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Условие Отговори Решение

USAPhO 2004, Полуфинал — Задача 6. B2

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Всички материали от USAPhO 2004

AAPT UNITED STATES PHYSICS TEAM AIP 2004 — 2004 Semi-Final Exam · 50 т.

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.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

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.
Do not discuss the contents of this exam with anyone until after the submission deadline.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

A very long solenoid has radius $b$ , $n$ turns per unit length, and carries current $I$ as shown in the accompanying cross-sectional view. The solenoid is firmly fixed in place and cannot rotate.

a. What is the magnetic field $\vec{B}$ inside the solenoid, far from its ends? [5 т.]

b. What is the magnetic energy density $u_\mathrm{B}$ stored in the solenoid? [5 т.]

There are two long cylindrical shells of length $h$ and mass $m$ coaxial to the solenoid – one, inside the solenoid at a radius $a$ , carries a charge $+Q$ , uniformly distributed over its surface; the other, outside the solenoid at radius $c$ , carries charge $–Q$ . $h \gg c$ . Use $r$ to represent the distance from the cylinders’ axis. The cylinders are free to rotate about their axes.

Cross-sectional view with x and y axes: three concentric circles; the inner one of radius a carries +Q, the middle one of radius b is the solenoid with a curved arrow showing counterclockwise current I, the outer one of radius c carries –Q. — cross-sectional view

c. What is the electric field $\vec{E}$ between the cylindrical shells, far from their ends? [5 т.]

d. What is the electric energy density $u_\mathrm{E}$ stored between the shells? [5 т.]

The current in the solenoid is gradually reduced from $I$ to zero at a constant rate, $\Delta I/\Delta t$ . The cylindrical shells begin to rotate. Assume the magnetic field generated by the rotating cylindrical shells is negligible.

e. Calculate the gain in angular momentum of each cylinder about its axis when the current in the solenoid is reduced from $I$ to zero. [15 т.]

f. In addition to storing energy, fields can also store linear momentum and angular momentum. The density of linear momentum stored in electromagnetic fields is

\varepsilon_0\vec{E} \times \vec{B}.

Calculate the change in angular momentum of the electromagnetic fields about the cylinders’ axis as the current in the solenoid is reduced from $I$ to zero. [15 т.]

$\{$ Suggested by a problem in Introduction to Electrodynamics, 3rd Ed., by David J. Griffiths (Prentice Hall, Upper Saddle River, New Jersey, 1999), page 359. $\}$

Отговори

Решение

B2. a. Using $\hat{z}$ to represent a unit vector in the $z$ -direction, $\vec{B} = \mu_0 nI\hat{z}$ for $r < b$ . (It is not necessary to derive this result from Ampere’s Law.)

u_B = \frac{1}{2\mu_0}B^2 = \frac{\mu_0 n^2 I^2}{2}.

It is also possible to derive this result from $U = \frac{1}{2}LI^2$ .

If edge effects are ignored $U = u_B hA$

where $h$ is the length and $A$ is the cross-sectional area of the solenoid. Combining this with

LI = N\Phi = (nh)(BA)

yields

u_B hA = \tfrac{1}{2}nhBAI,

u_B = \tfrac{1}{2}nBI = \tfrac{1}{2}\mu_0 n^2 I^2

c. The charge per unit length on the inner cylindrical shell is $\lambda = +Q/h$ .

Letting $\hat{r}$ represent a unit vector in the $r$ -direction, Gauss’s Law for cylindrical symmetry yields

\vec{E} = +\frac{Q}{2\pi\varepsilon_0 hr}\hat{r}.

u_E = \tfrac{1}{2}\varepsilon_0 E^2 = \frac{Q^2}{8\pi^2\varepsilon_0 h^2 r^2}.

e. Use Faradays Law to find the induced electric field at the location of each charged cylinder.

\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}.

$\vec{E}$ is constant around a circular path, the cross-section of the cylinder, $\oint \vec{E} \cdot d\vec{l} = E2\pi r$ .

The magnetic flux is $\Phi_B = BA$

where $A$ is the enclosed area that has a magnetic field. Thus

E2\pi r = -A\frac{dB}{dt} = -A\mu_0 n\frac{dI}{dt}.

Since $I$ is decreasing at a constant rate

E2\pi r = -A\mu_0 n\frac{\Delta I}{\Delta t}

For the inner cylinder, $r = a$ and $A = \pi a^2$ , and

E_a 2\pi a = -\pi a^2\mu_0 n\frac{\Delta I}{\Delta t}.

Therefore

E_a = -\frac{\mu_0 na}{2}\frac{\Delta I}{\Delta t}.

In order to oppose the decrease in $I$ , $E$ is in the same direction as $I$

\vec{E}_a = -\frac{\mu_0 na}{2}\frac{\Delta I}{\Delta t}\hat{\phi}

For the outer cylinder, $r = c$ and $A = \pi b^2$ , since the field $B = 0$ for $r > b$

and

E_c 2\pi c = -\pi b^2\mu_0 n\frac{\Delta I}{\Delta t}.

Therefore

E_c = -\frac{\mu_0 nb^2}{2c}\frac{\Delta I}{\Delta t}.

with

\vec{E}_c = -\frac{\mu_0 nb^2}{2c}\frac{\Delta I}{\Delta t}\hat{\phi}

Each cylinder experiences a torque caused by the electric force.

\frac{d\vec{L}}{dt} = \vec{\tau} = \vec{r} \times \vec{F}_E.

For the inner cylinder

\frac{d\vec{L}_a}{dt} = a\hat{r} \times Q\left(-\frac{\mu_0 na}{2}\frac{\Delta I}{\Delta t}\hat{\phi}\right) = -\frac{\mu_0 na^2Q}{2}\frac{\Delta I}{\Delta t}\hat{z}.

Since $I$ decreases at a constant rate, $L$ also changes at a constant rate.

\frac{\Delta\vec{L}_a}{\Delta t} = -\frac{\mu_0 na^2Q}{2}\frac{(0 - I)}{\Delta t}\hat{z}

\Delta\vec{L}_a = \tfrac{1}{2}\mu_0 nIa^2Q\hat{z}.

For the outer cylinder

\frac{d\vec{L}_c}{dt} = c\hat{r} \times (-Q)\left(-\frac{\mu_0 nb^2}{2c}\frac{\Delta I}{\Delta t}\hat{\phi}\right) = +\frac{\mu_0 nb^2Q}{2}\frac{\Delta I}{\Delta t}\hat{z}.

and

\frac{\Delta\vec{L}_c}{\Delta t} = +\frac{\mu_0 nb^2Q}{2}\frac{(0 - I)}{\Delta t}\hat{z}

\Delta\vec{L}_c = -\tfrac{1}{2}\mu_0 nIb^2Q\hat{z}.

f. $E = 0$ for $r < a$ and $B = 0$ for $r > b$ . Using $\vec{\pi}_{EM}$ to represent the linear momentum density

\vec{\pi}_{EM} = \varepsilon_0\vec{E} \times \vec{B} = \varepsilon_0\left(\frac{Q}{2\pi\varepsilon_0 hr}\right)\hat{r} \times (\mu_0 nI)\hat{z} = -\frac{Q\mu_0 nI}{2\pi hr}\hat{\phi} \quad \text{for } a < r < b

and

\vec{\pi}_{EM} = 0 \quad \text{for } r < a \text{ and } b < r

Using $\vec{\ell}_{EM}$ to represent the angular momentum density

\vec{\ell}_{EM} = \vec{r} \times \vec{\pi}_{EM}.

In the region $a < r < b$ ,

\vec{\ell}_{EM} = \vec{r} \times \vec{\pi}_{EM} = (r\hat{r}) \times \left(-\frac{Q\mu_0 nI}{2\pi hr}\hat{\phi}\right) = -\frac{Q\mu_0 nI}{2\pi h}\hat{z}.

The angular momentum density is constant in the region $a < r < b$ and zero outside. Therefore the total angular momentum originally stored in the fields is

\vec{L}_{EM} = (\text{volume containing field})\,\vec{\ell}_{EM}

\vec{L}_{EM} = \left[h\pi(b^2 - a^2)\right]\left(-\frac{Q\mu_0 nI}{2\pi h}\hat{z}\right) = -\tfrac{1}{2}Q\mu_0 nI(b^2 - a^2)\hat{z}

When the current is reduced to zero, this angular momentum goes to zero. Therefore the change in angular momentum is

\vec{L}_{EM} = -\tfrac{1}{2}Q\mu_0 nI(b^2 - a^2)\hat{z}.

Оригинал в Архива: Semifinal 2004.pdf · официални решения: Semifinal 2004 Solutions.pdf