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Условие Отговори Решение

USAPhO 2013, Полуфинал — Задача 6. Question B2

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Всички материали от USAPhO 2013

UNITED STATES PHYSICS TEAM 2013 — Semifinal Exam · 50 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 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 3 parts: the cover sheet (page 2), Part A (pages 3-5), and Part B (pages 7-8). 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, 2013.
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. A straight edge or ruler could also be useful.

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, 2013.

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 concerns three situations involving the transfer of energy into a region of space by electromagnetic fields. In the first case, that energy is stored in the kinetic energy of a charged object; in the second and third cases, the energy is stored in an electric or magnetic field.

In general, whenever an electric and a magnetic field are at an angle to each other, energy is transferred; for example, this principle is the reason electromagnetic radiation transfers energy. The power transferred per unit area is given by the Poynting vector:

\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}

In each part of this problem, the last subpart asks you to verify that the rate of energy transfer agrees with the formula for the Poynting vector. Therefore, you should not use the formula for the Poynting vector before the last subpart!

a. A long, insulating cylindrical rod has radius $R$ and carries a uniform volume charge density $\rho$ . A uniform external electric field $E$ exists in the direction of its axis. The rod moves in the direction of its axis at speed $v$ .

i. What is the power per unit length $\mathcal{P}$ delivered to the rod?

ii. What is the magnetic field $B$ at the surface of the rod? Draw the direction on a diagram.

iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.

b. A parallel plate capacitor consists of two discs of radius $R$ separated by a distance $d \ll R$ . The capacitor carries charge $Q$ , and is being charged by a small, constant current $I$ .

i. What is the power $P$ delivered to the capacitor?

ii. What is the magnetic field $B$ just inside the edge of the capacitor? Draw the direction on a diagram. (Ignore fringing effects in the electric field for this calculation.)

iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.

c. A long solenoid of radius $R$ has $\mathcal{N}$ turns of wire per unit length. The solenoid carries current $I$ , and this current is increased at a small, constant rate $\frac{dI}{dt}$ .

i. What is the power per unit length $\mathcal{P}$ delivered to the solenoid?

ii. What is the electric field $E$ just inside the surface of the solenoid? Draw its direction on a diagram.

iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.

Отговори

Решение

a. i. A length $l$ of the rod has charge $q = \pi R^2 l \rho$ ; the force on it is $F = qE$ and the power delivered is $P = Fv$ . Combining these,

P = \pi R^2 l \rho E v, \quad \mathcal{P} = \pi R^2 \rho E v.

ii. The length $l$ of the rod moves past a point in a time $t = \frac{l}{v}$ , so the current carried by the rod is

I = \frac{q}{t} = \pi R^2 \rho v.

Applying Ampere’s law to a loop of radius $R$ ,

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}

2\pi R B = \mu_0 \pi R^2 \rho v \quad \Rightarrow \quad B = \frac{1}{2}\mu_0 R \rho v

The field is circumferential as given by the right-hand rule.

iii. The electric and magnetic fields are perpendicular, so the Poynting vector has magnitude

S = \frac{1}{\mu_0} E B = \frac{1}{2} R \rho v E.

A quick application of the right hand rule indicates that it points inward along the surface of the cylinder, as it ought. The cylinder has area per unit length $2\pi r$ , so the rate of energy transfer per unit length is

\mathcal{P} = 2\pi r S = \pi R^2 \rho v E

in agreement with the previous result.

b. i. The capacitance is given by the standard parallel-plate capacitor formula,

C = \frac{\epsilon_0 \pi R^2}{d}.

The voltage on the capacitor is thus

V = \frac{Q}{C} = \frac{Qd}{\epsilon_0 \pi R^2}

and the power is

P = IV = \frac{IQd}{\epsilon_0 \pi R^2}.

Students may choose instead to apply the formula for the volume energy density,

\mathcal{U} = \frac{1}{2}\epsilon_0 E^2.

ii. Consider an Amperian loop encircling the edge of the capacitor, and use a flat Gaussian surface through the center of the capacitor. The electric field here is perpendicular to the surface and has magnitude

E = \frac{V}{d} = \frac{Q}{\epsilon_0 \pi R^2}.

The electric flux through the surface is thus

\phi_E = \pi R^2 E = \frac{Q}{\epsilon_0}.

This can also be determined directly using Gauss’s law and appropriate symmetries. There is no current through the surface, so from Ampere’s law

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}

2\pi R B = \mu_0 \frac{dQ}{dt} \quad \Rightarrow \quad B = \frac{\mu_0 I}{2\pi R}

The field is circumferential as given by the right-hand rule.

Note that we could instead use a curved Gaussian surface that avoids the center of the capacitor and intersects one of the charging wires! In this case we have directly

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I

and the calculation proceeds as before.

iii. The electric and magnetic fields are perpendicular, so again

S = \frac{1}{\mu_0} E B = \frac{IQ}{2\epsilon_0 \pi^2 R^3}.

A quick application of the right hand rule indicates that it points inward along the edge of the capacitor, as it ought. The area of this region is $2\pi R d$ , so the power delivered is

P = 2\pi R d S = \frac{IQd}{\epsilon_0 \pi R^2}

in agreement with the previous result.

c. i. Suppose that the solenoid has length $l$ . The inductance is

L = \mu_0 \mathcal{N}^2 \pi R^2 l.

Students may quote this formula directly, or derive it as follows. Consider an Amperian loop of length $d$ intersecting the solenoid. This loop encloses $\mathcal{N} d$ turns of wire, so from Ampere’s law (remembering that the magnetic field exists entirely within the solenoid)

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}

Bd = \mu_0 \mathcal{N} d I \quad \Rightarrow \quad B = \mu_0 \mathcal{N} I

There are $\mathcal{N} l$ loops, so the total flux is

\Phi = \mathcal{N} l B \pi R^2 = \mu_0 \mathcal{N}^2 I \pi R^2 l

and since $\Phi = LI$ ,

L = \mu_0 \mathcal{N}^2 \pi R^2 l

as quoted above.

The voltage across the inductor is thus

V = L \frac{dI}{dt} = \mu_0 \mathcal{N}^2 \pi R^2 l \frac{dI}{dt}

and the power delivered is

P = IV = \mu_0 \mathcal{N}^2 \pi R^2 l I \frac{dI}{dt}

or, dividing by $l$ ,

\mathcal{P} = \mu_0 \mathcal{N}^2 \pi R^2 I \frac{dI}{dt}.

Students may choose instead to apply the formula for the volume energy density,

\mathcal{U} = \frac{1}{2\mu_0} B^2.

ii. Consider an Amperian loop just inside the surface of the solenoid. From above, the magnetic field through this loop is $B = \mu_0 \mathcal{N} I$ , so

\oint \mathbf{E} \cdot d\mathbf{l} = \frac{d\phi_B}{dt}

2\pi R E = \mu_0 \mathcal{N} \pi R^2 \frac{dI}{dt} \quad \Rightarrow \quad E = \frac{1}{2}\mu_0 \mathcal{N} R \frac{dI}{dt}

The field is circumferential as given by Lenz’s law and the right-hand rule.

iii. The electric and magnetic fields are perpendicular, so again

S = \frac{1}{\mu_0} E B = \frac{1}{2}\mu_0 \mathcal{N}^2 R I \frac{dI}{dt}.

A quick application of the right hand rule indicates that it points inward towards the axis of the solenoid, as it ought. The area per unit length is just $2\pi R$ , so the power per unit length is

\mathcal{P} = 2\pi R S = \mu_0 \mathcal{N}^2 \pi R^2 I \frac{dI}{dt}

in agreement with the previous result.

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-2013.pdf · официални решения: USAPhO-2013-Solutions.pdf