Съдържание

Условие Отговори Решение

USAPhO 2012, Полуфинал — Задача 3. Question A3

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Всички материали от USAPhO 2012

UNITED STATES PHYSICS TEAM 2012 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

DO NOT DISTRIBUTE THIS PAGE

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.
Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets.
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

This problem inspired by the 2008 Guangdong Province Physics Olympiad

Two infinitely long concentric hollow cylinders have radii $a$ and $4a$ . Both cylinders are insulators; the inner cylinder has a uniformly distributed charge per length of $+\lambda$ ; the outer cylinder has a uniformly distributed charge per length of $-\lambda$ .

An infinitely long dielectric cylinder with permittivity $\epsilon = \kappa\epsilon_0$ , where $\kappa$ is the dielectric constant, has a inner radius $2a$ and outer radius $3a$ is also concentric with the insulating cylinders. The dielectric cylinder is rotating about its axis with an angular velocity $\omega \ll c/a$ , where $c$ is the speed of light. Assume that the permeability of the dielectric cylinder and the space between the cylinders is that of free space, $\mu_0$ .

Perspective drawing of the concentric cylinders cut off at the left by a zigzag break line: the large outer hollow cylinder, the shaded dielectric cylindrical shell with a curved arrow showing its rotation, and the thin inner cylinder protruding at the right; beside it an end-on cross-section with three concentric circles, the dielectric annulus shaded grey and an arrow along it indicating the sense of rotation.

a. Determine the electric field for all regions.

b. Determine the magnetic field for all regions.

Отговори

Решение

a. Consider a Gaussian cylinder of radius $r$ and length $l$ centered on the cylinder axis. The electric field is radial, so Gauss’s Law states that

\oint \mathbf{E}\cdot d\mathbf{A} = \frac{q_{\mathrm{in}}}{\epsilon_0} \quad\Rightarrow\quad 2\pi rEl = \frac{\lambda_{\mathrm{in}}l}{\epsilon_0}

where $\lambda_{\mathrm{in}}$ is the linear charge density enclosed in the cylinder, so

\mathbf{E} = \frac{\lambda_{\mathrm{in}}}{2\pi r\epsilon_0}\hat{\mathbf{r}}.

The field due to the hollow cylinders alone is therefore

\mathbf{E}_{\mathrm{applied}} = \frac{\lambda}{2\pi r\epsilon_0}\hat{\mathbf{r}} \times \begin{cases} 0 & r < a \\ 1 & a < r < 4a \\ 0 & r > 4a. \end{cases}

However, the field within the dielectric is reduced by a factor $\kappa$ , so that in total

\mathbf{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{\mathbf{r}} \times \begin{cases} 0 & r < a \\ 1 & a < r < 2a \\ 1/\kappa & 2a < r < 3a \\ 1 & 3 < r < 4a \\ 0 & r > 4a. \end{cases}

b. We can apply the results of the previous section to obtain the enclosed charge density $\lambda_{\mathrm{in}}$ as a function of radius,

\lambda_{\mathrm{in}} = \begin{cases} 0 & r < a \\ \lambda & a < r < 2a \\ \lambda/\kappa & 2a < r < 3a \\ \lambda & 3 < r < 4a \\ 0 & r > 4a. \end{cases}

Defining

\lambda_i = \left(1 - \frac{1}{\kappa}\right)\lambda

we conclude that a charge density $-\lambda_i$ exists on the inner surface of the dielectric, a charge density $\lambda_i$ exists on the outer surface, and there is no charge on the interior.

As with the case of a very long solenoid, we expect the magnetic field to be entirely parallel to the cylinder axis $\hat{\mathbf{z}}$ , and to go to zero for large $r$ . Consider an Amperian loop of length $l$ extending along a radius, the inner side of which is at radius $r$ and the outer side of which is at a very large radius. We have on this loop

\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{\mathrm{in}}.

Letting $B$ be the magnitude of the magnetic field at radius $r$ , we find

B = \frac{\mu_0 I_{\mathrm{in}}}{l}.

For $r > 3a$ , $I_{\mathrm{in}} = 0$ , since the charge on the hollow cylinders is not moving. For $2a < r < 3a$ , the loop now encloses the outer surface of the dielectric. In time $\frac{2\pi}{\omega}$ a charge $\lambda_i l$ passes through the loop, so the current due to the outer surface is

I_{\mathrm{out}} = \frac{\lambda_i l\omega}{2\pi}

and thus this is $I_{\mathrm{in}}$ for $2a < r < 3a$ . For $r < 2a$ , the loop now encloses both surfaces of the dielectric; the inner surface contributes a current that exactly cancels the outer one, so again $I_{\mathrm{in}} = 0$ . Putting this together,

\mathbf{B} = \frac{\mu_0\omega\lambda_i}{2\pi}\hat{\mathbf{z}} \times \begin{cases} 0 & r < 2a \\ 1 & 2a < r < 3a \\ 0 & r > 3a. \end{cases}

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