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USAPhO 2014, Полуфинал — Задача 6. Question B2

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Всички материали от USAPhO 2014

USA Physics Olympiad Exam — Part A / Part B · 50 т.

Внимание: Бележка към темата

The printed key is retained with its source defects: A3(a)(i) has a factor 1/2 in the expanded cubic term; A3(b)(ii) switches from question distance d to h; A4(b)(ii) prints r < b before describing the exterior field; B2(a)(i) drops the subscript 0 in one denominator. These are source discrepancies, not silent editorial corrections.

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 minutes 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 4 parts: the cover sheet (page 2), Part A (pages 3–5), Part B (pages 7–8), and several answer sheets for two of the questions in part A (pages 10–13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
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 15, 2014.
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.

USA Physics Olympiad Exam — Instructions

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 15, 2014.

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

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Условие

In parts a and b of this problem assume that velocities $v$ are much less than the speed of light $c$ , and therefore ignore relativistic contraction of lengths or time dilation.

a.

An infinite uniform sheet has a surface charge density $\sigma$ and has an infinitesimal thickness. The sheet lies in the $xy$ plane.

i. Assuming the sheet is at rest, determine the electric field $\vec E$ (magnitude and direction) above and below the sheet.

ii. Assuming the sheet is moving with velocity $\vec v=v\hat x$ (parallel to the sheet), determine the electric field $\vec E$ (magnitude and direction) above and below the sheet.

iii. Assuming the sheet is moving with velocity $\vec v=v\hat x$ , determine the magnetic field $\vec B$ (magnitude and direction) above and below the sheet.

iv. Assuming the sheet is moving with velocity $\vec v=v\hat z$ (perpendicular to the sheet), determine the electric field $\vec E$ (magnitude and direction) above and below the sheet.

v. Assuming the sheet is moving with velocity $\vec v=v\hat z$ , determine the magnetic field $\vec B$ (magnitude and direction) above and below the sheet.

b.

In a certain region there exists only an electric field $\vec E=E_x\hat x+E_y\hat y+E_z\hat z$ (and no magnetic field) as measured by an observer at rest. The electric and magnetic fields $\vec E'$ and $\vec B'$ as measured by observers in motion can be determined entirely from the local value of $\vec E$ , regardless of the charge configuration that may have produced it.

i. What would be the observed electric field $\vec E'$ as measured by an observer moving with velocity $\vec v=v\hat z$ ?

ii. What would be the observed magnetic field $\vec B'$ as measured by an observer moving with velocity $\vec v=v\hat z$ ?

c.

An infinitely long wire wire on the $z$ axis is composed of positive charges with linear charge density $\lambda$ which are at rest, and negative charges with linear charge density $-\lambda$ moving with speed $v$ in the $z$ direction.

i. Determine the electric field $\vec E$ (magnitude and direction) at points outside the wire.

ii. Determine the magnetic field $\vec B$ (magnitude and direction) at points outside the wire.

iii. Now consider an observer moving with speed $v$ parallel to the $z$ axis so that the negative charges appear to be at rest. There is a symmetry between the electric and magnetic fields such that a variation to your answer to part b can be applied to the magnetic field in this part. You will need to change the multiplicative constant to something dimensionally correct and reverse the sign. Use this fact to find and describe the electric field measured by the moving observer, and comment on your result. (Some familiarity with special relativity can help you verify the direction of your result, but is not necessary to obtain the correct answer.)

Решение

a. i.

By symmetry, the fields above and below the sheet are equal in magnitude and directed away from the sheet. By Gauss's Law, using a cylinder of base area $A$ ,

2EA=\frac{\sigma A}{\varepsilon_0}\quad\Rightarrow\quad E=\frac\sigma{2\varepsilon_0}

pointing directly away from the sheet in the $z$ direction, or

\mathbf E=\frac\sigma{2\varepsilon}\times\begin{cases}\hat z&\text{above the sheet,}\\-\hat z&\text{below the sheet.}\end{cases}

a. ii.

The motion does not affect the electric field, so the answer is the same as that of part (i).

a. iii.

Assuming $v>0$ , the right-hand rule indicates there is a magnetic field in the $-\hat y$ direction for $z>0$ and in the $+\hat y$ direction for $z<0$ . From Ampere's law applied to a loop of length $l$ normal to the $\hat x$ direction,

2Bl=\mu_0\sigma vl.

To get the right-hand side, note that in time $t$ , an area $vtl$ moves through the loop, so a charge $\sigma vtl$ moves through. Then the current through the loop is $\sigma vl$ .

Applying symmetry, we have

\mathbf B=\frac{\mu_0\sigma v}2\times\begin{cases}-\hat y&\text{above the sheet,}\\\hat y&\text{below the sheet.}\end{cases}

a. iv.

Again the motion does not affect the electric field, so the answer is the same as that of part (i).

a. v.

Applying Ampere's law and symmetry, there is no magnetic field above and below the sheet.

Interestingly, there's no magnetic field at the sheet either. Consider an Amperian loop of area $A$ in the $xy$ plane as the sheet passes through. The loop experiences a current of the form

A\sigma\delta(t).

But the loop also experiences an oppositely directed change in flux of the form

A\frac\sigma{\varepsilon_0}\delta(t),

so the right-hand side of Ampere's law, including the displacement current term, remains zero.

b. i.

In part (a), we showed that if the electric field was produced by a sheet of charge, then it was unaffected by the motion of an observer. Thus, in general,

\mathbf E'=\mathbf E.

b. ii.

No magnetic field was created by the motion of the sheet of charge in the direction of the electric field, so the magnetic field in the frame of reference of the moving observer should likewise not depend on the component of the electric field in the direction of motion. When the sheet of charge was moving in the $+\hat x$ direction, a magnetic field was created in the $-\hat y$ direction; the observer moving in the $+\hat x$ direction is equivalent to the sheet of charge moving in the $-\hat x$ direction, creating a magnetic field in the $+\hat y$ direction. That is, an electric field in the $\hat z$ direction causes an observer moving in the $\hat x$ direction to observe a magnetic field in the $\hat y=\hat z\times\hat x$ direction.

Furthermore, the magnitudes of the fields satisfied

B=\mu_0\varepsilon_0vE=\frac1{c^2}vE.

Combining this with the previous equation,

\mathbf B'=-\frac1{c^2}\mathbf v\times\mathbf E=\frac v{c^2}(E_y\hat x-E_x\hat y).

c. i.

The wire as a whole is neutral, so there is no electric field outside the wire.

c. ii.

The current in the wire is $\lambda v$ , so Ampere's law yields

B=\mu_0\frac{\lambda v}{2\pi r}

in the tangential direction. The current is in the $-\hat z$ direction, so by the right-hand rule, the circular $\mathbf B$ field lines would point clockwise looking in that direction.

c. iii.

The result of part (b) was

\mathbf B'=-\frac1{c^2}\mathbf v\times\mathbf E.

Exchanging the electric and magnetic fields, reversing the sign, and fixing the dimensions,

\mathbf E'=\mathbf v\times\mathbf B.

Taking the cross product yields an electric field vector that points outward, with magnitude

E'=v\mu_0\frac{\lambda v}{2\pi r}=\frac\lambda{2\pi\varepsilon_0r}\frac{v^2}{c^2}.

Physically, this can be explained by length contraction of the positive charges and inverse length contraction of the negative charges, which are now stationary. That is, we have derived a relativistic effect, second-order in $v/c$ , from the first-order field transformations!

Why wasn't this derivation used to discover relativity the moment Maxwell's equations were written down? We have implicitly assumed that Maxwell's equations are the same in all reference frames, but historically it was thought they were only valid in one frame, the reference frame of the ether. Assuming that Maxwell's equations are indeed the same in all frames yields an invariant speed, the speed of light, leaving inevitably to all of special relativity. Here we've taken one of many possible paths.

Оригинал в Архива: USAPhO-2014.pdf · официални решения: USAPhO-2014-Solutions.pdf