_xLw = \frac{n_Vq_e^2\mathcal{E}d}{\gamma} = \frac{\mathcal{E}d}{\rho_0}. \end{aligned}

2

(a) The current is $I = q_en_VaA_w$ , where $v$ is the speed of each electron. Then the angular momentum is

m_evR\hat{\mathbf{z}} = \frac{Im_eR}{q_en_VA_w}\hat{\mathbf{z}}.

Note that the angular momentum points into the page, because the electrons have negative charge.

(b) We compute the change in angular momentum associated with a change in flux $\Delta\Phi$ . By Faraday’s Law, there is an induced electric field resulting from the change in magnetic flux.

2\pi RE = -\frac{d\Phi}{dt}.

The electric field results in a force $q_eE$ on each electron, and thus we find that

2\pi R\frac{m_e}{q_e}\frac{dv}{dt} = -\frac{d\Phi}{dt}.

Then, the rate of change in angular momentum is

Rm_e\frac{dv}{dt} = \frac{q_e}{2\pi}\frac{d\Phi}{dt}.

Therefore,

\Delta L = -\frac{q_e}{2\pi}\Delta\Phi.

Then, for $L + Cq_e\Phi$ to be conserved,

C = \frac{1}{2\pi}.

(c) The circumference of such an orbit is $2\pi R$ . A de Broglie wavelength is given by

\lambda = \frac{hR}{L_{\mathrm{can}}}.

Then,

n = \frac{2\pi L_{\mathrm{can}}}{h} = \frac{L_{\mathrm{can}}}{\hbar},

or $L_{\mathrm{can}} = n\hbar$ where $\hbar = h/2\pi$ . Then the allowed mechanical momenta satisfy

L = n\hbar - \frac{q_e\Phi}{2\pi},

where $n$ is an integer.

(d) We require $\frac{|q_e|\Delta\Phi}{2\pi} = \hbar$ such that we get the same set of allowed mechanical angular momenta. Then,

\Delta\Phi = \frac{h}{|q_e|}.

Note that for a superconducting ring, the flux quantum is $h/2|q_e|$ because the electrons come in Cooper pairs, but this wasn’t part of this problem.

3

(a) The increasing flux induces a clockwise EMF. The Lorentz force on an electron is

q_e(\mathbf{E} + \mathbf{v}\times\mathbf{B}).

Then, $\mathbf{v}\times\mathbf{B}$ must oppose $\mathbf{E}$ . Therefore, $\mathbf{v}$ must point radially inward by the right hand rule. Therefore, electrons must be flowing radially inward.

(b) We have that

|I| = \frac{1}{R_H}\left|\frac{\mathrm{d}\Phi}{\mathrm{d}t}\right| \implies \Delta Q = \frac{1}{R_H}\Delta\Phi.

Therefore, the number of electrons that pass from one edge to another is

\frac{\Delta\Phi}{R_H|q_e|}.

(c) We set

|q_e|\,\nu = \frac{h}{R_H|q_e|} \implies R_H = \frac{h}{kq_e^2}.

This is known as the integer quantum Hall effect, and it applies to any material shape.

Оригинал в Архива: 2021-USAPhO_plus.pdf · официални решения: 2021-USAPhO_plus_Solutions_v2.pdf