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Условие Отговори Решение

USAPhO 2012, Полуфинал — Задача 6. Question B2

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Всички материали от USAPhO 2012

UNITED STATES PHYSICS TEAM 2012 — Semifinal Exam · 50 т.

Important Instructions for the Exam Supervisor

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This examination consists of two parts.
Part A has four questions and is allowed 90 minutes.
Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minute break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet.
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 1, 2012.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.
Please provide the examinees with graph paper for Part A.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time.
After you have completed Part A you may take a break.
Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time.
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Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

AAPT ID #

Doe, Jamie

A1 - 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 1, 2012.

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

$g = 9.8\ \mathrm{N/kg}$	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
$k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}$	$k_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}$
$c = 3.00 \times 10^{8}\ \mathrm{m/s}$	$k_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
$N_{\mathrm{A}} = 6.02 \times 10^{23}\ (\mathrm{mol})^{-1}$	$R = N_{\mathrm{A}}k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}$
$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}$	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
$1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$	$h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}$
$m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
$\sin\theta \approx \theta - \frac{1}{6}\theta^3$ for $\lvert\theta\rvert \ll 1$	$\cos\theta \approx 1 - \frac{1}{2}\theta^2$ for $\lvert\theta\rvert \ll 1$

Условие

For this problem, assume the existence of a hypothetical particle known as a magnetic monopole. Such a particle would have a “magnetic charge” $q_m$ , and in analogy to an electrically charged particle would produce a radially directed magnetic field of magnitude

B = \frac{\mu_0}{4\pi}\frac{q_m}{r^2}

and be subject to a force (in the absence of electric fields)

F = q_mB

A magnetic monopole of mass $m$ and magnetic charge $q_m$ is constrained to move on a vertical, nonmagnetic, insulating, frictionless U-shaped track. At the bottom of the track is a wire loop whose radius $b$ is much smaller than the width of the “U” of the track. The section of track near the loop can thus be approximated as a long straight line. The wire that makes up the loop has radius $a \ll b$ and resistivity $\rho$ . The monopole is released from rest a height $H$ above the bottom of the track.

Ignore the self-inductance of the loop, and assume that the monopole passes through the loop many times before coming to a rest.

a. Suppose the monopole is a distance $x$ from the center of the loop. What is the magnetic flux $\phi_B$ through the loop?

b. Suppose in addition that the monopole is traveling at a velocity $v$ . What is the emf $\mathcal{E}$ in the loop?

c. Find the change in speed $\Delta v$ of the monopole on one trip through the loop.

d. How many times does the monopole pass through the loop before coming to a rest?

e. Alternate Approach: You may, instead, opt to find the above answers to within a dimensionless multiplicative constant (like $\frac{2}{3}$ or $\pi^2$ ). If you only do this approach, you will be able to earn up to 60% of the possible score for each part of this question.

You might want to make use of the integral

\int_{-\infty}^{\infty}\frac{1}{(1 + u^2)^3}du = \frac{3\pi}{8}

or the integral

\int_0^\pi \sin^4\theta\ d\theta = \frac{3\pi}{8}

Отговори

Решение

a. The most direct way is to calculate the flux through a flat, circular surface bounded by the loop. However, by Gauss’s law the flux will remain the same if we deform this surface, as long as we keep its boundary the same and don’t hit the monopole. The easiest way is to deform the surface so it is part of a sphere of radius $r = \sqrt{b^2 + x^2}$ centered at the monopole.

Work in spherical coordinates, where $\theta$ is the angle to the $x$ -axis and $\phi$ is the azimuthal angle. The loop itself is at $\theta_0 = \tan^{-1}(b/x)$ , so

\Phi_B = \frac{\mu_0}{4\pi}\frac{q_m}{r^2}\int_0^{2\pi}d\phi\int_0^{\theta_0}r^2\sin\theta\,d\theta = \frac{\mu_0q_m}{2}(1 - \cos(\theta_0)).

We can easily check this by limiting cases: for small $x$ we have $\mu_0q_m/2$ , i.e. half of the total flux, while for large $x$ we have zero flux.

This form is acceptable, but we can also write it explicitly in terms of $x$ . By drawing a right triangle, we have

\sin\theta_0 = \frac{b}{r}, \quad \cos\theta_0 = \frac{x}{r}

which gives

\Phi_B = \frac{\mu_0q_m}{2}\left(1 - \frac{x}{\sqrt{b^2 + x^2}}\right).

We will use the variable $\theta_0$ in the parts below, dropping the subscript.

b. Since $\mathcal{E} = -d\Phi_B/dt$ , differentiating both sides of the above result gives

\mathcal{E} = \frac{\mu_0q_mv}{2}\frac{b^2}{(b^2 + x^2)^{3/2}} = \frac{\mu_0q_mv}{2b}\sin^3\theta

where we used the chain rule, with $v = dx/dt$ . Either form is acceptable.

c. Consider the force of the magnetic field produced by the current $I$ in the loop on the monopole. We use the Biot-Savart law,

\mathbf{B} = \frac{\mu_0I}{4\pi}\int\frac{d\mathbf{s}\times\mathbf{r}}{r^3}

where $\mathbf{r}$ is a vector from the monopole to a point on the rim, $d\mathbf{s}$ integrates along the loop, and $r = \sqrt{b^2 + x^2}$ as before. Since $d\mathbf{s}\times\mathbf{r} = r\sin\theta\,ds$ , the integral gives

B = \frac{\mu_0I}{4\pi}\frac{2\pi b\sin\theta}{r^2} = \frac{\mu_0I}{2b}\sin^3\theta.

The force on the monopole is $F = q_mB$ , so the acceleration is

a = \frac{q_mB}{m}.

The current $I$ is related to $\mathcal{E}$ by Ohm’s law, $\mathcal{E} = IR$ , where we’ll compute $R$ later. Then

a = \frac{\mu_0^2q_m^2v}{4b^2mR}\sin^6\theta.

The change in speed in one trip is

\Delta v = \int a\,dt \approx \frac{1}{v}\int_{-\infty}^{\infty}a\,dx = \frac{1}{v}\int_0^\pi a\frac{dx}{d\theta}\,d\theta = -\frac{b}{v}\int_0^\pi\frac{a}{\sin^2\theta}\,d\theta

where we approximated $v$ as constant throughout one trip and the bottom of the track as an infinite straight line. Plugging in our result for $a$ and using the second provided integral,

\Delta v = -\frac{\mu_0^2q_m^2}{4bmR}\int_0^\pi\sin^4\theta\,d\theta = -\frac{3\pi}{32}\frac{\mu_0^2q_m^2}{bmR}.

Finally, we need to find $R$ . Since $a \ll b$ , the loop is approximately a long cylindrical wire, so

R = \rho\frac{2\pi b}{\pi a^2}

which gives the final result

\Delta v = -\frac{3\pi}{64}\frac{\mu_0^2q_m^2a^2}{b^2m\rho}.

We haven’t kept track of the signs very carefully; all that matters is that $\Delta v$ is negative, as it must be by Lenz’s law.

The problem can also be solved by energy conservation, finding the energy dissipated in the loop every trip by integrating $P = \mathcal{E}^2/R$ over time, then relating that to the change in the monopole’s kinetic energy, $\Delta K \approx mv\Delta v$ .

If we had kept the integral above over $dx$ , we would instead have to use the first provided integral. Incidentally, these integrals aren’t too hard to derive. The first is related to the second by $u$ -substitution; for the second, note that

\int_0^\pi\sin^4\theta\,d\theta = \int_0^\pi\left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^4d\theta.

All of the terms in the expansion integrate to zero, by periodicity, except for the constant term. Then the integral is

\frac{1}{16}\binom{4}{2}\int_0^\pi d\theta = \frac{3\pi}{8}

as stated.

d. The initial speed is $\sqrt{2gH}$ by energy conservation, so

N = \frac{\sqrt{2gH}}{\Delta v} = \frac{64\sqrt{2}}{3\pi}\frac{b^2m\rho\sqrt{gH}}{\mu_0^2q_m^2a^2}.

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Оригинал в Архива: USAPhO-2012.pdf · официални решения: USAPhO-2012-Solutions.pdf